Elements of Propulsion: Gas Turbines and Rockets Jack D. Mattingly Professor Emeritus Department of Mechanical Engineering Seattle University, Seattle, Washington Foreword by
Hans von Ohain German Inventor of the Jet Engine
EDUCATION SERIES Joseph A. Schetz Series EditorinChief Virginia Polytechnic Institute and State University Blacksburg, Virginia
Published by the American Institute of Aeronautics and Astronautics, Inc. 1801 AlexanderBell Drive, Reston, Virginia 201914344
American Institute of Aeronautics and Astronautics, Inc., Reston, Virginia
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Library of Congress CataloginginPublication Data
Mattingly, Jack D. Elements of propulsion: gas turbines and rockets/Jack D. Mattingly; foreword by Hans yon Ohain. p. cm.(Education series) Includes bibliographical references and index. ISBN 1563477793 (alk. paper) 1. Aircraft gasturbines. 2. AirplanesJet propulsion. 3. Jet engines. 4. Rocket engines. I. Title. II. Series: AIAA education series. TL709.M388 2006 629.134'353dc22
2006017363
Copyright © 2006 by The American Institute of Aeronautics and Astronautics, Inc. All rights reserved. Printed in the United States. No part of this publication may be reproduced, distributed, or transmitted, in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. Data and information appearing in this book are for information purposes only. AIAA is not responsible for any injury or damage resulting from use or reliance, nor does AIAA warrant that use or reliance will be free from privately owned rights.
I have been blessed to share my life with Sheila, my best friend and wife. She has been my inspiration and helper, and the one w h o sacrificed the most to make this work possible. I dedicate this book and the accompanying software to Sheila. I would like to share with all the following passage I received from a very close friend about 30 years ago. This passage provides guidance and focus to my life. I hope it can be as much help to you.
FABRIC OF LIFE I want to say something to all of you Who have become a part Of the fabric of my life The color and texture Which you have brought into My being Have become a song And I want to sing it forever. There is an energy in us Which makes things happen When the paths of other persons Touch ours And we have to be there And let it happen. When the time of our particular sunset comes Our thing, our accomplishment Won't really matter A great deal. But the clarity and care With which we have loved others Will speak with vitality Of the great gift of life We have been for each other.
Gregory Norbert, O.S.B.
AIAA Education Series EditorinChief Joseph A. Schetz
Virginia Polytechnic Institute and State University
Editorial Board Takahira Aoki
Rakesh K. Kapania
University of Tokyo
Virginia Polytechnic Institute and State University
Edward W. Ashford Brian Landrum Karen D. Barker
Brahe Corporation
University of Alabama Huntsville
Robert H. Bishop
Timothy C. Lieuwen
University of Texas at Austin
Georgia Institute of Technology
Claudio Bruno
University of Rome Aaron R. Byerley
U.S. Air Force Academy Richard Colgren
University of Kansas
Michael Mohaghegh
The Boeing Company Conrad E Newberry
Naval Postgraduate School Mark A. Price
Queen's University Belfast
Kajal K. Gupta
NASA Dryden Flight Research Center
James M. Rankin
Ohio University
Rikard B. Heslehurst
David K. Schmidt
Australian Defence Force Academy
University of Colorado Colorado Springs
David K. Holger
Iowa State University
David M. Van Wie
Johns Hopkins University
Foreword to the Second Edition
We are very pleased to present the Second Edition of Elements of Propulsion: Gas Turbines and Rockets by Jack D. Mattingly. The original edition was a wellreceived, comprehensive, and indepth treatment of these important topics. This new edition has updated the material and expanded the coverage, and we anticipate that it will be equally well received. An interesting feature of the first edition was a very extensive foreword by the jet engine pioneer, Hans Von Ohain. Indeed, it is so extensive and comprehensive that it is more a historical overview and introduction than a foreword. We deem it so useful that it is included in this edition as well. This Second Edition has ten chapters, seven appendices and more than 850 pages. Jack Mattingly is extremely well qualified to write this book because of his broad expertise in the area. His command of the material is excellent, and he is able to organize and present it in a very clear manner. He is such a proficient author that he is the cowinner of the Martin Summerfield Book Award for another volume in this series. The A I A A Education Series aims to cover a very broad range of topics in the general aerospace field, including basic theory, applications, and design. A complete list of titles can be found at http://www.aiaa.org. The philosophy of the series is to develop textbooks that can be used in a university setting, instructional materials for continuing education and professional development courses, and also books that can serve as the basis for independent study. Suggestions for new topics or authors are always welcome.
Joseph A. Schetz EditorinChief AIAA Education Series
ix
Foreword to the First Edition
Background The first flight of the Wright brothers in December 1903 marked the beginning of the magnificent evolution of humancontrolled, powered flight. The driving forces of this evolution are the evergrowing demands for improvements in 1) Flight performance (i.e., greater flight speed, altitude, and range and better maneuverability); 2) Cost (i.e., better fuel economy, lower cost of production and maintenance, increased lifetime); 3) Adverse environmental effects (i.e., noise and harmful exhaust gas effects); 4) Safety, reliability, and endurance; and 5) Controls and navigation. These strong demands continuously furthered the efforts of advancing the aircraft system. The tight interdependency between the performance characteristics of aerovehicle and aeropropulsion systems plays a very important role in this evolution. Therefore, to gain better insight into the evolution of the aeropropulsion system, one has to be aware of the challenges and advancements of aerovehicle technology.
The Aerovehicle A brief review of the evolution of the aerovehicle will be given first. One can observe a continuous trend toward stronger and lighter airframe designs, structures, and materialsfrom wood and fabric to allmetal structures; to lighter, stronger, and more heatresistant materials; and finally to a growing use of strong and light composite materials. At the same time, the aerodynamic quality of the aerovehicle is being continuously improved. To see this development in proper historical perspective, let us keep in mind the following information. In the early years of the 20th century, the science of aerodynamics was in its infancy. Specifically, the aerodynamic lift was not scientifically well understood. Joukowski and Kutta's model of lift by circulation around the wing and Prandtl's boundarylayer and turbulence theories were in their incipient stages. Therefore, the early pioneers could not benefit from existent XV
xvi
FOREWORD
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scientific knowledge in aerodynamics and had to conduct their own fundamental investigations. The most desirable major aerodynamic characteristics of the aerovehicle are a low drag coefficient as well as a high lift/drag ratio LID for cruise conditions, and a high maximum lift coefficient for landing. In Fig. 1, one can see that the world's first successful glider vehicle by Lilienthal, in the early 1890s, had an LID of about 5. In comparison, birds have an LID ranging from about 5 to 20. The Wright brothers' first humancontrolled, powered aircraft had an LID of about 7.5. As the LID values increased over the years, sailplanes advanced most rapidly and now are attaining the enormously high values of about 50 and greater. This was achieved by employing ultrahigh wing aspect ratios and aerodynamic profiles especially tailored for the low operational Reynolds and Mach numbers. In the late 1940s, subsonic transport aircraft advanced to LID values of about 20 by continuously improving the aerodynamic shapes, employing advanced profiles, achieving extremely smooth and accurate surfaces, and incorporating inventions such as the engine cowl and the retractable landing gear. The continuous increase in flight speed required a corresponding reduction of the landing speed/cruise speed ratio. This was accomplished by innovative wing structures incorporating wing slots and wing flaps that, during the landing process, enlarged the wing area and increased significantly the lift coefficient. Today, the arrowheadshaped wing contributes to a high lift for landing (vortex lift). Also, in the 1940s, work began to extend the high LID value from the subsonic to the transonic flight speed regime by employing the sweptback wing and later, in 1952, the area rule of Whitcomb to reduce transonic drag rise. Dr. Theodore von Kfirmfin describes in his memoirs, The Wind and
FOREWORD
xvii
Beyond,* how the sweptback wing or simply swept wing for transonic and supersonic flight came into existence: The fifth Volta Congress in Rome, 1935, was the first serious international scientific congress devoted to the possibilities of supersonic flight. I was one of those who had received a formal invitation to give a paper at the conference from Italy's great Gugliemo Marconi, inventor of the wireless telegraph. All of the world's leading aerodynamicists were invited. This meeting was historic because it marked the beginning of the supersonic age. It was the beginning in the sense that the conference opened the door to supersonics as a meaningful study in connection with supersonic flight, and, secondly, because most developments in supersonics occurred rapidly from then on, culminating in 1946a mere 11 years laterin Captain Charles Yeager's piercing the sound barrier with the X1 plane in level flight. In terms of future aircraft development, the most significant paper at the conference proved to be one given by a young man, Dr. Adolf Busemann of Germany, by first publicly suggesting the sweptback wing and showing how its properties might solve many aerodynamic problems at speeds just below and above the speed of sound. Through these investigations, the myth that sonic speed is the fundamental limit of aircraft flight velocity, the sound barrier was overcome. In the late 1960s, the Boeing 747 with sweptback wings had, in transonic cruise speed, an LID value of nearly 20. In the supersonic flight speed regime, LID values improved from 5 in the mid1950s (such as LID values of the B58 Hustler and later of the Concorde) to a possible LID value of 10 and greater in the 1990s. This great improvement possibility in the aerodynamics of supersonic aircraft can be attributed to applications of artificial stability, to the area rule, and to advanced wing profile shapes that extend laminar flow over a larger wing portion. The hypersonic speed regime is not fully explored. First, emphasis was placed on winged reentry vehicles and lifting bodies where a high LID value was not of greatest importance. Later investigations have shown that the LID values can be greatly improved. For example, the maximum LID for a "wave rider" is about 6.* Such investigations are of importance for hypersonic programs.
The Aeropropulsion System At the beginning of this century, steam and internal combustion engines were in existence but were far too heavy for flight application. The Wright brothers recognized the great future potential of the internal combustion engine and developed both a relatively lightweight engine suitable for flight application and an efficient propeller. Figure 2 shows the progress of the propulsion systems over the years. The Wright brothers' first aeropropulsion system had a shaft power of 12 hp, and its power/weight ratio (ratio of power output to total propulsion system weight, including propeller and transmission) was about 0.05 hp/lb.
*Von K~irrn~n,T., The Windand Beyond, Little, Brown, Boston, 1967. *Heiser, W. H., and Pratt, D.J., HypersonicAirbreathing Propulsion, AIAA Education Series, AIAA, Washington, DC, 1994.
FOREWORD
xviii
50%
Wright brothers 1903: ~0.05 hp/lb • End WWII:0.8 hp/lb •
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Through the subsequent four decades of evolution, the overall efficiency and the power/weight ratio improved substantially, the latter by more than one order magnitude to about 0.8 hp/lb. This great improvement was achieved by engine design structures and materials, advanced fuel injection, advanced aerodynamic shapes of the propeller blades, variablepitch propellers, and engine superchargers. The overall efficiency (engine and propeller) reached about 28%. The power output of the largest engine amounted to about 5000 hp. In the late 1930s and early 1940s, the turbojet engine came into existence. This new propulsion system was immediately superior to the reciprocating engine with respect to the power/weight ratio (by about a factor of 3); however, its overall efficiency was initially much lower than that of the reciprocating engine. As can be seen from Fig. 2, progress was rapid. In less than four decades, the power/weight ratio increased more than 10fold, and the overall efficiency exceeded that of a diesel propulsion system. The power output of today's largest gas turbine engines reaches nearly 100,000 equivalent hp.
Impact on the Total Aircraft Performance The previously described truly gigantic advancements of stronger and lighter structures and greater aerodynamic quality in aerovehicles and greatly advanced overall efficiency and enormously increased power output/weight ratios in aeropropulsion systems had a tremendous impact on flight performance, such as on flight range, economy, maneuverability, flight speed, and altitude. The increase in flight speed over the years is shown in Fig. 3. The Wright brothers began with the first humancontrolled, powered flight in 1903; they continued to
FOREWORD
xix Jet era
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improve their aircraft system and, in 1906, conducted longer flights with safe takeoff, landing, and curved flight maneuvers. While the flight speed was only about 35 mph, the consequences of these first flights were enormous: 1) Worldwide interest in powered flight was stimulated. 2) The science of aerodynamics received a strong motivation. 3) The U.S. government became interested in power flight for potential defense applications, specifically reconnaissance missions. In 1909, the Wright brothers built the first military aircraft under government contract. During World War I, aircraft technology progressed rapidly. The flight speed reached about 150 mph, and the engine power attained 400 hp. After World War I, military interest in aircraft systems dropped, but aircraft technology had reached such a degree of maturity that two nonmilitary application fields could emerge, namely: 1) Commercial aviation, mail and passenger transport (first allmetal monoplane for passenger and mail transport, the Junkers F13, in 1919, sold worldwide); 2) Stunt flying leading to general aviation (sport and private transportation). In the period from 1920 to 1940, the speed increased from about 150 to 350 mph through evolutionary improvements in vehicle aerodynamics and engine technology, as discussed previously. At the end of World War II, the flight speed of propeller aircraft reached about 400450 mph, and the power output of the largest reciprocating engines was about 5000 hp. This constituted almost the performance limit of the propeller/reciprocating engine propulsion system. Today, the propeller/reciprocating engine survives only in smaller, lowerspeed aircraft used in general aviation.
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FOREWORD
In the late 1930s, jet propulsion emerged that promised far greater flight speeds than attainable with the propeller or piston engine. The first jetpropelled experimental aircraft flew in the summer of 1939 (the He178), and in early 1941, the first prototype jet tighter began flight tests (He280). In 1944, massproduced jet fighters reached a speed of about 550 mph (Me262). In the early 1950s, jet aircraft transgressed the sonic speed. In the mid1950s, the first supersonic jet bomber (B58 Hustler) appeared, and later the XB70 reached about Mach 3. Also during the 1950s, after more than 15 years of military development, gas turbine technology had reached such a maturity that the following commercial applications became attractive: 1) commercial aircraft, e.g., Comet, Caravelle, and Boeing 707; 2) surface transportation (land, sea); and 3) stationary gas turbines. In the 1960s, the highbypassratio engine appeared, which revolutionized military transportation (the C5A transport aircraft). At the end of the 1960s, based on the military experience with highbypassratio engines, the second generation of commercial jet aircraft came into existence, the widebody aircraft. An example is the Boeing 747 with a large passenger capacity of nearly 400. Somewhat later came the Lockheed L1011 and Douglas DC10. By that time, the entire commercial airline fleet used turbine engines exclusively. Advantages for the airlines were as follows: 1) Very high overall efficiency and, consequently, a long flight range with economical operation. 2) Overhaul at about 5 million miles. 3) Short turnaround time. 4) Passenger enjoyment of the very quiet and vibrationfree flight, short travel time, and comfort of smooth stratospheric flight. 5) Community enjoyment of quiet, pollutionfree aircraft. By the end of the 1960s, the entire business of passenger transportation was essentially diverted from ships and railroads to aircraft. In the 1970s, the supersonic Concorde with a flight speed of 1500 mph (the third generation of commercial transport) appeared with an equivalent output of about 100,000 hp.
Summary In hindsight, the evolution of aerovehicle and aeropropulsion systems looks like the result of a master plan. The evolution began with the piston engine and propeller, which constituted the best propulsion system for the initially low flight speeds, and had an outstanding growth potential up to about 450 mph. In the early 1940s, when flight technology reached the ability to enter into the transonic flight speed regime, the jet engine had just demonstrated its suitability for this speed regime. A vigorous jet engine development program was launched. Soon the jet engine proved to be not only an excellent transonic but also a supersonic propulsion system. This resulted in the truly exploding growth in flight speed, as shown in Fig. 3. It is interesting to note that military development preceded commercial applications by 1520 years for both the propeller engine and the gas turbine engine. The reason was that costly, highrisk, longterm developments conducted
FOREWORD
xxi
by the military sector were necessary before a useful commercial application could be envisioned. After about 75 years of powered flight, the aircraft has outranked all other modes of passenger transportation and has become a very important export article of the United States. The evolutions of both aerovehicle and aeropropulsion systems have in no way reached a technological level that is close to the ultimate potential! The evolution will go on for many decades toward capabilities far beyond current feasibility and, perhaps, imagination.
How Jet Propulsion Came into Existence The idea of airbreathing jet propulsion originated at the beginning of the 20th century. Several patents regarding airbreathing jet engines had been applied for by various inventors of different nationalities who worked independently of each other. From a technical standpoint, airbreathing jet propulsion can be defined as a special type of internal combustion engine that produces its net output power as the rate of change in the kinetic energy of the engine's working fluid. The working fluid enters as environmental air that is ducted through an inlet diffuser into the engine; the engine exhaust gas consists partly of combustion gas and partly of air. The exhaust gas is expanded through a thrust nozzle or nozzles to ambient pressure. A few examples of early airbreathing jet propulsion patents are as follows: 1) In 1908, Lorin patented a jet engine that was based on piston machinery (Fig. 4a). 2) In 1913, Lorin patented a jet engine based on ram compression in supersonic flight (Fig. 4b), the ramjet. 3) In 1921, M. Guillaume patented a jet engine based on turbomachinery; the intake air was compressed by an axialflow compressor followed by a combustor and an axialflow turbine driving the compressor (Fig. 4c). These patents clearly described the airbreathing jet principle but were not executed in practice. The reason lies mainly in the previously mentioned strong interdependency between aerovehicle and aeropropulsion systems. The jet engine has, in comparison with the propeller engine, a high exhaust speed (for example, 600 mph and more). In the early 1920s, the aerovehicle had a flight speed capability that could not exceed about 200 mph. Hence, at that time, the socalled propulsive efficiency of the jet engine was very low (about 3040%) in comparison to the propeller, which could reach more than 80%. Thus, in the early 1920s, the jet engine was not compatible with the tooslow aerovehicle. Also, in the early 1920s, an excellent theoretical study about the possibilities of enjoying jet propulsion had been conducted by Buckingham of the Bureau of Standards under contract with NACA. The result of this study was clearthe jet engine could not be efficiently employed if the aerovehicle could not greatly exceed the flight speed of 200 mph; a flight speed beyond 400 mph seemed to be necessary. The consequences of the results of this study were that the aircraft engine industry and the scientific and engineering community had no interest in the various jet engine inventions. Thus the early jet engine
xxii
FOREWORD
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1 Fig. 5 Whittle's turbojet patent drawing. (National Air and Space Museum.) concepts were forgotten for a long time. They were unknown to Sir Frank Whittle, to me, and to the British and German patent offices. In 1939, however, the retired patent examiner Gohlke found out about the early jet patents and published them in a synoptic review. The first patent of a turbojet engine, which was later developed and produced, was that of Frank Whittle, now Sir Frank (see Fig. 5). His patent was applied for in January 1930. This patent shows a multistage, axialflow compressor followed by a radial compressor stage, a combustor, an axialflow turbine driving the compressor, and an exhaust nozzle. Such configurations are still used today for small and mediumpower output engines, specifically for remotecontrolled vehicles.
Turbojet Development of Sir Frank Whittle Frank Whittle ~ was a cadet of the Royal Air Force. In 1928, when he was 21 years old, he became interested in the possibilities of rocket propulsion and
*Boyne,W. J., and Lopez,D. S., The JetAge, Forty Years of Jet Aviation, SmithsonianInst. Press, Washington,DC, 1979.
xxiv
FOREWORD
propeller gas turbines for aircraft, and he treated these subjects in his thesis. He graduated and became a pilot officer, continuously thinking about airbreathing jet propulsion. In 1929, he investigated the possibilities of a ducted fan driven by a reciprocating engine and employing a kind of afterburner prior to expansion of the fan gas. He finally rejected this idea on the basis of his performance investigations. The same idea was conceived later in Italy and built by Caproni Campini. The vehicle flew on August 28, 1940, but had a low performance, as predicted by Sir Frank in 1929. Suddenly, in December 1929, Frank Whittle was struck by the idea of increasing the fan pressure ratio and substituting a turbine for the reciprocating engine. This clearly constituted a compact, lightweight turbojet engine. He applied for a patent for the turbojet (Fig. 5) in January 1930. Frank Whittle discussed this idea with fellow officers and his superior officer. They were very impressed, and a meeting was arranged between him and officials of the British Air Ministry, Department of Engine Development. This department, in turn, sought advice from Dr. A. A. Griffith, who was interested in the development of a propeller gas turbine. Dr. Griffith expressed doubts about the feasibility of Whittle's turbojet concept from a standpoint of toohigh fuel consumption. Actually, in highspeedflight, the turbojet has great advantages over a propeller gas turbine due to the fact that the turbojet is much lighter than the propeller gas turbine and can fly faster because of the absence of the propeller. Whittle rightfully considered the turbojet as a fortunate synthesis or hybrid of the "propeller gas turbine" and "rocket" principles. As Sir Frank recalls, the department wrote a letter that, in essence, stated that any form of a gas turbine would be impractical in view of the long history of failure and the lack of turbine materials capable of withstanding the high stresses at high temperatures. Whittle' s outstanding and very important views were that the flying gas turbine had great advantages over a stationary gas turbine power plant due to the efficient ram pressure recovery, low environmental temperature in high altitude, and high efficiency of the jet nozzle. Unfortunately, these views were ignored by the department. Frank Whittle (Fig. 6) tried to interest the turbine industry in his concept of jet propulsion, but he did not succeed. Lacking financial support, Whittle allowed his patent to lapse. A long, dormant period was ahead for Frank Whittle's jet propulsion ideas. After five years, in mid1935, two former Royal Air Force (RAF) officers tried to revive Whittle's turbojet concept. Whittle was enthused and wrote, "the jet engine had, like the Phoenix, risen from its ashes. ''§ At that time, Whittle was under enormous pressure. He was preparing for the examination in mechanical sciences (Tripos); his goal was to graduate with "FirstClass Honors." Now, in addition, he had to design his first experimental jet engine in late 1935. In March 1936, a small company, Power Jets Ltd., was formed to build and test Whittle's engine, the W. U. (Whittle Unit). In spite of all the additional work, Whittle passed his exam in June 1936 with FirstClass Honors.
§Boyne,W. J., and Lopez,D. S., The JetAge, Forty Years ofJetAviation, SmithsonianInst. Press, Washington, DC, 1979.
FOREWORD
Fig. 6
xxv
Frank Whittle using slide rule to perform calculations. (Bettman.)
In April 1937, Whittle had his benchtest jet engine ready for the first test run. It ran excellently; however, it ran out of control because liquid fuel had collected inside the engine and started to vaporize as the engine became hot, thereby adding uncontrolled fuel quantities to the combustion process. The problem was easily overcome. This first test run was the world's first run of a benchtest jet engine operating with liquid fuel (Fig. 7). In June 1939, the testing and development had progressed to a point that the Air Ministry's Director of Scientific Research (D.S.R.) promised Frank Whittle a contract for building a flight engine and an experimental aircraft, the Gloster E28/29 (Gloster/Whittle). On May 15, 1941, the first flight of the Gloster/Whittle took place (Fig. 8). Senior ministry officials initially showed little interest, and a request for filming was ignored; however, during further flight demonstrations, interest in jet propulsion increased. Of particular interest was a performance demonstration given to Sir Winston Churchill. At that occasion, the Gloster/Whittle accelerated away from the three escorting fighters, one Tempest and two Spitfires.
xxvi
FOREWORD
Fig. 7 Whittle's first experimental engine after second reconstruction in 1938. (National Air and Space Museum.)
Several British aircraft engine corporations adapted the work of Frank Whittle. Specifically, RollsRoyce, due to the efforts of Sir Stanley G. Hooker, ¶ developed the first operational and the first production engine for the twoengine Gloster Meteor, Britain's first jet fighter. In March 1943, the Gloster Meteor prototype made its first flight, powered by two de Haviland (Hl) radial jet engines. In July 1944, the Meteor I, powered with two RollsRoyce Welland engines, became operational. Its only combat action (in World War II) was in August 1944 in a successful attack against the German V1 flying bomb; it was the only fighter with sufficient level speed for the purpose. Mass production began with the Meteor III powered by two RollsRoyce Dervents in 1945. The Meteor remained the RAF's firstline jet fighter until 1955. From the beginning of his jet propulsion activities, Frank Whittle had been seeking means for improving the propulsive efficiency of turbojet engines. ¶ He conceived novel ideas for which he filed a patent application in 1936, which can be called a bypass engine or turbofan. To avoid a complete new design, Whittle sought an interim solution that could be merely "tacked on" to a jet engine. This configuration was later known as the aft fan. Whittle's work on fan jets or bypass engines and aft fans was way ahead of his time. It was of greatest importance for the future or turbopropulsion.
¶Schlaifer, R., and Heron, S. D., Developmentof Aircraft Engines, and Fuels, PergamonPress, New York, 1970; reprint of 1950 ed.
FOREWORD
xxvii
Fig. 8 Gloster E28/29. (National Air and Space Museum.)
Whittle's Impact on U.S. Jet Development In the summer of 1941, U.S. Army Air Corp General Henry H. Arnold was invited to observe flight demonstrations of the Gloster/Whittle. He was very impressed and decided this technology should be brought over to the United States. In September 1941, an agreement was signed between U.S. Secretary of War Stimson and Sir Henry Self of the British Air Commission. The United States could have the engine W1X and a set of drawings of the Whittle W2B jet engine, provided that close secrecy was maintained and the number of people involved were held to a minimum. Under these conditions, open bids for the jet engine development were not possible. General Arnold chose General Electric for jet engine development because of the great experience this company had in the development of aircraft engine turbosuperchargers. The W2B engine was built and tested on March 18, 1942, under the name GE 1A. This engine had a static thrust of 12501b and weighed 1000lb. 3 In the meantime, the Bell Aircomet (XP59A) was being designed and built. On October 3, 1942, the Aircomet with two GE 1A engines flew up to 10,000 ft. This aircraft, while in the first tests seemed to have good performance characteristics, had an incurable "snaking" instability and so provided a poor gun platform for a fighter pilot. Also, another serious shortcoming was that the top speed was not sufficiently above that of an advanced propeller fighter. For these reasons, the Bell XP59A with two GE 1A engines (W2B) did not become a production fighter. From these experiences, it appeared that an engine of more than 40001b thrust was required for a singleengine fighter that would be capable of more than 500 mph operational speed. Lockheed was chosen to design a new jet fighter because when the project was discussed with the engineering staff, Lockheed's Kelly Johnson assured them a singleengine jet fighter in the "500 plus" mph class could be built on the basis of a 40001b thrust engine.
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General Electric developed the 40001b thrust engine, the 140 (an advanced version of Whittle's W2B engine), and Lockheed built the P80A Shooting Star, which flew on June 11, 1944. Although it did not enter combat during World War II, the Shooting Star became the United States' frontline fighter and outranked the Gloster Meteor with an international speed record (above 620 mph near the ground). By about 1945, Frank Whittle had successfully completed, with greatest tenacity under the most adverse conditions, the enormous task of leading Great Britain and the United States into the jet age.
Other Early Turbojet Developments in the United States Independent of European influence, several turbojet and propeller gas turbine projects had been initiated in the United States in 1939 and 1940. Although these projects had been terminated or prematurely canceled, they had contributed significantly to the knowhow and technology of aircraft gas turbines, specifically their combustor and turbomachinery components. One of these projects was the 2500hp Northrop propeller gas turbine (Turbodyne) and a highpressureratio turbojet under the excellent project leadership of Vladimir Pavelecka. Although the development goal of the large aircraft gas turbine engine was essentially met in late 1940, the project was canceled because the Air Force had lost interest in propeller gas turbines in view of the enormous advancement of the competitive jet engines. Westinghouse had developed outstanding axial turbojet engines. The first very successful test runs of the Westinghouse X 19A took place in March 1943. In the beginning of the 1950s, the Navy canceled the development contract, and top management of Westinghouse decided to discontinue work on turbojet engines. The Lockheed Corporation began to work on a very advanced turbojet conceived by an outstanding engineer, Nathan C. Price. This engine was so far ahead of its time that it would have needed a far longer development time than that provided by the contract. The development contract was canceled in 1941. Pratt & Whitney had started to work on its own jet propulsion ideas in the early 1940s but could not pursue these concepts because of the toostringent obligations during wartime for the development and production of advanced aircraft piston engines. After World War II, Pratt & Whitney decided to go completely into turbojet development using axialflow turbomachinery. The company began with the construction of a gigantic test and research facility. The government gave Pratt & Whitney a contract to build a large number of 50001b thrust RollsRoyce Nene engines with a radial compressor of the basic Whittle design. Subsequently, Pratt & Whitney developed its own large axialflow, dualrotor turbojet and later a fanjet with a small bypass ratio for the advanced B52.
Turbojet Development of Hans von Ohain in Germany My interest in aircraft propulsion began in the fall of 1933 while I was a student at the Georgia Augusta University of Gottingen in physics under
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Prof. R. Pohl with a minor in applied mechanics under Prof. Ludwig Prandtl. I was 21 years old and beginning my Ph.D. thesis in physics, which was not related to jet propulsion. The strong vibrations and noise of the propeller piston engine triggered my interest in aircraft propulsion. I felt the natural smoothness and elegance of flying was greatly spoiled by the reciprocating engine with propellers. It appeared to me that a steady, thermodynamic flow process was needed. Such a process would not produce vibrations. Also, an engine based on such a process could probably be lighter and more powerful than a reciprocating engine with a propeller because the steady flow conditions would allow a much greater mass flow of working medium per cross section. These characteristics appeared to me to be most important for achieving higher flight speeds. I made performance estimates for several steady flow engine types and finally chose a special gas turbine configuration that appeared to me as a lightweight, simple propulsion system with low development risks. The rotor consisted of a straightvane radial outflow compressor backtoback with a straight vane radial inflow turbine. Both compressor and turbine rotors had nearly equal outer diameters, which corresponds to a good match between them. In early 1935, I worked out a patent for the various features of a gas turbine consisting of radial outflow compressor rotor, combustor, radial inflow turbine, and a central exhaust thrust nozzle. With the help of my patent attorney, Dr. E. Wiegand, a thorough patent search was made. A number of interesting aeropropulsion systems without a propeller were found, but we did not come across the earlier patents of Lorin, Guillaume, and Frank Whittle. (I learned for the first time about one of Frank Whittle's patents in early 1937 when the German Patent Office held one of his patents and one patent of the Swedish corporation Milo, against some of my patent claims.) My main problem was finding support for my turbojet ideas. A good approach, it seemed to me, was to first build a model. This model should be able to demonstrate the aerodynamic functions at very lowperformance runs. The tip speed of this model was a little over 500 ft/s. Of course, I never considered highpower demonstration runs for two reasons. The cost for building such an apparatus could easily be a factor of 10 or 20 times greater than that for building a lowspeed model. Also, a test facility would be required for highperformance test or demonstration runs. I knew a head machinist in an automobile repair shop, Max Hahn, to whom I showed the sketches of my model. He made many changes to simplify the construction, which greatly reduced the cost. The model was built at my expense by Hahn in 1935 (Fig. 9). In mid1935, I had completed my doctoral thesis and oral examination and had received my diploma in November 1935. I continued working in Prof. Pohl's institute and discussed with him my project "aircraft propulsion." He was interested in my theoretical writeup. Although my project did not fit Pohl's institute, he was extremely helpful to me. He let me test the model engine in the backyard of his institute and gave me instrumentation and an electric starting motor. Because the combustors did not work, the model did not run without power from the starting motor. Long, yellow flames leaked out of the turbine. It looked more like a flame thrower than an aircraft gas turbine.
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Fig. 9
Max Hahn with model engine. (National Air and Space Museum.)
I asked Prof. Pohl to write me a letter of introduction to Ernst Heinkel, the famous pioneer of highspeed aircraft and sole owner of his company. Professor Pohl actually wrote a very nice letter of recommendation. I had chosen Heinkel because he had the reputation of being an unconventional thinker obsessed with the idea of highspeed aircraft. Intuitively, I also felt that an aircraft engine company would not accept my turbine project. I learned later that my intuition was absolutely right. Today, I am convinced no one except Heinkel (see Fig. 10) would have supported my jet ideas at that time. Heinkel invited me to his home on the evening of March 17, 1936, to explain the jet principle to him. He, in turn, gave me a view of his plan. He wanted the jet development to be apart from the airplane factory. For this purpose, he intended to construct a small, temporary building near the Warnow River. I was very enthusiastic about this idea since it gave me a feeling of freedom and independence from the other part of the company and an assurance of Heinkel's confidence in me. Also, he strongly emphasized that he himself wanted to finance the entire jet development without involvement of the German Air Ministry. Finally, he explained to me that he had arranged a meeting between me and his top engineers for the next morning. On March 18, 1936, I met with a group of 8 to 10 Heinkel engineers and explained my jet propulsion thoughts. Although they saw many problems, specifically with the combustion, they were not completely negative. Heinkel called me to a conference at the end of March. He pointed out that several uncertainties, specifically the combustion problems, should be solved before the gas
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Fig. 10 Ernst Heinkel (left) and Hans von Ohain (right). (National Air and Space Museum.)
turbine development could be started. He wanted me to work on this problem and to report to him all the difficulties I might encounter. He offered me a kind of consulting contract that stated that the preliminary work (combustor development) could probably be completed in about two months. If successful, the turbojet development would then be started, and I would receive a regular employment contract. I signed this contract on April 3, 1936, and would start working in the Heinkel Company on April 15. The first experiments with the model in early 1936 had convinced me that the volume of the combustion chambers was far too small for achieving a stable combustion. This was later substantiated in a discussion with combustion engineers at an industrial exhibit. I found a simple way to correct this condition. Cycle analysis of my model clearly showed that for high turbine inlet temperatures, such as 700°C and higher, a centrifugal compressor with a radial inflow turbine was most suitable as a basis for combustor development. My greatest problem was how to develop a functioning combustor in a few months. In my judgment, such a development would need at least six months, more likely one year, while Heinkel's estimate was two months. I had grave doubts whether Heinkel would endure such a long development time without seeing any visible progress, such as an experimental jet engine in operation. However, to avoid any combustor difficulties, I was considering a hydrogen combustor system with a nearly uniform turbine inlet temperature distribution. This hydrogen combustor system should be designed so that it could be built without any risk or need for preliminary testing. My idea was to separate the compressor and turbine on the rotor by a shaft and to employ an annular connecting duct from the exit of the compressor diffuser to
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the inlet of the turbine. Within this annular duct, I wanted to place a row of hollow vanes (about 60). These hollow vanes would have blunt trailing edges with many small holes through which hydrogen gas jets would be discharged into the air wakes behind the blunt trailing edges. In this way, the hydrogen combustion would be anchored at the blunt trailing edges of the hollow vanes. I was absolutely certain this combustor system would successfully function without any development or preliminary testing. I was also certain that no pretesting or development was necessary for the simple radialflow turbomachinery. Testing of the hydrogen demonstrator engine (Fig. 11) showed that my judgment was correct on both points. By midMay 1936 I had nearly completed the layout of the hydrogen demonstrator engine. To build this engine was, for me, most important, not only for quick achievement of an impressive demonstration of the jet principle, but also for very significant technical reasons: 1) One reason was to obtain a solid basis for the design of the flight engine and the development of the liquidfuel combustor, which should be started as a parallel development as soon as possible. 2) To achieve this solid basis, the hydrogen engine was the surest and quickest way when one does not have compressor and turbine test stands. 3) The anticipated stepbystep development approach: First testing the compressorturbine unit with the "norisk" hydrogen combustor and then
f ile
k_ RADIALTURBOJET(He Sl) WITHHYDROGEN (Built in 1936;tested in April 1937) Radius of rotor1 ft Thrust250 lb 10,000 rpm Fig. 11
Von Ohain's hydrogen demonstrator engine.
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using the tested turbomachine for exploring its interaction with the liquidfuel combustor system seemed to be good protection against timeconsuming setbacks. Now came the greatest difficulty for me: How could I convince Heinkel that first building a turbojet with hydrogen gas as fuel would be a far better approach than trying to develop a liquidfuel combustor under an enormous time pressure? According to my contract, of course, I should have worked on the liquidfuel combustor with the (impossible) goal of having this development completed by June 1936. I briefly explained to Heinkel my reasons for the hydrogen engine and emphasized this engine would be a full success in a short time. I was well prepared to prove my point in case Heinkel wanted me to discuss this matter in a conference with his engineers. Surprisingly, Heinkel asked only when the hydrogen demonstrator could run. My shortest time estimate was half a year. Heinkel was not satisfied and wanted a shorter time. I told him that I had just heard that Wilhelm Gundermann and Max Hahn would work with me, and I would like to discuss the engine and its time schedule with them. So, Heinkel had agreed with my reasons to build the hydrogen jet demonstrator first. About a week after my discussion with Heinkel, I joined Gundermann and Hahn in their large offÉce. I showed them the layout of the hydrogen engine. Gundermann told me he had attended my presentation to the group of Heinkel's leading engineers in March 1936. He was surprised that I departed from the liquidfueled turbojet program. I explained my reasons and also told about Heinkel's strong desire to have the hydrogen engine built in less than half a year. After studying my layout, both men came to the conclusion that it would not be possible to build this engine in less than six months, perhaps even longer. Gundermann, Hahn, and I began to work as an excellent team. The engine was completed at the end of February 1937, and the start of our demonstration program was in the first half of March, according to Gundermann's and my recollections. The first run is clearly engraved in my memory: Hahn had just attached the last connections between engine and test stand; it was after midnight, and we asked ourselves if we should make a short run. We decided to do it! The engine had a 2hp electric starting motor. Hahn wanted to throw off the beltconnecting starter motor and hydrogen engine if selfsupporting operation was indicated. Gundermann observed the exhaust side to detect possible hot spotsnone were visible. I was in the test room. The motor brought the engine to somewhat above 2000 rpm. The ignition was on, and I opened the hydrogen valve carefully. The ignition of the engine sounded very similar to the ignition of a home gas heating system. I gave more gas, Hahn waved, the belt was off, and the engine now ran selfsupporting and accelerated very well. The reason for the good acceleration probably was twofold: the relatively low moment of inertia of the rotor and the enormously wide operational range of the hydrogen combustion system. We all experienced a great joy that is difficult to describe. Hahn called Heinkel, and he came to our test stand about 20 minutes later, shortly before 1:00 a.m. We made a second demonstration run. Heinkel was enthused he congratulated us and emphasized that we should now begin to build the liquidfuel engine for flying.
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The next day and until the end of March, Heinkel began to show further demonstration runs to some of his leading engineers and important friends. The next day following our "night show," Heinkel visited us with Walter and Siegfried Guenther (his two top aerodynamic designers) for a demonstration run. They were very impressed and asked me about the equivalent horsepower per square meter. I replied, "A little less than 1000," but hastened to add that the flight engine would have more than 2500 h p / m 2 because of the much greater tip speed and greater relative flow cross sections. During April, we conducted a systematic testing program. After the first run of the hydrogen engine, Heinkel ordered his patent office to apply for patents of the hydrogen engine. Because of earlier patents, the only patentable item was my hydrogen combustion system. I became employed as division chief, reporting directly to Heinkel, and received an independent royalty contract, as I had desired. An enormous amount of pressure was now exerted by Heinkel to build the flight engine. During the last months of 1937, Walter and Siegfried Guenther began with predesign studies of the first jetpropelled aircraft (He178) and specified a static thrust of 1100 lb for the flight engine (He.S3). The aircraft was essentially an experimental aircraft with some provisions for armament. In late 1937, while I was working on different layouts of the flight engine, Max Hahn showed me his idea of arranging the combustor in the large unused space in front of the radialflow compressor. He pointed out that this would greatly reduce the rotor length and total weight. Hahn's suggestion was incorporated into the layout of the flight engine (see Fig. 12). In early 1938, we had a wellfunctioning annular combustor for gasoline. The design of the flight engine was frozen in the summer of 1938 to complete construction and testing by early 1939.
Fig. 12
1937 design of the He.S3 turbojet engine.
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In spring 1939, aircraft and engine were completed, but the engine performance was too low: about 8001b thrust, while a thrust of 10001100 lb was desirable to start the aircraft from Heinkel's relatively short company airfield. We made several improvements, mostly optimizing the easily exchangeable radial cascades of the compressordiffuser and turbine stator. In early August, we had reached 1000 lb of thrust. We made only several onehour test runs with the flight engine. However, upon suggestion of the Air Ministry, we completed a continuous 10hour test run with a rotor that was not used for flight tests. On August 27, 1939, the first flight of the He178 with jet engine He.S3B was made with Erich Warsitz as pilot (Fig. 13). This was the first flight of a turbojet aircraft in the world. It demonstrated not only the feasibility of jet propulsion, but also several characteristics that had been doubted by many opponents of turbojet propulsion: 1) The flying engine had a very favorable ratio of net power output to engine weightabout 2 to 3 times better than the best propeller/piston engines of equal thrust power. 2) The combustion chambers could be made small enough to fit in the engine envelope and could have a wide operational range from start to high altitude and from low to high flight speed.
Fig. 13 The world's first jetpowered aircraft, the Heinkel He178, was powered by the von Ohaindesigned He.S3B turbojet engine. (National Air and Space Museum.)
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The advantages of developing a flight demonstration turbojet in Heinkel's aircraft company were unique. Among the advantages were complete technical freedom, lack of importance attached to financial aspects, no government requirements, and no time delays; the aircraft was, so to speak, waiting for the engine. These great advantages were true only for the initial phases of jet engine development up to the first flight demonstrator. For making a production engine, however, enormous disadvantages included complete lack of experts in fabrication (turbomachinery, etc.), materials, research (turbines), accessory drives, control systems, no machine tools or component test stands, etc. Heinkel was very aware of this situation. His plan was to hire engineers from the aircraft engine field and to purchase an aircraft engine company.
Other Early Turbojet Developments in Germany The following events developed at the same time, which was of great importance for the early phases of the turbojet evolution: 1) Professor Herbert Wagner privately started an aircraft gas turbine development project. 2) The Air Ministry became aware of Heinkel's turbojet project in 1938 and exerted a strong influence on the engine industry to start turbo development projects. 3) Heinkel purchased an aircraft engine corporation and received a contract for development and production of a highperformance turbojet engine. In 1934, Wagner conceived the idea of an axialflow propeller gas turbine while he was a professor of aeronautics in Berlin and formed a corporation to pursue these ideas. (I heard about Wagner's project for the first time in spring of 1939.) By introducing a design parameter that was the ratio of propeller power input to total net power output, he had conceived a gas turbine engine that was a cross between a turbojet and a propeller gas turbine. Wagner first explored what would happen if the propeller power input was 50% of the total power output. This condition was favorable for longrange transport. Then in 1936, he investigated the "limiting case" of zero propeller power input, which constituted a turbojet. This engine was of great interest for highspeed aircraft because of its light weight. The unique feature of Wagner's design was the utilization of 50% reaction turbomachinery (or symmetric blading). A compressor with 50% reaction blading has the greatest pressure ratio and efficiency for a given blade approach Mach number; but the design is difficult because of the inherently strong threedimensional flow phenomena. This problem was solved by one of Wagner's coworkers, Rudolf Friedrich. At the time Wagner was working on the turbojet engine, in about 1936, he became technical director of the Junkers Airframe Corporation in Dessau. The jet engine work was conducted in the Junkers machine factory, which was located in Magdeburg. The head of his turbojet development was Max A. Mueller, his former "first assistant." In late fall, 1938, Wagner had decided to leave the Junkers Corporation, but he wanted to obtain funds from the Air Ministry for the continuation of his turbojet development work. The Air Ministry agreed to Wagner's request under the
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condition that the jet development be continued at the Junkers Aircraft Engine Company in Dessau. This seemed to be acceptable to Herbert Wagner. However, his team of about 12 very outstanding scientists and engineers (among them the team leader, Max A. Mueller, and the highly regarded Dr. R. Friedrich) refused to join the Junkers Aircraft Engine Company under the proposed working conditions. Heinkel made them very attractive work offers that convinced Wagner's former team to join the Heinkel Company. Heinkel added Wagner's axial turbojet to his development efforts (designated as the He.S30). Thus, in early 1939, Heinkel had achieved one goalto attract excellent engineers for his turbojet development. In early 1938, the Air Ministry had become aware of Heinkel's private jet propulsion development. The Engine Development Division of the Air Ministry had a small section for special propulsion systems that did not use propellers and piston engines, but rather used special rockets for shorttime performance boost or takeoff assistance. Head of this section was Hans Mauch. He asked Heinkel to see his turbojet development in early summer 1938, more than one year before the first flight of the He178. After he saw Heinkel's hydrogen turbojet demonstrator in operation and the plans for the flight engine, he was very impressed. He thanked Heinkel for the demonstration and pointed out that turbojet propulsion was, for him, a completely unknown and new concept. He soon became convinced that the turbojet was the key to highspeed flight. He came, however, to the conclusion that Heinkel, as an airframe company, would never be capable of developing a production engine because the company lacked engine test and manufacturing facilities and, most of all, it lacked engineers experienced in engine development and testing techniques. He wanted the Heinkel team to join an aircraft engine company (DaimlerBenz) and serve as a nucleus for turbojet propulsion development. Furthermore, he stated that Ernst Heinkel should receive full reimbursement and recognition for his great pioneering achievements. Heinkel refused. In the summer of 1938, Mauch met with Helmut Schelp, who was in charge of jet propulsion in the Research Division of the Air Ministry. Mauch invited Schelp to join him in the Engine Development Division. Schelp accepted the transfer because he saw far greater opportunities for action than in his Research Division. In contrast to Mauch, Schelp was very well aware of turbojet propulsion and was convinced about its feasibility. He was well versed in axial and radial turbomachinery and with the aerothermodynamic performance calculation methods of turbojet, ramjet, and pulse jet. Like Mauch, he was convinced of the necessity that the aircraft engine companies should work on the development of turbojet engines. However, Schelp did not see a necessity for Heinkel to discontinue his jet engine development. He saw in Heinkel's progress a most helpful contribution for convincing the engine industry to also engage in the development of turbojets, and for proving to the higher echelons of the Air Ministry the necessity of launching a turbojet development program throughout the aircraft engine companies. Schelp worked out the plans and programs for jet propulsion systems, decided on their most suitable missions, and selected associated aircraft types. Schelp's goal was to establish a complete jet propulsion program for the German aircraft engine industry. He also talked with Hans Antz of the Airframe Development Division of the Air Ministry to launch a turbojet fighter aircraft development as
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soon as possible. This became the Me262. To implement the program, Mauch and Schelp decided to visit aircraft engine manufacturersJunkers Motoren (Jumo), DaimlerBenz, BMW Flugmotorenbau, and Bradenburgische Motorenweke (Bramo). Mauch and Schelp offered each company a research contract to determine the best type of jet engine and its most suitable mission. After each study was completed and evaluated, a major engine development contract might be awarded. The industry's response to these proposals has been summed up by Schlaifer in Development of Aircraft Engines, and Fuels: "The reaction of the engine companies to Mauch's proposals was far from enthusiastic, but it was not completely hostile."** Anselm Franz and Hermann Oestrich were clearly in favor of developing a gas turbine engine. Otto Mader, head of engine development at Jumo, made two counter arguments against taking on turbojet propulsion developments. He said, first, that the highest priority of Jumo was to upgrade the performance of its current and future piston engines, and that this effort was already underpowered; and, second, Jumo did not have workers with the necessary expertise in turbomachine engine development! After several meetings between Mader and Schelp, however, Mader accepted the jet engine development contract and put Franz in charge of the turbojet project. At that time, Dr. Anselm Franz was head of the supercharger group. DaimlerBenz completely rejected any work on gas turbine engines at that time. Meanwhile, BMW and Bramo began a merger, and after it was finalized, Hermann Oestrich became the head of the gas turbine project for BMW. These developments show that the aircraft engine industries in Germany did not begin to develop jet engines on their own initiative, but rather on the initiative and leadership of Mauch, and specifically of Helmut Schelp of the technical section of the German Air Ministry. Without their actions, the engine companies in Germany would not have begun development work on turbojet propulsion. The net result of Schelp's planning efforts was that two important turbojet engine developments were undertaken by the German aircraft engine industry, the Junkers Engine Division and BMW. The Jumo 004 (shown in Fig. 14), developed under the leadership of Anselm Franz, was perhaps one of the truly unique achievements in the history of early jet propulsion development leading to mass production, for the following reasons: 1) It employed axialflow turbomachinery and straight throughflow combustors. 2) It overcame the nonavailability of nickel by aircooled hollow turbine blades made out of sheet metal. 3) The manufacturing cost of the engine amounted to about onefifth that of a propeller/piston engine having the equivalent power output. 4) The total time from the start of development to the beginning of largescale production was a little over four years (see Table 1). 5) It incorporated a variablearea nozzle governed by the control system of the engine, and model 004E incorporated afterburning.
**Schlaifer, R., and Heron, S. D., Developmentof Aircraft Engines, and Fuels, PergamonPress, New York, 1970; reprint of 1950 ed.
FOREWORD Starter motor
Airbleed
Compressor
Flamechamber
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Combustionchamber Turbinewithhollowblades
Fig. 14 Drawing of Jumo 004B turbojet engine showing air cooling system [thrust = 2000 lb, airflow = 46.6 lb/s, pressure ratio = 3.14, turbine inlet temperature = 1427°F, fuel consumption = 1.4 (lb/h)/h, engine weight = 1650 Ib, diameter = 30 in., length = 152 in., efficiencies: 78% compressor, 95% combustor, 79.5% turbine].
The preceding points reflect the design philosophy of Dr. A. Franz for the Jumo 004, which was lowest possible development risk, shortest development time, dealing with a complete lack of heatresistant materials, and minimizing manufacturing cost. From this design philosophy, it is understandable that the Jumo 004 engine, while fully meeting the requirements, did not have the highest overall performance compared to some contemporary experimental axialflow engines, such as the He.S30 and others. If it had been possible for the Jumo 004 to employ heat resistant materials, then the engine thrust, the thrust/weight ratio, and the efficiency would have been increased substantially. Also the engine life could have been drastically increased from about 25 h to well over 100 h. However, because the combat life of a German fighter was well below 25 h, the economical optimum could tolerate a short engine life and the avoidance of nickel. Furthermore, to avoid any development risk or time delay, the compressor type chosen for the Jumo 004 was one where essentially all the static pressure increase occurs in the rotor and none in the stator (a freevortex type of compressor having constant axial velocity over the blade span). Although such a compressor type does not have the best performance, at that time it was best understood. The previously described points show that the Jumo 004 represented an outstanding compromise between engine performance, the existent design constraints due to materials shortage, the need for short development time, and earliest possible production. Table 1
Jumo 004 development and production schedule
Start of development First test run First flight in Me262 Preproduction Beginning of production Introduction of hollow blades About 6000 engines delivered
Fall 1939 Oct. 11, 1940 July 18, 1942 1943 Early 1944 Late 1944 May 1945
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The BMW 003 turbojet engine, which was developed under the leadership of Hermann Oestrich, was also a resounding success. Because its thrust was smaller than that of the Jumo 004, it was ideally suited for the He162. After World War II, Oestrich and a group of prominent scientists and engineers from Germany went to France and helped lay the foundation for France's turbojet industry. Now I would like to go back to the end of 1939, when Heinkel began to make plans for buying an engine company. After the first flight of the He178 on August 27, 1939, Heinkel invited high officials of the Air Ministry to see a flight demonstration of the He178. This demonstration took place on November 1, 1939. At that occasion, Heinkel offered the development of a jet fighter, the He280, which had two outboard engines under the wing. Heinkel received a contract for this aircraft in early 1940. In addition, I believe Udet and Heinkel had made an agreement that Heinkel would get official permission to buy the Hirth Engine Company, if the first flight of the He280 could be demonstrated by April 1941. The He280 had severe restrictions with respect to distance of the engine nacelle from the ground. It actually was designed for the axial engine He.S30 (the Wagner turbojet engine). It appeared, however, unlikely that the He.S30 would be ready in time. On the other hand, it was impossible to use an engine of the He.S3B type, which had powered the He178, because the diameter of this engine type would have been far too large. Under these conditions, I could see only one possible solution for succeeding in time, and this solution had extreme high risk. I employed a radial rotor similar to that of the He.S3B and combined it with an axial (adjustable) vane diffuser and a straight throughflow annular combustor. The company designation of this engine was He.S8A. We had only about 14 months for this development, but we were luckyit worked surprisingly well, and Heinkel could demonstrate the first flight of the He280 on April 2, 1941. The government pilot was Engineer Bade. Earlier flights in late March were done by Heinkel's test pilot, Fritz Schaefer. A few days after this demonstration, Heinkel obtained permission to buy the Hirth Motoren Company in Stuttgart, which was known for its excellent small aircraft engines. This company had outstanding engineers, scientists, machinists, precision machine tools, and test stands. Heinkel relocated the development of the He.S30 to his new HeinkelHirth engine company to make use of the excellent test and manufacturing facilities. In the summer of 1942, the He.S30 was ready for testing. It performed outstandingly well. The continuous thrust was about 1650 lb. From a technical standpoint, this engine had by far the best ratio of thrust to weight in comparison to all other contemporary engines. The superiority of the He.S30 was, in large part, the result of its advanced 50% reaction degree axialflow compressor, designed by Dr. R. Friedrich. However, the success of the He.S30 came too late. The He280 had been thoroughly tested. While it was clearly superior to the best contemporary propeller/piston fighter aircraft, the He280 had considerably lower flight performance than the Me262 with respect to speed, altitude, and range. Also, the armament of the He280 was not as strong as that of the Me262. For these reasons, the He.S8 and the He.S30 were canceled in the fall of 1942; the He280 was officially
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Fig. 15 Messerschmitt Me262 jet fighter. (National Air and Space Museum.)
canceled in early 1943. The Me262 (Fig. 15) went on to become the first operational jet fighter powered by two Jumo 004B turbojet engines. The Air Ministry did give full recognition to the excellence of the 50% reaction degree compressor type as most suitable for future turbojet developments. Thus, in the fall of 1942, Heinkel had lost his initial leadership in jet aircraft and turbojet engines. Ironically, this happened when he had just reached his goal of owning an aircraft engine company with an outstanding team of scientists and engineers. It was the combined team of the original Hirth team, the Heinkel team, and Wagner's team. These conditions made Heinkel fully competitive with the existent aircraft engine industry. The Air Ministry had recognized the excellence of Heinkel's new team and facilities. Helmut Schelp, who in the meantime had become the successor of Hans Mauch, was in favor of the HeinkelHirth Company's receiving a new turbojet engine development contract. He clearly foresaw the need for a strong engine for the advanced Me262, the Arado234, the Junkers287, and others. This new engine was supposed to have a thrust of nearly 3000 lb with a growth potential to 4000 lb as well as a high pressure ratio for improving the fuel economy. We began working on the He.S011 engine in the fall of 1942. The He.S011 was an axialflow design with 50% reaction degree compressor blading, similar to that designed by Dr. Friedrich, and a twostage, axialflow, aircooled turbine. Note that Helmut Schelp not only had established the performance specifications, but also had contributed to the overall design with excellent technical input and suggestions pertaining to the advanced diagonal inducer stage, the twostage aircooled turbine, and the variable exhaust nozzle system. I was in charge of the He.S011 development, while the local director of the HeinkelHirth plant, Curt Schif, was in charge of He.S011 production. The top engineer of the Hirth Corporation was Dr. Max Bentele. He was well known for his outstanding knowledge in dealing with blade vibration problems. Upon special request of the Air Ministry, he had solved the serious turbine blade vibration problems of the Jumo 004 in the summer of 1943. Dr. Bentele was responsible for the component development of the He.S011. After considerable initial difficulties, he achieved excellent performance
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characteristics of compressor and turbine that made it possible, by the end of 1944, for the performance requirements of the 011 to be met or surpassed. He also contributed to the preparation for production of the 011, which was planned to start in June 1945. The end of World War II, of course, terminated the production plan before it had started. Only a few He.S011 engines are in existence today, and they are exhibited in several museums in the United States and Great Britain. In other countries, such as Russia and Japan, interesting developments in the field of propeller gas turbines and turbojets had also been undertaken. It is, however, not sufficiently known to what extent these developments were interrelated with, or influenced by, the previously described jet developments and what their development schedules had been. In summary, at the end of the 1930s and in the first half of the 1940s, turbojet propulsion had come into existence in Europe and the United States. It had been demonstrated that the turbojet is, for highspeed flight, uniquely superior to the propeller/piston engines because of the following two major characteristics: 1) The ratio of power output to weight of the early turbojets was at least 2 or 3 times greater than that of the best propeller/piston engines. This is one necessary condition of propulsion systems for highspeed flight and good maneuverability. 2) Because, in the turbojet, the propeller is replaced by ducted turbomachinery, the turbojet is inherently capable as a propulsion system of high subsonic and supersonic flight speeds. In the following, it will be discussed how the turbojet engine progressed to the performance capabilities of today.
Evolution of Airbreathing Turbopropulsion Systems to the Technology Level of Today The early turbojets were used as propulsion systems for highspeed fighter and reconnaissance aircraft. For these applications, the early turbojets (because of their superior power/weight ratios) were far more suitable propulsion systems than the traditional propeller/piston engines. However, the early turbojets were not suitable for those application areas where greatest fuel economy, highest reliability, and a very long endurance and service life were required. For making airbreathing turbojet propulsion systems applicable for all types of aircraft, ranging from helicopters to highspeed, longrange transports, the following development goals were, and still are, being pursued: 1) Higher overall efficiency (i.e., the product of thermodynamic and propulsion efficiencies), 2) Largerpoweroutput engines, 3) Larger ratios of power output to engine weight, volume, and frontal area, 4) Greater service life, endurance, and reliability, 5) Strong reduction of adverse environmental exhaust gases, 6) Reduced noise.
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To achieve these goals, parallel research and development efforts were undertaken in areas such as 1) Fundamental research in combustion processes, development, and technology efforts for increasing specific mass flow through combustors and reducing the total pressure drop; and for achieving nearly 100% combustion efficiency with a more uniform temperature profile at combustor exit. 2) Minimizing the excitation of vibrations (including aeroelastic effects) and associated fatigue phenomena. 3) Continuous improvement of the structural design and structural materials, such as composite materials, heat and oxidationresistant alloys and ceramics. 4) Increasing the turbine temperature capability by improving air cooling effectiveness. Also increasing the polytropic turbine efficiency. 5) Improvement of the compressor with respect to greater specific mass flow, greater stage pressure ratio, greater overall pressure ratio, and greater polytropic efficiency. 6) Advanced controllablethrust nozzles and their interactions with the aircraft. 7) Advanced control systems to improve operation of existing and new engines. All of these research and development areas were, and still are, of great importance for the progress in turbopropulsion systems; however, the compressor can perhaps be singled out as the key component because its advancement was a major determining factor of the rate of progress in turbo engine development. This will become apparent from the following brief description of the evolution of the turbopropulsion systems.
Development of HighPressureRatio Turbojets The first step to improve the early turbojets was to increase their overall efficiency. To do this, it was necessary to increase the thermodynamic cycle efficiency by increasing the compressor pressure ratio. The trend of compressor pressure ratio over the calendar years is shown in Table 2. In the early 1940s, it was well understood that a highpressureratio (above 6:1), singlespool, fixedgeometry compressor can operate with good efficiency only at the design point, or very close to it. The reason is that at the design Table 2
Trend in compressor pressure ratios
Calendar years
Compressor pressure ratio
Late 1930 to mid1940 Second half of 1940s Early 1950 Middle to late 1960s End of century (2000)
3:1 to about 5:1 5:1 and 6:1 About 10:1 20:1 to about 25:1 30:1 to about 40:1
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point, all compressor stages are matched on the basis of the compressibility effects. Consequently, under offdesign conditions where the compressibility effects are changed, the stages of a highpressureratio compressor are severely mismatched, resulting in a very low offdesign efficiency. For example, at an operational compressor rpm (which would be substantially below the design rpm), the compressibility effects are small. Under such offdesign conditions, the front stages tend to operate under stalled conditions, while the last stages tend to work under turbining conditions. Such compressor characteristics are unacceptable for the following reasons: 1) There are enormous difficulties in starting such an engine. 2) Very poor overall efficiency (or poor fuel economy) exists under partload operation. 3) Very low overall efficiency exists at high supersonic flight speed (because the corrected rpm is very low as a result of the high stagnation temperature of the air at the compressor inlet). By the end of the 1940s and early 1950s, excellent approaches emerged for the elimination of these shortcomings of simple highpressureratio compressors. Pratt & Whitney, under the leadership of Perry Pratt, designed a highpressureratio jet engine (the J57) with a dualrotor configuration. (In later years, triplerotor configurations were also employed.) The ratio of rpm values of the low and highpressure spools varies with the overall pressure ratio and, in this way, alleviates the mismatching effects caused by the changes in compressibility. At approximately the same time, Gerhard Neumann of General Electric conceived a highpressureratio, singlespool compressor having automatically controlled variable stator blades. This compressor configuration was also capable of producing very high pressure ratios because it alleviated the mismatching phenomena under offdesign operation by stator blade adjustment. In addition, the controlled variablestatorcompressor offers the possibility of a quick reaction against compressor stall. The variablestator concept became the basis for a new, highly successful turbojet engine, the J79, which was selected as the powerplant for many important supersonic Air Force and Navy aircraft. A third possibility to minimize mismatching phenomena was variable frontstage bleeding, which was employed in several engines. Before I continue with the evolution of the turbo engines toward higher overall efficiencies, a brief discussion about the research and development efforts on compressor bladings and individual stages is in order. Such efforts existed to a small degree even before the advent of the turbojet (in Switzerland, Germany, England, and other countries). However, in the mid1940s, those research efforts began to greatly intensify after the turbojet appeared. It was recognized that basic research and technology efforts were needed to continuously increase 1) stage pressure ratio and efficiency and 2) mass throughflow capability. Very significant contributions had been made by universities, research institutes, and government laboratories (NACA, later NASA; Aero Propulsion Laboratory, and others), and industry laboratories. Many outstanding research results were obtained from universities in areas of rotating stall, threedimensional and nonsteady flow phenomena and transonic flow effects, novel flow visualization techniques for diagnostics (which identified flow regions having improvement possibilities), understanding of noise origination, and many others.
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Two examples where government laboratories had achieved very crucial advancements that were later adapted by industry were the following: In the early 1950s, the NACA Lewis Research Center, under the leadership of Abe Silverstein, advanced compressor aerodynamics to transonic and supersonic flow, critical contributions for increasing the compressorstage pressure ratio. He initiated a large transonic and supersonic compressor research program. Many indepth studies furnished information for utilizing this new compressor concept. Later, the Air Force Propulsion Laboratory, under the leadership of Arthur Wennerstrom, conducted advanced compressor research and introduced a very important supersonic compressor concept, which was particularly applicable for front stages because it solved the most difficult combined requirements of high mass flow ratio, high pressure ratio, high efficiency, and broad characteristics. The continuous flow of research and technological contributions extended over the last four decades and is still going on. The total cost may have been several hundred million dollars. Some of the results can be summarized as follows: The polytropic compressor efficiency, which was slightly below 80% in 1943, is now about 92%; the average stage pressure ratio, which was about 1.15:1 in 1943, is now about 1.4:1 and greater; the corrected mass flow rate per unit area capability grew over the same time period by more than 50%. The improvement in compressor efficiency had an enormous impact on engine performance, specifically on the overall engine efficiency. The substantially increased stage pressure ratio and the increased corrected mass flow rate per unit area capability resulted in a substantial reduction of engine length, frontal area, and weightperpower output. Let us now return to the 1950s. As previously stated, the new turbojets employing a dualrotor configuration or controlled variablestator blades were capable of substantially higher pressure ratios than the fixedgeometry, singlespool engines of the 1940s. Engine cycle analysis showed that the high powerperunit mass flow rate needed for the advanced engines required higher turbine inlet temperatures. This requirement led to continuous major research efforts to achieve highefficiency combustion with low pollution, to increase the effectiveness of turbine bladecooling methods, and to improve the temperature capabilities of materials. Thus the turbojets of the 1950s had made substantial progress in thermodynamic efficiency and propulsive thrust per pound of inlet air per second. The latter characteristic means that the velocity of the exhaust gas jet had increased.
Development of HighBypassRatio Turbofans For supersonic flight speeds, the overall efficiency of these turbojets was outstanding. However, for high subsonic and transonic flight speeds (around 500600 mph), the velocity of the exhaust gas jet was too high to obtain a good propulsive efficiency. Under these conditions, the bypass engine (also called turbofan or fanjet) became a very attractive approach for improving the propulsive efficiency. The first fanjets had a relatively small bypass ratio of about 2:1. (Bypass ratio is the ratio of mass flow bypassing the turbine to mass flow passing through the turbine.) In the early 1960s, the U.S. Air Force had established requirements for military transports capable of an extremely long
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Fig. 16 The TF39 highbypassratio turbofan engine used on the Lockheed C5 transport. (Courtesy of General Electric Aircraft Engines.)
range at high subsonic cruise speeds. Such requirements could be met only by employing propulsion systems of highest possible thermodynamic and propulsive efficiencies, which led to the following engine characteristics: 1) Very high compressor pressure ratios between 20:1 and 30:1, which could be achieved by combining the concepts of variable stator and dual rotor, 2) Very high turbine inlet temperatures, 3) Very high bypass ratios, around 8:1. The first engine of this type was the TF39 (Fig. 16), a military transport engine developed by General Electric under the leadership of Gerhard Neumann. Four of the TF39 turbofan engines powered the Lockheed C5A. The Air Force Propulsion Laboratory, under the leadership of Cliff Simpson, had played a key role in the establishment of these requirements. Simpson also succeeded in convincing the highest Air Force and Department of Defense echelons of the significance of this new type of airbreathing propulsion system, which he considered a technological breakthrough. (At that time, it was difficult to generate interest in advanced airbreathing propulsion system concepts, since the nation's attention was focused on rocket propulsion for space exploration.) The advantages of the highbypassratio turbofan engines can be summarized as follows: 1) High overall efficiency, resulting in a long flight range, 2) Strong increase in propulsive thrust at low flight speeds, which is important for takeoff, climbing, and efficient partload operation, 3) Lower jet velocity, which leads to great noise reduction, and 4) Low fuel consumption, which reduces chemical emissions. These characteristics had been demonstrated in the late 1960s and early 1970s. They also became of great interest for commercial aircraft and led to the development of the widebody passenger aircraft.
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During the 1980s, the continuous flow of advanced technology of turbo engine components had brought a high degree of maturity to the various aircraft gas turbine engine types, such as the following: 1) the previously discussed highbypassratio engines, 2) the lowbypassratio engines with afterburning for supersonic flight, and 3) the pure turbojet for high supersonic flight Mach numbers of approximately 3. The small and medium aircraft shaftpower gas turbines also benefited enormously from the continuous improvement of turbomachinery technology. These small gas turbines are used in helicopters and subsonic propeller aircraft. The high power/weight ratio and the high thermodynamic efficiency of the advanced shaftpower gas turbine engines played a key role in the advancement of helicopters.
Future Potential of Airbreathing Jet Propulsion Systems In the future, major advancement of airbreathing jet propulsion systems can be expected from 1) Evolutionary improvements of the established largebypassratio turbofan engines for transonic flight speeds and the lowbypassratio turbofan, or pure turbojet engines, for supersonic flight speeds, 2) Improvements and new approaches to engineairplane integration, and 3) New approaches to airbreathing propulsion systems for high supersonic and hypersonic flight speeds. The evolutionary improvements of established engine types will result in greater fuel economy and better performance characteristics. By the end of the century, one can expect polytropic efficiencies of turbine and compressor of nearly 95%. Furthermore, one will see considerably increased singlestage pressure ratios; significantly higher turbine inlet temperatures resulting from better heat and oxidationresistant materials, and more effective bladecooling methods; and much lighter structural designs and materials (composite materials). This technological progress may result in an overall engine efficiency increase of about 20% and in a weight reduction for given horsepower output by probably a factor of 2 and higher. For the evolution of highbypassratio engines at cruise speeds between 500 and 600 mph, the following trend is important; the greater the turbine inlet temperature and the higher the polytropic efficiencies of the compressor and the turbine, the higher the optimum pressure ratio of the gas turbine engine and the bypass ratio of the fan. In the future, this trend will lead to larger bypass ratios; hence, the fan shroud will become relatively large in diameter and will contribute substantially to the weight and external drag of the propulsion system. Several solutions are conceivable to alleviate this problem, but at this time it is not possible to predict the most promising approach. One way is to eliminate the fan shroud by using an unshrouded fan (having a multiplicity of sweptback fan blades, see Fig. 17), also called a propfan. This configuration is currently in an experimental state and may become very important in the future for improving fuel economy. Another possibility
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Fig. 17 PWAllison 578DX propulsion system. (Courtesy of Allison Gas Turbine Division.)
may lie in the development of a transonic airframe configuration in which the large fan shrouds have a dual function: They contribute, in part, to the stability of the aircraft (horizontal and vertical stabilizer surfaces) while serving at the same time as a shroud for the fan. Finally, it may be conceivable that, in the future, a practical airframe and wing configuration can be developed that will be capable of extending a high lift/drag ratio to a flight speed regime that is very close to the speed of sound (a kind of wellknown "supercritical" airframe configuration proposed by Whitcomb of NASA). If such an airframe configuration could be developed, the bypass ratio at these relatively high flight speeds would be substantially lower than that for flight speeds around 500 mph and, therefore, a shrouded fan would be most applicable. Perhaps a most interesting question is: Can one expect a future supersonic passenger transport that is economically feasible in view of the progress of future transonic passenger transports? The general trend of the airplane lift/drag ratio is decreasing with increasing flight Mach number, while the overall efficiency of aeropropulsion systems increases with increasing flight Mach number. Currently, the lift/drag ratio of the Boeing 747 compared to that of the supersonic Concorde is about 3:1. In the future, the corresponding ratio may decrease to about 2:1, and the structural weight of the supersonic aircraft will greatly be improved. Such improvements may be the result of advancements in the aerodynamic shape, the structure, the structural airframe materials, and the best use of artificial stability. The overall efficiency of the supersonic flight engine can also be improved, including its capability of cruising subsonically with only one
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engine. This is important in case one of the two engines fails. Under this condition, the mission must be completed at subsonic speed without requiring an additional fuel reserve. It appears that advanced supersoniccruise aircraft systems have the potential to achieve these conditions and, thus, become economically acceptable in the future.
New Approaches to EngineAirplane Integration In general, the investigations of engineairplane integration have the goal to avoid or minimize losses due to adverse interface phenomena between the engine inlet stream or exhaust jet and the air vehicle. However, many favorable effects can be achieved by properly integrating functions of the propulsion system with functions of the airplane. Historically, Prof. Ackeret (University of Zurich, Switzerland) had first suggested, in 1921, reaccelerating the boundarylayer air near the trailing edge of the wing. He showed that this method of producing the propulsive thrust can result in substantial gains in overall efficiency of the propulsion system. In the 1970s, various investigations had shown that similar results can be obtained by using momentum exchange of the highpressure bypass air with the boundarylayer air. Other uses of bypass air are to energize the boundary layer at proper locations of the wing to prevent boundarylayer separation and to increase the circulation around the wing (often called supercirculation). This method may become important for advanced shorttakeoffandlanding (STOL) applications. For future V/STOL applications, new methods of thrust vectoring and thrust augmentation may have very attractive possibilities. Also, bypass air may be used for boundarylayer suction and ejection in advanced laminar systems.
Airbreathing Propulsion Systems for High Supersonic and Hypersonic Speeds In airbreathing propulsion systems, the combined compression by ram and turbo compressor is of great benefit to the thermodynamic propulsion process up to flight Mach numbers approaching 3. When the flight Mach number increases further, the benefits of the turbo compressor begin to decrease and the engine begins to operate essentially as a ramjet. When the flight Mach number exceeds about 3.5, any additional compression by a turbo compressor would be a disadvantage. Thus, if the engine operates best as a pure subsonic combustion ramjet, it fits in a flight Mach number regime from about 3.5 to 5. Beyond flight Mach numbers of about 6, the pressure and temperature ratios would be unfavorably high if the engine continued to operate as a subsonic combustion ramjet. The reasons are as follows: 1) High degree of dissociation of the combustor exhaust flow, reducing the energy available for exhaust velocity, 2) Pressures far too high for Brayton cycle operations or for the structure to withstand.
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For these reasons, the cycle will be changed from a subsonic to a supersonic combustion ramjet, and hydrogen will be used as fuel because hydrogen has the greatest 1) combustion heat and fuelair concentration range, 2) diffusion speed and reaction speed, and 3) heatsink capabilities. This supersonic combustion ramjet cycle is characterized by a reduction of the undisturbed hypersonic flight Mach number to a somewhat lower hypersonic Mach number with an increase in entropy, which should be as low as possible. The deceleration process must be chosen in such a manner that the increases in static pressure and entropy correspond to a highperformance Brayton cycle. The internal thrust generated by the exhaust gas must be larger than the external drag forces acting on the vehicle. To minimize the parasitic drag of the ramjet vehicle systems, various external ramjet vehicle configurations have been suggested. However, theoretical and experimental investigations will be necessary to explore their associated aerothermochemical problems. For the experimental research, one may consider investigations using fleeflight models or hypersonic wind tunnels with true temperature simulation. Historically, it may be of interest that many suggestions and investigations had been made as to how to achieve clean hypersonic airflows with true stagnation temperatures. I remember much work and many discussions with Dr. R. Mills, E. Johnson, Dr. Frank Wattendorf, and Dr. Toni Ferri about "air accelerator" concepts, which aimed to avoid flow stagnation and the generation of ultrahigh static temperatures and chemical dissociation. Since the Wright brothers, enormous achievements have been made in both lowspeed and highspeed airbreathing propulsion systems. The coming of the jet age opened up the new frontiers of transonic and supersonic flight (see Fig. 3). While substantial accomplishments have been made during the past several decades in the field of high supersonic and hypersonic flight (see Ref. 2), it appears that even greater challenges lie ahead. For me, being a part of the growth in airbreathing propulsion over the past 60 years has been both an exciting adventure and a privilege. This foreword has given you a view of its history and future challenges. The following book presents an excellent foundation in airbreathing propulsion and can prepare you for these challenges. My wish is that you will have as much fun in propulsion as I have.
Hans von Ohain German Inventor of the Jet Engine December 14, 1911March 13, 1998
Preface
This undergraduate text provides an introduction to the fundamentals of gas turbine engines and jet propulsion. These basic elements determine the behavior, design, and operation of the jet engines and chemical rocket motors used for propulsion. The text contains sufficient material for two sequential courses in propulsion: an introductory course in jet propulsion and a gas turbine engine components course. It is based on one and twocourse sequences taught at four universities over the past 25 years. The author has also used this text for a course on turbomachinery. The outstanding historical foreword by Hans yon Ohain (the German inventor of the jet engine) gives a unique perspective on the first 50 years of jet propulsion. His account of past development work is highlighted by his early experiences. He concludes with predictions of future developments. The text gives examples of existing designs and typical values of design parameters. Many example problems are included in this text to help the student see the application of a concept after it is introduced. Problems are included at the end of each chapter that emphasize those particular principles. Two extensive design problems for the preliminary selection and design of a gas turbine engine cycle are included. Several turbomachinery design problems are also included. The text material is divided into five parts: 1) Introduction to aircraft and rocket propulsion (Chapter 1), 2) Review of fundamentals (Chapter 2), 3) Rocket propulsion (Chapter 3), 4) Analysis and performance of airbreathing propulsion systems (Chapters 48), 5) Analysis and design of gas turbine engine components (Chapters 9 and 10). Chapter 1 introduces the types of airbreathing and rocket propulsion systems and their basic performance parameters. Also included is an introduction to aircraft and rocket systems that reveals the influence that propulsion system performance has on the overall system. This material facilitates incorporation of a basic propulsion design problem into a course, such as new engines for an existing aircraft. The fundamental laws of mass conservation, momentum, and thermodynamics for a control volume (open system), the properties of perfect gases, and compressible flow properties are reviewed in Chapter 2. New material on
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analysis of onedimensional gas dynamics [enthalpykinetic energy (HK) diagram] is presented in Chapter 2 to help student understanding of engine cycles from scramjets to afterburning turbofans. Fundamentals of thermochemistry are presented for later use in analysis of chemical rockets in Chapter 3. Chapter 3 presents the fundamentals of rocket propulsion. Emphasis in this chapter is on chemical propulsion (liquid and solid) with coverage of rocket motor performance. The chapter also covers requirements and capabilities of rocket propulsion. The analysis of gas turbine engines begins in Chapter 4 with the definitions of installed thrust, uninstalled thrust, and installation losses. This chapter also reviews the ideal Brayton cycle, which limits gas turbine engine performance. Two types of analysis are developed and applied to gas turbine engines in Chapters 58: parametric cycle analysis (thermodynamic design point) and performance analysis. The text uses the cycle analysis methods introduced by Frank E. Marble of the California Institute of Technology and further developed by Gordon C. Oates (deceased) of the University of Washington and Jack L. Kerrebrock of the Massachusetts Institute of Technology. The steps of parametric cycle analysis are identified in Chapter 5 and then used to model engine cycles from the simple ramjet to the complex, mixedflow, afterburning turbofan engine. Families of engine designs are analyzed in the parametric analysis of Chapters 5 and 7 for ideal engines and engines with losses, respectively. Chapter 6 develops the overall relationships for engine components with losses. The performance analysis of Chapter 8 models the actual behavior of an engine and shows why its performance changes with flight conditions and throttle settings. The results of the engine performance analysis can be used to establish component performance requirements. To keep the size of the new edition down, the analysis of some engine cycles has been moved to the electronic Supporting Materials available with this book. Chapter 9 covers both axialflow and centrifugalflow turbomachinery. Included are basic theory and meanline design of axialflow compressors and turbines, quick design tools (e.g., repeatingrow, repeatingstage design of axialflow compressors), example multistage compressor designs, flow path and blade shapes, turbomachinery stresses, and turbine cooling. Example output from the COMPR and TURBN programs is included in several example problems. Inlets, exhaust nozzles, and combustion systems are modeled and analyzed in Chapter 10. The special operation and performance characteristics of supersonic inlets are examined and an example of an external compression inlet is designed. The principles of physics that control the operation and design of main burners and afterburners are also covered. The appendices contain tables with properties of standard atmosphere and properties of combustion products for air and ( C H 2 ) n. There is also material on turbomachinery stresses as well as useful data on existing gas turbine engines and liquidpropellant rocket engines. Eight computer programs are provided for use with this textbook: 1) AFPROP, properties of combustion products for air and (CH2),, 2) ATMOS, properties of the atmosphere 3) COMPR, axialflow compressor meanline design analysis 4) EQL, chemical equilibrium analysis for reactive mixtures of perfect gases
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5) GASTAB, gas dynamic tables 6) PARA, parametric engine cycle analysis of gas turbine engines 7) PERF, engine performance analysis of gas turbine engines 8) TURBN, axialflow turbine meanline design analysis The AFPROP program can be used to help solve problems of perfect gases with variable specific heats. The ATMOS program calculates the properties of the atmosphere for standard, hot, cold, and tropical days. The EQL program calculates equilibrium properties and process end states for reactive mixtures of perfect gases, for different problems involving hydrocarbon fuels. GASTAB is equivalent to traditional compressible flow appendices for the simple flows of calorically perfect gases. It includes isentropic flow; adiabatic, constant area frictional flow (Fanno flow); frictionless, constant area heating and cooling (Rayleigh flow); normal shock waves; oblique shock waves; multiple oblique shock waves; and PrandtlMeyer flow. The PARA and PERF programs support the material in Chapters 58. PARA is very useful in determining variations in engine performance with design parameters and in limiting the useful range of design values. PERF can predict the variation of an engine's performance with flight condition and throttle (Chapter 8). Both are very useful in evaluating alternative engine designs and can be used in design problems that require selection of an engine for an existing airframe and specified mission. The COMPR and TURBN programs permit preliminary design of axialflow turbomachinery based on meanline design. As already suggested, the author has found the following two courses and material coverage very useful for planning.
Introductory Course Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter
1, all 2, as needed 3, all 4, all 5, Sections 5.159 6, Sections 6.16.9 7, Sections 7.17.4 8, Sections 8.18.5
Gas Turbine Engine Components Course Chapter 2, as needed Chapter 8, Sections 8.18.3 Chapter 9 and Appendix E, all Chapter 10, all The material in Chapters 2 and 9 along with Appendix E of this text have also been used to teach the major portion of an undergraduate turbomachinery course.
Jack D. Mattingly April 2006
Acknowledgments I am deeply indebted to Professor Gordon C. Oates (deceased) and Dr. William H. Heiser. Professor Oates taught me the basics of engine cycle analysis during my doctoral studies at the University of Washington and later until his untimely death in 1986. While Dr. Heiser was a Distinguished Visiting Professor at the U.S. Air Force Academy from 1983 to 1985, we developed an outstanding engine design course and wrote the first edition of the textbook Aircraft Engine Design. We have continued to work and teach together over the past 23 years, and he has been my friend and mentor. Special thanks to Brigadier General Daniel H. Daley (USAF, retired, and former Head, Department of Aeronautics, U.S. Air Force Academy), who sponsored my studies under Professor Oates and guided me during the compilation of the teaching notes "Elements of Propulsion." These notes were used for over a decade in the Academy's propulsion courses before the publication of Elements of Gas Turbine Propulsion in 1996 and are the basis for this textbook. Most recently General Daley helped me develop Chapters 2 and 3 and the general content of this new textbookI couldn't have done this book without his help. I also thank all of the students who have been in my propulsion courses and my coworkers at the Department of Aeronautics, U.S. Air Force Academy, Colorado; the Air Force Institute of Technology (AFIT), WrightPatterson Air Force Base, Ohio; the Air Force Aero Propulsion and Power Laboratory, WrightPatterson Air Force Base, Ohio; Seattle University; and University of Washington. Many of these students and coworkers provided insight and guidance in developing this material. I acknowledge the hard work and determination of Scott Henderson, who worked all of the problems and design problems and checked my solutions. He made significant improvements to the quality and clarity of these problems. I would like to thank the following reviewers and users for their many helpful comments and suggestions: John Anderson, University of Maryland; Saeed Farokhi, University of Kansas; Afshin Ghajar, Oklahoma State University; Frank Redd, Utah State University; Lt. Col. Brenda Haven, WrightPatterson AFB, Ohio; Aaron Byerley, U.S. Air Force Academy, Colorado; and David T. Pratt, University of Washington.
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Table of Contents
Foreword
to the Second Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Foreword to the First Edition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1. 1.1 1.2 1.3 1.4 1.5 1.6
Chapter 2. 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ix xv li Iv lvii
1
Propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Units and Dimensions .............................. Operational E n v e l o p e s a n d Standard A t m o s p h e r e . . . . . . . . . . . . . Airbreathing E n g i n e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Aircraft Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Rocket Engines .................................. Problems ......................................
1 2 4 5 29 49 56
Review of Fundamentals ...................... Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equations o f State a n d C o n s e r v a t i o n o f M a s s . . . . . . . . . . . . . . . Steady Flow E n e r g y E q u a t i o n . . . . . . . . . . . . . . . . . . . . . . . . . Steady Flow Entropy E q u a t i o n . . . . . . . . . . . . . . . . . . . . . . . . . Steady Flow M o m e n t u m Equation . . . . . . . . . . . . . . . . . . . . . . Perfect G a s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C o m p r e s s i b l e Flow Properties . . . . . . . . . . . . . . . . . . . . . . . . . O n e  D i m e n s i o n a l G a s D y n a m i c s   F i n i t e Control Volume A n a l y s i s a n d the H  K D i a g r a m . . . . . . . . . . . . . . . . . . . . . . . N o z z l e D e s i g n a n d N o z z l e Operating Characteristics . . . . . . . . . . O n e  D i m e n s i o n a l G a s D y n a m i c s   D i f f e r e n t i a l Control Volume Analysis ................................. C h e m i c a l Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems ......................................
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65 65 68 75 76 82 93 107 121 134 139 147
xii Chapter 3. 3.1 3.2 3.3 3.4 3.5
Chapter 4. 4.1 4.2 4.3 4.4 4.5 4.6
Chapter 5. 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12
Chapter 6. 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10
Rocket Propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . R o c k e t Propulsion R e q u i r e m e n t s and Capabilities . . . . . . . . . . . . R o c k e t Propulsion E n g i n e s . . . . . . . . . . . . . . . . . . . . . . . . . . Types o f Rocket N o z z l e s . . . . . . . . . . . . . . . . . . . . . . . . . . . Parameters for C h e m i c a l R o c k e t s . . . . . . . . . . . . . . . . . . . . . . Problems ......................................
161 161 166 176 189 194 228
Aircraft Gas Turbine Engine . . . . . . . . . . . . . . . . . . .
233
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T h r u s t Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N o t e o n Propulsive Efficiency . . . . . . . . . . . . . . . . . . . . . . . . G a s Turbine E n g i n e C o m p o n e n t s . . . . . . . . . . . . . . . . . . . . . . B r a y t o n Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Aircraft E n g i n e D e s i g n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems ......................................
233 233 243 244 252 257 258
Parametric Cycle Analysis of Ideal Engines . . . . . . . . .
261
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D e s i g n Inputs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Steps o f Engine Parametric Cycle A n a l y s i s . . . . . . . . . . . . . . . . A s s u m p t i o n s o f Ideal Cycle Analysis . . . . . . . . . . . . . . . . . . . . Ideal R a m j e t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ideal Turbojet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ideal Turbojet with Afterburner . . . . . . . . . . . . . . . . . . . . . . . Ideal Turbofan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ideal Turbofan with O p t i m u m B y p a s s Ratio . . . . . . . . . . . . . . . Ideal Turbofan with O p t i m u m Fan Pressure Ratio . . . . . . . . . . . . Ideal Pulse Detonation E n g i n e . . . . . . . . . . . . . . . . . . . . . . . . Problems ......................................
261 262 264 264 266 266 278 291 302 325 332 341 344
Component Performance
355
.....................
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Variation in G a s Properties . . . . . . . . . . . . . . . . . . . . . . . . . . C o m p o n e n t Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . Inlet a n d Diffuser Pressure Recovery . . . . . . . . . . . . . . . . . . . . C o m p r e s s o r and Turbine Efficiencies . . . . . . . . . . . . . . . . . . . . B u r n e r Efficiency a n d Pressure L o s s . . . . . . . . . . . . . . . . . . . . E x h a u s t Nozzle L o s s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M e c h a n i c a l Efficiency o f Power Shaft . . . . . . . . . . . . . . . . . . . S u m m a r y o f C o m p o n e n t Figures o f Merit ( C o n s t a n t Cp Values) . . . . . . . . . . . . . . . . . . . . . . . . . . . . C o m p o n e n t Performance with Variable cp . . . . . . . . . . . . . . . . . Problems ......................................
355 355 357 358 360 370 371 371 371 373 378
xiii C h a p t e r 7. 7.1 7.2 7.3 7.4
Parametric Cycle A n a l y s i s o f Real E n g i n e s
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Turbojet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Turbojet with Afterburner . . . . . . . . . . . . . . . . . . . . . . . . . . . TurbofanSeparate Exhaust Streams . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.........
381 381 399 404 427
C h a p t e r 8. 8.1 8.2 8.3 8.4 8.5
E n g i n e Performance A n a l y s i s
437
..................
381
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Gas Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Turbojet Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Turbojet Engine with Afterbuming . . . . . . . . . . . . . . . . . . . . . Turbofan EngineSeparate Exhausts and Convergent Nozzles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
437 447 464 486
C h a p t e r 9. 9.1 9.2 9.3 9.4 9.5 9.6
Turbomachinery
537
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Euler's Turbomachinery Equations . . . . . . . . . . . . . . . . . . . . . AxialFlow Compressor Analysis . . . . . . . . . . . . . . . . . . . . . . CentrifugalFlow Compressor Analysis . . . . . . . . . . . . . . . . . . AxialFlow Turbine Analysis . . . . . . . . . . . . . . . . . . . . . . . . . CentrifugalFlow Turbine Analysis . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
537 537 539 600 607 668 674
C h a p t e r 10. 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8
I n l e t s , N o z z l e s , and C o m b u s t i o n S y s t e m s . . . . . . . . . . Introduction to Inlets and Nozzles . . . . . . . . . . . . . . . . . . . . . . Inlets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subsonic Inlets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Supersonic Inlets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exhaust Nozzles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Introduction to Combustion Systems . . . . . . . . . . . . . . . . . . . . Main Burners . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Afterburners . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
685
499 524
...........................
685 685 686 695 726 744 757 769 779
A p p e n d i x A.
Altitude Tables . . . . . . . . . . . . . . . . . . . . . . . . . . .
785
A p p e n d i x B.
Gas Turbine E n g i n e D a t a . . . . . . . . . . . . . . . . . . . .
793
A p p e n d i x C.
D a t a for S o m e L i q u i d  P r o p e l l a n t R o c k e t E n g i n e s . . . .
801
A p p e n d i x D.
A i r and (CH2)n P r o p e r t i e s at L o w P r e s s u r e
803
A p p e n d i x E.
T u r b o m a c h i n e r y Stresses and Materials
.......
..........
821
xiv
Appendix F. About the Software . . . . . . . . . . . . . . . . . . . . . . . . 835
.
Appendix G References Index
Answers to Selected Problems . . . . . . . . . . . . . . . . . . 841
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .845
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .851
1 Introduction
1.1 Propulsion The Random House College Dictionary defines propulsion as "the act of propelling, the state of being propelled, a propelling force or impulse" and defines the verb propel as "to drive, or cause to move, forward or onward. ''1 From these definitions, we can conclude that the study of propulsion includes the study of the propelling force, the motion caused, and the bodies involved. Propulsion involves an object to be propelled plus one or more additional bodies, called propellant. The study of propulsion is concerned with vehicles such as automobiles, trains, ships, aircraft, and spacecraft. The focus of this textbook is on the propulsion of aircraft and spacecraft. Methods devised to produce a thrust force for the propulsion of a vehicle in flight are based on the principle of jet propulsion (the momentum change of a fluid by the propulsion system). The fluid may be the gas used by the engine itself (e.g., turbojet), it may be a fluid available in the surrounding environment (e.g., air used by a propeller), or it may be stored in the vehicle and carried by it during the flight (e.g., rocket). Jet propulsion systems can be subdivided into two broad categories: airbreathing and nonairbreathing. Airbreathing propulsion systems include the reciprocating, turbojet, turbofan, ramjet, turboprop, and turboshaft engines. Nonairbreathing engines include rocket motors, nuclear propulsion systems, and electric propulsion systems. We focus on gas turbine propulsion systems (turbojet, turbofan, turboprop, and turboshaft engines) in this textbook. The material in this textbook is divided into three parts: 1) Basic concepts and onedimensional gas dynamics, 2) Analysis and performance of airbreathing propulsion systems, and 3) Analysis of gas turbine engine components. This chapter introduces the types of airbreathing and rocket propulsion systems and the basic propulsion performance parameters. Also included is an introduction to aircraft and rocket performance. The material on aircraft performance shows the influence of the gas turbine engine on the performance of the aircraft system. This material also permits incorporation of a gas turbine engine design problem such as new engines for an existing aircraft. Numerous examples are included throughout this book to help students see the application of a concept after it is introduced. For some students, the material on basic concepts and gas dynamics will be a review of material covered in other
2
ELEMENTS OF PROPULSION
courses they have already taken. For other students, this may be their first exposure to this material, and it may require more effort to understand.
1.2
Units and Dimensions
Since the engineering world uses both the metric SI and English unit system, both will be used in this textbook. One singular distinction exists between the English system and S I   t h e unit of force is defined in the former but derived in the latter. Newton's second law of motion relates force to mass, length, and time. It states that the sum of the forces is proportional to the rate of change of the momentum (M = mV). The constant of proportionality is 1/gc:
~  ~ F   1 d ( m V ) _ 1 dM gc dt gc dt
(1.1)
The units for each term in the preceding equation are listed in Table 1.1 for both SI and English units. In any unit system, only four of the five items in the table can be specified, and the latter is derived from Eq. (1.1). As a result of selecting gc = 1 and defining the units of mass, length, and time in SI units, the unit of force is derived from Eq. (1.1) as kilogrammeters per square second (kg. m/s2), which is called the newton (N). In English units, the value of g~ is derived from Eq. (1.1) as
gc = 32.174 ft. lbm/(lbf s 2) Rather than adopt the convention used in many recent textbooks of developing material or use with only SI metric units (gc = 1), we will maintain g~ in all our equations. Thus g¢ will also show up in the equations for potential energy (PE) and kinetic energy (KE): PE  mgz gc
KE=
mV 2 2go
The total energy per unit mass e is the sum of the specific internal energy u, specific kinetic energy ke, and specific potential energy pe:
V 2 gz e = u + k e + pe = u + x    +  zgc gc There are a multitude of engineering units for the quantities of interest in propulsion. For example, energy can be expressed in the SI unit of joule Table 1.1
Unit system SI English
Units and dimensions
Force
gc
Mass
Length
Time
Derived Poundforce, lbf
1 Derived
Kilogram, kg Poundmass, Ibm
Meter, m Foot, ft
Second, s Second, s
INTRODUCTION
3
(1 J = 1 N . m ) , i n B r i t i s h t h e r m a l u n i t s ( B t u ) , o r i n f o o t  p o u n d f o r c e ( f t  l b f ) . O n e m u s t b e a b l e to u s e t h e a v a i l a b l e d a t a i n t h e u n i t s p r o v i d e d a n d c o n v e r t t h e u n i t s w h e n r e q u i r e d . T a b l e 1.2 is a u n i t c o n v e r s i o n t a b l e p r o v i d e d to h e l p y o u in y o u r e n d e a v o r s .
Table 1.2 Unit Length
Area Volume Time Mass Density Force Energy
Power
Pressure (stress)
E n e r g y per unit m a s s Specific heat Temperature
Temperature change Specific thrust Specific p o w e r T h r u s t specific fuel c o n s u m p t i o n (TSFC) P o w e r specific fuel consumption S t r e n g t h / w e i g h t ratio (o/p)
Unit conversion table Conversion
1 m = 3.2808 ft = 39.37 in. 1 k m = 0.621 m i l e 1 mile = 5280 ft = 1.609 krn 1 nm = 6 0 8 0 ft 1.853 k m 1 m 2 = 10.764 ft 2 1 c m 2 = 0.155 in. 2 1 gal = 0.13368 ft 3 = 3.785 liter 1 liter = 10  3 m 3 = 61.02 in. 3 1 h = 3600 s = 60 m i n 1 k g = 1000 g = 2.2046 I b m = 6.8521 x 10  2 slug 1 slug = 1 lbf. s2/ft = 32.174 I b m 1 s l u g / f t 3 = 512.38 k g / m 3 1 N = 1 kg. m/s 2 1 lbf = 4.448 N 1 J = 1 N. m = lkg. m2/s 2 1 B t u = 778.16 ft. lbf = 252 cal = 1055 J 1 cal = 4.186 J 1 kJ = 0.947813 B t u = 0.23884 kcal 1 W = 1 J / s  I k g . n2/S 3 1 hp = 550 ft l b f / s = 2545 B t u / h = 745.7 W 1 k W = 3412 B t u / h = 1.341 hp 1 a t m = 14.696 lb/in. 2 or psi = 760 torr = 101,325 Pa 1 a t m = 30.0 i n H g = 407.2 i n H 2 0 1 ksi = 1000 psi 1 m m H g = 0 . 0 1 9 3 4 psi = 1 ton" 1 Pa = 1 N / m 2 1 i n H g 3376.8 Pa 1 k J / k g = 0.4299 B t u / l b m 1 k J / ( k g . °C) = 0.23884 B t u / ( l b m . °F) 1 K = 1.8°R K = 273.15 + °C °R = 459.69 + °F 1 °C = 1.8°F 1 l b f / ( l b m / s ) = 9.8067 N / ( k g / s ) 1 h p / ( l b m / s ) = 1.644 k W / ( k g / s ) 1 l b m / ( l b f  h) = 28.325 m g / ( N , s) 1 l b m / ( h p  h ) 168.97 m g / ( k W , s) 1 k s i / ( s l u g / f t 3) = 144 ft2/s 2 = 13.38 m 2 / s 2
4
ELEMENTS OF PROPULSION
1.3 Operational Envelopes and Standard Atmosphere Each engine type will operate only within a certain range of altitudes and Mach numbers (velocities). Similar limitations in velocity and altitude exist for airframes. It is necessary, therefore, to match airframe and propulsion system capabilities. Figure 1.1 shows the approximate velocity and altitude limits, or corridor of flight, within which airlift vehicles can operate. The corridor is bounded by a lift limit, a temperature limit, and an aerodynamic force limit. The lift limit is determined by the maximum levelflight altitude at a given velocity. The temperature limit is set by the structural thermal limits of the material used in construction of the aircraft. At any given altitude, the maximum velocity attained is temperaturelimited by aerodynamic heating effects. At lower altitudes, velocity is limited by aerodynamic force loads rather than by temperature. The operating regions of all aircraft lie within the flight corridor. The operating region of a particular aircraft within the corridor is determined by aircraft design, but it is a very small portion of the overall corridor. Superimposed on the flight corridor in Fig. 1.1 are the operational envelopes of various powered aircraft. The operational limits of each propulsion system are determined by limitations of the components of the propulsion system and are shown in Fig. 1.2. The analyses presented in this text use the properties of the atmosphere to determine both engine and airframe performance. Since these properties vary with location, season, time of day, etc., we will use the U.S. standard atmosphere 2 to give a known foundation for our analyses. Appendix A gives the properties of the U.S. standard atmosphere, 1976, in both English and SI units. Values of the pressure P, temperature T, density p, and speed of sound a are given in dimensionless ratios of the property at altitude to its value at sea level (SL), the reference value. The dimensionless ratios of pressure, temperature, and density
Lift (stall) limit
\
/ Fig. 1.10a
Schematic d i a g r a m of a t u r b o p r o p .
A ~r ; n l ~ f
Threestage axial flnw
Prop, drive: "~
 " Combustion chamber
. . . . . v. . . . . . . . turbine
~enuuugai compressor
Fig. 1.10b C a n a d i a n P r a t t & W h i t n e y PT6 turboshaft. (Courtesy of P r a t t & Whitney of Canada.)
Fig. 1.10c
Allison T56 turboshaft. (Courtesy of Allison Gas T u r b i n e Division.)
14
ELEMENTS OF PROPULSION
the turbofan engine will have a better aerodynamic performance than the turboprop since the turbofan is essentially a ducted turboprop. Putting a duct or shroud around a propeller increases its aerodynamic performance. Examples of a turboshaft engine are the Canadian Pratt & Whitney PT6 (Fig. 1.10b), used in many small commuter aircraft, and the Allison T56 (Fig. 1.10c), used to power the C130 Hercules and the P3 Orion.
1.4.5 Ramjet The ramjet engine consists of an inlet, a combustion zone, and a nozzle. A schematic diagram of a ramjet is shown in Fig. 1.11. The ramjet does not have the compressor and turbine as the turbojet does. Air enters the inlet where it is compressed and then enters the combustion zone where it is mixed with the fuel and burned. The hot gases are then expelled through the nozzle, developing thrust. The operation of the ramjet depends on the inlet to decelerate the incoming air to raise the pressure in the combustion zone. The pressure rise makes it possible for the ramjet to operate. The higher the velocity of the incoming air, the greater the pressure rise. It is for this reason that the ramjet operates best at high supersonic velocities. At subsonic velocities, the ramjet is inefficient, and to start the ramjet, air at a relatively higher velocity must enter the inlet. The combustion process in an ordinary ramjet takes place at low subsonic velocities. At high supersonic flight velocities, a very large pressure rise is developed that is more than sufficient to support operation of the ramjet. Also, if the inlet has to decelerate a supersonic highvelocity airstream to a subsonic velocity, large pressure losses can result. The deceleration process also produces a temperature rise, and at some limiting flight speed, the temperature will approach the limit set by the wall materials and cooling methods. Thus when the temperature increase due to deceleration reaches the limit, it may not be possible to burn fuel in the airstream. In the past few years, research and development have been done on a ramjet that has the combustion process taking place at supersonic velocities. By using a supersonic combustion process, the temperature rise and pressure loss due to deceleration in the inlet can be reduced. This ramjet with supersonic combustion is known as the scramjet (supersonic combustion ramjet). Figure 1.12a shows the schematic of a scramjet engine similar to that proposed for the National
Fuel sprayring
•
~nlet Fig. 1.11
=[=
Flameholder
Combustionzone
=[ Nozzle~
Schematic diagram of a ramjet.
INTRODUCTION :,~Fuselage torebody
Fig. 1.12a
Inlet comp.ression
15
Fuse)age
Schematic diagram of a scramjet.
AeroSpace Plane (NASP) research vehicle, the X30 shown in Fig. 1.12b. Further development of the scramjet for other applications (e.g., the Orient Express) will continue if research and development produces a scramjet engine with sufficient performance gains. Remember that since it takes a relative velocity to start the ramjet or scramjet, another engine system is required to accelerate aircraft like the X30 to ramjet velocities.
1.4.6 Turbojet/RamjetCombinedCycle Engine Two of the Pratt & Whitney J58 turbojet engines (see Fig. 1.13a) are used to power the Lockheed SR71 Blackbird (see Fig. 1.13b). This was the fastest aircraft
Fig. 1.12b
Conceptual drawing of the X30. (Courtesy of Pratt & Whitney.)
ELEMENTS OF PROPULSION
16
Fig. 1.13a
Pratt & Whitney J58 turbojet. (Courtesy of Pratt & Whitney.)
(Mach 3+) when it was retired in 1989. The J58 operates as an afterburning turbojet engine until it reaches high Mach level, at which point the six large tubes (Fig. 1.13a) bypass flow to the afterburner. When these tubes are in use, the compressor, burner, and turbine of the turbojet are essentially bypassed, and the engine operates as a ramjet with the afterburner acting as the ramjet's burner.
1.4.7 Aircraft Engine Performance Parameters This section presents several of the airbreathing engine performance parameters that are useful in aircraft propulsion. The first performance parameter is the thrust of the engine that is available for sustained flight (thrust = drag), accelerated flight (thrust > drag), or deceleration (thrust < drag).
Fig. 1.13b
Lockheed SR71 Blackbird. (Courtesy of Lockheed.)
INTRODUCTION
17
As derived in Chapter 4, the uninstalled thrust F of a jet engine (single inlet and single exhaust) is given by F =
(,:no + ;nD Ve  ,hoVo gc
+ ( P e  P o ) A e
(1.5)
where rho, rhf = mass flow rates of air and fuel, respectively Vo, V~ = velocities at inlet and exit, respectively Po, Pe = pressures at inlet and exit, respectively It is most desirable to expand the exhaust gas to the ambient pressure, which gives Pe = Po. In this case, the uninstalled thrust equation becomes F =
(In 0 1 1;nf ) We  i r l o g 0
gc
for Pe = Po
(1.6)
The installed thrust T is equal to the uninstalled thrust F minus the inlet drag and minus the nozzle drag Dnoz, or
Dinle t
T = F

Oinlet 
Dnoz
(1.7)
Dividing the inlet drag D i n l e t and nozzle drag Dnoz by the uninstalled thrust F yields the dimensionless inlet loss coefficient ~binlet and nozzle loss coefficient ~noz, or Oinlet ~inlet  
F Dnoz 4~noz F
(1.8)
Thus the relationship between the installed thrust T and uninstalled thrust F is simply r = F(1

~inlet 
~noz)
(1.9)
The second performance parameter is the thrust specific fuel consumption (S and TSFC). This is the rate of fuel use by the propulsion system per unit of thrust produced. The uninstalled fuel consumption S and installed fuel consumption TSFC are written in equation form as
s= ~ F TSFC = m~ T where F S
= uninstalled thrust = uninstalled thrust specific fuel consumption
(1.1o) (1.11)
18
ELEMENTS OF PROPULSION
T = installed engine thrust TSFC = installed thrust specific fuel consumption rhf = mass flow rate of fuel The relation between S and TSFC in equation form is given by S = TSFC(1

t~inlet

(~noz)
(1.12)
Values of thrust F and fuel consumption S for various jet engines at sealevel static conditions are listed in Appendix B. The predicted variations of uninstalled engine thrust F and uninstalled thrust specific fuel consumption S with Mach number and altitude for an advanced fighter engine 3 are plotted in Figs. 1.14a 1.14d. Note that the thrust F decreases with altitude and the fuel consumption S also decreases with altitude until 36 kft (the start of the isothermal layer of the atmosphere). Also note that the fuel consumption increases with Mach number and that the thrust varies considerably with the Mach number. The predicted partialthrottle performance of the advanced fighter engine is shown at three flight conditions in Fig. 1.14e. The takeoff thrust of the JT9D highbypassratio turbofan engine is given in Fig. 1.15a vs Mach number and ambient air temperature for two versions. Note the rapid falloff of thrust with rising Mach number that is characteristic of this engine cycle and the constant thrust at a Mach number for temperatures of 86°F and below (this is often referred to as a flat rating). The partialthrottle performance of both engine versions is given in Fig. 1.15b for two combinations of altitude and Mach number. 50,000
SL
40,000
J
30,000 E
J2°kft 30kft 6 kft
.~.
20,000
40 kft 50 kft
10,000
I
0.4
I
I
0.8 1.2 Mach number
I
1.6
Fig. 1.14a Uninstalled thrust F of an advanced afterburning fighter engine at m a x i m u m power setting, afterburner on. (Extracted from Ref. 3.)
INTRODUCTION
19
2.20 Alt (kft) /
2.15
SL
/ 10
~ 2.10 ~ 2.05
E ~ 2.00
,
2o

5/"
30 50 36 40
"
~ 1.95 N .~ 1.90
e~
1.85
I
1.80
I
0.4
I
I
0.8 1.2 Mach number
1.6
Fig. 1.14b Uninstalled fuel consumption S of an advanced afterburning fighter engine at m a x i m u m power setting, afterburner on. (Extracted from Ref. 3.)
25,000
SL 20,000
A ~ ~
15,000
10
~
20_t30
e
4O
.~ 10,000
5O
5,000
I
0.4
I
I
0.8 1.2 Mach number
I
I
1.6
2
Fig. 1.14c Uninstalled thrust F of a n advanced afterburning fighter engine at military power setting, afterburner off. (Extracted from Ref. 3.)
20
ELEMENTS OF PROPULSION 1.15 Alt (kft) ~
^ u
20
30 50 / a ~
/ / / . ~ / 
~v
1.10 /
/
"~ 1.05
•=
1.00
~ 0.95
•~ 0.90
! .,~ 0 . 8 5
0.80
0.75
I
0.4
0.8
I
I
I
1.2
1.6
2
Mach number
Fig. 1.14d Uninstailed fuel consumption S of an advanced afterburning fighter engine at military power setting, afterburner off. (Extracted from Ref. 3.)
2,0
e~ 1.8 
36 /
36kf/
0.8
e~ r~
e
1.6
1.4
8 1.2
.3e~
1.0
0.8
0.6
I 5
I 10
I 15
I 20
I 25
I 30
I 35
Uninstalled thrust F (1000 lbf)
Fig. 1.14e Partialthrottle performance of an advanced fighter engine. (Extracted from Ref. 3.)
INTRODUCTION
50,000
21
\\ \,,\
....
"
"%
45,000 \

"

T9070
x
\~
"x
\
,~ 40,000 2e, 86°F and below
35,000 [
IO0°F 30,000 r
120°F I
I
0.1
0.2
I
I
I
I
0.3 0.4 0.5 Flight Machnumber
I
Fig. 1.15a JT9D70/70A turbofan takeoff thrust. (Courtesy of Pratt & Whitney.)
Although the aircraft gas turbine engine is a very complex machine, the basic tools for modeling its performance are developed in the following chapters. These tools are based on the work of Gordon Oates. 4 They permit performance calculations for existing and proposed engines and generate performance curves similar to Figs. 1.14al.14e and Figs. 1.15a and 1.15b.
0.80
• Maximumclimb rating k
• Maximumcruise rating
0.75 e~
O "5 e'~
0.70
",
~'..
.~
• 0.9
',,~
~3O,ooorto
8
.
_...a
0.65 I
4~0
I
6000
I
I
8000
I
I
I
I
10,000 12,000 Thrust (lbf)
I
I
14,000
I
I
16,0~
Fig. 1.15b JT9D70/70A turbofan cruisespecific fuel consumption. (Courtesy of Pratt & Whitney.)
ELEMENTS OF PROPULSION
22 Table 1.3
Typical aircraft engine thrust installation losses
MI
Aircraft type
~/~inlet
~)noz
(~inlet
(~noz
Fighter Passenger/cargo Bomber
0.05 0.02 0.03
0.01 0.01 0.01
0.05
0.03
0.04
0.02
The value of the installation loss coefficient depends on the characteristics of the particular engine/airframe combination, the Mach number, and the engine throttle setting. Typical values are given in Table 1.3 for guidance. The thermal efficiency ~/r of an engine is another very useful engine performance parameter. Thermal efficiency is defined as the net rate of organized energy (shaft power or kinetic energy) out of the engine divided by the rate of thermal energy available from the fuel in the engine. The fuel's available thermal energy is equal to the mass flow rate of the fuel rnf times the fuel lowerheating value hpR. Thermal efficiency can be written in equation form as '~T m
Wout
(1.13)
in where ~Tr = thermal efficiency of engine l~'out = net power out of engine Qin = rate of thermal energy released
OhfhpR)
Note that for engines with shaft power output, l~out is equal to this shaft power. For engines with no shaft power output (e.g., turbojet engine), Wout is equal to the net rate of change of the kinetic energy of the fluid through the engine. The power out of a jet engine with a single inlet and single exhaust (e.g., turbojet engine) is given by
Woot =
1
[(,h0 + 'hS)Ve  'h0 V0 I
The propulsive efficiency r/p of a propulsion system is a measure of how effectively the engine power Wout is used to power the aircraft. Propulsive efficiency is the ratio of the aircraft power (thrust times velocity) to the power out of the engine Wout. In equation form, this is written as qp _ where r/p T
= propulsive efficiency of engine = thrust of propulsion system
TVo
. Wout
(1.14)
INTRODUCTION Vo
23
= velocity of aircraft net power out of engine
Wou t =
For a jet engine with a single inlet and single exhaust and an exit pressure equal to the ambient pressure, the propulsive efficiency is given by
2(1 
~binlet 
np =
~bnoz)[(fh 0 q
('no +
I?lf)Ve /~/oVo]Vo
hS)Ve 
(1.15)
hoVo2
For the case when the mass flow rate of the fuel is much less than that of air and the installation losses are very small, Eq. (1.15) simplifies to the following equation for the propulsive efficiency:
~lp  Ve/Vo + 1
(1.16)
Equation (1.16) is plotted vs the velocity ratio V~/Vo in Fig. 1.16 and shows that high propulsive efficiency requires the exit velocity to be approximately equal to the inlet velocity. Turbojet engines have high values of the velocity ratio Ve/Vo with corresponding low propulsive efficiency, whereas turbofan engines have low values of the velocity ratio Ve/Vo with corresponding high propulsive efficiency. The thermal and propulsive efficiencies can be combined to give the overall efficiency ~1o of a propulsion system. Multiplying propulsive efficiency b y 100
90
80
70
o
60
50
40
30
1.0
Fig. 1.16
I
I
I
I
I
I
1.5
2.0
2.5
3.0
3.5
4.0
Propulsive efficiency vs velocity ratio (Ve/VO).
24
ELEMENTS OF PROPULSION
thermal efficiency, we get the ratio of the aircraft power to the rate of thermal energy released in the engine (the overall efficiency of the propulsion system): (1.17)
Tlo = r/er/r TV0
(1.18)
0o m .
Qin Several of the preceding performance parameters are plotted for general types of gas turbine engines in Figs. 1.17a, 1.17b, and 1.17c. These plots can be used to obtain the general trends of these performance parameters with flight velocity for each propulsion system. Since Oin : t;rtfhpR, Eq. (1.18) can be rewritten as TV0 '17°  lhfhpR
With the help of Eq. (1.11), this equation can be written in terms of the thrust specific fuel consumption as Vo ~7o  TSFC • hpR
(1.19)
Turbojet 90 80

~
_
_
/
~
B
P
R
turbofan
70 60 o
50
40 2 ~ 30 20
Advanced_~~prop~
~ .
Turboprop
10 Conventionalprop I
0.5
I
I
1.0 1.5 AircraflMachnumber
I
2.0
I
2.5
Fig. 1.17a Specific thrust characteristics of typical aircraft engines. (Courtesy of Pratt & Whitney.)
INTRODUCTION
25
1.3
1.1
0.9 =< 
/
/ /
~
0.7
HighBPR turbofan
Conventional prop 7 / J r A
0.5
/
'
z Z
vanced prop
/
0.3
0.1
I
I
0.5
I
1.0 1.5 Aircraft Mach number
I
I
2.0
2.5
Fig. 1.17b Thrustspecific fuel consumption characteristics of typical aircraft engines. (Courtesy of Pratt & Whitney.)
Otrall efficiency 11o = 0.7 
0.1
0 \
thrust x aircraft velocity heat added ~
%
~
~
Year 2000 +
0.6 0.5II
0.4
Turt
b ~= 0 . 3 
0.2Subsonic flight
0.100
I
I
I
I
I
0.1
0.2
0.3
0.4
0.5
I
I
I
I
0.6 0.7 0.8 0.9 thrust Propulsive efficiency t/p =  x aircraft velocity core power
I
1.0
Fig. 1.17c Efficiency characteristics of typical aircraft engines. (Courtesy of Pratt & Whitney.)
26
ELEMENTS
OF
PROPULSION
Using Eqs. (1.17) and (1.19), we can write the following for TSFC:
Vo
TSFC =
(1.20)
rle rlTheR
Example 1.1 An advanced fighter engine operating at Mach 0.8 and 10km altitude has the following uninstalled performance data and uses a fuel with hpg = 42,800 kJ/kg: F=50kN
rh0=45kg/s
and
rnf=2.65kg/s
Determine the specific thrust, thrust specific fuel consumption, exit velocity, thermal efficiency, propulsive efficiency, and overall efficiency (assume exit pressure equal to ambient pressure).
Solution: F rho
50kN   45 k g / s
1.1111 k N / ( k g / s ) =
1111.1 m / s
S  rhf _ _ _ 2 "k6g5/ s _ 0.053 ( k g / s ) / k N = 53 m g / N • s F 50kN
(a0)
VO = Moao = Mo  
\are f /
aref = 0.8(0.8802)340.3 = 239.6 m / s
From Eq. (1.6) we have
re
Fgc +/noVo
50,000 x 1 + 45 x 239.6
rh0 + rnf
45 + 2.65
Wo~
= 1275.6 m / s
(~ho + ,h~)Ve~  ,~oVo~
2gcthfheR
~'/T = Qin
Wout = (~h0 + ~hl)Ve~  ~h0Vo~
2gc _0in
m_
47.65 x 1275.62  4 5
x 239.62
2xl
 37.475 x 10 6 W
/nfhpR = 2.65 x 42,800 = 113.42 x 106 W li¢out
37.475 X 10 6
"qT = Qin  113.42 X 106
FVo r/p = W o u t
FVo r/° = Qin =
= 33.04%
50,000 x 239.6 37.475 x 106 = 31.97% 50,000 x 239.6 113.42 x 106 = 10.56%
INTRODUCTION
27
1.4.8 Specific Thrust vs Fuel Consumption For a jet engine with a single inlet and single exhaust and exit pressure equal to ambient pressure, when the mass flow rate of the fuel is much less than that of air and the installation losses are very small, the specific thrust F//no can be written as F &o
VeVo gc
(1.21)
Then the propulsive efficiency of Eq. (1.16) can be rewritten as 2 Fgc/(&oVo) + 2
rip
(1.22)
Substituting Eq. (1.22) into Eq. (1.20) and noting that TSFC = S, we obtain the following very enlightening expression: S 
Fgc//no + 2Vo 2~qThpR
(1.23)
Aircraft manufacturers desire engines having low thrust specific fuel consumption S and high specific thrust F//no. Low engine fuel consumption can be directly translated into longer range, increased payload, and/or reduced aircraft size. High specific thrust reduces the crosssectional area of the engine and has a direct influence on engine weight and installation losses. This desired trend is plotted in Fig. 1.18. Equation (1.23) is also plotted in Fig. 1.18 and shows that fuel consumption and specific thrust are directly proportional. Thus the aircraft manufacturers have to make a tradeoff. The line of Eq. (1.23) shifts in the desired direction when there is an increase in the level of technology (increased thermal efficiency) or an increase in the fuel heating value.
Eq. (1.23) r.~
g
t
e
c
h
n
o
l
o
g
y
SpecificthrustF/rho Fig. 1.18
Relationship between specific thrust and fuel consumption.
28
ELEMENTS OF PROPULSION
Another very useful measure of merit for the aircraft gas turbine engine is the thrust/weight ratio F/W. For a given engine thrust F, increasing the thrust/weight ratio reduces the weight of the engine. Aircraft manufacturers can use this reduction in engine weight to increase the capabilites of an aircraft (increased payload, increased fuel, or both) or decrease the size (weight) and cost of a new aircraft under development. Engine companies expend considerable research and development effort on increasing the thrust/weight ratio of aircraft gas turbine engines. This ratio is equal to the specific thrust F/&o divided by the engine weight per unit of mass flow W//no. For a given engine type, the engine weight per unit mass flow is related to the efficiency of the engine structure, and the specific thrust is related to the engine thermodynamics. The weights per unit mass flow of some existing gas turbine engines are plotted vs specific thrust in Fig. 1.19. Also plotted are lines of constant engine thrust/weight ratio F/W. The engine companies, in conjunction with the U.S. Department of Defense and NASA, are involved in a large research and development effort to increase the engine thrust/weight ratio F / W and decrease the fuel consumption while maintaining engine durability, maintainability, etc. An earlier program was called the integrated highperformance turbine engine technology (IHPTET) initiative (see Refs. 5 and 6).
25t
F/W=2
3
4
5
20
15
10
50
100 F/&o
Fig. 1.19 Engine thrust/weight ratio F/W.
130
INTRODUCTION 1.5
29
Aircraft Performance
This section on aircraft performance is included so that the reader may get a better understanding of the propulsion requirements of the aircraft. 7 The coverage is limited to a few significant concepts that directly relate to aircraft engines. It is not intended as a substitute for the many excellent references on this subject (see Refs. 811).
1.5.1 Performance Equation Relationships for the performance of an aircraft can be obtained from energy considerations (see Ref. 12). By treating the aircraft (Fig. 1.20) as a moving mass and assuming that the installed propulsive thrust T, aerodynamic drag D, and other resistive forces R act in the same direction as the velocity V, it follows that [T  (D + R)]V =
rate of mechanical energy input
W d ( ~ ~2) dt ÷ g ~ storage storage rate of rate of potential kinetic energy energy W dh
(1.24)
Note that the total resistive force D + R is the sum of the drag of the clean aircraft D and any additional drags R associated with such proturberances as landing gear, external stores, or drag chutes. By defining the energy height Ze as the sum of the potential and kinetic energy terms
V2 Ze  h +  
2g
(1.25)
Eq. (1.24) can now be written simply as [T  (D + R)]V = W dze dt
(1.26)
By defining the weight specific excess p o w e r Ps as Ps ~ 
dze dt
Aircraft velocity V *C
Fig. 1.20
Forces on aircraft.
(1.27)
30
ELEMENTS OF P R O P U L S I O N
Eq. (1.26) can now be written in its dimensionless form as
T(D+R)_Ps_ w
1 d (h+~g
~\
v
(1.28)
This is a very powerful equation that gives insight into the dynamics of flight, including both the rate of climb dh/dt and acceleration dV/dt.
1.5.2
Lift and Drag
We use the classical aircraft lift relationship
L = nW = CLqSw
(1.29)
where n is the load factor or number of g perpendicular to V (n = 1 for straight and level flight), CL is the coefficient of lift, Sw is the wing planform area, and q is the dynamic pressure. The dynamic pressure can be expressed in terms of the density p and velocity V or the pressure P and Mach number M as 1
V2
1
V2
q = "2P~c = 2 0"Orefgc
(1.30a)
or
q=~
T pM 2
_T~prefMo2 0 =2
(1.30b)
where 6 and ~r are the dimensionless pressure and density ratios defined by Eqs. (1.2) and (1.4), respectively, and y is the ratio of specific heats (y = 1.4 for air). The reference density Pref and reference pressure Pref of air are their sealevel values on a standard day and are listed in Appendix A. We also use the classical aircraft drag relationship
D CDqSw
(1.31)
Figure 1.21 is a plot of lift coefficient CL vs drag coefficient Co, commonly called the liftdrag polar, for a typical subsonic passenger aircraft. The drag coefficient curve can be approximated by a secondorder equation in CL written as
co = K~C~ + X2CL + Coo
(1.32)
where the coefficients K1, K2, and CDO are typically functions of flight Mach number and wing configuration (flap position, etc.). The Coo term in Eq. (t.32) is the zero lift drag coefficient that accounts for both frictional and pressure drag in subsonic flight and wave drag in supersonic flight. The K 1 and K2 terms account for the drag due to lift. Normally K2 is very small and approximately equal to zero for most fighter aircraft.
INTRODUCTION
31
1.0
0.8
0.6
cL 0.4
0.2
I
0.01
I
0.02
I
0.03
I
0.04
I
0.05
I
0.06
I
0.07
I
0.08
Co
Fig. 1.21 Typical liftdrag polar.
Example 1.2 For all the examples given in this section on aircraft performance, two types of aircraft will be considered.
a) Fighter aircraft (HF1). An advanced fighter aircraft is approximately modeled after the F22 Advanced Tactical Fighter shown in Fig. 1.22. For convenience, we will designate our hypothetical fighter aircraft as the HF1, having the following characteristics: Maximum gross takeoff weight WTo = 40,000 lbf (177,920 N) Empty weight = 24,000 lbf (106,752 N) Maximum fuel plus payload weight = 16,000 lbf (71,168 N) Permanent payload = 1600 lbf (7117 N, crew plus return armament) Expended payload = 2000 lbf (8896 N, missiles plus ammunition) Maximum fuel capacity = 12,400 lbf (55,155 N) Wing area Sw = 720 ft 2 (66.9 m 2) Engine: lowbypassratio, mixedflow turbofan with afterburner Maximum lift coefficient CLmax = 1.8 Drag coefficients given in Table 1.4
b) Passenger aircraft (HP1). An advanced 253passenger commercial aircraft approximately modeled after the Boeing 787 is shown in Fig. 1.23. For convenience, we will designate our hypothetical passenger aircraft as the HP 1, having the following characteristics: Maximum gross takeoff weight WTO = 1,645,760 N (370,000 lbf) Empty weight = 822,880 N (185,500 lbf)
32
ELEMENTS OF PROPULSION
Fig. 1.22 F22, Advanced Tactical Fighter. (Photo courtesy of Boeing Defense & Space Group, Military Airplanes Division.)
Maximum landing weight = 1,356,640 N (305,000 lbf) Maximum payload = 420,780 N (94,600 lbf, 253 passengers plus 196,000 N of cargo) Maximum fuel capacity 2 716,706 N (161,130 lbf) Wing area Sw = 282.5 m (3040 ft 2) Engine: highbypassratio turbofan Maximum lift coefficient CLmax = 2.0 Drag coefficients given in Table 1.5.
Table 1.4
Drag coefficients for hypothetical fighter aircraft (HF1)
Mo
Ki
K2
CDO
0.0 0.8 1.2 1.4 2.0
0.20 0.20 0.20 0.25 0.40
0.0 0.0 0.0 0.0 0.0
0.0120 0.0120 0.02267 0.0280 0.0270
INTRODUCTION
Fig. 1.23
33
Boeing 787. (Photo courtesy of Boeing.)
Table 1.5 Drag coefficients for hypothetical passenger aircraft (HP1)
Mo
KI
K2
CDO
0.00 0.40 0.75 0.83
0.056 0.056 0.056 0.056
 0.004  0.004  0.008  0.008
0.0140 0.0140 0.0140 0.0150
Example 1.3 Determine the drag polar and drag variation for the HF1 aircraft at 90% of m a x i m u m gross takeoff weight and the HP1 aircraft at 95% of m a x i m u m gross takeoff weight.
a) Fighter aircraft (HF1). The variation in Coo and KI with Mach number for the HF1 are plotted in Fig. 1.24 from the data of Table 1.4. Figure 1.25 shows the drag polar at different Mach numbers for the HF1 aircraft. Using these drag data and the preceding equations gives the variation in aircraft drag with subsonic Mach number and altitude for level flight (n = 1), as shown in Fig. 1.26a. Note that the minimum drag is constant for Mach numbers 0 to 0.8 and then increases. This is the same variation as Coo. The variation of drag with load factor n is shown in Fig. 1.26b at two altitudes. The drag increases with increasing
34
ELEMENTS OF PROPULSION /£2=0
0.030
I
I
0.40
0.35
0.025
CD0
0.020
0.30
Cvo
K1 0.015
0.25
0.010
0.20
0.005
0.15
0.000
I
0.0
Fig. 1.24
0.4
I
0.8 1.2 Mach number
I
1.6
2.0
Values of K] and Coo for HF1 aircraft.
1.0
~3~" ~.~k/.~/_~ ~ 1.0
0.8
~
0.6
G 0.4
0.2
0.00.0
I 0.04
Fig. 1.25
0.08
CD
0.12
0.16
Liftdrag p o l a r for HF1 aircraft.
I 0.2
9.10
INTRODUCTION
SL
8OOO 30 36 40 7000
35
10
20
50 kft
li 20 / / ~
30
/
36
60~
500(
40 50
400( Alt=SL 10kft 20kft 300(] 0
I
L
0.3
0.4
I
30kft 36kft I
i
0.5 0.6 0.7 Machnumber
I
I
0.8
0.9
I
1.0
Fig. 1.26a Drag for level flight (n = 1) for HF1 aircraft.
40,000
30,000
20,000
Sea /
10,000
i
0.4 Fig. 1.26b
i
I
0.8 1.2 Machnumber
I
1.6
i
2.0
Drag of HF1 aircraft at sea level and 36 kft for n = 1 and n = 5.
36
ELEMENTS OF PROPULSION K 1 = 0.056 0.0152
I
0.0150
0.0148
0.0146
Coo 0.0144
i
?/ I
0.0142
0.0090
i
0.0080
0.0070
0.0060
 0.0050
 0.0040
Coo
 0.0030
0.0140
0.0138 0.0
I 0.2
I 0.4
I 0.6
I 0.8
0.0020 1.0
Mach number
Fig. 1.27
Values of Kz and Coo for HP1 aircraft.
load factor, and there is a flight Mach number that gives minimum drag for a given altitude and load factor. b) Passenger aircraft (liP1). The variation in CDo and K2 with Mach number for the HP1 is plotted in Fig. 1.27 from the data of Table 1.5. Figure 1.28 shows the drag polar at different Mach numbers for the HP1 aircraft. Using these drag data and the preceding equations gives the variation in aircraft drag with subsonic Mach number and altitude for level flight (n = 1), as shown in Fig. 1.29. Note that the minimum drag is constant for Mach numbers 0 to 0.75 and then increases. This is the same variation as CDo.
Example 1.4 Calculate the drag at Mach 0.8 and 40kft altitude of the HF1 aircraft at 90% of maximum gross takeoff weight with load factors of 1 and 4. Solution: We begin by calculating the dynamic pressure q: q=~ T
8PrefM0 2 =
0.7 X 0.1858 X 2116 X 0.8 2 = 176.1 lbf/ft 2
From Fig. 1.24 at M = 0.8, Coo = 0.012, K1 = 0.20, and K2 = 0.
INTRODUCTION 0.8
37

0.7 
3
~
0.6 0.5
cL 0.4 0.3 0.2 0.1 \ i 0.016
0.0 0.012
i 0.02
i 0.024
i 0.028
i 0.032
i i 0 . 0 3 6 0.04
co Fig. 1.28
L i f t  d r a g p o l a r for H P  1 aircraft. 3
SL
100,000 
6
SL
95,000
90,000
z ~ 85,000
80,000
/
75,000 
70,000 0.2
Fig. 1.29
! SL
" 11 km 3km
I 0.3
9
0.4
km
I
I
0.5 0.6 Mach number
9km
I 0.7
I 0.8
D r a g for level flight (n = 1) for H P  1 aircraft.
I 0.9
38
ELEMENTS OF PROPULSION Case 1:
n = 1
nW qSw
CL
1 x 0.9 x 40,000 176.1 x 720 = 0.2839
Co = K1C2L 4 K2CL 4 Coo = 0.2(0.28392) I 0.012 = 0.0281 D = CoqSw = 0.0281 x 176.1 x 720 = 3563 lbf Case 2:
n = 4
CL .
nW . qSw
4 x 0.9 x 40,000 . . 176.1 x 720
1.136
Co = K~C~ + KaCL + Coo = 0.2(1.1362) + 0.012 = 0.2701 D = CoqSw = 0.2701 x 176.1 x 720 = 34,247 lbf Note that the drag at n = 4 is about 10 times that at n  1.
1.5.3 Stall, Takeoff, and Landing Speeds Stall is the flight condition when an aircraft's wing loses lift. It is an undesirable condition since vehicle control is lost for a time. During level flight (lift = weight), stall will occur when one tries to obtain a lift coefficient greater than the wing's m a x i m u m CLmax. The stall speed is defined as the level flight speed that corresponds to the w i n g ' s maximum lift coefficient, or
2gc W V pCLmax Sw
=,[ Wstall
(1.33)
To keep away from stall, aircraft are flown at velocities greater than Vstan. Takeoff and landing are two flight conditions in which the aircraft velocity is close to the stall velocity. For safety, the takeoff speed VTo o f an aircraft is typically 20% greater than the stall speed, and the landing speed at touchdown VTD is 15% greater: VTO = 1.20Vstall
(1.34)
VTD = 1 . 1 5 Vstal I
Example 1.5 Determine the takeoff speed of the HP1 at sea level with maximum gross takeoff weight and the landing speed with m a x i m u m landing weight.
INTRODUCTION
39
From Appendix A we have p = 1.255 k g / m 3 for sea level. From Example 1.2b we have CLmax = 2.0, W = 1,645,760 N, Sw 282.5 m 2, and
V /
Wstall ~"
2 × 1 1,645,760 69.0 m / s 1.225 x 2.0 282.5 =
Thus VTO = 1.20Vstall = 8 2 . 8
m/s (~185 mph)
For landing, W = 1,356,640 N, and
5 Vstall •
2_x_l 1,356,640 = 62.6 m / s V 1.225 X 2.0 282.5
Thus VTD ~ 1.15Vstall =
72.0 m / s (~161 mph)
1.5.4 Fuel Consumption The rate of change of the aircraft weight d W / d t is due to the fuel consumed by the engines. The mass rate of fuel consumed is equal to the product of the installed thrust T and the installed thrust specific fuel consumption. For constant acceleration of gravity go, we can write dW
.  m y g o = _T(TSFC)(g0"~ = wf = gc ,gcJ
This equation can be rewritten in dimensionless form as dW _ W
T(TSFC)(go~d t w \go~
(1.35)
1.5.4.1 E s t i m a t e of TSFG. Equation (1.35) requires estimates of installed engine thrust T and installed TSFC to calculate the change in aircraft weight. For many flight conditions, the installed engine thrust T equals the aircraft drag D. The value of TSFC depends on the engine cycle, altitude, and Mach number. For preliminary analysis, the following equations (from Ref. 12) can be used to estimate TSFC in units of (lbm/h)/lbf, and 0 is the dimensionless temperature ratio T/Tref:
1) Highbypassratio turbofan TSFC = (0.4 + 0 . 4 5 M 0 ) ~
(1.36a)
40
ELEMENTS OF PROPULSION 2) Lowbypassratio, mixedflow turbofan Military and lower power settings: TSFC = (1.0 t 0 . 3 5 M 0 ) ~
(1.36b)
Maximum power setting: TSFC = (1.8 + 0.30Mo)x/O
(1.36c)
3) Turbojet Military and lower power settings: TSFC = (1.3 + 0 . 3 5 M o ) ~
(1.36d)
Maximum power setting: TSFC = (1.7 + 0 . 2 6 M o ) ~
(1.36e)
TSFC = (0.2 + 0.9Mo)VrO
(1.36f)
4) Turboprop
1.5.4.2 Endurance. For level unaccelerated flight, thrust equals drag (T = D) and lift equals weight (L = W). Thus Eq. (1.35) is simply dW W
CL (TSFC)
dt
(1.37)
We define the endurance factor (EF) as EF 
CL gc CD(TSFC) go
(1.38)
Then Eq. (1.37) becomes dW W
dt EF
m
(1.39)
Note that the minimum fuel consumption for a time t occurs at the flight condition where the endurance factor is maximum. For the case when the endurance factor is constant or nearly constant, Eq. (1.39) can be integrated from the initial to final conditions and the following expression obtained for the aircraft weight fraction:
ws
/
t \
= e x p /  ==}
(1.40a)
INTRODUCTION
41
or
Wf = e x p [ _ ~D_(TSFC)tgo] Wi k CL gc_l
(1.40b)
1.5.4.3 Range. For portions of aircraft flight where distance is important, the differential time dt is related to the differential distance ds by ds = V dt
(1.41)
Substituting into Eq. (1.37) gives
dW _ W
CD TSFC go ds CL V gc
(1.42)
CL V gc Co TSFC go
(1.43)
We define the range factor (RF) as RF 
Then Eq. (1.42) can be simply written as dW W

ds RF
(1.44)
Note that the minimum fuel consumption for a distance s occurs at the flight condition where the range factor is maximum. For the flight conditions where the RF is constant or nearly constant, Eq. (t .42) can be integrated from the initial to final conditions and the following expression obtained for the aircraft weight fraction:
WI = e x p (  ~ )s
(1.45a)
or
Wf  _ _ exp ( Wi
CD TSFC x s go] CL V gc/
(1.45b)
This is called the Breguet range equation. For the range factor to remain constant, Ct./Co and V/TSFC need to be constant. Above 36kft altitude, the ambient temperature is constant, and a constant velocity V will correspond to constant Mach and constant TSFC for a fixed throttle setting. If CL is constant, CL/Co will remain constant. Since the aircraft weight W decreases during the flight, the altitude must increase to reduce the density of the ambient air and produce the required lift (L = W) while maintaining CL and velocity constant. This flight profile is called a cruise climb.
42
ELEMENTS OF PROPULSION
Example 1.6 Calculate the endurance factor and range factor at Mach 0.8 and 40kft altitude of hypothetical fighter aircraft HF1 at 90% of maximum gross takeoff weight and a load factor of 1. Solution: q
= ~6PrefM0 2 =
2
0.7 x 0.1858 x 2116
x 0.8 2 =
176.1 l b f / f t 2
From Fig. 1.24 at M = 0.8, Coo = 0.012, K1 = 0.20, and K2 = 0: nW CL = qSw
1 x 0.9 x 40,000 176.1 x 720
0.2839
CD = K1C2 + K2CL + Coo = 0.2(0.28392) + 0.012 0.0281 Using Eq. (1.36b), we have TSFC = (1.0 + 0.35M0)q'0 (1.0 + 0.35 x 0 . 8 ) ~
= 1.110 (lbm/h)/lbf
Thus
EF=
CL g__f_~= 0.2839 32.174__= 9.102 h CD(TSFC)g0 0.0281 x 1.11032.174
RF
CL V gc CDTSFCgo 0.28390.8 x 0.8671 x l l l 6 f l / s 3 6 0 0 s / h 32.174 1.110 (lbm/h)/lbf 6080 ft/nm 32.174 0.0281
= 4170 nm
Example 1.7 Determine the variation in endurance factor and range factor for the two hypothetical aircraft HF 1 and HP 1. a) Fighter aircraft (HFI). The endurance factor is plotted vs Mach number and altitude in Fig. 1.30 for our hypothetical fighter aircraft HF1 at 90% of maximum gross takeoff weight. Note that the best endurance Mach number (minimum fuel consumption) increases with altitude, and the best fuel consumption occurs at altitudes of 30 and 36 kft. The range factor is plotted vs Mach number and altitude in Fig. 1.31 for the HF1 at 90% of maximum gross takeoff weight. Note that the best cruise Mach number (minimum fuel consumption) increases with altitude, and the best fuel consumption occurs at an altitude of 36 kft and Mach number of 0.8.
INTRODUCTION 10
43
[[
Alt (kft) 36
1
8
i , 0
0.3
0.4
Fig. 1.30
50 I
I
0.5 0.6 0.7 Mach number
0.8
0.9
1.0
Endurance factor for HF1 aircraft.
5000
4500 Alt (kft)
40
50
4000
3500 2C
v 3000 10 2500 SL
20
2000 I0
1500 1000 0
50 0.3
Fig. 1.31
0.4
0.5 0.6 0.7 Mach number
0.8
Range factor for HF1 aircraft.
0.9
1.0
44
ELEMENTS OF PROPULSION 38
SL
Alt (km)
36 6 34 9
32 SL 30
28 0.2
I 0.3
Fig. 1.32
I
0.4
~3 0.5 0.6 Mach number
/1\ 0.7
/9 0.8
I 0.9
Endurance factor for HP1 aircraft.
b) Passenger aircraft (HPI). The endurance factor is plotted vs Mach number and altitude in Fig. 1.32 for our hypothetical passenger aircraft HP1 at 95% of maximum gross takeoff weight. Note that the best endurance Mach number (minimum fuel consumption) increases with altitude, and the best fuel consumption occurs at sea level. The range factor is plotted vs Mach number and altitude in Fig. 1.33 for the HP1 at 95% of maximum gross takeoff weight. Note that the best cruise Mach number (minimum fuel consumption) increases with altitude, and the best fuel consumption occurs at an altitude of 11 km and Mach number of about 0.83. Since the weight of an aircraft like the HP1 can vary considerably over a flight, the variation in range factor with cruise Mach number was determined for 95 and 70% of maximum gross takeoff weight (MGTOW) at altitudes of 11 and 12km and is plotted in Fig. 1.34. If the HP1 flew at 0.83 Mach and 12km altitude, the range factors at 95% MGTOW and at 70% MGTOW are about the same. However, if the HP1 flew at 0.83 Mach and l l  k m altitude, the range factor would decrease with aircraft weight, and the aircraft's range would be less than that of the HP1 flown at 0.83 Mach and 12km altitude. One can see from this discussion that the proper cruise altitude can dramatically affect an aircraft's range. 1.5.4.4 Maximum CL/CD. For flight conditions requiring minimum fuel consumption, the optimum flight condition can be approximated by that
INTRODUCTION
45
28,000 Alt (km)
26,000
11
24,000
22,000 20,000 6
18,000 
7/6/9
16,000 14,000 
I
12,000 0.2
Fig. 1.33
/3
0.3
I 0.4
I
I
I~,L
0.5 0.6 Mach number
0.7
,
I
0.8
0.9
Range factor for HP1 aircraft for various altitudes.
26,400 95% @ 11 km
26,000 i
25,600
~ 5 ~x~'~ ~ ~
25,200
24,800
24,000 0.71
Fig. 1.34
I
0.79
I
I
0.80 0.81 Mach number
R a n g e f a c t o r f o r H P  1 a i r c r a f t a t 70 and
I
I
0.82
0.83
95%
MGTOW.
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INTRODUCTION
47
90,000
32
80,000
30
70,000
28 ~
60,000
26
50,000
24
40,000
22
_v
CdCD 20 ~
~ 30,000
18
20,000 I 0.65
I I 0.70 0.75 Mach number
Fig. 1.36 Comparison of drag HP1 at l l  k m altitude.
CL/CD,endurance
10,000 0.60
I 0.80
16 0.85
factor, and range factor for the
the HP1 is a maximum at the same Mach number that CL/Co is maximum due to the lower TSFC of the highbypassratio turbofan engine [see Eq. (1.36a)]. The Mach number for an altitude giving a maximum range factor is called the best cruise Mach (BCM). The best cruise Mach normally occurs at a little higher Mach than that corresponding to (Cc/CD)*. This is because the velocity term in the range factor normally dominates over the increase in TSFC with Mach number. As a first approximation, many use the Mach number corresponding to (CL/CD)* for the best cruise Mach.
Example 1.8 Calculate the Mach giving maximum CIJCD at 20kft altitude for the HF1 aircraft at 90% of maximum gross takeoff weight and a load factor of 1.
Solution:
From Fig. 1.24 at M0 < 0.8, Coo = 0.012, K1 = 0.20, and K2 = 0: C~ = V K1
V~
= 0.2449
w 0.9 x 40,000 _ 204.16 lbf/ft 2 q  CLSw  0.2449 X 720 M0=~
q ~0 204.16 (y/2)6Pref .7 × 0.4599 × 2116 = 0.547
48
ELEMENTS OF PROPULSION
1.5.4.5 A c c e l e r a t e d flight. For flight conditions when thrust T is greater than drag D, an expression for the fuel consumption can be obtained by first noting from Eq. (1.28) that
T W
m
Ps V[1  (D + R)/T]
~
We define the ratio of drag D + R to thrust T as D+R u   T
(1.49)
The preceding equation for thrust to weight becomes
T
P~
(1.50)
W  V(1  u) Now Eq. (1.35) can be rewritten as dW W Since P~ dt forms as
=
m
TSFC go Ps dt V(1  u) gc
dZe, the preceding equation can be expressed in its most useful
dW TSFC gOdze W V(iu)gc
V2]
TSFC go ( V~U)~c d h+~g /
(1.51)
The term 1  u represents the fraction of engine power that goes to increasing the aircraft energy ze, and u represents that fraction that is lost to aircraft drag D + R. Note that this equation applies for cases when u is not unity. When u is unity, either Eq. (1.39) or Eq. (1.44) is used. To obtain the fuel consumption during an acceleration flight condition, Eq. (1.51) can be easily integrated for known flight paths (values of V and Ze) and known variation of TSFC/[V(1  u)] with ze.
1.5.5 Aerospace Vehicle Design~A Team Effort Aeronautical and mechanical engineers in the aerospace field do many things, but for the most part their efforts all lead to the design of some type of aerospace vehicle. The design team for a new aircraft may be divided into four principal groups: aerodynamics, propulsion, structures, and flight mechanics. The design of a vehicle calls on the extraordinary talents of engineers in each group. Thus the design is a team effort. A typical design team is shown in Fig. 1.37. The chief engineer serves as the referee and integrates the efforts of everyone into the vehicle design. Figure 1.38 shows the kind of aircraft design that might result if any one group were able to dominate the others.
INTRODUCTION
49
Chief engineer vehicle integration
Armament and [ avionics
I
Aerodynamics group
I Aerodynamic
IJ I
heating
Fig. 1.37
Aerospace ground equipment (AGE)
Propulsion group
Engine
Structures group
I
Inlet
I
Stress
J rI
Flight mechanics group
I
I
Weights Performance Stability and control
Organization of a typical vehicle design team.
Controls group
Aerodynamic group
Power plant group
Stress group
Fig. 1.38 Aircraft designs.
1.6
Rocket Engines
Nonairbreathing propulsion systems are characterized by the fact that they carry both fuel and the oxidizer within the aerospace vehicle. Such systems thus may be used anywhere in space as well as in the atmosphere. Figure 1.39
50
ELEMENTS OF PROPULSION Fuel
~S~
Fuel pump I
Oxidizer ~ Vq
~
~
Gas turbine Oxidizer pump
Combustion chamber
Fig. 1.39 Liquidpropellant rocket motor.
shows the essential features of a liquidpropellant rocket system. Two propellants (an oxidizer and a fuel) are pumped into the combustion chamber where they ignite. The nozzle accelerates the products of combustion to high velocities and exhausts them to the atmosphere or space. A solidpropellant rocket motor is the simplest of all propulsion systems. Figure 1.40 shows the essential features of this type of system. In this system, the fuel and oxidizer are mixed together and cast into a solid mass called the grain. The grain, usually formed with a hole down the middle called the perforation, is firmly cemented to the inside of the combustion chamber. After ignition, the grain burns radially outward, and the hot combustion gases pass down the perforation and are exhausted through the nozzle. The absence of a propellant feed system in the solidpropellant rocket is one of its major advantages. Liquid rockets, on the other hand, may be stopped and later restarted, and their thrust may be varied somewhat by changing the speed of the fuel and oxidizer pumps.
1.6.1 Rocket Engine Thrust A natural starting point in understanding the performance of a rocket is the examination of the static thrust. Application of the momentum equation developed in Chapter 2 will show that the static thrust is a function of the propellant flow rate mp, the exhaust velocity Ve and pressure Pe, the exhaust a r e a Ae, and the ambient pressure Pa. Figure 1.41 shows a schematic of a stationary rocket
I~/
Sol'id;ropel'lant . grain . . Perforation
I
Nozzle
Fig. 1.40 Solidpropellant rocket motor.
INTRODUCTION
51
••
Propellant tank(s)
•
Yt ~"~~,
X
,!
~,
f
Ve
Ae, Pe
~
(Y
1" l
(Pe Pa)Ae
I"
t Forces on controlvolume
!
f
t
I
~Ve
Momentumflux for controlvolume
Fig. 1.41
Schematic diagram of static rocket engine.
to be considered for analysis. We assume the flow to be onedimensional, with a steady exit velocity Ve and propellant flow rate rhp. About this rocket we place a control volume ~r whose control surface intersects the exhaust jet perpendicularly through the exit plane of the nozzle. Thrust acts in the direction opposite to the direction of Ve. The reaction to the thrust F necessary to hold the rocket and control volume stationary is shown in Fig. 1.41. The momentum equation applied to this system gives the following: 1) Sum of forces acting on the outside surface of the control volume:
y~Fx=F
(PePa)Ae
2) The net rate of change of momentum for the control volume: A(momentum) = 3;/out  inpVe
gc Since the sum of the forces acting on the outside of the control volume is equal to the net rate of change of the momentum for the control volume, we have
F  (Pe  Pa)Ae  #;rtpVe gc
(1.52)
52
ELEMENTS OF PROPULSION
If the pressure in the exhaust plane Pe is the same as the ambient pressure Pa, the thrust is given by F = rhpVe/gc. The condition Pe = Pa is called ondesign or optimum expansion because it corresponds to maximum thrust for the given chamber conditions. It is convenient to define an effective exhaust velocity C such that C ~ Ve ]
(Pe  Pa)Aegc
(1.53)
Thus the static thrust of a rocket can be written as
F = inpC gc
(1.54)
1.6.2 Specific Impulse The specific impulse lsp for a rocket is defined as the thrust per unit of propellant weight flow:
/sp  
F ~Vp

F gc lhp g 0
(1.55)
where go is the acceleration due to gravity at sea level. The unit Of/sp is the second. From Eqs. (1.54) and (1.55), the specific impulse can also be written as C Isp =  go
Example
(1.56)
1.9
Find the specific impulse of the space shuttle main engine (SSME) shown in Fig. 1.42a that produces 470,000 lbf in a vacuum with a propellant weight flow of 1030 lbf/s. By using Eq. (1.55), we find that the SSME has a specific impulse Isp of 456 s ( = 470,000/1030) in vacuum. An estimate of the variation in thrust with altitude for the space shuttle main engine is shown in Fig. 1.42b. The typical specific impulses for some rocket engines are listed in Table 1.6. Other performance data for rocket engines are contained in Appendix C.
1.6.3 Rocket Vehicle Acceleration The mass of a rocket vehicle varies a great deal during flight due to the consumption of the propellant. The velocity that a rocket vehicle attains during powered flight can be determined by considering the vehicle in Fig. 1.43. The figure shows an accelerating rocket vehicle in a gravity field. At some time, the mass of the rocket is m and its velocity is V. In an infinitesimal time dt, the rocket exhausts an incremental mass dmp with an exhaust velocity Ve
INTRODUCTION
53
Fig. 1.42a Space shuttle main engine (SSME).
relative to the rocket as the rocket velocity changes to V + dV. The net change in momentum o f the control volume cr is composed of the momentum out o f the rocket at the exhaust plus the change of the momentum of the rocket. The momentum out o f the rocket in the V direction is  Ve dmp, and the change in the momentum of the rocket in the V direction is m dV. The forces acting on the control volume o" are composed of the net pressure force, the drag D, and the gravitational force. The sum o f these forces in the V direction is
) ~Fv
=
(Pe

ea)Ae

D  m__ggcos 0 gc
The resultant impulse on the rocket ()~F v ) d t must equal the momentum change of the system A(momentum) = (  Ve dmp + m dV)/gc. Thus
(Pe

Pa)Ae  D  mg   cos O] dt =  V e dmp + m dV gc J gc
54
ELEMENTS OF PROPULSION 470
f
460 450 44O E430
420 410 400
I 20
Fig. 1.42b
I I 40 60 Altitude (kft)
I 80
100
Rocket thrust variation with altitude.
F r o m the preceding relationship, the momentum change of the rocket (m dV) is
m dV  [(Pe  Pa)Ae  D gc mg Since drop = rnp dt =
(dm/dt)dt,
cos
then Eq. (1.57) can be written as
md_Vgc [(PePa)Ae + thpVeDmgcosOldtgc Table 1.6
O]dt~ Vegc
gc
_1
Ranges of specific impulse lsp for typical rocket engines
Fuel/oxidizer
Isp, s
Solid propellant Liquid O2: kerosene (RP) Liquid 02:H2 Nuclear fuel: H2 propellant
250 310 410 840
(1.57)
INTRODUCTION
55
mg
E
Fig. 1.43
R o c k e t v e h i c l e in flight.
By using Eq. (1.53), this relationship becomes = gc
\gc
cos
dt
gc
or d V =  C dm m
Dgc dt _ g c o s O d t
(1.58)
m
The velocity of a rocket along its trajectory can be determined from the preceding equation if C, D, g, and 0 are known. In the absence of drag and gravity, integration of Eq. (1.58) gives the following, assuming constant effective exhaust velocity C: AV : C fmt mf
(1.59)
where AV is the change in velocity, m i is the initial mass of the rocket system, and m f i s the final mass. Equation (1.59) can be solved for the mass ratio as mi

my
= exp
AV C
(1.60)
56
ELEMENTS OF PROPULSION
Example 1.10 We want to estimate the mass ratio (final to initial) of an H2O2 (C = 4000 m/s) rocket for an Earth orbit (AV = 8000 m/s), neglecting drag and gravity. Using Eq. (1.59), we obtain mf/mt = e 2 = 0.132, or a singlestage rocket would be about 13% payload and structure and 87% propellant.
Problems 1.1
Calculate the uninstalled thrust for Example 1.1, using Eq. (1.6).
1.2
Develop the following analytical expressions for a turbojet engine: (a) When the fuel flow rate is very small in comparison with the air mass flow rate, the exit pressure is equal to ambient pressure, and the installation loss coefficients are zero, then the installed thrust T is given by
T = __/n°(Ve  Vo) gc (b)
For the preceding conditions, the thrust specific fuel consumption is given by TSFC 
(C)
(d)
Tgc//no + 2V0 2"OThpR
For Vo = 0 and 500 ft/s, plot the preceding equation for TSFC [in (lbm/h)/lbf] vs specific thrust T/rho [in lbf/(lbm/s)] for values of specific thrust from 0 to 120. Use ~ r = 0.4 and hpR = 18,400 Btu/lbm. Explain the trends.
1.3
Repeat 1.2c, using SI units. For Vo = 0 and 150 m/s, plot TSFC [in (mg/ s)/N] vs specific thrust T/rho [in N/(kg/s)] for values of specific thrust from 0 to 1200. Use ~T = 0.4 and hpR = 42,800 kJ/kg.
1.4
A J57 turbojet engine is tested at sealevel, static, standardday conditions (Po = 14.696 psia, To = 518.7°R, and Vo = 0). At one test point, the thrust is 10,200 lbf while the airflow is 164 lbm/s and the fuel flow is 8520 lbm/h. Using these data, estimate the exit velocity Ve for the case of exit pressure equal to ambient pressure (Po = Pe).
1.5
The thrust for a turbofan engine with separate exhaust streams is equal to the sum of the thrust from the engine core Fc and the thrust from the bypass stream Fs. The bypass ratio of the engine a is the ratio of the mass flow through the bypass stream to the core mass flow, or a =/nM/nc. When the exit pressures are equal to the ambient pressure,
INTRODUCTION
57
the thrusts of the core and bypass stream are given by I
Fc =   [(rhc + Fnf)Vce  thcVo] gc rnB FB =   ( V B e  VO)
gc
where Vce and Vse are the exit velocities from the core and bypass, respectively, Vo is the inlet velocity, and rnf is the mass flow rate of fuel burned in the core of the engine. Show that the specific thrust and thrust specific fuel consumption can be expressed as
F_
l~lo
1 C+inf/th c
gc
ol
)
"1~'~ Wce "q'" ]  ' ~ wBe  VO
s ='hs ,ns/,hc F (F/rh0)(1 + c0 where rno = rhc + rh~. 1.6
The CF6 turbofan engine has a rated thrust of 40,000 lbf at a fuel flow rate of 13,920 lbm/h at sealevel static conditions. If the core airflow rate is 225 lbm/s and the bypass ratio is 6.0, what are the specific thrust [lbf/ (Ibm/s)] and thrust specific fuel consumption [(lbm/h)/lbf]?
1.7
The JT9D highbypassratio turbofan engine at maximum static thrust (Vo = 0) on a sealevel, standard day (Po = 14.696 psia, To = 518.7°R) has the following data: the air mass flow rate through the core is 247 Ibm/s, the air mass flow rate through the fan bypass duct is 1248 Ibm/s, the exit velocity from the core is 1190 ft/s, the exit velocity from the bypass duct is 885 ft/s, and the fuel flow rate into the combustor is 15,750 lbm/h. Estimate the following for the case of exit pressures equal to ambient pressure (Po Pe): (a) The thrust of the engine (b) The thermal efficiency of the engine (heating value of jet fuel is about 18,400 Btu/lbm) (c) The propulsive efficiency and thrust specific fuel consumption of the engine =
1.8
Repeat Problem 1.7, using SI units.
1.9
One advanced afterburning fighter engine, whose performance is depicted in Figs. 1.14a1.14e, is installed in the HF1 fighter aircraft. Using the aircraft drag data of Fig. 1.26b, determine and plot the variation of weight specific excess power (Ps in feet per second) vs flight Mach number for
58
ELEMENTS OF PROPULSION level flight (n = 1) at 36kft altitude. Assume the installation losses are constant with values of ~binle t = 0.05 and ~bnoz= 0.02.
1.10
Determine the takeoff speed of the HF1 aircraft.
1.11
Determine the takeoff speed of the HP1 aircraft at 90% of maximum gross takeoff weight.
1.12
Derive Eqs. (1.47) and (1.48) for maximum CL/Co. Start by taking the derivative of Eq. (1.46) with respect to CL and finding the expression for the lift coefficient that gives maximum CL/Co.
1.13
Show that for maximum given by
CL/CD,the corresponding drag coefficient CD is
CD = 2CDo+ K2 C~ 1.14
An aircraft with a wing area of 800 f t 2 is in level flight (n = 1) at maximum CL/Co. Given that the drag coefficients for the aircraft are Coo = 0.02, K2 = 0, and K1 = 0.2, find (a) The maximum CL/CDand the corresponding values of CL and CD (b) The flight altitude [use Eqs. (1.29) and (1.30b)] and aircraft drag for an aircraft weight of 45,000 lbf at Mach 0.8 (c) The flight altitude and aircraft drag for an aircraft weight of 35,000 lbf at Mach 0.8 (d) The range for an installed engine thrust specific fuel consumption rate of 0.8 (lbm/h)/lbf, if the 10,0001bf difference in aircraft weight between parts b and c is due only to fuel consumption
1.15
An aircraft weighing 110,000 N with a wing area of 42 m 2 is in level flight (n = 1) at the maximum value of CL/Co. Given that the drag coefficients for the aircraft are CDo = 0.03, K2 = 0, and g l = 0.25, find the following: (a) The maximum CL/CDand the corresponding values of CL and Co (b) The flight altitude [use Eqs. (1.29) and (1.30b)] and aircraft drag at Mach 0.5 (c) The flight altitude and aircraft drag at Mach 0.75
1.16
The Breguet range equation [Eq. (1.45b)] applies for a cruise climb flight profile with constant RF. Another range equation can be developed for a level cruise flight profile with varying RF. Consider the case where we keep CL, Co, and TSFC constant and vary the flight velocity with aircraft
INTRODUCTION
59
weight by the expression
/2gcW
v= vCwcSw Using the subscripts i and f for the initial and final flight conditions, respectively, show the following: (a) Substitution of this expression for flight velocity into Eq. (1.42) gives dW
"v/Wi 
(b)
Wi
(d)
1.17
1.18
ds
Integration of the preceding between the initial i and final f conditions gives
, (c)

gFi
s]2 2(/~Fi)
For a given weight fraction Wf/Wi, the maximum range s for this level cruise flight corresponds to starting the flight at the maximum altitude (minimum density) and maximum value of vrCL/Co. For the drag coefficient equation of Eq. (1.32), maximum ~'CZ/Co corresponds to CL = (1/6K1)(x/12K1Coo + K~  K2).
An aircraft begins a cruise at a wing loading W/Sw of 100 lbf/ft 2 and Mach 0.8. The drag coefficients a r e K 1 = 0 . 0 5 6 , K 2 =  0 . 0 0 8 , and Coo = 0.014, and the fuel consumption TSFC is constant at 0.8 (lbm/h)/lbf. For a weight fraction Wf/Wi of 0.9, determine the range and other parameters for two different types of cruise. (a) For a cruise climb (maximum CL/CD) flight path, determine CL, Co, initial and final altitudes, and range. (b) For a level cruise (maximum ~CZ/Co) flight path, determine CL, Co, altitude, initial and final velocities, and range. An aircraft weighing 70,000 lbf with a wing area of 1000 ft 2 is in level flight (n = 1) at 30kft altitude. Using the drag coefficients of Fig. 1.24 and the TSFC model of Eq. (1.36b), find the following: (a) The maximum CL/CD and the corresponding values of CL, Co, and Mach number (Note: Since the drag coefficients are a function of Mach number and it is an unknown, you must first guess a value for the Mach number to obtain the drag coefficients. Try a Mach number of 0.8 for your first guess.) (b) The CL, CD, CL/CD, range factor, endurance factor, and drag for flight Mach numbers of 0.74, 0.76, 0.78, 0.80, 0.81, and 0.82
60
ELEMENTS OF PROPULSION (c) (d)
The best cruise Mach (maximum RF) The best loiter Mach (maximum EF)
1.19
An aircraft weighing 200,000 N with a wing area of 60 m 2 is in level flight (n = 1) at 9kin altitude. Using the drag coefficients of Fig. 1.24 and TSFC model of Eq. (1.36b), find the following: (a) The maximum CL/CD and the corresponding values of CL, CD, and Mach number (Note: Since the drag coefficients are a function of the Mach number and it is an unknown, you must first guess a value for the Mach number to obtain the drag coefficients. Try a Mach number of 0.8 for your first guess.) (b) The CL, CD, CL/CD, range factor, endurance factor, and drag for flight Mach numbers of 0.74, 0.76, 0.78, 0.80, 0.81, and 0.82 (c) The best cruise Mach (maximum RF) (d) The best loiter Mach (maximum EF)
1.20
What is the specific impulse in seconds of the JT9D turbofan engine in Problem 1.7?
1.21
A rocket motor is fired in place on a static test stand. The rocket exhausts 100 lbm/s at an exit velocity of 2000 ft/s and pressure of 50 psia. The exit area of the rocket is 0.2 ft2. For an ambient pressure of 14.7 psia, determine the effective exhaust velocity, the thrust transmitted to the test stand, and the specific impulse.
1.22
A rocket motor under static testing exhausts 50 kg/s at an exit velocity of 800 m / s and pressure of 350 kPa. The exit area of the rocket is 0.02 m 2. For an ambient pressure of 100 kPa, determine the effective exhaust velocity, the thrust transmitted to the test stand, and the specific impulse.
1.23
The propellant weight of an orbiting space system amounts to 90% of the system gross weight. Given that the system rocket engine has a specific impulse of 300 s, determine: (a) The maximum attainable velocity if all the propellant is burned and the system's initial velocity is 7930 m / s (b) The propellant mass flow rate, given that the rocket engine thrust is 1,670,000 N
1.24
A chemical rocket motor with a specific impulse of 400 s is used in the final stage of a multistage launch vehicle for deepspace exploration. This final stage has a mass ratio (initial to final) of 6, and its single rocket motor is first fired while it orbits the Earth at a velocity of 26,000 ft/s. The final stage must reach a velocity of 36,700 ft/s to escape the Earth's gravitational field. Determine the percentage of fuel that must be used to perform this maneuver (neglect gravity and drag).
INTRODUCTION
61
Gas Turbine Design Problems 1.D1
Background (HP1 aircraft). You are to determine the thrust and fuel consumption requirements of the two engines for the hypothetical passenger aircraft, the HP1. The twinengine aircraft will cruise at 0.83 Mach and be capable of the following requirements: 1) Takeoff at maximum gross takeoff weight WTo from an airport at 1.6km pressure altitude on a hot day (38°C) uses a 3650m (12kft) runway. The craft is able to maintain a 2.4% singleengine climb gradient in the event of engine failure at liftoff. 2) It transports 253 passengers and luggage (90 kg each) over a stillair distance of 11,120 km (6000 n mile). It has 30 min of fuel in reserve at end (loiter). 3) It attains an initial altitude of 11 km at beginning of cruise (Ps = 1.5 m/s). 4) The singleengine craft cruises at 5km altitude at 0.45 Mach (Ps = 1.5 m/s). All of the data for the HP1 contained in Example 1.2 apply. Preliminary mission analysis of the HP1 using the methods of Ref. 12 for the l l,120km flight with 253 passengers and luggage (22,770kg payload) gives the preliminary fuel use shown in Table P1.D1. Analysis of takeoff indicates that each engine must produce an installed thrust of 214 kN on a hot day (38°C) at 0.1 Mach and 1.6km pressure altitude. To provide for reasonablelength landing gear, the maximum diameter of the engine inlet is limited to 2.2 m. Based on standard design practice (see Chapter 10), the maximum mass flow rate per unit area is given by A=rh231.8_~o~0°
(kg/s)/m2
Table P1.D1
Description Taxi Takeoff Climb and acceleration Cruise Descent Loiter (30 min at 9km altitude) Land and taxi
Distance, km
Fuel used, kg 200a 840a
330 10,650 140
600 a
11,120 aThese fuel consumptionscan be considered to be constant.
5,880 a
50,240 1,090a 2,350 61,200
62
ELEMENTS OF PROPULSION Thus on a hot day (38°C) at 0.1 Mach and 1.6km pressure altitude, 0 = (38 + 273.1)/288.2 = 1.079, 00 = 1.079 x 1.002 = 1.081, 8 = 0.8256, 6o = 0.8256 x 1.007 = 0.8314, and the maximum mass flow through the 2.2mdiam inlet is 704.6 kg/s.
Calculations (HP1 Aircraft). 1)
If the HP1 starts out the cruise at 11 km with a weight of 1,577,940N, find the allowable TSFC for the distance of 10,650 km for the following cases: (a) Assume the aircraft performs a cruise climb (flies at a constant Co/CL). What is its altitude at the end of the cruise climb? (b) Assume the aircraft cruises at a constant altitude of 11 km. Determine Co/CL at the start and end of cruise. Using the average of these two values, calculate the allowable TSFC. 2) Determine the loiter (endurance) Mach numbers for altitudes of 10, 9, 8, 7, and 6 km when the HP1 aircraft is at 64% of WTO. 3) Determine the aircraft drag at the following points in the HP1 aircraft's l l,120km flight based on the fuel consumptions just listed: (a) Takeoff, M = 0.23, sea level (b) Start of cruise, M = 0.83, 11 km (c) End of cruise climb, M = 0.83, altitude = ? ft (d) End of 11km cruise, M = 0.83, 11 km (e) Engine out (88% of WTO), M = 0.45, 5 km
1 .D2
Background (HF1 Aircraft). You are to determine the thrust and fuel consumption requirements of the two engines for the hypothetical fighter aircraft HF1. This twinengine fighter will supercruise at 1.6 Mach and will be capable of the following requirements: 1) Takeoff at maximum gross takeoff weight WTO from a 1200ft (366m) runway at sea level on a standard day. 2) Supercruise at 1.6 Mach and 40kft altitude for 250 nm (463 km) at 92% of WTO. 3) Perform 5g tums at 1.6 Mach and 30kft altitude at 88% of WTO. 4) Perform 5g tums at 0.9 Mach and 30kft altitude at 88% of WTO5) Perform the maximum mission listed in the following. All of the data for the HF1 contained in Example 1.2 apply. Preliminary mission analysis of the HF1 using the methods of Ref. 12 for the maximum mission gives the preliminary fuel use shown in Table P1.D2: Analysis of takeoff indicates that each engine must produce an installed thrust of 23,500 lbf on a standard day at 0.1 Mach and sealevel altitude. To provide for optimum integration into the airframe, the maximum area of the engine inlet is limited to 5 ft2. Based on standard design practice (see Chapter 10), the maximum mass flow rate per unit
INTRODUCTION
63
Table P1.D2
Description
Distance, nm
Fuel used, Ibm
35 12 203 0 0 0 23 227
1,800 a
Warmup, taxi, takeoff Climb and acceleration to 0.9 Mach and 40 kft Accelerate from 0.9 to 1.6 Mach Supercruise at 1.6 Mach and 40 kft Deliver payload of 2000 lbf Perform one 5g turn at 1.6 Mach and 30 kft Perform two 5g turns at 0.9 Mach and 30 kft Climb to best cruise altitude and 0.9 Mach Cruise climb at 0.9 Mach Loiter (20 min at 30kft altitude) Land
700 a
500
700 a
4,400 0a 1,000 a 700 a 400 a 1,600 1,100 0a 12,400
~These fuel c o n s u m p t i o n s c a n be considered to be constant.
area for subsonic flight conditions is given by
rn A
47.5~
(lbm/s)/ft 2
Thus at 0.1 Mach and sealevel standard day, 0 = 1.0, 0o = 1.002, 6 = 1.0, 6 0   1 . 0 0 7 , and the m a x i m u m mass flow through the 5ft 2 inlet is 238.9 l b m / s . For supersonic flight conditions, the m a x i m u m mass flow rate per unit area is simply the density of the air p times its velocity V.
Calculations (HF1 Aircraft). 1)
2) 3)
I f the HF1 starts the supercruise at 4 0 k f t with a weight of 36,800 lbf, find the allowable T S F C for the distance o f 203 nm for the following cases: (a) Assume the aircraft performs a cruise climb (flies at a constant Co/CL). What is its altitude at the end of the cruise climb? (b) Assume the aircraft cruises at a constant altitude of 40 kft. Determine CD/CL at the start and end of cruise. Using the average of these two values, calculate the allowable TSFC. Find the best cruise altitude for the subsonic return cruise at 0.9 Mach and 70.75% of WTo. Determine the loiter (endurance) Mach numbers for altitudes of 32, 30, 28, 26, and 24 kft when the HF1 aircraft is at 67% of WTo.
64
ELEMENTS OF PROPULSION
4) Determine the aircraft drag at the following points in the HF1 aircraft's m a x i m u m mission based on the fuel consumptions just listed: (a) Takeoff, M = 0.172, sea level (b) Start o f supercruise, M = 1.6, 40 kft (c) End o f supercruise climb, M = 1.6, altitude = ? ft (d) End o f 40kft supercruise, M = 1.6, 40 kft (e) Start o f subsonic cruise, M = 0.9, altitude  best cruise altitude (f) Start o f loiter, altitude = 30 kft
2 Review of Fundamentals
2.1
Introduction
The operation of gas turbine engines and of rocket motors is governed by the laws of mechanics and thermodynamics. The field of mechanics includes the mechanics of both fluids and solids. However, since the process occurring in most propulsion devices involves a flowing fluid, our emphasis will be on fluid mechanics or, more specifically, gas dynamics. Understanding and predicting the basic performance of gas turbine engines and rocket motors requires a closed set of governing equations (e.g., conservation of mass, energy, momentum, and entropy). For transparency, understanding, and capturing the basic physical phenomena, we model the gas as a perfect gas and the flow as onedimensionalflow, in which the fluid properties are constant across the flow and vary only in the direction of flow (axial direction). This chapter is intended as a review of thermodynamics and onedimensional gas dynamics that are the foundation of this textbook. We include new material from Hypersonic Airbreathing Propulsion by Heiser and Pratt 13 on the graphical solution and description of onedimensional gas dynamics in Section 2.8 to improve the reader's comprehension of this fundamental material. This chapter concludes with a short section on chemical reactions to provide the fundamentals needed to understand reacting flows (e.g., combustion). With this material, we will predict the equilibrium state of reacting flows and the performance of liquidfuel rocket motors. For further study, the student is directed to references such as Ref. 14.
2,2
Equations of State and Conservation of Mass
The state of a system is described by specifying the values of the properties of the system. Pressure P, temperature T, specific internal energy e, and density p or specific volume v = 1/p are some basic thermodynamic properties. The specification of any two independent intensive properties will fix the thermodynamic state of a simple (meaning in absence of motion and force fields) system and, therefore, the values of all other thermodynamic properties of the system. The values of the other properties may be found through equations of state. The most common working fluid addressed in this book is a gas that is modeled using the perfect gas equations of state. 65
66
ELEMENTS OF PROPULSION
A function relating one dependent and two independent thermodynamic properties of a simple system of unit mass is called an equation of state. When the three properties are P, v, and T as in f(P,v, T)=0
(2.1)
the equation is called the thermal equation o f state. In general, we cannot write the functional relationship equation (2.1) in the form of an equation in which specified values of the two properties will allow us to determine the value of the third. Although humans may not know what the functional relation equation (2.1) is for a given system, one does exist and nature always knows what it is. When the solution set of Eq. (2.1) cannot be determined from relatively simple equations, tables that list the values of P, v, and T (elements of the solution set) satisfying the function may be prepared. This has been done for water (in all of its phases), air, and most c o m m o n gases. The functional relation between the internal energy e of a simple system of unit mass and any two independent properties for the set P, v ( = l / p ) , T is called the energy equation of state. This equation can be written functionally as e = e(T, v)
or
e = e(P, v)
or
e = e(P, T)
(2.2)
As with the thermal equation of state, we may not be able to write an analytical expression for any of the functional relations of Eq. (2.2). The important thing is that energy is a property; hence, the functional relations exist. If the solution sets of the thermal and energy equations of state of a simple system of unit mass are known, all thermodynamic properties of the system can be found when any two of the three properties P, v, T are specified. From the solution set, we can form a tabulation of v and e against specified values of P and T for all states of the system. From these known values of P, T, v, and e, we can determine any other property of the simple system. For example, the value of the property enthalpy h is found for any state of the system by combining the tabulated values o f P, v, and e for that state by h = e + Pv
(2.3)
Four other definitions are listed here for use in the later sections of this chapter: specific heat at constant volume c~, specific heat at constant pressure Cp, ratio of specific heats % the speed of sound a, and the Mach number M:
c~ 
(2.4) u
ce =
(2.5) P
3' = cp/c~
(2.6)
R E V I E W OF F U N D A M E N T A L S
2.2.1
67
a2=_ gc(OP'~
(2.7)
M = V/a
(2.8)
Definition of Steady Flow
Consider the flow of fluid through the control volume o shown in Fig. 2.1. If the properties of the fluid at any point i in the control volume do not vary with time, the flow is called steady flow. For such flows we may conclude that for any property R within odR~ = 0 dt
in steady flow
(2.9)
2.2.2 Definition of OneDimensional Flow If the intensive stream properties at a permeable control surface section normal to the flow directions are uniform, the flow is called onedimensional. Many flows in engineering may be treated as steady onedimensional flows. The term onedimensional is synonymous in this use with uniform and applies only at a control surface section. Thus the overall flow through a control volume may be in more than one dimension and still be uniform (onedimensional flow) at permeable sections of the control surface normal to the flow direction. The flow in Fig. 2.2 is called onedimensional flow because the intensive properties, such as velocity, density, and temperature, are uniform at sections 1 and 2.
2.2.3
Conservation of Mass Equation
The law of mass conservation for any control volume system tr is simply dm~ dt
=
thin  thou t
(2. lOa)
I pVi dA
(2.10b)
where /n =
A
O"
i
~. v I i_ .
.
.
.
.ff~
~v 2
.
(1) Fig. 2.1
(2) Control volume for steady flow.
68
ELEMENTS OF PROPULSION
V/ ' ~
.~.~2)
(1) Fig. 2.2 Onedimensional flow through a convergent duct. The flow is uniform at sections 1 and 2, hence onedimensional, even though the flow direction may vary elsewhere in the flow.
and V± is the velocity component normal to area A. This equation is known as the conservation of mass equation. For steady flows through any control volume, Eq. (2.10a) simplifies to (2.11)
thout ~ thin
If the flow is steady and onedimensional through a control volume with a single inlet and exit such as shown in Fig. 2.2, then by Eq. (2.10b)
th = pAV± which is called the
(2.12a)
onedimensional massflow equation, and (2.12b)
plA1V±~ = P2A2V_L2
2.3 Steady Flow Energy Equation We consider steady onedimensional flow of a fluid through a control volume and surface or (Fig: 2.3). Fluid crosses or at the in and out stations only. A shaftwork interaction Wx and heat interaction Q occur at the boundary of tr. If the energy within the control volume does not change with time (steady flow), the first law of thermodynamics can be written as •
Q
V2
V2
Wx
where the dimensions of Q,l~/x, etc., are those of power or energy per unit time. Dividing by th gives V2
qWx=(h+~gc
+gz)
~c/out
(h
V2
"~ ~gc "Jr~cZ)in
(2.14)
REVIEW OF FUNDAMENTALS
•• Vin
i
I
1
v°ut
)out(
Zin (iln) ~
Fig. 2.3
~
69
~ Controlsurfaceo l
Zout
Steady flow through control volume ~r.
where q and Wx are the heat and shaftwork interactions per unit mass flow through 0. All of the terms in this equation have units of energy per unit mass.
Example 2.1 The first step in the application of the steady flow energy equation is a clear definition of a control surface 0. This is so because each term in the equation refers to a quantity at the boundary of a control volume. Thus, to use the equation, one needs only to examine the control surface and identify the applicable terms of the equation. In the application of Eq. (2.14) to specific flow situations, many of the terms are zero or may be neglected. The following example will illustrate this point. The perfect gas equation (presented in detail later in this chapter) is used to model the relationship between change in enthalpy and temperature of the gas. Consider a turbojet aircraft engine as shown in Fig. 2.4a. We divide the engine into the control volume regions: o1: inlet 02: compressor 03: combustion chamber
04: turbine 05: nozzle
Let us apply the steady flow energy equation to each of these control volumes. In all cases, the potential energy change (gz/gc)out  (gz/gc)in is zero and will be ignored. It is advisable in using the steady flow energy equation to make two sketches of the applicable control surface 0, showing the heat and shaft work interactions (q, Wx) in one sketch and the fluxes of energy [h, V2/(2gc)] in the second sketch. The term [h + V2/(2gc)]out is a flux per unit mass flow of internal energy
70
ELEMENTS OF PROPULSION rhf
,
t tU ~ ~ ~
(1)
(o)
(~
i
~.
O2 • .
Inlet
(a)
"~
~~i
I
., ~ e
G4
a3
Compressor . _ Combustor _ _ Turbine_ _ Nozzle .
(2)
Fig. 2.4a
Fuel tank

I
I
(3)
(4)
(5)
(9)
(e)
Control v o l u m e for a n a l y z i n g each c o m p o n e n t of a turbojet engine.
e, kinetic energy vZ/(2gc), and flow w o r k Pv. W e will use the energy to include the flow work flux also.
expression flux of
a) Inlet and nozzle: or1 and 05. There are no shaft work interactions at control surfaces crI and ~r5. Heat interactions are n e g l i g i b l e and m a y be taken as zero. Therefore, the steady flow energy equation, as depicted in Fig. 2.4b, for the inlet or n o z z l e control surfaces gives the result
)ou,
0 = ( h + V2
Nozzle: numerical example Let the gases flowin~g through the nozzle control v o l u m e o5 be perfect with Cpgc=6OOOft/(s°R). D e t e r m i n e V9 for T5 = 1800°R, V5 = 400 ft/s, and T9 = 1200°R. q = 0, Wx= 0 I I
,
a 1, %
I
(In) hin
,
(e)
__ ]
Interactions 0
Fig. 2.4b
(Out) "1I ] , hout
I
I I I
,,
~gc in
~ "
I, I
a~,%
I
,
I
l
~gc out
Net energy flux ( h + V2 ~gc )oat 
(h
V2 + ~gc )in
E n e r g y equation applied to control v o l u m e s o"1 and ors.
REVIEW OF FUNDAMENTALS
71
Solution: From the steady flow energy equation with 5 and 9 as the in and out stations, respectively, we have h9 q ~ c : h5 q  2go and
V9 = v/2gc(h5  h9)t V2 = v/2Cpgc(T5  T9)t V2 or
=~/2(6000)(1800  1200) + 4002 SO
= 2700 ft/s
b) Compressor and turbine: o2 and o4. The heat interactions at control surfaces o2 and o4 are negligibly small. Shaft work interactions are present because each control surface cuts a rotating shaft. The steady flow energy equation for the compressor or for the turbine is depicted in Fig. 2.4c and gives V2
V2
Wx: (h~gc)out
(h'q'~gc)in
Compressor and turbine: numerical example For an equal mass flow through the compressor and turbine of 185 lb/s, determine the compressor power and the turbine exit temperature T5 for the following conditions: Cpgc : 6000 ft2/(s2°R) Compressor
Turbine
T2 = 740°R, T3 = 1230°R
T4 = 2170°R, T5 = ?
V2 = V3
V4 = V5
Wx< 0 for c 2 (compressor) Wx> 0 for 134 (turbine)
(In) I
"1 I
°2' °4
Interactions
Fig. 2.4c
[ ,
Wx V2
I'
q=0
wx
hin
[ ~
(G)i ~
(Out) I 1 I •
[ n
02' 04
'I '
hout
I
v2
II
' (G)o
ut
N e t e n e r g y flux
(h+~~c)out (h+ V2
~gc ) i n
E n e r g y e q u a t i o n a p p l i e d to control v o l u m e s 0.2 and 0"4.
72
ELEMENTS OF PROPULSION
Solution:
The compressor power Wc = (inWx)~2 is, with V2 = V3, ~VVc= th(h3  h2)  thcp(T3  T2) 6000 (ft/s) 2 =  ( 1 8 5 lbm/s) 32.174 ftlbf/(lbms z ) (1230  740)   1 6 . 9 × 106 fllbf/s x
lhp 550 ftlbf/s
= 30,700hp The minus sign means the compressor shaft is delivering energy to the air in tr2. The turbine drives the compressor so that the turbine power Wt = (rhWx)~4 is equal in magnitude to the compressor power. Thus Wt = Wc, where, from the energy equation, ff't = rh(h5

h4)
and
Wc   t h ( h 3

h2)
Thus ;ncp(T5 
T4) =  & c p ( T 3
 T2)
and =
T4  ( ~

T2)
= 2170°R  (1230°R  740°R) = 1680°R = 1220°F
c) Combustion chamber: ~r3. Let us assume that the fuel and air entering the combustion chamber mix physically in a mixing zone (Fig. 2.4d) to form what we will call reactants (denoted by subscript R). The reactants then enter a combustion zone where combustion occurs, forming products of combustion (subscript P) that leave the combustion chamber. We apply the steady flow energy equation to combustion zone ira. Because the temperature in the combustion zone is higher than that of the immediate surroundings, there is a heat interaction between o'3 and the surroundings that, per unit mass flow of reactants, is negligibly small (q < 0 but q = 0). Also the velocities of the products leaving and of the reactants entering the combustion zone are approximately equal. There is no shaft work interaction for o'3. Hence the steady flow energy equation, as depicted in Fig. 2.4d, reduces to hR3 = he4
(2.15 )
We must caution the reader about two points concerning this last equation. First, we cannot use the relation cpAT for computing the enthalpy difference between
REVIEW OF FUNDAMENTALS Fuel
(3) I
Air
Mixing zone
IiJ' I
S~3
rhR
(In)
(Out) I
I
hR ~ I 2g'c
q=O, wx=O
Interactions
mp
I
I I I j
o3
(4)
Combustion zone
"1
I I I I
73
=
I , hp
I
l
2gc
vp=vR Net energy flux
_
+ 2ggc)outFig. 2.4d
)in
Energy equation applied to control v o l u m e s o'3.
two states of a system when the chemical aggregation of the two states differs. Second, we must measure the enthalpy of each term in the equation relative to the same datum state. To place emphasis on the first point, we have introduced the additional subscripts R and P to indicate that the chemical aggregations of states 3 and 4 are different. To emphasize the second point, we select as our common enthalpy datum a state d having the chemical aggregation of the products at a datum temperature Td. Then, introducing the datum state enthalpy (he)d into the last equation, we have hR3  hpd ~ hP4  hpa
(2.16)
Equation (2.16) can be used to determine the temperature of the products of combustion leaving an adiabatic combustor for given inlet conditions. If the combustor is not adiabatic, Eq. (2.16), adjusted to include the heat interaction term q on the lefthand side, is applicable. Let us treat the reactants and products as perfect gases and illustrate the use of Eq. (2.16) in determining the temperature of the gases at the exit of a turbojet combustion chamber via an example problem.
Combustion chamber: numerical example For the turbojet engine combustion chamber, 45 Ibm of air enters with each 1 Ibm of JP8 (kerosene) fuel. Let us assume these reactants enter an adiabatic combustor at 1200°R. The heating value her of JP8 is 18,400 Btu/lbm of fuel at 298 K. [This is also called the
74
ELEMENTS OF PROPULSION
lower heating value (LHV) of the fuel.] Thus the heat released (AH)298K by the fuel per 1 Ibm of the products is 400 Btu/lbm (18,400/46) at 298 K. The following data are known: Cpe = 0.267 Btu/(lbm°R)
and
Cpl¢ = 0.240 Btu/(lbm°R)
Determine the temperature of the products leaving the combustor. Solution: A plot of the enthalpy equations of state for the reactants and the products is given in Fig. 2.5. In the plot, the vertical distance hR  he between the curves of hR and he at a given temperature represents the enthalpy of combustion AH of the reactants at that temperature (this is sometimes called the heat o f combustion). In our analysis, we know the enthalpy of combustion at T d 298 K (536.4°R). ="
States 3 and 4, depicted in Fig. 2.5, represent the states of the reactants entering and the products leaving the combustion chamber, respectively. The datum state d is arbitrarily selected to be products at temperature Td. State d' is the reactants' state at the datum temperature Td. In terms of Fig. 2.5, the lefthand side of Eq. (2.16) is the vertical distance between states 3 and d, or hR3  
hed = hR3  hRa, + hRd,  hpa
h,
R
I
I
I
Td
T3
T4
p
j.
T
Fig. 2.5 Enthalpy vs temperature for reactants and products treated as perfect gases.
REVIEW OF FUNDAMENTALS
75
and since A h l ¢ = cpRAT
and
hRd,  hpd = (AH)ra
then hR3  hpa = cpR(T3  Td) + ( A H ) ~
(i)
Similarly, the righthand side of Eq. (2.16) is hp 4  hea = cpp(T4  Td)
(ii)
Substituting Eqs. (i) and (ii) in Eq. (2.16), we get CpR(T3  Td) + (AH)Td = Cpp(T4  Ta)
(2.17)
We can solve this equation for T4, which is the temperature of the product gases leaving the combustion chamber. Solving Eq. (2.17) for T4, we get T4
=
cpR(T3  Ta) + (AH)Ta
Cpp
~Ta
_ 0.240(1200  536.4) + 400 + 536.4 0.267 = 2631°R (2171°F) This is the socalled adiabatic flame temperature of the reactants for a 45 : 1 mixture ratio of air to fuel weight. For the analysis in portions of this book, we choose to sidestep the complex thermochemistry of the combustion process and model it as a simple heating process. The theory and application of thermochemistry to combustion in jet engines are covered in many textbooks, such as the classic text by Penner (see Ref. 14).
2.4 Steady Flow Entropy Equation From the second law of thermodynamics, we have for steady State
Sout  Sin > ~ Tsource
(2.18)
where S = rhs, Q is the rate of heat interaction into the control volume ~r, and Ts..... is the heat source temperature as shown in Fig. 2.3. For steady state and adiabatic flow through a control volume, this reduces to the statement that the entropy flux out is greater than or equal to the entropy flux in
Sour ~> Sin
76
ELEMENTS OF PROPULSION
For one outlet section (2) and one inlet section (1), this and the continuity condition [Eq. (2.11)] yield
(2.19)
S2 > S 1
2.5 Steady Flow Momentum Equation Newton's second law of motion for a control volume o~is ~  ' ~ F ~ = 1 ( dM~ ) g~ T + n o u t  Win
(2.20)
where M~ = f~ V dm is the momentum of the mass within the control volume oand Mout = lout Vdrh is the momentum flux leaving the control volume. In words, Eq. (2.20) says that the net force acting on a fixed control volume ois equal to the time rate of increase of momentum within oplus the net flux of momentum from o. This very important momentum equation is in fact a vector equation, which implies that it must be applied in a specified direction to solve for an unknown quantity. Applying control volume equations to a steady flow problem gives useful results with only a knowledge of conditions at the control surface. Nothing needs to be known about the state of the fluid interior to the control volume. The following examples illustrate the use of the steady onedimensional flow condition and the momentum equation. We suggest that the procedure of sketching a control surface (and showing the applicable fluxes through the surface and the applicable forces acting on the surface) be followed whenever a control volume equation is used. This situation is similar to the use of free body diagrams for the analysis of forces on solid bodies. We illustrate this procedure in the solutions that follow.
Example 2.2 Water (p = 1000 k g / m 3) is flowing at a steady rate through a convergent duct as illustrated in Fig. 2.6a. For the data given in the figure, find the force of the fluid F~o acting on the convergent duct D between stations 1 and 2. Solution: We first select the control volume o" such that the force of interest is acting at the control surface. Because we want the force interaction between D and the flowing water, we choose a control surface coincident with the inner wall surface of D bounded by the permeable surfaces 1 and 2, as illustrated in Fig. 2.6a. By applying the steady onedimensional continuity equation [Eq. (2.11)], as depicted in Fig. 2.6b, we find V1 as follows:
plAI VI = P2A2V2 A2 1/2 V1 =All = 3 m/s
(Pl = P 2 )
REVIEW OF FUNDAMENTALS
77
O
v~
II
,
~
(1)
v2
PlA1VI I, I .I . . . . .
.2
Mass flux in = mass flux out
(2)
P1 = 137,900 Pa A1 = 0 . 2 m 2 Pl = P2 = 1000 kg/m 3
 I P2A2V2
P2 = 101,325 Pa A2 = 0.1 m 2 V2 = 6 m/s b) Continuity equation
a) Given data 0
P1 A1
,I I
VI(PlA1V1)'~I , ~
It _ _l
"~1V2(y2A2V2)
P2 A2
x forces on c
Net x momentum from G c) Momentum equation
Fig. 2.6
Flow through a convergent duct.
W i t h V 1 determined, we can apply the m o m e n t u m Eq. (2.20) to o and find the force of the duct walls on or, d e n o t e d by FDo (FoD = Foo.). B y s y m m e t r y , Foo is a horizontal force, and so the horizontal x c o m p o n e n t s o f forces and m o m e n t u m fluxes will be considered. T h e x forces acting on o are d e p i c t e d in Fig. 2.6c a l o n g with the x m o m e n t u m fluxes through o. F r o m Fig. 2.6c, w e h a v e m o m e n t u m :
P1AI  FDo.  P2A2 = 1 [(p2A2V2)V 2 _ (PlA 1Vl)Vl]
A n d by Fig. 2.6b, continuity:
plA1V1 = p2A2V2 = C o m b i n i n g the continuity and m o m e n t u m equations, w e obtain
in P1A1  Foo  P2A2 = ~ ( V 2  V1) gc
78
ELEMENTS OF PROPULSION
or
rh V
Foa = P1A1  P2A2    ( 2  
Vl)
gc
With rh = plA1V!  1000 k g / m 3 x 0.2 m 2 x 3 m / s = 600 kg/s, we have
FD~ = 137,000 N / m 2 x 0.2 m 2  101,325 N / m z x 0.1 m 2  600 kg/s x (6  3) m / s or
Fo~ = 27,580 N  10, 132 N  1800 N and  15,648 N
acts to left in assumed position
Finally, the force of the water on the duct is acts to fight
F~rD = FDcr = 15,648 N
Example 2.3 Figure 2.7 shows the steady flow conditions at sections 1 and 2 about an airfoil mounted in a wind tunnel where the frictional effects at the wall are negligible. Determine the section drag coefficient Cd of this airfoil.
Solution: Since the fluid is incompressible and the flow is steady, the continuity equation may be used to find the u n k n o w n velocity VB as follows: (pAg)l
that is, rhL = rh2
= (pAV)2
or
(2
1 )
P2
and
A1 = A 2
2
1V
p l A I V a = P 2 ~VBq~Vc A2 but Pl =
Va ~ "~ VB "~ 5
C
thus VB _  3"~ VA  ~1V C = 31.5 m / s
REVIEW OF FUNDAMENTALS
79
G / / / / / l V£Y 30m/s 3Om,
I VB=7 ]//" 0.1 m
I 0.3m
.I
I /
I
~
.
~
Wake
p = 0.618 kg/m3 = constant
I "1
O . l m VC = 27 m/s
I
~ ~ I I
I
VB=? t_
//////,(/?,
/ / / / /
(1) P1 = 74,730 Pa Tunnel area = 0.1 m 2 Chord = 0.15 m
Fig. 2.7
O.lm
vB VA
(2) P2  74,700 Pa
W i n d tunnel d r a g d e t e r m i n a t i o n for an airfoil section.
The momentum equation may now be used to find the drag on the airfoil. This drag force will include both the skin friction and pressure drag. We sketch the control volume o with the terms of the momentum equation as shown in Fig. 2.8. Taking forces to right as positive, we have from the sketch
Z F~ = PIA1 
P2A2 + FDo= plalV2a
a//1 = ( P l A 1 V A ) V A
M2=
p2(~A2)V~ +p2(1Az)V~ 2
2
=p2A2( 2V21Bt~V2"~c) '
P1 AI
I
~tl
I
P2 A2
" ~t2
[ Momentum equation sketch ZF o = ~/2  ~'/1 F i g . 2.8
S k e t c h for m o m e n t u m
e q u a t i o n for airfoil section.
80
ELEMENTS OF PROPULSION
For p : Pl = P2 and A : A1 = A2,
or
7400)01+0618 01i0 = 3.0 N  0.278 N  F o a = 2.722 N
.'. F o a
acts to left
Fan = drag force for section : Foa /~ .
FaD . . b
Cd F b' qc
and
FaD
2.722 N . . 0.333m and
acts to left
8.174 N / m q = pV2 2gc
where
V~=Va
Fb 8.174 = 0.196 Cd = [(pVZ)/(2gc)]C  (0.618 × 302/2)0.15
Example 2.4 Figure 2.9 shows a test stand for determining the thrust of a liquidfuel rocket. The propellants enter at section 1 at a mass flow rate of 15 k g / s , a velocity of 30 m / s , and a pressure of 0.7 MPa. The inlet pipe for the propellants is very flexible, and the force it exerts on the rocket is negligible. At the nozzle exit, section 2, the area is 0.064 m 2, and the pressure is 110 kPa. The force read by the scales is 2700 N, atmospheric pressure is 82.7 kPa, and the flow is steady.
Propellantn] = 15 kg/s
,
~0~
~
//;,
0.8m
,
~ !
~///////~ Fig. 2.9

, ~ Combustion = / chamber Nozzle~ ()
0/()
U//////////////////// Liquidfuel rocket test setup.
REVIEW OF FUNDAMENTALS
81
Determine the exhaust velocity at section 2, assuming onedimensional flow exists. Mechanical frictional effects m a y be neglected.
Solution: First, determine the force on the lever by the rocket to develop a 2700 N scale reading. This may be done by summing moments about the fulcrum point 0 (see Fig. 2.10a). Sum the horizontal forces on the rocket engine as shown in Fig. 2.10b. W e note that the unbalanced pressure force on the exterior of the rocket engine is PaA2, and the interior forces (pressure and friction) are contained within the force Fc~. Next, draw an internal volume ~r around a) °
0.2m
Y_,Mo= 2700 x 0.8FR× 0.2 = 0 ... FR = 2700 x 0.8 = 10,800 N 0.2
2700 N
b)
_J
P~2
Fc~ FR
1
x
c)
y
I I
f °
\
S
Fcc
PzA2 £'outx = k2v2
¢ •
Fig. 2.10
"%..
X
Momentum
e q u a t i o n s k e t c h ~ F ~ = (~/2 
h41)/gc.
82
ELEMENTS OF PROPULSION
the fluid within the rocket engine as shown, and indicate the horizontal forces and momentum flux (see Fig. 2.10c). Summing the forces on the rocket engine as shown in Fig. 2.10b, we obtain FR + PaA2 = Fc~r
Applying the momentum equation to the control volume oshown in Fig. 2.10c, we obtain Foc  P2A2 =
m2V2 gc
Combining these two equations to remove Foc gives FR  (/)2  Pa)A2 =
k2V2 gc
which is the same as Eq. (1.52). Because the flow is steady, the continuity equation yields rhl = th2 = 15 kg/s. Therefore, v2 =
FR  (P 2


Pa)A2
tn2/gc
10,800  [(110  82.7) x 103 N/m2](0.064 m 2) 15kg/s = 603.5 m/s
2.6 Perfect Gas 2.6.1 General Characteristics The thermodynamic equations of state for a perfect gas are P = pRT
(2.21)
e = e(T)
(2.22)
where P is the thermodynamic pressure, p is the density, R is the gas constant, Tis the thermodynamic temperature, and e is the internal energy per unit mass and a function of temperature only. The gas constant R is related to the universal gas constant ~ and the molecular weight of the gas M by 7¢ R = M
(2.23)
Values of the gas constant and molecular weight for typical gases are presented in Table 2.1 in several unit systems; 7~ = 8.31434kJ/(kmolK)= 1.98718 Btu/ (mol°R).
REVIEW OF FUNDAMENTALS Table 2.1
Gas Air Argon Carbon dioxide Carbon monoxide Hydrogen Ni~ogen Oxygen Sulfur dioxide W~er vapor
83
Properties of ideal gases at 298.15 K (536.67°R)
Molecular %, Cp, R, R, (ft Ibm)/ weight kJ/(kg. K) Btu/(lbm. °R) kJ/(kg. K) (Ibm. °R) 28.97 39.94 44.01 28.01 2.016 28.02 32.00 64.07 18.016
1.004 0.523 0.845 1.042 14.32 1.038 0.917 0.644 1.867
0.240 0.125 0.202 0.249 3.42 0.248 0.217 0.154 0.446
0.286 0.208 0.189 0.297 4.124 0.296 0.260 0.130 0.461
53.34 38.69 35.1 55.17 766.5 55.15 48.29 24.1 85.78
y 1.40 1.67 1.29 1.40 1.40 1.40 1.39 1.25 1.33
From the definition of enthalpy per unit mass h of a substance [Eq. (2.3)], this simplifies for a perfect gas to h=e+RT
(2.24)
Equations (2.24) and (2.3) combined show that the enthalpy per unit mass is also only a function of temperature h = h(T). Differentiating Eq. (2.24) gives dh = de + R dT
(2.25)
The differentials dh and de in Eq. (2.25) are related to the specific heat at constant pressure and specific heat at constant volume [see definitions in Eqs. (2.4) and (2.5)], respectively, as follows: dh = Cp d T de = Cv d T
Note that both specific heats can be functions of temperature. These equations can be integrated from state 1 to state 2 to give T2 i*
e2  el = I C v d T
(2.26)
d
TI T2
hz  hi = ] Cp d T
(2.27)
Tl
Substitution of the equations for dh and de into Eq. (2.25) gives the relationship between specific heats for a perfect gas cp=cv+R
(2.28)
84
ELEMENTS OF PROPULSION
and y is the ratio of the specific heat at constant pressure to the specific heat at constant volume, or, as in Eq. (2.6), 3' = Cp/Cv
The following relationships result from using Eqs. (2.28) and (2.6): R
 y
1
(2.29)
Cv
R
y1   3'
cp
(2.30)
The Gibbs equation relates the entropy per unit mass s to the other thermodynamic properties of a substance. It can be written as ds =
de + P d(1/p) T
=
dh  ( l / p ) dP T
(2.31)
For a perfect gas, the Gibbs equation can be written simply as dT . nd(1/P) ds = cv ~t t~
(2.32)
dT dP ds = Cp f  R ~ 
(2.33)
These equations can be integrated between states 1 and 2 to yield the following expressions for the change in entropy s2  sl: T2
Cv~qRf~ pl
S2   S 1 =
(2.34)
P2
Tj
T2
I $2   S1 =
dT Cp ~   
E~ P2 g
Tl
pm i
(2.35)
If the specific heats are known functions of temperature for a perfect gas, then Eqs. (2.26), (2.27), (2.34), and (2.35) can be integrated from a reference state and tabulated for further use in what are called gas tables. The equation for the speed of sound in a perfect gas is easily obtained by use of Eqs. (2.7) and (2.21) to give a Tx/Y~cT
2.6.2
(2.36)
Calorically Perfect Gas
A calorically perfect gas is a perfect gas with constant specific heats (Cp and Cv). In this case, the expressions for changes in intemal energy e, enthalpy h, and
REVIEW OF FUNDAMENTALS
85
entropy s simplify to the following: (2.37)
ezel =cv(TzTl) h2  hi = cp(T2 
T1)
(2.38)
T2
::¢P2
f~ T2
Pl ~ P2
S2  Sl = Cv f~c~11 R

(2.39) (2.40)
Equations (2.39) and (2.40) can be rearranged to give the following equations for the temperature ratio Tz/T1: T2
P2
~R/cv
$2  s1
~1 = \ ~ /
eXPcv
~=
exp
11
',P1.]
cp
From Eqs. (2.29) and (2.30), these expressions become
T2 (p2"]71 s z  sl TT= \pT/ e X P   c v
(2.41)
T2 {P2"~(~1)/~ sz sl T~ ~PII) exp Cp
(2.42)

2.6.3 Isentropic Process For an isentropic process (S 2 following equations:
=
S1) ,
T2
Eqs. (2.41), (2.42), and (2.21) yield the
(P2~ (y1)/~' = \PTJ
T2
eT
=
(p2] ~1
(2.43) (2.44)
=
Note that Eqs. (2.43), (2.44), and (2.45) apply only to a calorically perfect gas undergoing an isentropic process.
86
ELEMENTS OF PROPULSION
Example 2.5 Air initially at 20°C and 1 atm is compressed reversibly and adiabatically to a final pressure of 15 atm. Find the final temperature.
Solution: Because the process is isentropic from initial to final state, Eq. (2.43) can be used to solve for the final temperature. The ratio of specific heats for air is 1.4: /'15\0'4/1"4
( e( r2)1 ) / r ~ Tz = T1
= (20 + 2 7 3 . 1 5 ) ~ T )
= 293.15 × 2.1678 = 635.49 K (362.34°C)
Example 2.6 Air is expanded isentropically through a nozzle from T1 = 3000°R, V1 = 0, and P1 = 10 atm to V2 = 3000 ft/s. Find the exit temperature and pressure.
Solution: Application of the first law of thermodynamics to the nozzle gives the following for a calorically perfect gas: cpT1 "JI~gc = opT2 ~ 2gc This equation can be rearranged to give T2: T 2 = Tl
V ~  V~ _ 3000
2gcCp
30002 2 x 32.174 x 0.240 x 778.16
= 3000  748.9 = 2251.1°R Solving Eq. (2.43) for
P2 gives
[T2\~'/(~1) 1 /2251"1\3"s P2 = P, ~~I) = 0~ 3~~) = 3 . 6 6 a t m
2.6.4
Mollier Diagram for a Perfect Gas
The Mollier diagram is a thermodynamic state diagram with the coordinates of enthalpy and entropy s. Because the enthalpy of a perfect gas depends on temperature alone, dh = Cp dT temperature can replace enthalpy has the coordinate of a Mollier diagram for a perfect gas. When temperature T and entropy s are the coordinates of a Mollier diagram, we call it a Ts diagram. We can construct lines of constant pressure and density in the Ts diagram by using Eqs. (2.34) and (2.35). For a calorically
REVIEW OF FUNDAMENTALS
87
[*
u
0 Fig. 2.11
0.4
0.8 1.2 Entropy [kJ/(kg • K)]
1.6
.I 2
A Ts diagram for air as a calorically perfect gas.
perfect gas, Eqs. (2.39) and (2.40) can be written between any state and the entropy reference state (s = 0) as T s = c v ~ee~ref S = Cp ~  
T
p R ~¢ Pref P
 R ~
Zref
Pref
where Tref, Pref, and Pref are the values o f temperature, pressure, and density, respectively, when s = 0. Because the most c o m m o n working fluid in gas turbine engines is air, Fig. 2.11 has been constructed for air by using the preceding equations with these data: Cp = 1.004 k J / ( k g . K ) ,
Tref = 288.2 K
R = 0.286 k J / ( k g . K ) ,
2.6.5
Pref = 1.225 k g / m 3
Pref = 1 atm = 101,325 Pa
Mixtures of Perfect Gases
W e consider a mixture of perfect gases, each obeying the perfect gas equation: P V = NTCT
88
ELEMENTS OF PROPULSION
where N is the number of moles and 7~ is the universal gas constant. The mixture is idealized as independent perfect gases, each having the temperature T and occupying the volume V. The partial pressure of gas i is
Pi = NiT~.
T
According to the GibbsDalton law, the pressure of the gas mixture of n constituents is the sum of the partial pressures of each constituent: n
P= ~
(2.46)
Pi
i=1
The total number of moles N of the gas is n
(2.47)
N : ~ Ni i=1
The ratio of the number of moles of constituent i to the total number of moles in the mixture is called the mole fraction Xi. By using the preceding equations, the mole fraction of constituent i can be shown to equal the ratio of the partial pressure of constituent i to the pressure of the mixture:
Ni
Pi
Xi = ~ = ~
(2.48)
The GibbsDalton law also states that the internal energy, enthalpy, and entropy of a mixture are equal, respectively, to the sum of the internal energies, the enthalpies, and the entropies of the constituents when each alone occupies the volume of the mixture at the mixture temperature. Thus we can write the following for a mixture of n constituents. Energy: n
n
E = Z Ei = Z m i e i i=1
(2.49)
i=1
Enthalpy: n
mihi
(2.50)
S:~Si=~~misi
(2.51)
H = Z Hi = i=1
i=1
Entropy:
i=1
where m i is the mass of constituent i.
i=1
REVIEW OF FUNDAMENTALS
89
The specific heats of the mixture follow directly from the definitions of Cp and c~ and the preceding equations. For a mixture of n constituents, the specific heats are n
n
E micpi
E micoi
i=1 Cp   
i=l cv   
and
m
(2.52)
m
where m is the total mass of the mixture.
2.6.6
Gas Tables
In the case of a perfect gas with variable specific heats, the specific heat at constant pressure Cp is normally modeled by several terms of a power series in temperature T. This expression is used in conjunction with the general equations presented and the new equations that are developed next to generate a gas table for a particular gas (see Ref. 15). For convenience, we define T
(2.53)
h = I Cp d T Tref T
~b ~
I
cp ~
(2.54)
Tref
P r ~
Vr 
e x p (~ 
(/)ref R
exp ( 1 ~
i cv a T ) r~f
(2.55)
(2.56)
where Pr and or are called the r e d u c e d p r e s s u r e and r e d u c e d v o l u m e , respectively. Using the definition of ~b from Eq. (2.54) in Eq. (2.35) gives P~ $2  S1 = ~2  ~1  R E~e% P1 For an isentropic process between states 1 and 2, Eq. (2.57) reduces to P~ ~2 ~1 ~g~v  ~ P1 which can be rewritten as P2 =  ~b1 _ exp (~b2/R) (~'l)s=cons t exp~b2~ exp(~bl/R)
(2.57)
90
ELEMENTS OF PROPULSION
Using Eq. (2.55), we can express this pressure ratio in terms of the reduced pressure Pr a s
(Pq P1 ]
_Pr2
(2.58)
Prl
s=const
Likewise, it can be shown that (v2~ /) 1 ,/s=const
=
Vr2
(2.59)
/)rl
For a perfect gas, the properties h, Pr, e, Vr, and ~bare functions of T, and these can be calculated by starting with a polynomial for Cp. Say we have the seventhorder polynomial
Cp = Ao + AIT + A2T2 FA3T3 + A4T4 +A5T 5 + A6T6 t A7T7
(2.60)
The equations for h and ~bas functions of temperature follow directly from using Eqs. (2.53) and (2.54): A1 ,re +~As T3 +~A3 T4 + A4 A5 T6 + ~ 6 T7 +~Z7T8 h=href+ZoT+'~l ~ T5 +~
(2.61)
A2,,,2 A4T4 15 A5T5"F~ A6T6}~T7 q~= 6ref+ a° E~T+ alT'}21q~~ T3 +'4
(2.62)
After we define reference values, the variations of P~ and Eqs. (2.55) and (2.56), and the preceding equations.
2.6.7
Vr follow from
Comment for Appendix D
Typically, air flows through the inlet and compressor of the gas turbine engine whereas products of combustion flow through the engine components downstream of a combustion process. Most gas turbine engines use hydrocarbon fuels of composition ( C H 2 ) n. We can use the preceding equations to estimate the properties of these gases, given the ratio of the mass of fuel burned to the mass of air. For convenience, we use the fuel/air ratio f, defined as f 
mass of fuel mass of air
(2.63)
The maximum value of f is 0.0676 for the hydrocarbon fuels of composition (CH2)n. Given the values of Cp,h, and 4) for air and the values of combustion products, the values of Cp, h, and ~b for the mixture follow directly from the mixture
REVIEW OF FUNDAMENTALS Table 2.2
Constants for air and combustion products used in Appendix D and program AFPROP (from Ref. 16) Combustion products of air and (CH2),, fuels
Air alone Ao A1 A2 A3 A4
As A6 A7 hre f
~ref
91
2.5020051 5.1536879 6.5519486 6.7178376   1.5128259 7.6215767  1.4526770 1.0115540  1.7558886 0.0454323
x 10  l x 10 5 x 10 8
Ao A1 A2
x 10 12
A3
× 10 14 x 10 18 x 10 21
A5
X 10 25
A7
A4 A6
Btu/lbm Btu/(lbm. °R)
hre f
~ref
7.3816638 x 10 2 1.2258630 x 10 3  1.3771901 x 1 0  6 9.9686793 x 10 lO 4.2051104 x 10 13 1.0212913 X 1 0  1 6  1.3335668 x 10 20 7.2678710 × 10 25 30.58153 Btu/lbm 0.6483398 Btu/(lbm • °R)
equations [Eqs. (2.49)(2.52)] and are given by R =
1.9857117 B t u / ( l b m • °R) 28.97  f
x 0.946186
Cp air '~fCp prod Cp =
h
 hair
(2.64a) (2.64b)
1 +f
+ f hprod 1 +f
t~   tflair d  f t ~ p r ° d
(2.64c)
(2.640)
1 +f Appendix D is a table of the properties h and Pr a s functions of the temperature and f u e l / a i r r a t i o f f o r air and combustion products [air with hydrocarbon fuels o f composition (CH2)n] at low pressure (perfect gas). These data are based on the preceding equations and the constants given in Table 2.2, which are valid over the temperature range of 3 0 0  4 0 0 0 ° R . These constants come from the gas turbine engine modeling work of Capt. John S. M c K i n n e y (U.S. Air Force) while assigned to the A i r F o r c e ' s Aero Propulsion Laboratory, 16 and they continue to be widely used in the industry. Appendix D uses a reference value o f 2 for Pr at 600°R and f = 0.
2.6.8 Comment for Computer Program AFPROP The computer program A F P R O P was written by using the preceding constants for air and products o f combustion from air with ( C H 2 ) n. The program can calculate the four primary thermodynamic properties at a state (P, T, h, and s) given the fuel/air ratio f and two independent thermodynamic properties (say, P and h).
92
ELEMENTS OF PROPULSION
To show the use of the gas tables, we will resolve Examples 2.5 and 2.6, using the gas tables of Appendix D. These problems could also be solved by using the computer program AFPROP.
Example 2.7 Air initially at 20°C and 1 atm is compressed reversibly and adiabatically to a final pressure of 15 atm. Find the final temperature.
Solution: Since the process is isentropic from initial to final state, Eq. (2.58) can be used to solve for the final reduced pressure. From Appendix D at 20°C (293.15 K) and f 0, Pr = 1.2768 and Pr2 P2 15 Prl   P 1 P r 2 
15 × 1.2768 = 19.152
From Appendix D for Pr2 = 19.152, the final temperature is 354.42°C (627.57 K). This is 7.9 K lower than the result obtained in Example 2.5 for air as a calorically perfect gas.
Example 2.8 Air is expanded isentropically through a nozzle from T1 = 3000°R, V1 = 0, and P1 = 10 atm to V2 = 3000 ft/s. Find the exit temperature and pressure.
Solution: Application of the first law of thermodynamics to the nozzle gives the following for a calorically perfect gas: V2 V22 hi + ~ g c = h 2 +  2 g c From Appendix D at f = 0 and T1 = 3000°R, hi790.46 Btu/lbm and Prl = 938.6. Solving the preceding equation for h2 gives h2
=
hi
V22  V2  790.46 2go
30002
2 × 32.174 × 778.16
= 790.46  179.74 = 610.72 Btu/lbm For h = 6 1 0 . 7 2 B t u / I b m and f = 0, Appendix D gives 7"2 = 2377.7°R and Pr2 = 352.6. Using Eq. (2.58), we solve for the exit pressure
Pr_..~2_~ 1 0 / 3 5 2 . 6 \ _ 3.757 atm P2 P1pr 1 ~ )
REVIEW OF FUNDAMENTALS
93
These results for temperature and pressure at station 2 are higher by 126.6°R and 0.097 atm, respectively, than those obtained in Example 2.6 for air as a calorically perfect gas.
2.7 Compressible Flow Properties For a simple compressible system, we learned that the state of a unit mass of gas is fixed by two independent intensive properties such as pressure and temperature. To fully describe the condition and thus fix the state of this same gas when it is in motion requires the specification of an additional property that will fix the speed of the gas. Thus three independent intensive properties are required to fully specify the state of a gas in motion. At any given point in a compressible fluid flowfield, the thermodynamic state of the gas is fixed by specifying the velocity of the gas and any two independent thermodynamic properties such as pressure and temperature. However, we find that to specify the velocity directly is not always the most useful or the most convenient way to describe onedimensional flow. There are other properties of the gas in motion that are dependent on the speed of the gas and that may be used in place of the speed to describe the state of the flowing gas. Some of these properties are the Mach number, total pressure, and total temperature. In this section we define these properties and describe briefly some of the characteristics of compressible flow.
2. 7.1
Total Enthalpy and Total Temperature
The steady flow energy equation in the absence of gravity effects is V2
g2
q
+ ~ g c ) in
For a calorically perfect gas this becomes V2
V2
q  Wx= cp(Tl 2~pcCp)outCp(Tl 2~cCp)in The quantities h + V2/(2gc) and T ÷ V2/(2gcCp) in these equations are called the stagnation or total enthalpy ht and the stagnation or total temperature Tt, respectively. Total enthalpy:
V2 2g¢
(2.65)
V2 Tt  r +  
(2.66)
ht = h + 
Total temperature:
2gcCp
94
ELEMENTS OF PROPULSION
The temperature T is sometimes called the static t e m p e r a t u r e to distinguish it from the total temperature Tt. W h e n V = 0, the static and total temperatures are identical. From these definitions, it follows that, for a calorically perfect gas, Aht = cpATt
Using these new definitions, we see that the steady flow energy equation in the absence of gravity effects becomes q  Wx = htout  htin
(2.67)
or, for a calorically perfect gas, q  Wx
= cp(Ttout

(2.68)
Ttin)
If q  Wx = 0, we see from Eqs. (2.67) and (2.68) that htout = htin and that, for a calorically perfect gas, Trout = Ttin. Consider an airplane in flight at a velocity Va. To an observer riding with the airplane, the airflow about the wing of the plane appears as in Fig. 2.12. W e mark out a control volume ~r as shown in the figure between a station far upstream from the wing and a station just adjacent to the wing's leading edge stagnation point, where the velocity o f the airstream is reduced to a negligibly small magnitude.
Stagnationpoint
Streamlines
a) Airflow over a wing
//•///
/
[ / IWin~ I
Coincidentwith ~ / / / / 7 / / / /streamlines .. ~ / ~  ' ~ / // / ~" k : V 7 / / Y,. j Stagnationpoint ~" ~ /  " '  = ~ _ _ ~~'   ~1" 
(a)
~o"
(1)
VI = 0
b) Enlarged view of flow in the neighborhood of the stagnation point Fig. 2.12 Control volume ~r with freestream inlet and stagnation exit conditions (reference system at rest relative to wing).
REVIEW OF FUNDAMENTALS
95
Applying the steady flow energy equation to the flow through or of Fig. 2.12, we have
qWx ( h +
1
h+~g c a
or
O=CpTlCp
T+ a
From this equation, we find that the temperature of the air at the stagnation point of the wing is T1 = Ta_ ~
V2a
__ Tta
2 gcCp Thus we see that the temperature, which the leading edge of the wing "feels," is the total temperature Tta. At high flight speeds, the freestream total temperature Tta is significantly different from the freestream ambient temperature Ta. This is illustrated in Fig. 2.13, where Tta  Ta is plotted against V~ by using the relation
(rt:r)a
2 gcCp
, 12, 000
(Vo)2o R
\H0/
with Va expressed in ft/s. Because the speed of sound at 25,000 ft is 1000 ft/s, a Mach number scale for 25,000 ft is easily obtained by dividing the scale for V~ in Fig. 2.13 by 1000. (Mach number M equals Va divided by the local speed of sound a.) Therefore, Mach number scales are also given on the graphs. Referring to Fig. 2.13, we find that at a flight speed of 800 ft/s corresponding to a Mach number of 0.8 at 25,000ft altitude, the stagnation points on an airplane experience a temperature that is about 50°R higher than ambient temperature. At 3300 ft/s (M = 3.3 at 25,000 ft), the total temperature is 900°R higher than ambient! It should be evident from these numbers that vehicles such as the X15 airplane and reentry bodies experience high temperatures at their high flight speeds. These high temperatures are produced as the kinetic energy of the air impinging on the surfaces of a vehicle is reduced and the enthalpy (hence, temperature) of the air is increased a like amount. This follows directly from the steady flow energy equation, which gives, with q = wx = 0, h + V2
)in
or
V2
96
ELEMENTS OF PROPULSION
80
10,000
60
~" 1,000
o.... I
~ 40 20
100
1o
0 0
200
400
600 V a (ft/s)
800
1000
I
o
100
1000
10,000 Vc, fit/s)
I
I
I
I
I
I
I
I
I
0.2
0.4
0.6
0.8
1.0
0. l
1.0
l0
M @ 25,000 ft
t
00,000
100
M @ 25,000 fl
Fig. 2.13 Total temperature minus ambient temperature vs flight speed and vs flight Mach number at 25,000 ft [gcCp is assumed constant at 6000 ft2/(s2°R); therefore, these are approximate curves].
and
AT
AV 2
2gcCp
Thus a decrease in the kinetic energy of air produces a rise in the air temperature and a consequent heat interaction between the air and the surfaces of an air vehicle. This heat interaction effect is referred to as aerodynamic heating.
Example 2.9 The gas in a rocket combustion chamber is at 120 psia and 1600°R (Fig. 2.14). The gas expands through an adiabatic frictionless (isentropic) nozzle to 15 psia. What are the temperature and velocity of the gas leaving the nozzle? Treat the gas as the calorically perfect gas air with y = 1.4 and gcCp = 6000 ft2/(s 2. °R).
Solution: Locate the state of the combustion chamber gas entering control volume ~r on a Ts diagram like Fig. 2.11 in the manner depicted in Fig. 2.15. Then, from the diagram, find Sl. Because the process is isentropic, s2 = Sl. The entropy at 2, along with the known value of P 2 , fixes the static state of 2. With 2 located in the Ts diagram, we can read T2 from the temperature scale as 885°R, and we can verify this graphical solution for T2 by using the isentropic relation (2.43) with "),=1.4. Thus Tz=(1600°R)(15/120)°zs6=885°R (checks).
REVIEW OF FUNDAMENTALS (1)
Combustion chamber
97 (2)
a
I I ,~ V1=0
~ V2=?
I I
P1 = 120 psia T] = 1600°R Fig. 2.14
P2 = 15 psia T2=?
Rocket exhaust nozzle.
If, in addition to P2 and T2, the total temperature Tt2 of the flowing gas at 2 is known, then the state of the gas at 2 is completely fixed. For with P2 and T2 specified, the values of all thermodynamic properties independent of speed (the s t a t i c properties) are fixed, and the speed of the gas is determined by T2 and Tt2. From the steady flow energy equation (Fig. 2.15), we find that Tt2 = Tta = T t and, hence, V2 from the relation
2 &Cp
= ~t2   T 2
W e see from this equation that the vertical distance Tt2  T2 in the T  s diagram is indicative o f the speed of the gas at 2, which is V~ = 2(6000)(1600  885)ft2/s 2. Thus V2  2930 ft/s.
PI = (120 psia) (1)
~1
(1600°R) T1
(15 psia)
I J
5/
i 
,' i
(885 °R) T2 /
(2)
q = wx = 0
V] = 0
V~
2
tW__ _"
0=h:+
"2 ha 2g c
S 1 = S2
he2 = htl = h 1
v~ = r ~  r 2 2geCp
Fig. 2.15
, h2
 .........
V2
I
~
Process plot for example rocket nozzle.
"1
98
ELEMENTS OF PROPULSION
The series of states through which the gas progresses in the nozzle as it flows from the combustion chamber (nozzle inlet) to the nozzle exit is represented by path line a in the Ts diagram. The speed of the gas at any intermediate state y in the nozzle is represented by the vertical distance on the path line from 1 to the state in question. This follows from the relations Try =
2. 7.2
and
T 1
V2Y 2gcCp
  Try   T y =
T 1  Ty
S t a g n a t i o n or Total P r e s s u r e
In the adiabatic, noshaftwork slowing of a flowing perfect gas to zero speed, the gas attains the same final stagnation temperature whether it is brought to rest through frictional effects (irreversible) or without them (reversible). This follows from the energy control volume equation applied to o" of Fig. 2.16 for a calorically perfect gas. Thus, from
V2y  V~ q   Wx : Cp(ry  rl) +  2 gc with
q =
Wx =
and Vy =
Ty becomes
O,
Vl
T y = T o = TI +  2gcCp Because the energy control volume equation is valid for frictional or frictionless flow, Ty = To is constant and independent of the degree of friction between 1 and y as long as q = Wx = Vy = O.
Ptl = Pa
(Y) T
(Yr)
(Yi)
PI
~tl,Vy=0
Vl
/
(1) ~1 S 1 = Sy r
Fig. 2.16
Definition of total pressure.
I~ Syi = S 1
s
REVIEW OF FUNDAMENTALS
99
Although the gas attains the same final temperature To in reversible or irreversible processes, its final pressure will vary with the degree of irreversibility associated with the slowing down process. The entropy state and control volume equations for the flow through oare T,, P,, Cp (~._z_~ _ R E ~ = Sy T1 P1
 S 1 >

(2.69)
Since Ty = To = const from Eq. (3.5), the final value of Py depends on the entropy increase Sy  Sl, which in turn is a measure of the degree of irreversibility between 1 and y. When the slowing down process between 1 and y is reversible, with sy  Sl = 0, the final pressure is defined as the total pressure Pt. The final state is called the total state tl of the static state 1. Using this definition of total pressure, we have, from Eq. (2.43), _ [Tt\
p, = t ' t ~ )
'y/(~' 1)
(2.70)
These ideas are illustrated in the Ts diagram of Fig. 2.16. Let us imagine the flowing gas at station 1 to be brought to rest adiabatically with no shaft work by means of a duct diverging to an extremely large area at station y, where the flow velocity is zero. If the diverging duct is frictionless, then the slowing down process from 1 to y is isentropic with the path line C~rin the Ts diagram of the figure. If the diverging duct is frictional, then the slowing down process from 1 to y is irreversible and adiabatic (Syi > Sl) to satisfy the entropy control volume equation for adiabatic flow and is shown as the path line ai, in the Ts diagram. The total pressure of a flowing gas is defined as the pressure obtained when the gas is brought to rest isentropically. Thus the pressure corresponding to state Yr of the Ts diagram is the total pressure of the gas in state 1. The state point Yr is called the total or stagnation state tl of the static point 1. The concepts of total pressure and total temperature are very useful, for these two properties along with the third property (static pressure) of a flowing gas are readily measured, and they fix the state of the flowing gas. We measure these three properties in flight with pitotstatic and total temperature probes on modern highspeed airplanes, and these properties are used to determine speed and Mach number and to provide other data for many aircraft subsystems. Consider a gas flowing in a duct in which P and T may change due to heat interaction and friction effects. The flow total state points tl and t2 and the static state points 1 and 2, each of which corresponds to flow stations 1 and 2, are located in the Ts diagram of Fig. 2.17. By definition, the entropy of the total state at any given point in a gas flow has the same value as the entropy of the static state properties at that point. Therefore, st~ = $1 and st2 = s2.
100
ELEMENTS OF PROPULSION Pt2 Ptl
7~t2~11 ,
vl
v2
I P2 I

I /P1 (1)
(2) Jl L I
Fig. 2.17
s
I,t
$2S 1
•
]•
St2  Stl
•
Entropy change in terms of the stagnation properties Tt and P t .
From the entropy equation of state of a perfect gas, the entropy change between 1 and 2 is $2S1 = Cp~¢¢~11  R ( ~ The entropy change between total state points t~ and t 2 is Tt2
st2  s,l = C p & w 
ltl
 R~,~
Pt2
Pt7
(2.71a)
Since Stl ~ S 1 and st2 ~ $2, we have St2  Stl ~ S2  S 1 Therefore, the change of entropy between two states of a flowing gas can be determined by using total properties in place of static properties. Equation (2.71a) indicates that in an adiabatic and noshaftwork constantTt flow (such as exists in an airplane engine inlet and nozzle, or flow through a shock wave), we have Pt2 s2  sl =  R ~ , ~  Ptl
(2.71b)
By virtue of this equation and the entropy control volume equation for adiabatic flow, s2  sl > 0. Thus, in a constantTt flow, et2
m %. Once that point has been reached, it is physically impossible to add any more energy without forcing the upstream conditions to change, and the flow is said to be t h e r m a l l y c h o k e d . The conditions necessary for thermal choking can and do occur in practice, especially when the inlet Mach number is not far from 1, and should not be considered merely as an intellectual curiosity. Finally, it is also possible to show that the entropy always increases as the flow is heated. This is best done by substituting the preceding results into the Gibbs equation and rearranging to show that
Cp
\MiJ
\1 + 7Me2]
]
(2.110)
Since the grouping
m2
(1 + ~M2)~ has a maximum when M is 1, and since heating always drives M toward 1, it follows that heating is always accompanied by increased entropy. The GASTAB software is included to help solve problems. As an example, consider air (y = 1.4) flowing at Mach 0.5 in a constant area duct heated from 300 to 400 K. From Rayleigh flow tab of GASTAB, the maximum total temperature of the flow (Tt* = Tt%) is 433.9 K (=300/0.6914). For the exit temperature of 400 K (Tt/Tt* = 400/433.9 = 0.9219), the exit Mach number is 0.7197 and the total pressure ratio is 0.9314[Pt2/Ptl = (Pt2/Pt*)/(Ptl/Pt*) = 1.0377/1.1141]. Other example problems are included within the Supporting Material.
Example Case 2.5:
Constant Area Friction and Frictional Choking
Consider a onedimensional flow of constant area without energy interactions and frictional force F f with the surroundings, i.e., q = w,: = 0
Ai = Ae
Under these circumstances, which correspond to classical Fanno line flow, Eq. (2.12b) becomes [Eq. (2.101)] piVi = PeV~
REVIEW OF FUNDAMENTALS
117
the energy equation, Eq. (2.14), reduces to [Eq. (2.81)] h i + V2i = h e 3  V2 = ht 2 gc 2 gc and Eq. (2.20) becomes  F u = A ( P i + piV~2/gc)  A ( P e 3 PeV2/gc) = Ii  le
(2.11 1)
Taking the working fluid to be calorically perfect, Eq. (2.81) becomes [Eq. (2.82)] V2 = cpTe + v 2 = cpT, cpTi + ~gc 2go which can be written simply as [Eq. (2.85)] H+K=I where [Eq. (2.86)] H = cpT cp Tti
and
V2 K =  2gcCp Tti
and Eq. (2.111) becomes Ffgc = Vi(1 + gcRTi/V 2)  Ve(1 3 gcRTe/V 2) = Sai  Sae {n
(2.112)
where [Eq. (2.94)] S a   Igc & _ V(1 + g c R T / V 2) Using the dimensionless stream thrust function dp becomes Ffgc/th _ ~i  ~e
gv cA:,i
[Eq. (2.96)], Eq. (2.112)
(2.113)
which means that the change in stream thrust function is directly proportional to the frictional force Ff. Graphical solution of Eqs. (2.85) and (2.103) is presented in Fig. 2.27. Consider flow entering at Mi = 0.5 and dPi = 1.20 (point d in Fig. 2.27). If the frictional force decreases the dimensionless stream thrust function to dPe = 0.99, then the exit flow chokes with Me = 1 (point c in Fig. 2.27). There is no solution
118
ELEMENTS OF PROPULSION 1.4
1.2
V/\
,•
1.0
~
0.8
(
"~ hi
0.6
t ,\
x,
M
3.0"
0.4
0.2
0.0 0.0
0.2
0.4 0.6 0.8 1.0 Dimensionless Kinetic Energy
1.2
K = V2/(2gcCpTtl)
Fig. 2.27 Graphical solution for the frictionless, constant area frictional flow based on Eqs. (2.85) and (2.113). The straight lines emanating from the origin are lines of constant Mach number, increasing from left to right as indicated by their labels.
for further decreases of • with the same inlet conditions. Similarly, flow entering at M / = 2.74 and (I) i = 1.20 (point u in Fig. 2.27) will choke when • is decreased to 0.99. HK diagram. The great utility of graphically displaying flow processes in terms of dimensionless static enthalpy vs dimensionless kinetic energy has been amply demonstrated by the elementary example cases already considered. This method of presentation will prove even more valuable in explaining and illustrating the more complex internal flow behavior of airbreathing engines. To ease communications, this diagram will hereinafter simply be called the HK diagram (H for dimensionless static enthalpy and K for dimensionless kinetic energy). Please note that the HK diagram, for all its other virtues, is not a state diagram because only one axis is an intensive thermodynamic property. In other words, there is no necessary relationship between a point on the HK diagram and the other intensive thermodynamic properties of the fluid, such as static pressure or static entropy. Nevertheless, the HK diagram will provide more than enough information to reveal the things we really need to know about the flow. Also, under some frequently encountered conditions, such as onedimensional flow with known rh, A, and T,, the HK diagram is a state diagram.
REVIEW OF FUNDAMENTALS
119
An especially useful generalization for moving about the H  K diagram, based on the assemblage of preceding example cases, is that heating, friction, and area decrease all act separately and together to drive the Mach number toward 1. They may therefore in an intuitive sense be said to block, constrict, obstruct, restrict, or occlude the flow. Similarly, cooling, reverse friction (i.e., any streamwise force), and area increase could be said to unblock, enlarge, relieve, open, or free the flow. You will find these mental images helpful in what follows. Figure 2.28 shows the H  K diagram with representative constantproperty isolines. The H  K diagram will be used in this text to improve understanding of the flow through airbreathing engines. The H  K diagram often provides a convenient means for visualizing, comprehending, and even analyzing the operation of propulsion devices, as illustrated in the following for the scramjet engine. 17 It is important to note that the dimensionless stream thrust function at a given Mach number is not some arbitrary value but is specified by Eq. (2.114) as rearranged into the Mach number form:
 I ÷ ~ M
1)
2 I+TM~
(2.114)
e~
DimensionlessKineticEnergy K = V2/(2gccpTtl) Fig. 2.28 The H  K diagram, depicting representative constantproperty isolines. Key: 0 = freestream reference state. Point c = choked condition at constant impulse. Points u and d denote end states of normal shock. Circled numbers denote isolines of constant property as follows: 1) static enthalpy, static temperature; 2) kinetic energy, velocity, pressure (for frictionless heating or cooling only); 3) Mach number; 4) total enthalpy, total temperature (adiabat), ~ = T t / T t o = 1; 5) postheat release adiabat, t > 1; 6) impulse function/stream thrust, area (for frictionless flow with heating or cooling only), case I = Io; 7) impulse function, case ~>~o.
120
ELEMENTS OF PROPULSION
Example Case 2.6: Scramjet The basic concept of the scramjet is that the flow must remain supersonic throughout to avoid the high static temperatures that reduce performance by causing chemical dissociation of the combustion products, as well as the high static pressures that cause structural problems. Consequently, in practical terms, when the freestream Mach number exceeds 5  6 , the flow is designed to enter the combustor at supersonic speeds, and the device is known as the supersonic combustion ramjet, or scramjet. Figure 2.29 shows the HK diagram for air with 3' = 1.40 being processed by a scramjet that is powering a vehicle at a freestream Mach number of 10.0, where Eq. (2.114) has been used to show that qb0 = 1.390. The air is first decelerated and compressed from the freestream condition (point 0) to the burner entry condition (point 1) by means of a combination of isentropic compression and oblique shock waves. The purposes of this compression are to provide a large enough static temperature ratio T1/To for satisfactory thermodynamic cycle efficiency (usually in the range of 6  8 , and 6.50 for this example) and to produce high enough values of P1 and Tl to support complete and stable combustion in the burner. Even when these criteria have been met, the burner entry M1 = 3.340 remains supersonic, as Fig. 2.29 clearly reveals. The air is then heated in a combustion process that releases the chemical energy of the fuel. The heating is represented in this type of analysis by an increasing total temperature, in this example case by a factor of 1.40. The precise path of this process depends on the philosophy of the burner design, and two of many possible different types are depicted in Fig. 2.29. The first,
] .4
1.40
'
'~
c
'
'
,Scram et, 1.2l,t~o ,/.,. ~,o \~.~o I \/ \\ \ 1.0 ~'. 1.O0 "/'N I,a,I{,,~ ~ ,>
08[
~
0.6
•
~
/
~.\ ~
TI/To = 6.50
\ \~
Mo=lO.O ,i,o=,.o
\
~e= "40
0.2 0.0 t ~  0.0 0.2
~
, ~, ~\, 0.4 0.6 0.8 1.0 Dimensionless Kinetic Energy
",1 1.2
1.4
K = V2/(2gccpTtO)
Fig. 2.29
The H  K diagram for an example scramjet.
REVIEW OF FUNDAMENTALS
121
joining point 1 to point 2, is frictionless, constant area heating, which is a Rayleigh line having qbl = 1.250. The second, joining point 1 to point 3, is frictionless, constant pressure heating, which is found in Problem 2.49d also to be a line of constant velocity. There is clearly no danger of reaching point c and thermal choking for either combustor in this scenario. The heated air is then accelerated and expanded from a burner exit condition such as point 2 or 3 to the freestream static pressure at point 4. Because there are total pressure losses in the scramjet, the Mach number at point 4 can never be quite as large as the freestream Mach number, but it can be large enough that the kinetic energy and velocity at point 4 exceed that of point 0, which means that the scramjet produces net thrust. As a corollary, the total pressure losses and therefore the precise location of point 4 also depend on the type of burner design. Nevertheless, the H  K diagram makes it clear that the potential thermodynamic performance is greater for constant area heating than for constant velocity heating because each increment of heat is added at a higher temperature in the former case. The readers are encouraged to expand their familiarity and facility with the H  K diagram by either explaining the impact of other heating processes (e.g., constant temperature or constant Mach number) on the potential thermodynamic performance of the scramjet, or by constructing the H  K diagram for the ramjet, for which the combustor Mach number is very subsonic.
2.9
Nozzle Design and Nozzle Operating Characteristics
Figure 2.30 shows flow through a nozzle from a subsonic flow (state 1) through the sonic conditions (state *) to supersonic flow (state 2). The flow is isentropic, and both the total temperature and total pressure are constant. For analysis, we consider accelerating a gas through a nozzle with mass flow rate rhc. Let the flow originate in a large storage chamber (Fig. 2.31) at chamber pressure
Thermometer T Tt

u~_!_ lot
(1)
x
(2)
r~ /
Tt and Pt are constant
Flow to a lowerpressure Fig. 2.30
(2)
Simple nozzle flow.
122
ELEMENTS OF PROPULSION
(In) I
(Out)
I Tt I
v o
T
cha Imber
~ (In)
t
P
ill//lUg" .
.
.
.
.
.
t(3
t_,.x I
r
Pc = Pt
V
Tc= rt Fig. 2.31
Control volume for simple nozzle flow.
Pc = Pt, and chamber temperature Tc = Tt. The stream properties at any station in the flow are related by the following equations: Pc = Pt = P ( T t / T ) "y/('y1)
(i)
Tc = Tt = T + V 2 / ( 2 gcCe)
(ii)
/nc = p A V = P A V / ( R T )
(iii)
For given chamber gas conditions Pc, To, R, y, and known rhc, there are four variables P, T, V, A in these three equations. We may select one variable as independent and find each of the remaining three in terms of this one. Practical problems generally fall into two categories: 1) Nozzle design. We wish to pass a given mass flow with minimum frictional losses between two regions of different pressure (storage chamber at Pc and exhaust region at Pa) with, say, some assumed variation in pressure between the two regions. 2) Nozzle operating characteristics. Given a nozzle, what mass rates of flow and pressure distribution will prevail through the nozzle for various nozzle pressure ratios (P, = Pc/Pa)? In case 1, our independent, or known, variable is pressure P, which is a function of position x. In case 2, our independent, or known, variable is area A, which is a function of position x. We will consider each case in turn.
2.9.1
Nozzle Design
We shall illustrate the design of a nozzle by example.
REVIEW OF FUNDAMENTALS
123
Example 2.11 A s s u m e we wish to e x p a n d gases at 28 l b m / s from a highaltitude secondstage rocket c o m b u s t i o n c h a m b e r to an a m b i e n t pressure of 0.618 psia (~70 kft altitude). Pertinent data simulating the A g e n a rocket engine are as follows: Pc = 206 psia,
Cpgc : 6000 ft2/(s2°R),
N o z z l e length = 30 in., Rgc = 1715 ft2/(s2°R),
T c = 5000°R
rhc = 28 l b m / s
Exit pressure = 0.618 psia
To solve the problem, we must determine the flow area, gas temperature, velocity, and M a c h n u m b e r (A, T, V, and/14) at each nozzle station. A s s u m e all sections of the nozzle are circular, and assume simple area flow. F i g u r e 2.32 provides the length of, and pressure in, the nozzle to be designed. Solution: F o r the given conditions, the constants of Eqs. (i) through (iii) are known. Thus Pt = Po = 206 psia, Rgc = 1715 ft2/(s2°R),
Tt = Tc = 5000°R,
rhc = 28 l b m / s
Cpgc = 6000 ft2/(s2°R),
3 / = 1.4
x(in) 20 22 23 24 25 26 28 32 42 50
f 200 
'~I
Combustion 20 chamber

~
150 v
100 8
50 0
30 I~
I 10
" t " 20
40 I
50 
I
Determine nozzle contour from station 20 to 50 where flow occurs from Pcto Pa
30
I 40
Station x (in) Fig. 2 . 3 2
P(psia) 200 183.5 174.8 144.5 108.9 50 15.2 5.6 1.34 0.618
Nozzle pressure distribution.
I 50
124
ELEMENTS OF PROPULSION
Rewrite Eqs. (i) through (iii) as T  Tt(P/Pt) (y)/'r
(iv)
V = ~/2gccp(Tt  T)
(v)
A = thRT/(PV)
(vi)
With P known (Fig. 2.32), use Eq. (iv) to find Tat any given station. Equation (v) will then give V. With T and V determined, and rh, R, and P known, Eq. (vi) gives the nozzle area A at the selected station. In this manner, the nozzle area and gas properties can be found at all stations. The Mach number follows from (vii)
M = V/v/yRgcT
The results of the computation outlined are plotted in Fig. 2.33. The curves and nozzle contour in this figure illustrate that in order to decrease P and increase V in a simple area flow: A converging nozzle contour is required in subsonic flow. A diverging nozzle contour is required in a supersonic flow.
Thus we find that the typical shape of a nozzle that is accelerating a gas from rest to supersonic speeds is convergentdivergent (CD). At the design operating point of a supersonic CD nozzle, the flow is subsonic up to the throat, sonic at the throat, and supersonic after the throat. The exit plane pressure Pe equals the exhaust region pressure Pa. We shall denote the nozzle pressure ratio for the
CL
1
C L 
1

~6 0
7
I
V 6" (103fffsec) 5p 4(10 3 slug/fi3) 3"
A 2(102 ft2) 
T (103 R)
\
/
M
\
o 2o 25 30 3'5 40 4'5 5'0 x Fig. 2.33
5
250
4
200
3
150 p (psia)
2
 100
1
5O
0 I
20 25 ~0 ~5 ~0 ~5 50 x
N o z z l e flow p r o p e r t i e s vs station for air.
REVIEW OF FUNDAMENTALS
125
7"/
P* M~li M=I
Fig. 2.34
Isentrope line ~ .
design point as P~, where
Ph=(Pn)design=(P'~a)design The path line of an isentropic flow is called an isentrope. The area variation required to progress along an isentrope in a given direction is shown in the Ts diagram of Fig. 2.34. To progress downward along the isentrope requires a converging area (dA < 0) in subsonic flow and diverging area (dA > 0) in a supersonic stream. To progress upward along the isentrope requires a converging area (dA < 0) in supersonic flow and diverging area (dA > 0) in subsonic flow. This is why the engine intakes on the Concorde and various other supersonic aircraft converge from the inlet entrance to a throat and then diverge to the compressor face. This design reduces the speed of the air entering the compressor. Since P, T, p, and V2/(2gcCp) can be displayed in the Ts diagram, the isentrope line oc properly interpreted summarizes most of the characteristics of isentropic flow. The stream area/velocity variations discussed can be explained on the basis of the continuity equation by examining how the gas density varies with velocity in an isentropic flow. Area, velocity, and density are related as follows by the onedimensional steady flow continuity equation: A = __rn
pV
(rh = const)
By reference to a Ts diagram, with lines of constant density thereon, we know that in an isentropic flow, p decreases as V increases. In subsonic flow, V increases faster than p decreases so that, according to the preceding equation, A must decrease. In supersonic flow, p decreases more rapidly than V increases, and therefore A must increase to satisfy the continuity equation. Last, we have the important result that M = 1 in the throat of a nozzle accelerating a gas. When M = 1 at the throat, the nozzle reaches maximum possible mass flow for the given chamber pressure and temperature, and the nozzle is
126
ELEMENTS OF PROPULSION
said to be choked, and M will equal 1 only at the nozzle throat. In a decelerating diffuser flow, on the other hand, the throat Mach number may be less than, equal to, or greater than 1.
2.9.2 Nozzle Operating Characteristics for Isentropic Flow Having designed a nozzle for a specific operating condition, we now examine its offdesign operating characteristics. We wish to answer the following question: Given a nozzle, what are the possible isentropic pressure distributions and mass flow rates through the nozzle? A simple way to investigate this question is to deal with a single equation that contains all the restrictions placed on the flow by the perfect gas state equations and the control volume equations. The governing equations may be combined into a single equation. We have Mass:
PAV in = pA V =  RT
(viii)
Energy: V2
Tt = T +  2 gcCp
(ix)
P Pt(T/Tt) ~'/~'l)
(x)
Entropy:
Equation (viii) can be written as
/n A
PV RT
(xi)
wherein
P=Pt
T = Tt
and
V = x/2gccp(Tt  T) =
2gccpTt 1  \PTtJ
J ]
Substituting these expressions for P, T, and V in Eq. (xi) and simplifying, we obtain a single equation representing the simultaneous solution of
REVIEW OF FUNDAMENTALS
127
Eqs. (viii), (ix), and (x):
~n
Ptl2gc
~/ I (£~ 2/Y_(£~ (T+I)/T
(2.115)
If Eq. (2.115) is satisfied at every station of the flow through a nozzle, it follows that the conditions imposed on the flow by the thermal state equation and the mass, energy, and entropy control volume equations are satisfied. With Pc and T, known in any given nozzle flow, we may effect a graphical solution of Eq. (2.115) by plotting Fn/A vs P/Pt. In a physical flow, P/Pt m a y vary from 1.0 in a storage chamber (P = Pc = Pt) to 0 in a vacuum (P0). A graph of Fn/A vs P/Pt is given in Fig. 2.35 for Pc206 psia and Tc 5000°R. Because there is a unique value of M for each P/Pt, we show a Mach number scale along with the P/Pt axis. Notice that M increases as P/Pt decreases from left to right in the figure. W e note that for a given value of th/A, there are two possible values of P/Pt in Fig. 2.35. In a particular problem, we can determine which value of P/Pt is applicable by examining the physical aspects of the flow. Assume the nozzle depicted in Fig. 2.36 is discharging air isentropically from a storage chamber with Pc = 206 psia and Tc = 5000°R. Let us plot the nozzle
1.8 1.41.2 1.0
2 0.8
Pc= 206 psia ] Tc= 5000°R
c
/

d'
d
_ _
_
_
e
0.6 0.4 0.2 i
I
0.8 I 1
Fig.
I 0.5
2.35
, / ,
,
/
0.6 0.4 Pressure ratio P/Pt
0.2
i
I
I
I
0.8 1.0 1.2 Mach number (M)
K~/Av s P/Pt a n d
Mach
number
I
I
li
1.6
2.0
3
for
air.
128
ELEMENTS OF PROPULSION
I
/ B
i D
C
1.0
Station B C D E
Area (in. 2) 21.6 14.4 17.3 28.8
/
e'
~ 0.5
c
II I
I B
C
D
Fig. 2.36
ie I E Station
Nozzle pressure distribution.
pressure distributions for various nozzle mass flows. W e shall determine the pressure distribution for m a x i m u m mass flow first. W i t h the chamber pressure and temperature known, the Fn/A vs P/Pt plot of Fig. 2.35 is made. Then, because for maximum mass flow M = 1 at the nozzle throat, ~hm~ is determined by the relation
/~/rnax ~
(A)
Athroat
M=I
= [1.55(lbm/s)/in.2](14.4 in. z) = 22.3 l b m / s With rh and the areas at nozzle stations B, C, D, and E of Fig. 2.36 known, we determine Fn/A at these stations. With these values of rh/A, we locate state
REVIEW OF FUNDAMENTALS
129
points b, c, d, and e, as shown in Fig. 2.35, and read values of P/Pt corresponding to stations B, C, D, and E. Beginning at the storage chamber, P/Pt = 1. Then as A decreases and th/A increases, P/Pt decreases from 1.0 to band c at the throat, as indicated in Figs. 2.35 and 2.36. After passing through the nozzle throat, A increases and /n/A decreases. Now there may physically exist either value of the ratio P/Pt, corresponding to a given/n/A with a continuous variation in pressure through the nozzle being maintained. Thus at section D, the pressure may be that corresponding to d or d' in Figs. 2.35 and 2.36. Whichever value exists depends on the nozzle pressure ratio Pn = P~/Pa.The isentropic nozzle pressure distributions for maximum mass flow are the solid lines labeled I and II in the graph of Fig. 2.36. The dashed line represents a nozzle pressure distribution for a mass flow less than maximum and, hence, subsonic flow through the nozzle. The nozzle pressure distribution corresponding to flow in I of Fig. 2.36 can be produced by nozzle pressure ratios other than the design value P~. However, the nozzle exit plane pressure Pe and the exhaust region pressure P~ are equal only for the design nozzle pressure ratio P~. At the offdesign pressure ratios producing flow I, Pe remains the same, as given by
Pe = Pc/P~ but is either greater than or less than the exhaust region pressure Pa. When Pe > Pa, the nozzle is said to be underexpanded. Under these conditions, the gas in the nozzle has not expanded down to the exhaust region pressure. Similarly, when P~ < Pa, the nozzle is said to be overexpanded because the gas in the nozzle has expanded to a value below the exhaust region pressure. We see from Fig. 2.36 that there are no solutions of the equation
which gives Pe = Pa for exhaust region Pa between Pe, and Pe. Physically, it is possible to have a discharge region pressure in this range. What happens when such an exhaust region pressure exists? To answer this question, let us discuss the operating characteristics of a nozzle that might be used as a highspeed wind tunnel.
2.9.3
Nozzle Flow and Shock Waves
Figure 2.37 shows a windtunnel nozzle that we shall use for purposes of discussion. The tunnel operates between an air storage chamber maintained at Pc and Tc and an evacuated receiver. The pressure of the receiver Pa increases as air flows from the storage chamber through the tunnel into the receiver. In this way, the nozzle pressure ratio Pn = Pc/Pa decreases from a very high value (due to a low Pa initially) to a value of 1 when the receiver pressure becomes equal to the storage chamber pressure and flow ceases. A rocket engine nozzle
130
ELEMENTS OF PROPULSION
j c S h t a ° r a b g ~ l
l 2
I ~'~
section
2
~ 1
"3a
Receiver Pe
No flow
1.0
7 6
0.8 0.6
m5
0.4
4 3b
0.2  
.13
1
I
3a
2
Tunnel station Fig. 2.37
Nozzle flow with shock waves.
descending through the atmosphere would experience a similar decrease in pressure ratio as P~ increases and Pc remains constant. During the operation of the tunnel, the air flowing into the evacuated receiver raises the pressure in the nozzle exhaust region Pa and decreases the nozzle pressure ratio Pn. As a result, seven distinct nozzle pressure ratio operating conditions are present. They are depicted in Figs. 2.37 and 2.38 and tabulated in Table 2.3. The coordinates of the operating diagram in Fig. 2.38 are the nozzle pressure ratio and nozzle area ratio e, where e is the ratio of the nozzle exit area Ae to the nozzle throat area At. If we assume the tunnel of Fig. 2.37 has an area ratio of e = 2, then the operating points of the tunnel all lie along the horizontal line e = 2 in the nozzle operating diagram of Fig. 2.38. The following conditions are possible: 1) Underexpanded, Pn > Pa. The pressure in the evacuated receiver is less than the nozzle exit plane pressure, so that Pe > Pa. The nozzle is operating underexpanded with P, > Ph. The flow inside the nozzle is the same as that corresponding to the design point pressure ratio Ph.The flow outside the nozzle does not correspond to that for P~ since P~ ¢ Pa. The transition of the nozzle exit plane pressure from Pe to the lower receiver pressure P, occurs in the exhaust region, as depicted in the underexpanded portion of the nozzle operating diagram (Fig. 2.38).
I
I
O
I
I
\
I
I
I
A
I 3 o!1~.I ~o.re o [ Z Z O N
Ill
I
%
I
% %
I
"7"
%
%
REVIEW OF FUNDAMENTALS
t
~V/)V=
%
\
\ \ \
°
131
u
132
ELEMENTS OF PROPULSION Table 2.3
Operand point 1) Underexpanded 2) Design 3) Overexpanded a) Regular reflection b) Mach reflection 4) Normal shock at exit 5) Normal shock in divergent section 6) Sonic at throat subsonic elsewhere 7) Subsonic everywhere
Nozzle operating points
Exit section pressure Pe, Pe = Pc/P~
Nozzle pressure ratio P, = Po/P~
Mass flow rate
Pe > P, Pe = Pa
Pn > Pa Pn = P~
Pe < Pa
Pn < P~
Maximum Maximum Maximum
(Pe)x < (Pe)y = Pa
Pn < Pn
Maximum
Pe = P.
Pn > Pa
Maximum
Pe = P~
P. < Pa
Maximum
Pe = Pa
Pn < P~
Less than maximum
2) Design expansion, Pn = P~. The pressure in the receiver has been increased to the nozzle exit plane pressure Pe = Pa by the air flowing into the receiver. This is the nozzle design operating point with Pn Pa. No pressure disturbances occur in the jet issuing from the nozzle. This operating point is on the design expansion line of the nozzle operating diagram. 3) Overexpanded, Pn < P~. The pressure in the receiver is greater than the nozzle exit plane pressure Pe < Pa" The nozzle is operating overexpanded with P~ < P~. The transition from Pe to the higher receiver pressure is produced by either an oblique shock wave system (regular reflection 3a) or a combined obliquenormal shock wave system (Mach reflection pattern 3b). This nozzle operating condition lies between the design expansion line and normalshockatexit line of the nozzle operating diagram. 4) Normal shock at exit. The pressure in the receiver has increased to a value that has moved the normal shock wave of condition 3b into the nozzle exit plane. The pressure of the gas entering the normal shock is (Pe)x, and the pressure leaving is (P~)y = Pa. The loci of this operating condition are on the normalshockatexit line of the nozzle operating diagram (Fig. 2.38). 5) Normal shock inside. The receiver pressure has increased to a value that has caused the normal shock to move into the diverging portion of the nozzle. Flow in the nozzle preceding the shock is unaffected. The flow through the shock is irreversible at constant Tt such that the total pressure decreases across the shock wave. The flow downstream of the shock is subsonic. This operating condition lies between the normalshockatexit and sonic limit lines of the nozzle operating diagram. 6) Sonic limit. The receiver pressure has reached a value that produces isentropic shockfree flow throughout the nozzle with sonic throat conditions and
REVIEW OF FUNDAMENTALS
133
a) Pn=l.5, normal shock, inside
b) Pn=2.5, Mach reflection, overexpansion
e) Pn=4.5, regular reflection, overexpansion
d) Pn=8.0, underexpansion
Fig. 2.39 Spark Schlieren photographs of nozzle exhaust flow patterns for an area ratio of 1.5. (Department of AeroMechanical Engineering, AFIT, WPAFB, Ohio.) subsonic flow elsewhere. The nozzle pressure ratios corresponding to this operating condition lie on the sonic limit line of the nozzle operating diagram. This nozzle flow corresponds to flow I of Fig. 2.36. 7) Subsonic flow. The receiver pressure has risen to a value producing subsonic flow throughout with a reduced mass flow. This operating condition is bounded by the sonic limit and the Pn  1 (noflow) lines of the nozzle operating diagram. Figure 2.39 is a picture of several nozzle exhaust flow patterns for a conical nozzle with an area ratio of e = 1.5. For Pn = 1.5, the normal shock is located well within the nozzle. As Pn is increased to 2.5, the flow is overexpanded in the Mach reflection regime, as evidenced by [he clearly defined normal shock internal to the flow. Further increase in Pn causes this normal shock in the Mach reflection pattern to move farther out and diminish in size. It disappears when the regular reflection pattern is obtained. For Pn = 4.5, the nozzle is overexpanded in the regular reflection regime. The photograph with P~ = 8.0 corresponds to an underexpanded operating point with the exhaust gas expanding down to the ambient pressure in the jet plume.
2.9.4 Nozzle Characteristics of Some Operational Engines The operating point of a nozzle is determined by the nozzle pressure ratio Pn and area ratio e. These ratios are presented in Table 2.4 along with other data for
134
ELEMENTS OF PROPULSION Table 2.4
Some nozzle characteristics of rocket and turbojet engines
Engine
Saturn F1 without extension F1 with extension J2 Atlas Booster Sustainer Subsonic airbreathing turbofan (40,000 ft) Supersonic airbreathing turbofan (40,000 ft)
Chamber pressure Pc, psia
Area ratio e
965 965 763
P,
(Pa)operate
10 16 27.5
140 275 610
Varies with altitude Pa
703 543 814
8 25 1.0
100 528 1.9
Varies with altitude P. 35
856
12
1.98
320
some operational nozzles. The data in Table 2.4 permit us to locate the operating points of the Saturn and Atlas engines at a given e in the nozzle operating diagram (Fig. 2.38). In the Saturn F1 engine, e = 16 and Pc = 965 psia. Thus at 46,000 ft with Pa = 2 psia, we have Pn = 965/2 = 482 and a nozzle operating point of (482, 16). This operating point (assuming , / = 1.4) indicates that the F1 nozzle is operating above the nozzle design point (underexpanded) at 46,000 ft. At sea level for the F1, Pn is about 65, and the operating point of the engine nozzle (assuming ,,/= 1.4) is in the overexpanded regular reflection region. The turbojet engines of highperformance, airbreathing, subsonic aircraft generally use convergent nozzles (e = 1.0) and operate with nozzle pressure ratios greater than the design value of 1.9 (assuming y = 1.4). These nozzles therefore operate in the underexpanded operating regime. Turbojet/turbofan engines of supersonic aircraft, however, have convergingdiverging nozzles.
2.10 OneDimensional Gas DynamicsDifferential Control Volume Analysis The steady onedimensional flow of a chemically inert perfect gas with constant specific heats is conveniently described and governed by the following definitions and physical laws.
2.10.1
Definitions
Perfect gas: P = pRT
(i)
M 2 = V2/(TRgcT)
(ii)
Mach number:
REVIEW OF FUNDAMENTALS
135
Total temperature: (iii) Total pressure: y
Pt : P(I + ~ M
1
2x~~/(~1)
)
(iv)
2.10.2 Physical Laws For onedimensional flow through a control volume having a single inlet and exit sections 1 and 2, respectively, we have Onedimensional mass flow:
PlA1 V! z P2A2V2
(v)
Momentum: Frrict =
A(PA + pA V2/gc)
(vi)
Energy equation (no shaft work):
q = cp(Tt2  Ttl)
(vii)
Entropy equation (adiabatic flow): (viii)
$2 _~> S1
w h e r e Ffrict is the frictional force of a solid control surface boundary on the flowing gas, and A is the flow crosssectional area normal to the velocity V. The frictional force acting on a length L can be written as
F~ict
=f{pVZ/(2gc)}
wetted perimeter x L
or
gfrJct
pV 2 4A ~f(2~gc)(O) z
where f = frictional coefficient D = hydraulic diameter =
4xA wetted perimeter
A = flow crosssectional area
(2.116)
136
ELEMENTS OF PROPULSION
Now consider the differential element of duct with length dx, as shown in Fig. 2.40. The independent variables are the area change, total temperature change, and frictional force. The dependent variables are P, T, p, V, M 2, and P~. The application of Eqs. (ivi) to flow in Fig. 2.40, having the presence of the simultaneous effects of area change, heat interaction, and friction, results in the following set of equations for the infinitesimal element dx. Perfect gas: dP dp dT F +  = 0 P p T
(2.117a)
Total temperature: dT+ [(3" 1)/21M 2 d M 2 = d T t T 1 q [(T  1)/2] M2 M2 Tt
(2.117b)
Onedimensional mass flow: dp + ~ p
+
dV
(2.117c)
V
Total pressure: dP (3"/2)M 2 dM 2 dPt + = P 1 q  [ ( 3 '  1)/2] M2 M2 Pt

dFfrict = f [pV ~/ (2gc)](4A/D) dx
A
rI
_
dq = cp dTt /
7
P
T
I
Controlsurface
I
P I V I M2 I Pt
1~ I
D = hydraulic diameter Fig. 2.40
dx
(2.117d)
A+dA P +dP T+ dT p+dp V+dV
M2+dM 2 Pt + dPt
f = friction coefficient
I n d e p e n d e n t a n d d e p e n d e n t v a r i a b l e s for o n e  d i m e n s i o n a l flow.
REVIEW OF FUNDAMENTALS
137
Momentum: dP
dV
'fi~ yM2v+
2 ' m 2 dx T~ f ~ = 0
(2.117e)
Mach number: 2 dV V
dT T

dM 2 M2

(2.117f)
With the exception of the momentum equation, these equations are obtained by taking the derivative of the natural log of Eqs. (iv). In these equations, heat interaction effects are measured in terms of the total temperature change according to Eq. (vii). The entropy condition of Eq. (viii) is also applicable if dTt = 0. If dTt is not zero, then the entropy requirement is ds = dq/T. The six dependent variables M 2, V, P, p, T, and Pt in the preceding set of six linear algebraic equations may be expressed in terms of the three independent variables A, Tt, and 4f dx/D. The solution is given in Table 2.5. Note that D is the hydraulic diameter ( = 4 x flow crosssectional area/wetted perimeter). General conclusions can be made relative to the variation of the stream properties of the flow with each of the independent variables by the relations of Table 2.5. As an example, the relationship given for dV/V at the bottom of the table indicates that, in a constantarea adiabatic flow, friction will increase the stream velocity in subsonic flow and will decrease the velocity in supersonic flow. Similar reasoning may be applied to determine the manner in which any dependent property varies with a single independent variable.
Example 2.12 Consider the onedimensional flow of a perfect gas in a channel of circular cross section. The flow is adiabatic, and we wish to design the duct so that its area varies with x such that the velocity remains constant. 1) Show that in such a flow, the temperature must be a constant, hence the Mach number must be a constant. 2) Show that the total pressure varies inversely with the crosssectional area. 3) Show that 2 dA/C = yM2fdx, where C is the circumference. 4) For a constantfand a circular channel of diameter D, show that the diameter must vary in accordance with the following equation to keep the velocity constant:
D = Dl + yMZfx Solution: This problem requires that we apply the relationships of Table 2.5 and other fundamental relationships. 1) Because the flow is adiabatic, the total temperature is constant. From the definition of the total temperature, we have Tt = T + V2/(2Cpgc). Since Tt and
138
ELEMENTS OF PROPULSION T a b l e 2.5
I n f l u e n c e coefficients f o r s t e a d y o n e  d i m e n s i o n a l f l o w
Independent dA A
Dependent
2(1 q  ~
dM 2
M2
1
dV
1
V
I  M 2
dP
TM2
P
1  M2
~m2(l @ ~  ~ m 2 ) 1
2
1+ ~

M
2
M2 2(1
'yM2(1 q  ~ l M 2 ) 1
M2 1 M2
dT
(y_ 1)M2
T
1  M2
M
(1 q ~
_TM2[I +(T1)M 2] 2(1
]M2)
2(1
1  M z

M 2)
TM2
1  M 2
o
TM2   M 2)
2
(IyM2)(I+T~IM
M2
1 
1  M2
p
Pt
4fdx D
M2 ) M
dp
dPt
dR Tt
2)
M 2)
_y(y_1)M
2(1
 TM 2
 TM 2
2
2
4
M 2)
This table is read: d VV= V
_~
A ] IM 2
Tt
"yM2 4fdx 2 ( I  M 2) D
V are constant in the duct, the preceding equation requires that the static temperature remain constant. With the static temperature of the gas constant, the speed of sound will be constant (a = ~ / ~ c T ) . With constant velocity and speed of sound, the Mach number will be constant. 2) Application of the continuity equation to the constantvelocity flow gives A i / A = P/Pi. For a perfect gas with constant static temperature, we have P/Pi = P / P i and from Eq. (iv) for constantMach flow, we get Pt/Pti = P/Pi. Thus Pt/Pti = Ai/A. 3) To obtain this relationship, we need to get a relationship between two independent properties to keep the dependent property of velocity constant. Table 2.5 gives the basic relationships between dependent and independent properties for onedimensional flow. We are interested in the case where velocity is constant,
REVIEW OF FUNDAMENTALS
139
and thus we write the equation listed as an example at the bottom of Table 2.5 for the case where both dV and dTt are zero:
0=
1M 2
4042(1M
2) D
The area of the duct is equal to the circumference C times one quarter of the diameter D. Thus A = CD/4, and the preceding relationship reduces to
2 dA/C =
,M2f dx
4) For a circular cross section, A = IrD2/4 and C = ~D. Thus 2dA/C =dD. Substitution of this relationship into the preceding equation and integration give the desired result (at x = 0, D = Di):
D = Di + yM2fx
2.11
2.11.1
Chemical Reactions
General Characteristics
A chemical reaction from reactants A and B to products C and D is generally represented by
NAA + NsB + NcC + NDD where NA and NB are the number of moles of reactants A and B, respectively. Likewise Nc and ND are the number of moles of products C and D. An example reaction between one mole of methane (CH4) and two moles of oxygen (02) can be written as CH4 + 202 ~
C 0 2 4 2H20
and represents complete combustion between the fuel, methane, and oxidizer, oxygen. Combustion is complete when all of the carbon in the fuel burns to CO2 and the hydrogen burns to H20. The combustion is incomplete when there is any unburned fuel or compounds such as carbon, hydrogen, CO, or OH. This theoreticalprocess of complete combustion is very useful in our analysis because it represents the number of moles of each reactant that are needed for complete combustion to the number of moles of each product. The theoretical or complete combustion process is also referred to as a stoichiometric combustion process and is represented as VcH4CH4 4 v0202 ~ vc02CO2 4 vttzoH20 where vcH4 and v02 are the number of moles of reactants CH 4 and O2, and vc02 and vn20 are the number of moles of products CO2 and H20. The vi in the
140
ELEMENTS OF PROPULSION
preceding equation are called the stoichiometric coefficients for the theoretical reaction. Thus the stoichiometric coefficients for the preceding reaction of methane and oxygen are I)CH 4 :
1
vo2 : 2
vco2 = 1
IJH20 :
2
The stoichiometric coefficients give the number of moles of each constituent required for complete combustion. Note that the units of both the mole number Ni and the stoichiometric coefficients vi can also be regarded as moles per unit mass of the gas mixture. The actual process between reactants normally has different amounts of each reactant than that of the theoretical (stoichiometric) process. For example, consider one mole of methane reacting with three moles of oxygen. The stoichiometric coefficients of the theoretical reaction indicates that only two moles of oxygen are required for each mole of methane. Thus one mole of oxygen will not react and remains on the products side of the chemical equation as shown: CH4 + 302 ~
2.11.2
CO2 + 2H20 + 02
Chemical Equilibrium
The chemical reactions represented so far were shown as complete reactions. In reality, the reactions go both forward and backward and the actual process is denoted as NAA + NBB ~ NcC + NDD
Chemical equilibrium is reached when the forward rate of reaction of reactants A and B equals the backward rate of reaction of products C and D. It can be shown that chemical equilibrium corresponds to the m i n i m u m value of the Gibbs function (G = H  TS) for the entire gas mixture as shown in Fig. 2.41. Chemical equilibrium also corresponds to m a x i m u m entropy for an adiabatic system.
Gibbs function G
I [
100% reactants Fig. 2.41
Chemical equilibrium
Ib 100% products
Chemical equilibrium.
REVIEW OF FUNDAMENTALS Chemical equilibrium for a stoichiometric (theoretical represented as
141 reaction can be
liAA q liBB ~  l i c C b tODD
(2.118)
For solution of equilibrium problems, we define the equilibrium constant of a perfect gas as vc vo K p _ P ~ PD PaRR
VB
(2.119)
where PA, PB, Pc, and Po are the partial pressures of components A, B, C, and D; lia, V,~, lic, and Vo are the stoichiometric coefficients shown in Eq. (2.118). For a perfect gas, Kp is only a function of temperature. From Eq. (2.46), the partial pressure of constituent i can be written as
Pi =
N/ P
(2.120)
Ntotal
where Ntota I is the total number of moles present in the reaction chamber. Using Eq. (2.120), Eq. (2.119) for the equilibrium constant can be rewritten as
vcvD( P ]Av
Kp
N~N~ 
(2.121)
where Av = Vc + vo  liA  lOB.Equation (2.121) can be used to verify equilibrium results from complex computer programs as will be done later in this section.
Example 2.13 One mole of hydrogen and onehalf mole of oxygen are in equilibrium at a temperature of 2000 K and a pressure of 10 atm. Determine the number of moles of H20, H2, and O2 in the products (neglect other products such as OH, H, O, etc.).
Solution: We start by writing the chemical reaction equation for both the stoichiometric reaction and the actual reaction being modeled. Stoichiometric:
1
H2 +702 ~
H20
(thus liH20
=
1, v142= 1, v02 = 1/2)
Actual: H2 q ~1 0 2 >xH2 q yO2 +zH20
142
ELEMENTS OF PROPULSION
Because there are the three unknowns (x, y, and z), three equations are required. Two equations come from the atom balance of each element (H and O) and the third from the chemical equilibrium coefficient (Kp): A t o m balance: H:2 = 2x+2z
orz=
l x
ory=x/2
O:1 2y+z
Now the actual reaction equation can be written with just one unknown: 1 H2 ÷~O2 + xH2 q~x O 2 + (1 x)H20
From the J A N N A F tables in the Supporting Material, log Kp = 3.540 at 2000 K for the reaction 1
H2 q~O2 + H20 Thus
gp
103.540 3467.4 ~
~
Ke
__
Nc t~H2
t~O 2
Na N B
~Vtotal/
where NA = x, N8 = x/2, N c = 1  x, Av =  1/2, and Ntotal = 1 + x/2. Thus (1x)
Kp  xl
{
10 ,~1/2
xv/~ \1 + x / 2 J
= 3467.4
or (1  x)~/1 + x / 2
X3/2
3 4 6 7 . 42x/~v/ /77,
7753.3
Solving for x gives x = 0.002550, and the reaction can be written as 1 H2 ÷ ~O2 + 0.002550H2 + 0.00127502 ÷ 0.997450H20
Thus there are 0.002550 moles of hydrogen, 0.001275 moles of oxygen, and 0.997450 moles of water.
REVIEW OF FUNDAMENTALS
143
2.11.3 EQL Software The included EQL software program by David T. Pratt uses the NASA Glenn thermochemical data and the GordonMcBride equilibrium algorithm (see Ref. 18). This equilibrium calculation is based on finding the mixture composition that minimizes the Gibbs function, which allows rapid solution of reactions resulting in many products.
Example 2.14 The EQL software program gives the following result for one mole of hydrogen and onehalf mole of oxygen at a temperature of 2000 K and a pressure of 10 atm: H 2 + ~02 1 + 0.002717H2 + 0.00111802 I 0.9952H20 + 0.000974OH +2.668 x 1 0  5 H + 7 . 0 1 6 x 1 0  6 0 + 5 . 9 6 3 x 107HO2 + 1.992 x 107H202 The 0.0022 moles less of H20 are produced in this equilibrium mixture than Example 2.13, and these molecules show up in increases in the number of moles of H2 and O2. The equilibrium constant can be determined from these data and Eq. (2.121):
K p _ N cvH2o ( P ~ = Av v.2 8 vo2 \N~otal/ NAN
0.9952 0.00271 x ~ 1 8
(
10
~,0.~1
~1/2 = 3471
or log Kp = 3.540, which agrees with the JANNAF tables.
2.11.4 Enthalpy of Chemical Component, Enthalpy of Formation, Heat of Reaction, and Adiabatic Flame Temperature For reacting systems, the enthalpy of each component must be written in a form that has the same reference state. The enthalpy of chemical components (products or reactants) can be calculated using Eq. (2.50) with h for a component written as T
(2.122) Td
where h] is the enthalpy offormation (also called the heat of formation) at the reference state (datum) of 25°C (298 K), 1 atm (h  h°) is the enthalpy change due to the temperature change from the reference s t a t e (Td), and 0p is the specific heat at constant pressure per mole. Typical values of hf are given in Table 2.6 and values (h  h ) for typical gases are given in the JANNAF tables of the Supporting Material.
ELEMENTS OF PROPULSION
144 Table 2.6
Enthalpy of formation hf for some reactants and product gases at datum temperature 536°R/298 K (Refs. 12, 13)
Gas
Btu/lbmol
kJ/kgmol
Methane, CH4 Ethane, C2H6
 32,192  36,413  71,784  89,600  152,981  47,520  169,18l 93,717 0  103,966 107,139 0 16,967 203,200 0 35,275 38,817 14,228
 74,883  84,701  166,978  208,421  355,853 a  110,537  393,538 217,997 0  241,838 249,218 0 39,467 472,668 0 82,053 90,293 33,096
H e x a n e , C6H14
Octane, CsH1s JetA, C12H23 Carbon monoxide, CO Carbon dioxide, CO2 Atomic hydrogen, H Hydrogen, H2 Water vapor, H20 Atomic oxygen, O Oxygen, O2 Hydroxyl, OH Atomic nitrogen, N Nitrogen, N2 Nitrous oxide, N20 Nitric oxide, NO Nitrogen dioxide, NO2
aFor heating value hpR = 18,400Btu/lbm = 42,800 kJ/kg.
The enthalpy of both products and reactants in a reaction can be plotted vs temperature as shown in Fig. 2.41. A reaction typically causes a change in temperature. The heat of reaction AH is defined as the positive heat transfer to the products that is required to bring them back to the original temperature of the reactants. For ideal (perfect) gases, the heat of reaction AH at the standard reference temperature can be calculated using np
tl R o
A H = He  HI~ = ~.~ NiAh/i  ~ i=1
j
NjAhj)
(2.123)
1
where Ni and Nj are the number of moles of the products and reactants, respectively, and Ahj is the heat of formation per mole of species i and j, AH is the vertical difference between the enthalpy line of the reactants and that of the products in Fig. 2.42. Consider the adiabatic flow through a combustion chamber like that of Fig. 2.4d. The reactants enter at a temperature denoted by 1 in Fig. 2.42 and the products leave at a temperature denoted by 2. Neglecting changes in kinetic energy, the energy balance for this process yields H~ //2
or
Hp = HR
(2.124)
The temperature at state 2 is called the adiabatic flame temperature. This temperature can be calculated by determining the equilibrium state with the
REVIEW OF FUNDAMENTALS R
145 p
H
,
..o..oo...o.
k/":
/ i
/
l
I
r. Fig. 2.42
I
I
r,
r2
Enthalpy states for a reaction.
same enthalpy H and pressure P as state 1. To facilitate calculation, we rewrite Eq. (2.124) in terms of the enthalpy difference (h  h °) and the heat of reaction at the datum (reference) state ([AH]r ~. First we write H2  [He]r.= H1  [HR]rd+[AH]r~ Since np
j=l
and nR
H1  [HR]ra= E
Ni(h  h°)i
i=1
then np
H2  [Hplr~ = Z j=l
/2R
Nj(/t  / t ° ) j = Z N i ( h
 h°)i + [AHlr a
(2.125)
i=1
The following steps are used to calculate the adiabatic flame temperature: 1) Assume a final temperature T2. 2) Calculate the mole fraction of the products for the resulting Kp at T2. 3) Calculate Y~':~I Ni(h  h°)i + [AH]v~. 4) Calculate ~ j ~ l Nj(/~  / t ° ) j . If its value is greater than that of step 3, reduce the value of T2 antiperform steps 2  4 again.
146
ELEMENTS
OF
PROPULSION
This calculation by hand is tedious and has been programmed for rapid calculation using computers. The EQL software calculates the adiabatic flame temperature for given reactants and inlet temperature T and pressure P as shown in Example 2.16.
Example 2.15 Calculate the heat of reaction for the reaction of Example 2.13.
Solution: We have H2 q ~1 0 2 + 0.002024H2 + 0.00101202 + 0.997976H20 at 2000 K. From the JANNAF tables in the Supporting Materials, at 2000 K we have Ah)H2  o = O, Ahjo2 o = O, and AhjH2o o = 60.150 kcal/mole. Eq. (2.78) gives 

AH = H p HR l'lp
1~R
= Z NiAhfi  Z NjAh~j = 0.997976 x 60.150 = 60.028 kcal i=1
j=l
Thus 60.028 kcal (251.28 kJ) of energy must be removed during the reaction to keep the temperature at 2000 K.
Example 2.16 Using the EQL software, determine the adiabatic flame temperature for one mole of oxygen and onehalf mole of methane with the reactants at a temperature of 500 K and pressure of 800 kPa.
Solution: We first enter the reactants and inlet pressure and temperature into the opening screen of EQL. Next we select the Equilibrium Processes Tab, which displays the numerous possible combustion processes. We select 'Adiabatic Flame Temperature' and perform the calculations. The resulting adiabatic flame temperature is 3349.85 K, and the resulting principal products are listed below: Compound H20 CO CO2 OH 02 H2 H O
Mole fraction 0.4093 0.1540 0.1190 0.0989 0.0792 0.0681 0.0382 0.0330
REVIEW OF FUNDAMENTALS
147
Problems 2.1
Consider Fig. P2.1. A stream of air with velocity of 500 ft/s and density of 0.07 lbm/ft 3 strikes a stationary plate and is deflected 90 deg. Select an appropriate control volume and determine the force Fp necessary to hold the plate stationary. Assume that atmospheric pressure surrounds the jet and that the initial jet diameter is 1.0 in.
/A
,h
Fig. P2.1
2.2
Consider Fig. P2.1. An airstream with density of 1.25 k g / m 3 and velocity of 200 m / s strikes a stationary plate and is deflected 90 deg. Select an appropriate control volume and determine the force Fp necessary to hold the plate stationary. Assume that atmospheric pressure surrounds the jet and that the initial jet diameter is 1.0 cm.
2.3
Consider the flow shown in Fig. P2.2 of an incompressible fluid. The fluid enters (station l) a constantarea circular pipe of radius r0 with uniform velocity Va and pressure P1 The fluid leaves (station 2) with the parabolic velocity profile V2 given by
V2=Vmax 1and uniform pressure P2 Using the conservation of mass and momentum equations, show that the force F necessary to hold the pipe in place can be V1
F
1
V2 = Vmax [ 1
2 Fig. P2.2
r
2
]
148
ELEMENTS OF PROPULSION expressed as
(
F:~r 2 P1P2+~g~j 2.4
Consider the flow of an incompressible fluid through a twodimensional cascade as shown in Fig. P2.3. The airfoils are spaced at a distance s and have unit depth into the page. Application of the conservation of mass requires Vt cos/3t = Ve cos/3e" (a) From the tangential momentum equation, show that rn Fo =  (Vt gc
sin/3t  Ve sin fie)
(b) From the axial momentum equation, show that Fz = s(Pe  Pt)
(c) Show that the axial force can be written as Fz=
~/
~
Streamlines
Fig. P2.3 2,5
When a freejet is deflected by a blade surface, a change of momentum occurs and a force is exerted on the blade. If the blade is allowed to move at a velocity, power may be derived from the moving blade. This is the basic principle of the impulse turbine. The jet of Fig. P2.4, which is initially horizontal, is deflected by a fixed blade. Assuming the same pressure surrounds the jet, show that the horizontal (Fx) and vertical forces (Fy) by the fluid on the blade are given by
Fx =
/1/(/'/1/'12COS/3) gc
and
/1'/~2sin/3
Fy  
gc
REVIEW OF FUNDAMENTALS
149
lade rh
~.
Fig. P2.4 Calculate the force Fy for a mass ul = u2 = 2000 ft/s, and/3 = 60 deg.
2.6
flow
rate
of
100ibm/s,
One method of reducing an aircarft' s landing distance is through the use of thrust reversers. Consider the turbofan engine in Fig. P2.5 with thrust reverser of the bypass airstream. It is given that 1500 lbm/s of air at 60°F and 14.7 psia enters the engine at a velocity of 450 ft/s and that 1250 lbm/s of bypass air leaves the engine at 60 deg to the horizontal, velocity of 890 ft/s, and pressure of 14.7 psia. The remaining 250 lbm/s leaves the engine core at a velocity of 1200ft/s and pressure of 14.7 psia. Determine the force on the strut F~ Assume the outside of the engine sees a pressure of 14.7 psia. F~
1500 lbngs 450 ft/s
l ore
250 lbm/s 1200 ft/s
Fig. P2.5 2.7
Consider Fig. P2.6. Air with a density of 0.027 lbm/ft 3 enters a diffuser at a velocity of 2470 ft/s and a static pressure of 4 psia. The air leaves the diffuser at a velocity of 300 ft/s and a static pressure of 66 psia. The entrance area of the diffuser is 1.5 ft 2, and its exit area is 1.7 ft 2. Determine the magnitude and direction of the strut force necessary to hold the diffuser stationary when this diffuser is operated in an atmospheric pressure of 4 psia.
150 2.8
ELEMENTS OF PROPULSION Consider Fig. P2.6. It is given that 50 k g / s of air enters a diffuser at a velocity of 750 m / s and a static pressure of 20 kPa. The air leaves the diffuser at a velocity o f 90 m / s and a static pressure of 330 kPa. The entrance area of the diffuser is 0.25 m 2, and its exit area is 0.28 m 2. Determine the magnitude and direction of the strut force necessary to hold the diffuser stationary when this diffuser is operated in an atmospheric pressure of 20 kPa.
L[ 1
Diffuser
Fig. P2.6
2.9
Consider Fig. P2.7. It is given that 100 l b m / s of air enters a nozzle at a velocity of 600 ft/s and a static pressure of 70 psia. The air leaves the nozzle at a velocity of 4000 f t / s and static pressure of 2 psia. The entrance area of the nozzle is 14.5 ft 2, and its exit area is 30 ft 2. Determine the magnitude and direction of the strut force necessary to hold the nozzle stationary when this nozzle is operated in an atmospheric pressure of 4 psia.
Nozzle
Fig. P2.7
2.10
Consider Fig. P2.7. Air with a density of 0.98 k g / m 3 enters a nozzle at a velocity of 180 m / s and a static pressure of 350 kPa. The air leaves the nozzle at a velocity of 1200 m / s and a static pressure of 10 kPa. The entrance area of the nozzle is 1.0 m 2, and its exit area is 2.07 m2. Determine the magnitude and direction of the strut force necessary to hold the nozzle stationary when this nozzle is operated in an atmospheric pressure of 10 kPa.
2.11
For a calorically perfect gas, show that P + pV2/gc can be written as P(1 + yM2). Note that the Mach number M is defined as the velocity V divided by the speed of sound a.
REVIEW OF FUNDAMENTALS
151
2.12
Air at 1400 K, 8 atm, and 0.3 Mach expands isentropically through a nozzle to 1 atm. Assuming a calorically perfect gas, find the exit temperature and the inlet and exit areas for a mass flow rate of 100 kg/s.
2.13
It is given that 250 lbm/s of air at 2000°F, 10 atm, and 0.2 Mach expands isentropically through a nozzle to 1 arm. Assuming a calorically perfect gas, find the exit temperature and the inlet and exit areas.
2.14
Air at 518.7°R is isentropically compressed from 1 to 10 atm. Assuming a calorically perfect gas, determine the exit temperature and the compressor's input power for a mass flow rate of 150 lbm/s.
2.15
It is given that 50 kg/s of air at 288.2 K is isentropically compressed from l to 12 atm. Assuming a calorically perfect gas, determine the exit temperature and the compressor's input power.
2.16
Air at  55°F, 4 psia, and M = 2.5 enters an isentropic diffuser with an inlet area of 1.5 ft 2 and leaves at M = 0.2. Assuming a calorically perfect gas, determine: (a) The mass flow rate of the entering air (b) The pressure and temperature of the leaving air (c) The exit area and magnitude and direction of the force on the diffuser (assume outside of diffuser sees 4 psia)
2.17
Air at 225 K, 28 kPa, and M = 2.0 enters an isentropic diffuser with an inlet area of 0.2 m 2 and leaves at M = 0.2. Assuming a calorically perfect gas, determine: (a) The mass flow rate of the entering air (b) The pressure and temperature of the leaving air (c) The exit area and magnitude and direction of the force on the diffuser (assume outside of diffuser sees 28 kPa)
2.18
Air at 1800°F, 40 psia, and M = 0.4 enters an isentropic nozzle with an inlet area of 1.45 ft 2 and leaves at 10psia. Assuming a calorically perfect gas, determine: (a) The velocity and mass flow rate of the entering air (b) The temperature and Mach number of the leaving air (c) The exit area and magnitude and direction of the force on the nozzle (assume outside of nozzle sees 10 psia)
2.19
Air at 1500 K, 300 kPa, and M = 0.3 enters an isentropic nozzle with an inlet area of 0.5 m 2 and leaves at 75 kPa. Assuming a calorically perfect gas, determine: (a) The velocity and mass flow rate of the entering air (b) The temperature and Mach number of the leaving air (c) The exit area and magnitude and direction of the force on the nozzle (assume outside of nozzle sees 75 kPa)
152
ELEMENTS OF PROPULSION
2.20
It is given that 100 lb/s of air enters a steady flow compressor at 1 atm and 68°F. It leaves at 20 atm and 800°F. If the process is adiabatic, find the input power, specific volume at exit, and change in entropy. Is the process reversible? (Assume a calorically perfect gas.)
2.21
It is given that 50 kg/s of air enters a steady flow compressor at 1 atm and 20°C. It leaves at 20 atm and 427°C. If the process is adiabatic, find the input power, specific volume at exit, and change in entropy. Is the process reversible? (Assume a calorically perfect gas.)
2.22
It is given that 200 lb/s of air enters a steady flow turbine at 20 atm and 3400°R. It leaves at 10 atm. For a turbine efficiency of 85%, determine the exit temperature, output power, and change in entropy. (Assume a calorically perfect gas.)
2.23
It is given that 80 kg/s of air enters a steady flow turbine at 30 atm and 2000 K. It leaves at 15 atm. For a turbine efficiency of 85%, determine the exit temperature, output power, and change in entropy. (Assume a calorically perfect gas.)
2.24
Rework Problem 2.13 for variable specific heats, using Appendix D or the program AFPROP. Compare your results to Problem 2.13.
2.25
Rework Problem 2.15 for variable specific heats, using Appendix D or the program AFPROP. Compare your results to Problem 2.15.
2.26
Rework Problem 2.16 for variable specific heats, using Appendix D or the program AFPROP. Compare your results to Problem 2.16.
2.27
Rework Problem 2.19 for variable specific heats, using Appendix D or the program AFPROP. Compare your results to Problem 2.19.
2.28
Rework Problem 2.20 for variable specific heats, using Appendix D or the program AFPROP. Compare your results to Problem 2.20.
2.29
Rework Problem 2.23 for variable specific heats, using Appendix D or the program AFPROP. Compare your results to Problem 2.23.
2.30
It is given that 100 lbm/s of air at total pressure of 100 psia, total temperature of 40°F, and static pressure of 20 psia flows through a duct. Find the static temperature, Mach number, velocity (ft/s), and flow area (ft2).
2.31
Products of combustion (3' = 1.3) at a static pressure of 2.0 MPa, static temperature of 2000 K, and Mach number of 0.05 are accelerated in an isentropic nozzle to a Mach number of 1.3. Find the downstream static pressure and static temperature, if the mass flow rate is 100 kg/s and the gas constant R is 286 J/(kg. K), use the mass flow parameter (MFP) and find the flow areas for M = 0.5 and M = 1.3.
REVIEW
OF
FUNDAMENTALS
153
2.32
Data for the JT9D highbypassratio turbofan engine are listed in Appendix B. If the gas flow through the turbines (from station 4 to 5) is 251 lbm/s with the total properties listed, what amount of power (kW and hp) is removed from the gas by the turbines? Assume the gas is calorically perfect with 7 = 1.31 and Rgc  1716 ft2/(s 2 °R).
2.33
At launch, the space shuttle main engine (SSME) has 1030 lbm/s of gas leaving the combustion chamber at Pt = 3000 psia and T, = 7350°R. The exit area of the SSME nozzle is 77 times the throat area. If the flow through the nozzle is considered to be reversible and adiabatic (isentropic) with R& = 3800 ft2/(s 2. °R) and 3' = 1.25, find the area of the nozzle throat (in. 2) and the exit Mach number. Hint: Use the mass flow parameter to get the throat area and Eq. (3.14) to get the exit Mach number.
2.34
The experimental evaluation of a gas turbine engine's performance requires the accurate measurement of the inlet air mass flow rate into the engine. A bellmouth engine inlet (shown schematically in Fig. P2.8) can be used for this purpose in the static test of an engine. The freestream velocity Vo is assumed to be zero, and the flow through the bell mouth is assumed to be adiabatic and reversible. See Fig. P2.8. Measurements are made of the freestream pressure Pro and the static pressure at station 2 (P2), and the exit diameter of the inlet D 2. (a) For the bellmouth inlet, show that the Mach number at station 2 is given by
M2 =
Lkp, ° _
1
and the inlet mass flow rate is given by
_
4
eTo )
k eT }
outh inlet
Tto
Pro
ue
;iiii
I AP=PtoP2
Fig. P2.8
]
ELEMENTS OF PROPULSION
154 (b)
For the following measured data, determine the inlet air mass flow rate, Mach number M2, static temperature T2, and velocity V2:
Pro = 77.80kPa
Tto = 27°C
AP = 3.66kPa 2.35
D2
=
0.332m
An ideal ramjet (see Fig. P2.9) is operated at 50,000ft altitude with a flight Mach number of 3. The diffuser and nozzle are assumed to be isentropic, and the combustion is to be modeled as an ideal heat interaction at constant Mach number with constant total pressure. The crosssectional area and Mach number for certain engine stations are given in Table P2.1. The total temperature leaving the combustor Tr4 is 4000°R. Assume ambient pressure surrounding the engine flow passage. (a) Determine the mass flow rate of air through the engine (Ibm/s). (b) Complete the table with flow areas, static pressures, static temperatures, and velocities. (c) Find the thrust (magnitude and direction) of the diffuser, combustor, and nozzle. (d) Find the thrust (magnitude and direction) of the ramjet.
Diffuser ~ :
~.Nozzle ~ i< Combustor i< Centerline
1
Fig. P2.9
2.36
If the flow enters the diffuser of Problem 2.7 at  5 5 ° F , is the process isentropic ?
2.37
If the flow enters the nozzle of Problem 2.10 at 1250 K, is the process isentropic?
Table P2.1
Station Area (ft 2) Mach P (psia) T (°R) V fit/s)
1
2
3
4
5
6
4.235 3
1
0.15
0.15
1
3
REVIEW OF FUNDAMENTALS 2.38
2.39
Air at 20 kPa, 260 K, Determine: (a) Total temperature (b) Total temperature (c) Static temperature Air upstream
155
and Mach 3 passes through a normal shock. and pressure upstream of the shock and pressure downstream of the shock and pressure downstream of the shock
of a normal
shock has
the
following properties:
Pt = 100 psia, Tt = 100°F, and M = 2. Find the upstream static temperature, static pressure, and velocity (ft/s). Find the downstream total temperature, Mach number, total pressure, static temperature, static pressure, and velocity (ft/s). 2.40
If the diffuser of the ideal ramjet in Problem 2.35 has a normal shock in front of the inlet, determine: (a) The total and static pressures and the static temperature downstream of the shock (b) The mass flow rate through the engine (Ibm/s) (assume choked flow at diffuser throat) (c) The thrust of the engine
2.41
Air at a total pressure of 1.4 MPa, total temperature of 350 K, and Mach number of 0.5 is accelerated isentropically in a nozzle (see Fig. P2.10) to a Mach number of 3 (station x), passes through a normal shock (x to y), and then flows isentropically to the exit. Given a nozzle throat area of 0.05 m 2 and the exit area of 0.5 me: (a) Find the area at the shock (b) Find the static pressure and static temperature upstream of the shock (station x) (c) Find the Mach number and the total and static pressures and temperatures downstream of the shock (station y) (d) Find the Mach number, static pressure, and static temperature at the exit
Inlet
Throat
••Ex•t Shock
Centerline
x
y
Fig. P2.10 2.42
Air Tt = area (a)
flows through an isentropic nozzle with inlet conditions of 2000°R and Pt = 100 psia. The throat area is 2 ft 2, and the exit is 10.32 ft 2. If the flow is choked at the throat, find: Mass flow rate through the nozzle
156
ELEMENTS OF PROPULSION (b) (c)
Mach number, static temperature, and static pressure at exit without a shock Mach number, static temperature, and static pressure at exit with a shock in the divergent section where the flow area is 4.06 ft 2 (see Fig. P2.10)
2.43
Air at a total pressure of 400 psia, total temperature of 500°R, and subsonic Mach number is accelerated isentropically in a nozzle (see sketch in Fig. P2.10) to a supersonic Mach number (station x), passes through a normal shock (x to y), and then flows isentropically to the exit. Given a nozzle inlet area of 3 ft 2, throat area of 1 ft 2, and exit area of 6 ft2: (a) Find the Mach number at the inlet (b) Find the Mach number upstream of the shock (station x) (c) Find the static pressure and static temperature upstream of the shock (station x) (d) Find the Mach number and the total and static pressures and temperatures downstream of the shock (station y) (e) Find the Mach number, static pressure, and static temperature at the exit
2.44
A 20deg wedge (see Fig. P2.11) is to be used in a wind tunnel using air with test conditions of M = 3, Tt = 500°R, and Pt = 100 psia. Determine the angle of the oblique shocks and the downstream total and static properties (pressure and temperature). Use the GASTAB program.
10 ° Fig. P2.11 2.45
A 15deg ramp is used on a supersonic inlet at M = 3.5 and an altitude of 20 km. Use the GASTAB program. (a) Determine the angle of oblique shock and flow properties (Mach number, total and static temperature, total and static pressure) upstream and downstream of the shock (b) At what Mach number does the shock detach from the ramp?
2.46
Draw an HK diagram for the J79 afterburning turbojet engine of Fig. 1.6. Assume c), is constant and Tto is 518.7°R.
2.47
Draw an HK diagram for the F100 afterburning turbofan engine of Table B.4. Assume Cp is constant and Tto is 518.7"R.
2.48
Show that dP/P is a constant for Example 2.12.
REVIEW OF FUNDAMENTALS 2.49
A calorically perfect gas undergoes an ideal heat interaction in a duct. The area of the duct is varied to hold the static pressure constant. Using the influence coefficients of Table 2.5, show the following: (a) The area variation required to hold the static pressure constant is given by
dA a (b)
(l+~_mz)
(c)
1+
M2 dTt T,
The relationship for the area of the duct in terms of the Mach number is given by _
(d)
dTt T~
The relationship between the Mach number and total temperature, for the preceding area variation, is given by
M2 
2.50
157
M1
2
where subscripts refer to states 1 and 2. Show V is CONSTANT.
You are required to design a nozzle to pass a given mass flow rate of air with m i n i m u m frictional losses between a storage pressure chamber Po and an exhaust region P,. with a given variation in pressure between the two regions. The design conditions are rn = 1000 l b m / s
Po = 3000 psia
Tc = 3700°R
Table P2.2
Stationx, in.
P, psia
0 2 4 6 8 10 (throm) 12 16 24 41 60
2918 2841 2687 2529 2124 1585 727 218 19.8 3.21 1.17
P/Pt M T/Tt A/A* T, °R a, ft/s V, ft/s A, in. 2 D, in.
158
ELEMENTS OF PROPULSION (a) (b) (c) (d)
Using the design conditions, complete Table P2.2 Make a plot of the nozzle contour (see Fig. 2.33) Calculate the nozzle design pressure ratio P~ and the nozzle area ratio e Using the altitude table, determine the design altitude for this nozzle
(P,. = Pa) (e)
Determine the thrust of a rocket motor using this nozzle at its design altitude
2.51
A perfect gas enters a constantarea heater at a Mach number of 0.3, total pressure of 600 kPa, and total temperature of 500 K. A heat interaction of 500 kJ/kg into the gas occurs. Using the GASTAB software, determine the Mach number, total pressure, and total temperature after the heat interaction for the following gases: (a) 3' = 1.4 and Cp = 1.004 kJ/(kg K) (b) 3' = 1.325 and Cp = 1.171 k J / ( k g . K)
2.52
A perfect gas enters a constantarea heater at a Mach number of 0.5, total pressure of 200 psia, and total temperature of 1000°R. A heat interaction of 100 Btu/lbm into the gas occurs. Using the GASTAB software, determine the Mach number, total pressure, and total temperature after the heat interaction for the following gases: (a) 3' = 1.4 and Cp = 0.24 Btu/(lbm. °R) (b) y = 1.325 and cp = 0.28 Btu/(lbm. °R)
2.53
A convergentonly nozzle is to be used on an afterburning gas turbine engine as shown in Fig. P2.12. Model the afterburner (station 6 to station 7) as a constantarea duct (A6 = A7) with simple heat interaction qin into the air. The flow through the nozzle (station 7 to station 8) is isentropic. The exit area of the nozzle As is varied with the afterburner setting Tt7 tO keep sonic (M = 1) flow at station 8 and the inlet conditions (mass flow rate, Pt, and Tt) constant at station 6. (a) Using the mass flow parameter, show that the area ratio A8/A 6 is given by
A8 _ Pt6 T / ~ MFP(M6) et6 T/~(A*~ A6 P t T ~ t 6 M F P ~ ~ 1)PtTVTt6\A,]M6
/ qin : Nozzle : • :
Afterburner
Fig. P2.12
:
~,
REVIEW OF FUNDAMENTALS (b)
159
For M6 = 0.4, determine the area ratio As~A6 for the following values of Tt7/Tt6: 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, and Tt7 = Tt*
2.54
One mole of hydrogen and one mole of oxygen are in equilibrium at a temperature of 1500 K and a pressure of 5 atm. Determine the number of moles of H20, H2, and Oz in the products (neglect other products such as OH, H, O, etc.). Compare your results with that of the EQL program.
2.55
One mole of hydrogen and 0.45 moles of oxygen are in equilibrium at a temperature of 2200 K and a pressure of 20 atm. Determine the equilibrium constant from the results of the EQL program and compare to the JANNAF tables.
2.56
Calculate the heat of reaction for the reaction of Problem 2.54.
2.57
Using the EQL software, determine the adiabatic flame temperature for one mole of hydrogen and onehalf mole of oxygen with the reactants at a temperature of 300 K and a pressure of 10 atm.
3 Rocket Propulsion
3.1
Introduction
The types of rocket propulsion and basic parameters used in performance evaluation of rocket propulsion are described in this chapter. First, the concepts of thrust, effective exhaust velocity, and specific impulse are presented for a gaseous continuum leaving a rocket engine. These concepts are covered in more detail than that contained in Chapter 1. To determine the requirements of rocket vehicles, some introductory mechanics of orbital flight are investigated along with rocket mass ratios, rocket acceleration, and multistage rockets. This section on requirements and capabilities is followed by a general discussion of the various sources of rocket propulsion and a summary of their capabilities. Following this general introduction to rocket propulsion and rocket flight is a discussion of rocket nozzle types, a study of the detailed performance parameters for rocket engines expelling a continuum, detailed example problems, and discussion of liquid and solidpropellant rockets.
3.1.1
Rocket Thrust
The rocket thrust studied here will be applicable to rockets that eject a gaseous continuum material from a nozzle at a constant rate. Chemical and nuclearH2 rockets are examples of such rocket propulsion systems. The thrust of an ionic rocket that expels discrete ion particles at high speeds, on the other hand, is not related to properties of a continuum such as pressure and density. The thrust equation to be developed will not apply, therefore, to ionic rockets. The thrust of ion rockets, however, still depends on the same basic principle of operation as chemical or nuclearH2 rockets, i.e., the reaction force resulting from imparting momentum to a mass.
3.1.2
Ideal Thrust Equation
The ideal rocket thrust is defined as the force required to hold a rocket at rest as the rocket ejects a propellant rearward under the following assumed conditions: 1) Atmospheric pressure acts everywhere on the rocket's external surface. (There is no interaction between the jet issuing from the rocket nozzle and the air in contact with the external surface of the rocket.) 161
162
E L E M E N T S OF P R O P U L S I O N
2) The propellant flows from the rocket at a constant rate in uniform onedimensional flow at the nozzle exit section. 3) The rate of change of the momentum within the rocket is negligibly small and may be neglected. The rocket of Fig. 3. la is held in place by a reaction strut and is ejecting a propellant to the rear under the preceding conditions. The shear force at section BB of the reaction strut is equal in magnitude to the ideal rocket thrust. As shown by the freebody diagram in Fig. 3.1a, the thrust is produced by the internal force on the rocket shell. In a vacuum, the thrust would be identical with Fint, but is less than the internal force in the atmosphere because of the ambient pressure forces acting on the external surface of the rocket. The external force is due to atmospheric pressure forces only and is given by Next =
PaAe
If we assume internal isentropic flow, the interior force is due to pressure forces only. It is the sum of all of the pressure forces exerted by the propellant on the interior wall surfaces. The internal force is readily evaluated in terms of propellant flow properties by application of the momentum equation to the control surface indicated in Fig. 3.1b, and by using the fact that the force of the propellant on
Thrust = iR x I = Fint  Next F~xt
f r e e b o d y ~ diagram
iRx[
a) Forces on rocket shell
control /E/_ surface • . .............
control
~¢/surface ."" ...........
PeAe <s
Forces on control surface b) Evaluation of Fig. 3.1
=
 ,,,~ g/~c ""
Momentum flux through control surface
[Fint[by momentum equation Ideal rocket thrust.
I
•
ROCKET PROPULSION
163
the internal walls equals, in magnitude, the force of the walls on the propellant. For the magnitude of the internal force, we obtain 1
1;nge
2
IFintl = PeAe +PeAeV{ = PeAe +  gc gc
The difference between Fin t and Fex t is the ideal rocket thrust, Fs, f i = f i n t   Next
Fi 
~riVe g,.
+ (Pe  Pa)Ae
(3.1)
This equation gives the thrust in terms of the propellant flow properties, the exit area, and the ambient pressure. It is well to remember that only the term P, Ae on the right side of Eq. (3.1) represents a real force acting on the rocket shell.
3.1.3 Optimum Ideal Thrust Let us consider a rocket ejecting a given propellant supersonically from a combustion chamber at a fixed temperature and pressure. Under these conditions, the mass flow through the rocket nozzle is constant at a value given by
PtA
2 gc R
Y \PtJ
\PtJ
J
(3.2)
Let us further assume that the nozzle exit area is adjustable and that the ambient pressure Pa is constant. Now notice that as we vary Ae, the pressure P~ and the velocity Ve will vary, but mass flow rate m and P~ will remain constant. The following question arises: For what value of Ae will the corresponding values of Ve and Pe make the thrust of Eq. (3.1) a maximum? The simplest way to demonstrate the optimum thrust condition is to deal with the real forces acting on the rocket shell, and to determine directly under what conditions these produce maximum thrust. First, we observe that as Ae is varied by adding a further divergent section on the nozzle exit, any change in thrust comes about by the net force acting on this added section. This follows from the fact that forces acting on the original surfaces of the rocket are not changed by the added section because atmospheric pressure still acts on the original external surfaces, and the original internal pressure distribution remains the same as long as supersonic flow exists at the nozzle exit. Because any thrust change, therefore, comes about by the action of the forces acting on the added surfaces introduced in changing Ae, we need only to examine the forces associated with the added surface to maximize thrust. Figure 3.2 shows an enlarged view of the pressure distribution on the external and internal surfaces near the exit of a rocket nozzle. The external
164
ELEMENTS OF PROPULSION z
enlarged
P>Pa
Fig. 3.2
Forces acting on rocket nozzle wall for optimum thrust consideration.
pressure is constant at Pa" The internal pressure decreases from a value greater than Pa at x to Pa at y, and to a value lower than Pa at Z, corresponding to an overexpanded nozzle. Now, as Ae is increased by adding surfaces up to station y, we find the thrust is increased because the internal pressure on these added surfaces is greater than Pu. If Ae is increased beyond that value corresponding to y, the thrust is decreased because there is a net drag force acting on the surfaces beyond y because the internal pressure is less than Pa between y and z. To eliminate the wall surface producing a net drag and to not eliminate any wall surface producing a net thrust, the nozzle should be terminated at y where Pe = P,. The exit velocity corresponding to this condition will be denoted (V~)opt so that
(Fi)op t 
m(Ve)op,  
g,.
(3.3a)
If the nozzle of Fig. 3.2 is at a higher altitude, the ambient pressure is reduced to a value of Pa', indicated by the dashed line of Fig. 3.2. The pressure at this higher altitude is such that the internal pressure acting on the nozzle wall from y to z is greater than the new ambient pressure. At this higher altitude, then, the nozzle surface from y to z is a thrustproducing surface and should be retained. This means a higher nozzle area ratio at higher altitudes for optimum thrust.
3.1.4
Vacuum Ideal Thrust
In a vacuum (very high altitudes), Fex t = 0. Thus (Fi)vac = Fint, where Fint c a n be evaluated by ( F i ) v a c = Fint =
;nVe + PeAe gc
(3.3b)
ROCKET PROPULSION
165
Thrust F1
r
Altitude
0 Fig. 3.3
3.1.5
Ideal thrust variation with altitude.
Thrust Variation with Altitude
To understand the variation of thrust with altitude, we consider two rockets, A and B. Rocket A has infinitely adjustable area ratio, whereas rocket B's area ratio is fixed (constant) at a value that gives optimum expansion at sea level. For each rocket, Fi Fiat  Fext. At higher than sealevel altitudes, rocket A's thrust increases because Fin t remains the same, and Fext decreases to zero a s Pa approaches zero. The thrust of rocket A exceeds that of rocket B at altitudes since Fext decreases the same as for rocket B. In addition, rocket A has thrustproducing surfaces (such as segment x  y of Fig. 3.2) added on to it as it climbs and expands to lower ambient pressures. The variation of thrust for rockets A and B is sketched in Fig. 3.3.
3.1.6 Effective Exhaust Velocity C and Specific Impulse (Isp) The effective exhaust velocity C is defined in Chapter 1 by Eq. (1.53) as C ~
g e t
(Pe  Pa)Ae /n / gc
(3.4)
Using this definition of effective exhaust velocity in Eq. (3.1), the ideal thrust is simply F;
mC gc
(3.5)
The values of a rocket's exhaust velocity Ve and exhaust pressure Pe depend on the type of rocket engine and its design and operation. For launch from the
166
ELEMENTS OF PROPULSION
surface of the Earth to Earth orbit, the atmospheric pressure Pa varies from a low of zero (vacuum) to a high of that at sea level. For a rocket engine where the mass flow rate and combustion chamber conditions are constant, the effective exhaust velocity C and ideal thrust Fi will vary from a low value for sealevel operation to a high value at vacuum conditions. Figure 1.42b shows the predicted variation of thrust with altitude for the space shuttle main engine (SSME) under these conditions. The variation of effective exhaust velocity C will be the same as the thrust times a constant. Even though the effective exhaust velocity C can vary, an average value of the effective exhaust velocity may be used for ease of analysis or calculation. The specific impulse Isp for a rocket is defined as the thrust per unit of propellant weight flow: Isp

F w

F gc rh go
(3.6)
where go is the acceleration of gravity at sea level. Specific impulse has units of seconds. Using Eq. (3.5), the effective exhaust velocity is directly related to the specific impulse by C = lspg0
(3.7)
As just discussed, the effective exhaust velocity can vary during operation. From Eq. (3.7), the specific impulse Isp for a rocket engine will vary in direct proportion to its effective exhaust velocity C. For ease of analysis or calculation, an average value of specific impulse may be used.
Example 3.1 Appendix C lists the F1 engine used on the Saturn V rocket as having a vacuum thrust of 1,726,000 lbf and vacuum l~p of 305 s. Determine the mass flow rate at the vacuum conditions.
Solution:
Solving Eq. (3.6) for the mass flow rate, we have rn .
F gc 1,726,000 32.174 . . . . 305 32.174
lsp go
5659 lbm/s
3.2 Rocket Propulsion Requirements and Capabilities 3.2.1 Requirements The propulsion requirements for space flight missions are usually expressed in terms of speed increment that the propulsion system must supply to the space vehicle. When the appropriate increment in speed is supplied to a vehicle, the vehicle's guidance system must adjust the direction of the speed to attain the orbit desired. As a simple example, the ideal speed increment to place a
ROCKET PROPULSION
167 r
(R m go
Re
2
e
Fig. 3.4
Forces acting on nearEarthorbiting mass.
vehicle in orbit about the Earth is that speed that will give a balance between the centrifugal and gravitational forces acting on the body in orbit. Considering the balance of these forces, we can determine the needed velocity increment. The force of gravity acting on a mass m at a distance r from the Earth's center is mg~," (~)2, where Re is the Earth's radius (3959 miles, 6370 km), and go is the acceleration of gravity on the surface of the Earth (32.174 ft/s 2, 9.807 m/s2). Referring to the force diagram of Fig. 3.4, we have
mV,2 _ m gO_(Re) 2_ g~r gc \ r / Solving for the satellite velocity V, at a height h above the surface of the Earth (r = Re + h) gives vs =
h
(3.8)
For a circular Earth orbit at 160 km (100 miles), Eq. (3.8) gives a satellite velocity of 7800 m/s (25,600 ft/s). Thus a rocket must increase the velocity from zero at launch to 7800 m/s at 160 km. The propulsion system's fuel is the original source of the kinetic energy and the higher potential energy possessed by the orbital mass of Fig. 3.4. The kinetic energy (KE) of the orbital mass is simply given by {mV~2/(2gc)}, and the change in potential energy (APE) is given by Re+h
APE=
J m ggcd r = m e Re
Re+h
2g° egc J ~dr = m ggco ( R R ~ h ) Re
(3.9)
168
ELEMENTS OF PROPULSION
The sum of the changes in potential and kinetic energies per unit mass {A(pe + ke)} of the orbiting mass relative to sealevel launch is given by A(pe + ke)
_
m

gc !~Re+h]
+ 
2 gc
(3.10)
To place a mass in circular Earth orbit at 100 miles requires that the sum of the potential and kinetic energies be increased by 13,760 Btu/lbm (32 MJ/kg). This energy change is equal to the kinetic energy of a mass at a velocity of 26,250 ft/s (8000 m/s). In addition to this energy, the propulsion system must supply the energy that is delivered to the viscous atmosphere by the ascending vehicle due to aerodynamic drag forces. Thus the equivalent velocity increment requirement of a propulsion system to attain a near Earth orbit is greater than 26,250 ft/s (8000 m/s) due to aerodynamic drag. Taking these items into account gives an equivalent velocity increment requirement of approximately 30,000ft/s (9140 m/s) for a near Earth orbit. Table 3.1 gives this velocity increment and those required for certain other space missions. To escape the Earth's gravitational field (h + oo), solution of Eq. (3.9) Lives the change in potential energy as Earth escape: Ape = g°Re g,
(3.11)
This potential energy change is equal to the kinetic energy corresponding to the escape velocity Vescapegiven by Vescape =
~/2 goRe
(3.12)
Comparison of Eq. (3.12) for the escape velocity to Eq. (3.8) for the satellite velocity evaluated for low Earth orbit gives Wescape: ~/2V~.
(3.13)
This equation explains why the velocity increment of 42,000 ft/s (12,800 m/s) required for Earth escape (see Table 3.1) is 12,000 ft/s (3660 m/s) higher than that required for low Earth orbit. T a b l e 3.1
Mission
equivalent
Mission Low Earth orbit Earth escapelunar hit High Earth orbit Lunar orbit Mars, Venus Probe Lunar soft landing Lunar round trip Escape from solar system
velocity increment
requirements
AVequiv (m/s)
AVequiv (ft/s)
9,140 12,800 13,720
30,000 42,000 45,000
15,240 18,280
50,000 60,000
ROCKET PROPULSION
169
The total mass of a rocket vehicle is the sum of its parts. For convenience, the vehicle mass is considered to be divided into three different masses: the payload mass mpl , the propellant mass mp, and the remaining mass, which will be called the dead weight mass mdw. The mass of the structure includes the rocket engines, guidance and control system, tankage, rocket structure, etc. The initial mass of a rocket vehicle, mo, can be expressed as
(3.14)
mo = mpl q mp q mdw
For convenience, we define the payload mass ratio A and dead weight mass ratio as A=mpl
and
6=mdw
mo
(3.15)
mo
If all of the propellant is consumed during firing, the resulting vehicle mass is defined as the burnout mass mbo, or mbo
z
(3.16)
mo  mp mpl + mdw
In addition to the mass ratios of Eq. (3.15), the vehicle mass ratio (MR) is defined as the initial mass mo divided by the burnout mass mbo, or MR =
mo
(3.17)
mbo
Using Eq. (3.16) and the mass ratio defined in Eq. (3.15), the M R can be written as M R 
3.2.2
1
(3.18)
Equation of Motion for an Accelerating Rocket
In this section, the velocity increment AV of a rocket is related to the propulsion system's burning time tb, the ejected mass velocity V~, and the space vehicle's MR. The equation of motion for an accelerating rocket (see Fig. 3.5) rising from the Earth under the influence of gravity and drag as it expels matter was developed in Ch~;:,~r 1. The analysis resulted in the general expression for the differential velocity change given by Eq. (1.58), or
m

(L
)
+ g cos 0 dt
(3.19)
where C is the effective exhaust velocity. For launch from the surface of the Earth, the acceleration of gravity g can be written in terms of the radius from the center of the Earth, r, the acceleration of gravity at sea level, go, and the
170
ELEMENTS OF PROPULSION
Fig. 3.5 radius of the Earth,
Accelerating rocket.
Re, as follows: {Re 2
Substitution of this expression into Eq. (3.19) yields
m In this equation, dVis the velocity increment supplied to the vehicle of mass m in time dt as mass dm is expelled rearward at effective velocity C relative to the vehicle. Notice that dm is a negative quantity for leaving mass and hence C(dm/m) becomes a positive number exceeding the D
R
cos0}d,
2
when values of C are sufficiently high to produce a positive increment in V. When dm is zero (no burning), dV becomes a negative until the kinetic energy possessed by the vehicle at burnout is all converted into potential energy of the vehicle at zero velocity and maximum height. After this condition is reached, the vehicle begins to descend, and its potential energy is converted back into kinetic energy that, in turn, is dissipated into the atmosphere during reentry. To avoid this simple yoyo motion, we assume that a guidance system has the rocket follow a specified trajectory. The trajectory is such that the vehicle's velocity vector is perpendicular to the radius r at the instant of burnout, and a near Earth orbit is established corresponding to the velocity of 26,000 ft/s.
ROCKET PROPULSION
171
Integration of Eq. (3.20) from liftoff to burnout gives us the desired relation between the vehicle's velocity increment AV and the propulsion system's burning time tb, the effective exhaust velocity C, the drag D, the trajectory, and the vehicle's MR. Thus g~
mbo
AV= I d V = 
tbo
I C dm go m
][D/g@g~+ (~£)2co s 0 } dt
(3.21)
t=0
mo
where subscript bo denotes burnout conditions. Evaluation of this equation requires detailed knowledge of the trajectory and vehicle drag. Using Eq. (3.21), the effective velocity increment Ageffective , the drag velocity increment AVarag, and gravity velocity increment Aggravity are defined by mbo
AVeffective ~ 
I C
dm_m
No tbo
t=0
Aggravity ~ g o
bjol(;cos} 0
dt
t=0
Note that the effective velocity increment Ageffective is the specific vehicle impulse (engine thrust per unit mass of the vehicle integrated over time), or mbo
tbo
cdm =_
AVe~'ective = 
m
mo
tbo
/nC dt ~ gc
 dt
m
t=0
m
(3.23)
t=0
Equation (3.21) for the vehicle's velocity increment can be rewritten as (3.24)
A V = Ageffective  Agdrag  Aggravity
This equation shows the effect of drag and gravity on effective velocity increment that was discussed earlier in this section, namely, that the effective velocity increment Ageffective is greater that the vehicle's velocity increment AV due to vehicle drag losses and gravity. Assuming constant effective exhaust velocity C, Eq. (3.22) gives the following expression for effective velocity increment: Areffective
=
( ~ \mbo/
(
C{~ mo = C(~(MR) = C(,~z ~
1 )
(3.25)
Equation (3.25) is plotted in Fig. 3.6, which shows the exponential relationship between the mass ratio and the effective velocity ratio (AVen~c,ive/C). Both Fig. 3.6 and Eq. (3.25) show the extreme sensitivity of the mass ratio to the effective exhaust velocity for a given high energy mission (large AVeffective ).
172
E L E M E N T S OF P R O P U L S I O N 100
.
.
.
.
i
,
,
,
,
i
,
,
,
'
i
.
.
.
.
i
,
,
MR 10
1
Fig. 3.6
2 3 AVeffective/C
4
Ideal rocket vehicle mass ratio.
Many space missions require multiple Ageffective. Consider a mission from Earth orbit to the surface of the moon and return. This mission will require at least six AV: one to exit Earth orbit and reach transit speed; a second to enter lunar orbit; a third to land on the moon; a fourth to reach lunar orbit again; a fifth to exit lunar orbit and reach transit speed; and a sixth to enter Earth orbit. Additional AV will be required for trajectory corrections, vehicle orientation, etc. Each AV expends vehicle mass, and the mass ratio (mo/mbo) for the total mission of a singlestage vehicle can be obtained using Eq. (3.25) where A Veffective is the sum of all of the effective velocity changes (A Ve~'~ctive) in the mission. For a near Earth orbit, Re/r is close to unity (Re = 3959 miles, r = 3959 t 100 = 4059 miles, and Re/r = 0.975), and AVgravitycan be approximated as AVgravity ~ g0tbo. Assuming constant effective exhaust velocity C and negligible drag (AVdrag = 0), Eqs. (3.24) and (3.25) give A V = Vbo ~ C [ j c ( m ° ~ _ g0tbo
(3.26)
\ mbo /
This simple equation provides a means of comparing the performance of different propulsion systems in terms of vehicle gross mass morequired to put a given mass mbo into a low Earth orbit requiting a vehicle velocity increment of Vbo.
Example 3.2 To demonstrate the importance of a high propellant exhaust velocity Ve, let us determine the orbital payload capabilities of various propulsion systems for a fixed initial vehicle gross weight. Table 3.2 gives approximate representative values of effective exhaust velocity C for four propulsion systems. The dead
ROCKET PROPULSION Table 3.2
173
Performance of highthrust rocket propulsion systems Effective exhaust velocity, C
Propulsion system fuel and propellant
m/s
ft/s
Assumed dead weight ratio (6)
Solid Liquid Ozkerosene (RP) Liquid O2 and H 2 Nuclear fuel using H2 propellant
2440 3050 4110 8230
8000 10,000 13,500 27,000
0.03 0.03 0.06 0.10
weight ratio 6 listed consists of tankage, pumps, and structural members excluding the payload. As a reference point, we determine first the gross weight at liftoff for a 20001bin payload placed in a near Earth orbit by a liquid Ozkerosene propulsion system. At burnout, the payload, m p l , and the rocket dead weight, maw, have attained the velocity Vbo and both will orbit. Therefore, mbo = mpl ÷ mdw = mpl + ~mo = mpl ÷ 0.03 mo W e will assume for each propulsion system that the burning time is 100 s. W e have, from Eq. (3.26), mo
mo
mbo
mpl + 0.03 mo
M R  mo mbo 
/
c
/
mo [30,000 + 3217 / 2000 + 0.03 mo = e x p [ 10~() ] = exp{3.32} ~ 28
Thus mo = 28{2000 + 0.03 mo} = 56,000 + 0.84mo mo = 56,000/0.16 = 350,000 lbm
Thus to place a 20001bm payload in a near Earth orbit requires a sealevel liftoff gross weight of 350,000 lbf. With this gross weight as a reference point, let us evaluate the orbiting payload capabilities of the other propulsion systems of Table 3.2. For the propulsion system using liquid 02 and Hz, we find a pad weight of 350,000 lbf is capable of placing 8890 lbm into a near Earth orbit. The computation follows:
mpl ~
mbo  mdw = mo exp
mpl = 350,000 exp  ~ , mo = 88901bm
 0.06
= 350,000(0.0854  0.06)
174
ELEMENTS OF PROPULSION
Table 3.3
NearEarth orbital payload for 350,000 Ibm (158,760 kg) launch mass
Solid propellant C (ft/s)l(m/s) 6 (dead weight ratio) Payload (lbm)l(kg) A (payload ratio)
800012440 0.03 Nothing 0.0142
Liquid O2RP 10,00013050 0.03 2000]907 0.0057
Liquid H2O2
NuclearH2
13,50014115 0.06 8890]4040 0.0250
27,00018230 0.10 67,000130,390 0.1914
Similar calculations give the results shown in Table 3.3. The solidpropellant propulsion system's performance is not sufficient to place any mass in orbit under the assumed conditions. The results in Table 3.3 indicate that a high value of effective exhaust velocity is extremely beneficial, if not mandatory. An increase in C by a factor of 2.7, in going from LOXRP to nuclearH> permits a 33fold increase in the payload even though the dead weight of the nuclearH2 system is three times that of the LOXRP propulsion unit. Obviously, it is advantageous to have a high exhaust velocity.
3.2.3
Rocket Vehicles in Free Space
When rockets are fired in "free space," there are no drag or gravitational penalties. Integration of Eq. (3.21 ) under these conditions and assuming constant effective exhaust velocity C yields
AV/C=~(MR)=~( 1 "] \A+O/
(3.27a)
or 1
MR   A+6
exp(AV/C)
(3.27b)
For rocket vehicles in free space, the effective velocity increment Ageffective equals the vehicle's velocity increment AV, and thus Fig. 3.6 is also a plot of Eq. (3.27b).
3.2.4
MultipleStage Rocket Vehicles
The launch of a payload using singlestage rocket vehicles was investigated in the previous section. When large energy changes are required (large AV) for a given payload, the initial vehicle mass of a singlestage vehicle and its associated cost become very enormous. We consider a liquid H2O2 chemical rocket ( C = 4115 m/s, 13,500ft/s) launching a 900kg (2000Ibm) payload on a mission requiring a AV of 14,300 m / s (46,900 ft/s). From Fig. 3.6 or Eq. (3.27b), the required vehicle MR is 32.30. This high mass ratio for a payload of 900 kg and 3% dead weight results in an initial vehicle mass of 938,000 kg mass and
ROCKET PROPULSION
175
a vehicle dead weight mass of about 29,070 kg. This large initial vehicle mass can be dramatically reduced by using a multistage vehicle. For a multistage vehicle with N stages, the payload of a lower stage, [mpl]i , is the mass of all higher stages, [mo]i+l, or
[mp,]i [mo]i+l
for i < N
From Eq. (3.27b), the initial mass of stage i and above (denoted by given by
(3.28)
[mo]i) is
1
[moJi exp{fAV/C)i }  8i [mpl]i
(3.29a)
and the payload ratio of stage i (denoted by A~) is given by ~i =
exp{(AV/C)i}
 (3i
(3.29b)
Using Eq. (3.27a), the total velocity change, AVtotal, for an Nstage rocket can be written as ota, =
(3.30)
The overall payload ratio Ao for a multistage rocket is the product of all the ~i and can be expressed as
N Ao = U Ai
(3.31)
iI Equations (3.28) and (3.29) can be used to obtain the mass ratio required for each stage i of a multistage rocket.
Example 3.3 Consider both a twostage vehicle and a threestage vehicle for the launch of the 900kg (2000Ibm) payload. Each stage uses a liquid H2O2 chemical rocket (C = 4115 m/s, 13,500 ft/s), and the AVtouz of 14,300 m / s (46,900 ft/s) is split evenly between the stages. Because each stage has the same AV and effective exhaust velocity C, the payload ratio Ai of the stages of a multistage vehicle are the same. The payload ratio for a stage A i is 0.1460 for a twostage vehicle and 0.2840 for a threestage vehicle. Table 3.4 summarizes the results for one, two, and threestage vehicles. Note that a very large reduction in initial vehicle mass is obtained when going from a singlestage vehicle to a twostage vehicle (a twostage vehicle is 4.5% of the mass of the singlestage vehicle), and only a slight reduction in mass is obtained going from a twostage vehicle to a threestage vehicle (a threestage vehicle is 93% of the mass of the twostage vehicle).
ELEMENTS OF PROPULSION
176 Table 3.4
Multistage vehicle sizes for a 900kg payload at a AV of 14,300 m / s with liquid H2O2
Singlestage vehicle
Twostage vehicle
Threestage vehicle
14,300 m/s 0.03 0.0009590 938,480 kg 28,170 kg 900 kg
7150 m/s 0.03 0.02130 42,250 kg 1,270 kg 6,170 kg 6,170 kg 185 kg 900 kg
4,767 m/s 0.03 0.02290 39,300 kg 1,180 kg 11,200 kg 11,200 kg 336 kg 3,180 kg 3,180 kg 95 kg 900 kg
Stage 5V Stage 6i Stage )to Vehicle mo 1st Stage mdw 1st Stage mpj 2nd Stage mo 2nd Stage mdw 2nd Stage mp! 3rd Stage mo 3rd Stage maw 3rd Stage rap!
3.3
Rocket Propulsion Engines
Figure 3.7 shows the basic features of any propulsion system. The function of the propulsion system is to combine energy with matter to produce a directed stream of highspeed particles. That portion of the propulsion system that performs this function may be called the accelerator. In most systems, it is necessary to transport the propellant to the accelerator, and this implies pumps and a certain amount of plumbing. In some systems, it is necessary to supply the propellant in a special form. For example, the ion accelerator requires the propellant in the form of an ionized gas. We lump all of these functions under the heading Propellant Feed System in Fig. 3.7. In all propulsion systems, it is necessary to convert the energy stored onboard the vehicle into a form that is compatible with the particular mechanisms used to accelerate the exhaust particles. Thus there must be an energy conversion system as shown in Fig. 3.7. The energy conversion system may range from a single combustion chamber, wherein the energy of chemical bonds is released, to a nuclearelectric power plant. For the purposes of further discussion, we divide propulsion systems into two broad categories: 1) thermal systems and 2) electric systems.
~
lantFeed Propellant __~[ Propel System I Accelerator ~ Energy _~[ EnergyConversio~n Source lSystem l
I
Fig. 3.7
" "
Basic features of a rocket propulsion system.
ROCKET PROPULSION
177
3.3.1 ThermalPropulsion Systems The basic principle of a thermal propulsion system is the paragon of simplicity. Energy from a chemical reaction, or from a nuclear reactor, or even from an electric arc discharge, is used to elevate the temperature of the propellant. The individual particles of the propellant thereby obtain considerable kinetic energy in the form of random thermal motion. The propellant is then expanded through a convergentdivergent nozzle whose purpose is to convert the random thermal energy of the propellant into a more or less unidirectional stream of highspeed particles. In this context, a thermal system is a hot gas generator with a nozzle for an accelerator. The nozzleaccelerator simply tries to make an ordered state of affairs out of the chaos of random thermal motion. A key parameter in the analysis of propulsion systems in general is specific impulse Isp. Thermodynamic analysis indicates, under some ideal assumptions, that the specific impulse of a thermal system is proportional to the quantity To~ 34, where T~ is the propellant combustion chamber temperature and M is its molecular weight. To produce a high specific impulse in a thermal system, it is then desirable to have a high operating temperature in connection with a low molecular weight of exhaust products. The extent to which these two desirable objectives can be achieved is a rough basis for comparing thermal propulsion systems.
3.3.1.1 Chemicalpropulsionsystems. Chemical propulsion systems are a subclass of thermal systems that use the energy released by exothermic chemical reaction in a combustion chamber. Chemical systems may be further divided into liquidbipropellant engines, solidpropellant engines, and hybrid engines. Figure 3.8a shows the essential features of a liquid rocket system. Two propellants (an oxidizer and a fuel) are pumped into the combustion chamber. The hybrid engine is shown in Fig. 3.8b, and the propellant is a solid fuel around the combustion chamber. We may summarize this system as follows: 1) Energy conversion systemcombustion chamber, 2) Propellant feed systemfuel pump, oxidizer pump, and gas turbine (not required in hybrid system), 3) Acceleratornozzle. Figure 3.9 shows the essential features of a solidpropellant propulsion system. In this case, the fuel and oxidizer are mixed together and cast into a
Fuel
Fuel Pump
Gas Turbine
4, Oxidizer
Oxidizer Pump
Combustion Chamber
Fig. 3.8a Liquidbipropellant rocket engine.
Nozzle
178
ELEMENTS OF PROPULSION
Regulator Liquid ~ ~ ~ I ~ L "
oxidizer
~ ~ Valve inj~tm Fig. 3.8b Hybrid propulsion rocket engine.
solid mass called the grain. The grain is usually formed with a hole down the middle called the perforation and is firmly cemented to the inside of the combustion chamber. After ignition, the grain burns radially outward, and the hot combustion gases pass down the perforation and are exhausted through the nozzle. We may summarize a solidpropellant system as follows: l) Energy conversion systemcombustion chamber 2) Propellant feed systemnone 3) Acceleratornozzle The absence of a propellant feed system in the solidpropellant chemical rocket is one of its outstanding advantages. Table 3.5 gives the pertinent characteristics of several different chemical propellants. It is important to note that the combustion temperature Tc and the molecular weight of exhaust products, A4, are largely determined by the specific chemical reaction taking place in the combustion chamber. It is possible to select different combinations and thereby achieve an increase in specific impulse, but there are limitations because the products of chemical combustion inherently tend to have relatively high molecular weights.
Fig. 3.9
Solidpropellant rocket engine.
ROCKET PROPULSION Table 3.5
179
Characteristics of chemical propellants
Oxidizer and fuel Ozkerosene O2H2 FzH2 Niwocellulose
T~, K
Jg
~p,S
3400 2800 3100 23003100
22 9 7.33 2228
260 360 400 200230
This discussion serves to point out an important characteristic of chemical propulsion systems, namely, the oxidizer and fuel serve as both the source of energy and the source of particles for the propulsion system. Thus the process of supplying energy to the accelerator and the process of supplying particles to the accelerator are closely related. These fundamental facts place an upper bound of about 400 s on the specific impulse I~p obtainable by chemical means.
3.3.1.2 Nuclear heat transfer propulsion systems. Figure 3.10 shows the essential features of the nuclear heat transfer propulsion system. As the name suggests, this system contains a nuclear reactor that serves two important purposes: 1) it produces the energy necessary to heat the propellant gas to a high temperature; 2) it transfers this energy to the propellant gas. Thus the reactor serves both as an energy source and a heat exchanger. Once heated in the reactor, the hot propellant gas is expanded through a nozzle in the manner characteristic of all thermal systems. The major components of this system are: 1) Energy conversion systemsolid core nuclear reactor 2) Propellant feed systempump and turbine 3) Acceleratornozzle An important feature of the nuclear heat transfer system is that the source of exhaust particles and the source of energy are independent. The propellant supplies the particles and the reactor supplies the energy. Thus the propellant need not be selected on the basis of its energy content, but can be selected on the basis of its suitability as exhaust material for a thermal propulsion system. In other words, the primary consideration in the selection of a propellant is low molecular weight.
Nuclear Reactor
Propellant
~;~ N~:~g~ N®~NNI
Nozzle
l ur ineL Fig. 3.10
Nuclear heat transfer propulsion system.
ELEMENTSOF PROPULSION
180 Table 3.6
Characteristics of propellants for nuclear propulsion
Propellant
To, K
./fL
Max lsp, S
H2 C3H8 (propane) NH3 (ammonia)
2800 2800 2800
2 5.8 8.5
1000 530 460
Table 3.6 lists some important characteristics of different propellants for nuclear propulsion. Note that hydrogen produces the highest specific impulse because it has the lowest molecular weight. This fact is the primary reason that hydrogen is most often mentioned as the propellant for nuclear systems. Note also that the temperature Tc is the same for all three propellants listed in the table. The temperature Tc is determined primarily by the structural limitations of the reactorheat exchanger and not by the nature of the propellant.
3.3. 1.3 Electrothermal propulsion systems. Electrothermal propulsion systems use electric power to heat a propellant gas. One method of doing this is to pass the propellant around an electric arc as suggested by Fig. 3.11. The resulting propulsion system is sometimes called a thermal arcjet. A second possibility is to use tungsten heater elements (temperature increased by electrical energy) as a mechanism for imparting thermal energy to the propellant. In either case, the propellant can be heated to a rather high temperature, perhaps as high as 10,000 K. The temperature could be made higher, but beyond about 10,000 K, a significant portion of the input energy is used up in ionization and disassociation of the atoms of the propellant. This energy (called 'frozen flow' losses) is, therefore, not available for the basic purpose of the device, which is to impart kinetic energy to a directed stream of particles. We should also add that at very high temperatures, the exhaust stream would contain significant pieces of the electrodes and nozzle. These two effects alone tend to place an upper bound on the temperature to which the propellant may be heated, and hence there is an upper bound (~2500 s) on the specific impulse of the electrothermal devices. The major components of the system are: 1) Energy conversion systemelectric plant and arc chamber 2) Propellant feed systempropellant pump 3) Acceleratornozzle
:tric Jer [ Propellant it Fig. 3.11
Electrode~ Thermal arcjet propulsion system.
ROCKET PROPULSION
181
The electrothermal system is also characterized by the fact that the propellant plays no part in the process of supplying energy to the accelerator. The system also requires an electric power plant in the energy conversion system, and this feature is an important consideration in the design of vehicles propelled by electrothermal means. Because of the electric power plant, this system is often classified as an electric propulsion system. We have called it a thermal system because of the characteristic pattern of a high temperature gas plus expansion through a nozzle.
3.3.2 Electric Propulsion Systems Electric propulsion systems are characterized by an accelerator that makes use of the interaction of electromagnetic fields and charged particles. In a rather crude sense, the electric charge on a particle provides a way for an electromagnetic field to regulate and direct the motion of the particle. In particular, it is possible to accelerate the particle to a high speed and eject it from a rocket vehicle. As should be clear by now, the ejection of highspeed particles is the very essence of rocket propulsion. This simple picture of an electric propulsion system conceals some rather basic subtleties. First, the propellant feed system for such a system must do more than merely transport the propellant to the accelerator. It must also operate on the propellant to produce significant amounts of ionized particles at the entrance to the accelerator. Second, we note that electromagnetic fields are capable of accelerating charged particles to a very high specific impulse. However, it is necessary to supply the energy to the accelerator in electric form, and hence it is necessary to carry an electric power plant along on the mission. The electric power plant is the key to the problem because estimates of the power level for propulsion run into the megawatt range. In other words, the energy conversion system for an electric propulsion system is quite likely to be a very large and massive device. We will have more to say about this shortly.
3.3.2.1 Electromagneticpropulsion systems. Electromagnetic systems use a magnetic field to accelerate a collection of gas, called a plasma, to a speed on the order of 50,000 m/s (150,000 ft/s). A plasma is a fully or partially ionized gas containing essentially equal numbers of electrons and ions. Because the plasma is a rather good conductor of electricity, it will interact with electromagnetic fields. Because the plasma is also a gas, it displays the properties of a continuum or fluid. As a result, the basic description of the interaction of a magnetic field and a plasma combines all of the pleasantries of electromagnetic field theory and fluid mechanics into a whole new field of study called magnetohydrodynamics (MHD), or sometimes, magnetofluidmechanics. In this new field, it is quite possible to write down the governing equations, but it is quite another matter to solve them. The factors involved in the study of MHD include the electrical and fluid properties of the plasma, the time history of the magnetic field (e.g., pulsed or continuous), and the geometry of the accelerator. The upshot of all of this is that there are MHD forces acting on the plasma. These
182
ELEMENTS OF PROPULSION
Plasmaor Propellant~ ~ Generat ~/////~///////~El ectric Power Plant Fig. 3.12
+ I Accelerat°r~
Plasma propulsion system.
forces may act on the boundary of the plasma or may be distributed throughout the volume. In any case, the basic problem is to use these forces to drive the plasma out of the vehicle and thereby produce useful propulsive thrust. Because of the variety of ways in which MHD forces can be made to act on a plasma, there is a bewildering array of plasma accelerators. The list of devices is constantly growing, and it is quite hopeless to try to classify them in any simple way that would do justice to the subject. In this case, we must be content with the schematic picture of the electromagnetic system shown in Fig. 3.12. We should recognize that the plasma generator and plasma accelerator shown in the figure represent a variety of possible devices. However, they are all united by the joint feature of using the action of a magnetic field on a macroscopically neutral plasma. The major components of the system are: 1) Energy conversion systemelectric power plant 2) Propellant feed systempump and plasma generator 3) Acceleratorplasma accelerator The names MHD propulsion and plasma propulsion are often associated with this system. Although the specific impulse range for this system is not definitely established, the range 2,00010,000 s is a reasonable consensus of the values given in open literature. Hydrogen, helium, and lithium are mentioned as propellants. There is an important feature of the plasma system that is well worth mentioning at this point. The exhaust stream of this system is electrically neutral, and hence the neutralization problems of the ion engine, which we discuss next, are avoided in this case.
3.3.2.2
Electrostatic propulsion systems. The operating principle of
this system is based on the static electric fields to accelerate and eject electrically charged particles. Figure 3.13 shows the main features of an ion engine. The propellant introduced into the engine is electrically neutral because charged particles cannot be stored in appreciable amounts. The ions are produced immediately before they are exposed to the accelerating field. This is accomplished by stripping off one or more electrons from the neutral propellant molecule, which is then a single or multiplecharged positive ion. The positive ion is then accelerated by the electric field and ejected from the engine at a high speed. During this process of acceleration, it is necessary to focus and shape the ion beam to minimize the number of collisions between the ions and the structural members of the engine. Finally, both ions and electrons must be ejected from the engine to keep it
ROCKET PROPULSION
183
Electron Flow Propellant
++ _ +
Ion / Source
Electron Injector Focusing Accelerating Electrode Electrode
Fig. 3.13
Decelerating Electrode
Sketch of ion engine (electrostatic accelerator).
electrically neutral. This is accomplished by gathering up the electrons available at the ion source and transporting them to the engine exit where they are injected into the ion stream, thus hopefully producing an electrically neutral propulsive beam. The three major parts of an ion engine are the ion source, accelerator section, and beam neutralizer (electron injector). In the simplest terms, a charged particle acquires a kinetic energy equal to its loss of electric potential energy when it "falls" through a difference in electric potential. This basic idea leads to the following expression for the exhaust speed on an ion engine: l
2
~ m i V e = qAv
(3.32)
where Av = q = mi = Ve =
net accelerating voltage, V electric charge on the ion, C mass of the ion, kg exhaust speed, m/s
The factors affecting the exhaust speed are the net accelerating voltage Av and the charge to mass ratio q/mi of the exhausted particles. To have a compact, efficient engine, it is desirable to have a rather high accelerating voltage, something on the order of several tens of thousands of volts. With this constraint in mind, we can investigate the influence of chargetomass ratio on the selection of the propellant.
Example 3.4 If we assume 1) that the ions are protons (disassociated and ionized hydrogen, q / m i = 0.91 × 108 C/kg) and 2) a reasonable accelerating voltage of 10,000 V, we obtain an exhaust speed of 1.34 × 10 6 m / s . This corresponds to a specific impulse of 134,000 s.
184
ELEMENTS OF PROPULSION
This value of specific impulse is several orders of magnitude higher than that obtainable from thermal systems. Thus it appears that the ion engine can produce incredibly high specific impulses. However, several other considerations enter the picture to modify this apparently rosy situation. Performance analysis indicates there is an upper bound on the specific impulse so that the system has the capability of accomplishing useful missions. The reasons for this upper bound and the various factors involved will not emerge until one takes up the discussion of rocket performance. It suffices to say, at this point, that the ion engine is capable of producing specific impulses that are too high to be useful, and that a reasonable upper bound on specific impulse is something on the order of 100,000 s. This situation points out one of the very profound differences between ion propulsion and thermal propulsion. In the latter, the propellant is often selected to make the specific impulse as high as possible. With ion propulsion, the problem is to select a propellant that will hold the specific impulse (at reasonable accelerator voltages) down to a useful value. For this reason, our interest centers on the heavier elements in the periodic table. An additional consideration is the fact that the propellant must be supplied to the accelerator in ionized form, and this involves the expenditure of a certain amount of energy that is, therefore, not available for the purpose of propulsion. The amount of energy required for this purpose is measured by the ionization potential of an atom. Table 3.7 gives some important properties of the alkali metals. The alkali metals are the most promising propellants, because they combine the desirable features of low ionization potential together with reasonable accelerating voltages at useful specific impulses. Figure 3.14 shows a schematic diagram of an electrostatic propulsion system. The major components may be identified as: 1) Energy conversion systemelectric power plant 2) Propellant feed systempump and ion source 3) Acceleratorion engine For obvious reasons, this system is also called an ion propulsion system. The preceding discussion points out there is an upper limit on the specific impulse for an ion system. It will also be difficult to obtain efficient operation at lower specific impulses as well. This lower limit arises due to a number of practical considerations such as electrode erosion at low accelerator voltages and Table 3.7
Characteristics of propellants for electrostatic propulsion
Propellant
First ionization potential, ev
Atomic weight
Accelerating voltage at l~p = l05 s
Sodium Rubidium Cesium
5.14 4.18 4.89
23 85.4 133.0
12,000 V 45,000 V 69,000 V
ROCKET PROPULSION
Propellant Electric Power Plant Fig. 3.14
~// ~
Source
///I
185
+ l IonEngine~
IIIIII//~
Major features of an electrostatic (ion) propulsion system.
radiation from the hot surfaces of the ion source. The lower limit appears to be about 7500 s.
3.3.2.3 Electric power plants for space propulsion. A major component in any electric propulsion system is the electric power plant, and so we examine the situation regarding the generation of power in space. Electrical power will be required for a variety of reasons, including communication, control, life support, and propulsion. There are a great many methods of power generation that could satisfy some of these needs. Our interest is the generation of electric p o w e r for the purposes of propulsion. The very stringent requirements of electric propulsion narrow the field of interest. One important factor is the power levels of interest in electric propulsion. For Earth satellite missions, such as altitude control, the power levels are on the order of 5 0  1 0 0 kW. More ambitious missions, such as lunar flight, require 1  3 MW. Manned interplanetary flight will require 2 0  1 0 0 M W of electric power. As a matter of perspective, we note that the installed capacity of Hoover D a m is 1200 MW. In other words, we are talking about electric power plants that are comparable to Earthbound station applications that must operate for long periods of time in the rather unfriendly environs of space. Our discussion will center on two parameters of space power plants: efficiency and specific mass c~,.. Specific mass is defined as the mass of the power plant per unit output (electrical) power, or mc
Cec =  
Pc
(3.33)
where ac = the specific mass, k g / W Pc = the electrical output power, W m,. = the total mass of the power plant, kg All space power plants consist of at least three basic elements: an energy source, a conversion device, and a waste heat radiator. The only two energy sources that seem to offer much hope of producing the power levels of interest are the sun and the nuclear fission reactor. Solar collectors for the megawatt range become extremely large and must be constructed in a very sophisticated (i.e., flimsy) manner. Such
186
ELEMENTS OF PROPULSION
devices could stand only very small accelerations, and meteorite erosion of polished surfaces would be a problem. Nuclear power reactors axe very powerful sources of energy. They have the important advantage that the power output is limited primarily by the capability of the reactor coolant to remove heat. They have a low specific mass and long life at high power levels. On the other hand, the reactor produces a radiation hazard for the crew and payload. Both the solar collector and the nuclear reactor produce thermal energy. Among the possible methods of converting thermal energy into electrical energy are the following: thermionic, thermoelectric, and turboelectric. These three methods of conversion use vastly different mechanisms, but all three have one common featurevery low efficiency. The inefficiency of power conversion schemes results in an increase in the specific mass of the power plant for two important reasons. First, the overall size and mass of the energy source must be increased to "supply" the losses. Second, the losses must somehow be dumped overboard by a waste heat rejection system. For the times of operation and power levels of interest in space propulsion, the only mechanism available for the rejection of heat waste is radiation into space. Because of the inherent low efficiency of the system, the waste heat radiator will be a very large and massive device. It will be susceptible to meteorite puncture and the attendant loss of working fluid from the system. For many designs, it appears that the waste heat radiator will be the dominant mass in the power plant for outputs exceeding a few hundred kilowatts. Table 3.8 gives some projected characteristics of space power plants. As we have noted, all of the systems have low efficiency. The values of specific mass given in the table should be taken as tentative, particularly in the 1  10 M W range, because systems of this power level have not been built. Authorities differ widely in their estimate of the situation, and there is considerable uncertainty in the figures. In all of this, we should not lose sight of the fact that the purpose of the propulsion system is to impart kinetic energy to a directed stream of particles. In electric propulsion, this energy first appears as thermal energy and then must be converted into electrical energy before it is in a form suitable for the basic task at hand. As we have suggested, the process of converting thermal energy into electricity in space is not noted for its efficiency. For the foreseeable future, it appears that electric propulsion will continue to pay a severe penalty tor the several conversion processes involved in the generation of electric power. It is no exaggeration to state that the use of electric propulsion will be paced by the development of lightweight, efficient power conversion systems.
Table 3.8
Characteristics of space power systems
Specific mass  c~c (kgm/W) System Turboelectric Thermionic Thermoelectric
Efficiency
10100 kW
1 10 MW
1020% 510% 510%
0.020.04 0.5  1.0 0.51.0
0.0030.004 0.003 0.004 
ROCKET PROPULSION
187
3.3.3 Summary of Propulsion Systems There are a number of ways in which propulsion systems can be classified to indicate certain fundamental features. In our discussion, we have used the classifications of thermal and electric. This classification is based on the mechanism used by the accelerator to impart kinetic energy to the exhaust stream. Thus propulsion systems may be summarized as follows.
Thermal: 1) Chemical 2) Nuclear heat transfer 3) Electrothermal (thermal arc jet)
Electric: 1) Electrostatic (ion) 2) Electromagnetic (plasma) There are other fundamental features of propulsion systems of no less importance, and consideration of these leads to other possible classifications. For example, in our discussion, we noted that in certain systems, the propellant does not supply energy to the accelerator. That is, the source of energy and the source of particles are separate. Such propulsion systems are called separately powered. On the basis of this feature, we have the following classification.
Separately powered: l) 2) 3) 4)
Nuclear heat transfer Electrothermal (thermal arc jet) Electromagnetic (plasma) Electrostatic (ion)
Nonseparately powered: 1) Chemical Note that the chemical propulsion system is termed nonseparately powered because the propellant supplies both energy and particles to the accelerator. Propulsion systems can also be classified according to their performance capabilities as measured by the two key parameters, /sp and thrusttoweight (sealevelweight) ratio. Table 3.9 presents typical values of these quantities for various propulsion systems. The data in Table 3.9 offer several new ways to classify propulsion systems. For example, specific impulse values range over a fairly continuous spectrum from 200 (the lower end of the chemical range) up to 100,000 (the upper end
Table 3.9
Propulsion system Chemical Nuclear heat transfer Electrothermal Electromagnetic Electrostatic
Rocket performance summary
l~p, s 300400 4001000 4002500 2000104 7500105
Thrusttoweight ratio up to up to up to up to up to
102 30 102 10 3 10 4
188
ELEMENTS OF PROPULSION
of the ion range). Thus the order in which the systems are listed in Table 3.9 is a classification based on magnitude of specific impulse. There is, of course, some overlap, but in large measure, each of the systems fills a gap in the specific impulse spectrum. A very important distinction between propulsion systems is found in Table 3.9. Note that there are several orders of magnitude difference between the thrusttoweight ratio of electric propulsion systems and the chemical or nuclear propulsion systems. This leads to a classification of systems as either high thrust or low thrust.
Highthrust systems: 1) Chemical 2) Nuclear heat transfer Lowthrust systems: 1) Electrothermal 2) Electromagnetic 3) Electrostatic Note that all of the lowthrust systems require an electric power plant as the energy conversion system. This fact accounts for the low values of thrusttoweight ratios for these systems. Significant increases in the thrusttoweight capabilities of these systems can be achieved only by reducing the weightperunit power of space power plants. Figure 3.15 is a summary of propulsion systems obtained by plotting the data in Table 3.9. Notice the very clear distinction between the highthrust and lowthrust systems. Notice also that there is a fairly continuous range of specific impulse capability from 200 to 100,000 s with some significant overlapping among the electric propulsion systems. 10 2
(high thrust)
101
o 10°
Chemical Nuclear
10 ]
~ 10_2 ~
10 3
[" 104
!•
(low thrust)
l therrnal l . I E l e c t r o _ ~
J
105 10 6 10 2
I'JmagneticJ"1 nectro~'~ 10 3
104
10 5
SpecificImpulse, Isp (s) Fig. 3.15
Thrusttoweight ratio vs specific impulse.
ROCKET PROPULSION
3.4
189
Types of Rocket Nozzles
The initial geometry of a circularsection rocket nozzle is fixed by the nozzle inlet, throat, and exit areas. The inlet area Ac is established by combustion chamber design considerations. The throat and exit areas are determined from a knowledge of the combustion chamber temperature and pressure, acceptable mean values of y and R, the propellant mass flow rate, and the designed nozzlepressure ratio P,'/Pa = P c / P e • With these data, the nozzle throat and exit areas can be obtained by application of the mass flow parameter (MFP). Having established A,:, At, and A e (or the radii Re, Rt, Re), the problem of the nozzle contour design remains. Because of weight and space limitations, determination of the nozzle shape reduces to the problem of choosing a contour providing minimum length (a measure of volume and weight) and yielding maximum thrust. In addition, the selected contour must be sufficiently simple to manufacture.
3.4.1
Conical Nozzle
The conical nozzle represents a compromise of the length, thrust, and ease of manufacturing design criteria weighted somewhat in favor of the last factor. A conical nozzle consists of two truncated cones (Fig. 3.16), joined top to top along their axis by a suitable radius to form the nozzle throat. The combustion chamber is similarly faired into the convergent nozzle section. The converging contour of the nozzle is not critical as regards the flow, and a rather rapid change in cross section is permissible here with a conical apex halfangle on the order of 400 commonly used. The divergence angle of the supersonic portion of the nozzle, however, is limited by flow separation considerations and must not exceed a value of about 150. For divergence angles too much greater than 15 deg, the flow will separate from the nozzle walls short of the exit even though the nozzle is operating at design altitude P, = Pe. Conversely, for a given divergence angle, the flow will separate if the nozzle backpressure ratio Pa/Pe is too high (Pa/Pe > 1 when a nozzle is operating at an altitude lower than the design altitude). When separation occurs, oblique shock wave A (Fig. 3.17) is located inside the nozzle, and the gas flow is contained within the jet boundary or slip line shown. Figure 3.17 depicts the real flow situation that occurs (other than under ideal laboratory conditions) in place of normal shocks near the exit section as shown in Fig. 2.37. Summerfield has established a criterion for judging2whether or not separation is likely to occur in the exit section of a conical nozzle. 0 The criterion is based on
~~Tf~//,
Rc ~
a< Rc
.Rb=2R t
+,/
. / . ,
40 °
Fig. 3.16
Conical convergentdivergent nozzle.
i
190
ELEMENTS OF PROPULSION
separation point Fig. 3.17
S e p a r a t i o n in o v e r e x p a n d e d nozzle.
an accumulation of experimental separation data points from many sources. The experimental data give the pressure ratios P , / P w e (Pwe = wall pressure at exit = Pe for 15deg conical nozzle) that produced flow separation in tests of 15deg conical nozzles for various nozzlepressure ratios. The data points are scattered throughout the crosshatched region of Fig. 3.18. The Summerfield criterion states that for backpressure ratios greater than 2.5, separation is likely to occur in the exit of a 15deg conical nozzle for nozzlepressure ratios in excess of 20. Current data on large rocket nozzles indicate that the critical backpressure ratio for these large nozzles is 3.5. This increase in the permissible backpressure ratio for no separation is to be expected as nozzle sizes increase because the boundary layer forms a smaller percentage of the total flow for larger nozzles. Using this modified Summerfield criterion value of 3.5, it should be noted that _
15"Conical Cone (P w,,=Pc)
•e p a r a ~ ~ ) / / / / ~ / ew
~mm~rfie~d
1
I
20
Fig. 3.18
I
~i~i~
I
I
I
40 60 80 IO0 Nozzle Pressure Ratio P,./~ S u m m e r f i e l d s e p a r a t i o n criterion.
120
ROCKET PROPULSION
191
g a) nonuniform
b) uniform
Fig. 3.19
Pressure profile at nozzle exit.
for large rocket nozzles, separation occurs when P, > 3.5 Pwe. For a 15deg conical nozzle, Pwe = Pe because the stream properties at the exit section are essentially uniform at the values determined by onedimensional flow analysis. In some nozzles, the streams are not uniform across the exit section and Pwe > Pe (Fig. 3.19). Separation will occur sooner (at lower values of Pa) in a conical nozzle than in those of the same area ratio for which Pwe > Pe.
3. 4.2
Bell Nozzle
The bellshaped nozzle of the Atlas sustainer engine (shown in Fig. 3.20) is designed to reduce the thrust and length disadvantages of a conical nozzle. To reduce length, a bellshaped nozzle employs a high divergence angle at the throat with a very rapid expansion of the gases from the throat. The flow is then turned rather abruptly back toward the axial direction. The comparative lengths of a bellshaped (/,8) and a conical (Lc) nozzle having the same expansion ratio are indicated in Fig. 3.20. The bellshaped contour is used on several current engines. The bellshaped nozzle on the HI Saturn 1B engine has a length 20%
~ ~ "~5deg nozzle
LB
3.5 Pwe) The remaining curves present Eq. (3.44) for various values of Pc/P,. Example
3.6
We can illustrate the use of the graph of Fig. 3.31 with an example. Let us find CFi for a rocket at lO0,000ft altitude with an area ratio of 25, y : 1.29, and a chamber pressure Pc = 500 psia. Because the value of y is not one of the specific values plotted in Figs. 3.30 and 3.323.35, use of Fig. 3.31 presents a convenient
2.3
' ' ' I
. . . .
I
. . . .
r
. . . .
I
. . . .
I
'
'
'
'
I
'
'
'
'
2.2 2.1 2.0
(cfi),,at19 1.8
_
~
1.7 1.6 1.5 1.0
. . . .
Fig. 3.31
I
1.1
. . . .
I
1.2
. . . .
1.3
I
,
,
~
,
1.4
V a c u u m thrust coefficient, (Ctl)v,c.
I
1.5
~
,
,
,
1.6
ROCKET PROPULSION
205
2.4 2.2 2.0
( CFi)opt 1.8
~ o ~ 5 o ' 6 o
1.6
1.4
~o loo
~8=3
1.2 l0
1O0
1000
10,000
Pc/Pe = Pc/P~ Fig. 3.32
way of finding w r i t t e n as
O p t i m u m thrust coefficient vs pressure ratio.
CFi w i t h o u t
i n t e r p o l a t i o n . First, w e n o t e t h a t Eq. (3.45) c a n b e
CFi = (CFi)vac
2.0
I
f



_
_
Pc At
i I I I I [
I
I
[
I I I I I
1000
1.8
Lineof
1.6
CF,
I
Pa Ae 
P
c
~
5OO
400
//.~~/1oo
1.4
t/20
1.2
,/
\~ 1. Stability requires that burning rate pressure exponent n be less than unity. If n is greater than unity, increases in chamber pressure above operating point will cause the propellant gas generation to increase faster than the nozzle flow rate. The result would lead to a rapid rise in chamber pressure with an accompanying explosion. From Eq. (i), the chamber pressure P,: can be expressed as
{
a(p s  pg)C* Ah [ ~~" go ZJ
Pc. :
(3.58)
This equation can be used to determine the chamber pressure when data are available for the terms on the righthand side of the equation. Because the density
Operating point
!
Nozzle flow rate nI SpecificationI~
:
[CustomerrequirementsI
Preliminarystudies: choiceofcycle typeof turbomachinery layout
less, g!y ii ............................................. [ ~ic
Iae;
co'me;o I
t
Fmo u
'
iuC;
IPe ° Lnc211:
inlet'nozzle'etc' ' ~
i...[ Mechanicaldesign: stressingofdisks, blades,casings; vibration,whirling, bearings
@
Detaildesign and manufacture
.
1
tem I dies ]
I
I Testanddevelopment I I
I
Production
[
~ Ii Fieldservice ]
Fig. 4.28 Typical aircraft gas turbine design procedure (from Ref. 29).
258
ELEMENTS OF PROPULSION
and offdesign performance be in the initial steps of design. The iterative nature of design is indicated in Fig. 4.28 by the feedback loops. Although only a few loops are shown, many more exist. Those items within the dashed lines of Fig. 4.28 are addressed within this textbook.
Problems 4.1
The inlet for a highbypassratio turbofan engine has an area A1 of 6.0 m 2 and is designed to have an inlet Mach number M~ of 0.6. Determine the additive drag at the flight conditions of sealevel static test and Mach number of 0.8 at 12km altitude.
4.2
An inlet with an area A 1 of 10 ft 2 is designed to have an inlet Mach number M~ of 0.6. Determine the additive drag at the flight conditions of sealevel static test and Mach number of 0.8 at 40kft altitude.
4.3
Determine the additive drag for an inlet having an area A ~ of 7000 in9 and a Mach number Mj of 0.8 while flying at a Mach number Mo of 0.4 at an altitude of 2000 ft.
4.4
Determine the additive drag for an inlet having an area A~ of 5.0 m 2 and a Mach number M~ of 0.7 while flying at a Mach number Mo of 0.3 at an altitude of 1 km.
4.5
A turbojet engine under static test (Mo = 0) has air with a mass flow of 100 k g / s flowing through an inlet area Aa of 0.56 m 2 with a total pressure of 1 atm and total temperature of 288.8 K. Determine the additive drag of the inlet.
4.6
In Chapter 1, the loss in thrust due to the inlet is defined by Eq. (1.8) as (])inlet = Dinlet/F. Determine (])inlet for the inlets of Example 4.1.
4.7
Determine the variation of inlet mass flow rate with Mach number Mo for the inlet of Example 4.2
4.8
In Chapter 1, the loss in thrust due to the inlet is defined by Eq. (1.8) as qSinLet= DinJet/F. For subsonic flight conditions, the additive drag Dadd is a conservative estimate of Dinlet. (a) Using Eq. (4.9) and isentropic ftow relations, show that (])inlet can be written as Dadd
~inlet F (b)
(Mo/M1)~(1
I "yM2)  (A,/Ao I yM o) (Fgc//no)( yMo/ao)
Calculate and plot the variation of q~inlet with flight Mach number Mo from 0.2 to 0.9 for inlet Mach numbers Mn of 0.6 and 0.8 with (fgc/tho)(y/ao) = 4.5.
AIRCRAFT GAS TURBINE ENGINE
259
4.9
A bellmouth inlet (see Fig. P2.8) is installed for static testing of jet engines. Determine the force on the bellmouth inlet with the data of Problem 2.34b for an inlet area that is 8 times the area at station 2 (assume that the inlet wall is a stream tube and that the outside of the bellmouth inlet sees a static pressure equal to Po.
4.10
The maximum power out of an ideal Brayton cycle operating between temperatures T2 and T4 is given by Eq. (4.17). By taking the derivative of the net work out of an ideal Brayton cycle with respect to the pressure ratio (PR) and setting it equal to zero, show that Eq. (4.16) gives the resulting compressor temperature ratio and Eq. (4.17) gives the net work out.
4.11
Show that Eq. (4.18) gives the thermal efficiency for the ideal Brayton cycle with regeneration.
4.12
For the ideal Brayton cycle with regeneration, regeneration is desirable when T9 > T3. Show that the maximum compressor pressure ratio PRmax for regeneration (T9 = ~ ) is given by
PRmax = \~22,]
5 Parametric Cycle Analysis of Ideal Engines
5.1
Introduction
Cycle analysis studies the thermodynamic changes of the working fluid (air and products of combustion in most cases) as it flows through the engine. It is divided into two types of analysis: parametric cycle analysis (also called designpoint or ondesign) and engine performance analysis (also called offdesign). Parametric cycle analysis determines the performance of engines at different flight conditions and values of design choice (e.g., compressor pressure ratio) and design limit (e.g., combustor exit temperature) parameters. Engine performance analysis determines the performance of a specific engine at all flight conditions and throttle settings. In both forms of analysis, the components of an engine are characterized by the change in properties they produce. For example, the compressor is described by a total pressure ratio and efficiency. A certain engine's behavior is determined by its geometry, and a compressor will develop a certain total pressure ratio for a given geometry, speed, and airflow. Because the geometry is not included in parametric cycle analysis, the plots of specific thrust F//no, and thrust specific fuel consumption S vs, say, Mach number or compressor pressure ratio are not portraying the behavior of a specific engine. Each point on such plots represents a different engine. The geometry for each plotted engine will be different, and thus we say that parametric cycle analysis represents a "rubber engine." Parametric cycle analysis is also called designpoint analysis or ondesign analysis because each plotted engine is operating at its socalled design point. 4'a2'28'29 The main objective of parametric cycle analysis is to relate the engine performance parameters (primarily thrust F and thrust specific fuel consumption S) to design choices (compressor pressure ratio, fan pressure ratio, bypass ratio, etc.), to design limitations (burner exit temperature, compressor exit pressure, etc.), and to flight environment (Mach number, ambient temperature, etc.). From parametric cycle analysis, we can easily determine which engine type (e.g., turbofan) and component design characteristics (range of design choices) best satisfy a particular need.
Supporting Material for this chapter is available electronically. See page 869 for instructions to download.
261
262
ELEMENTS OF PROPULSION
The value of parametric cycle analysis depends directly on the realism with which the engine components are characterized. For example, if a compressor is specified by the total pressure ratio and the isentropic efficiency, and if the analysis purports to select the best total pressure ratio for a particular mission, then the choice may depend on the variation of efficiency with pressure ratio. For the conclusions to be useful, a realistic variation of efficiency with total pressure ratio must be included in the analysis. The parametric cycle analysis of engines will be developed in stages. First the general steps applicable to the parametric cycle analysis of engines will be introduced. Next these steps will be followed to analyze engines where all engine components are taken to be ideal. Trends of these ideal engines will be analyzed, given that only basic conclusions can be deduced. The parametric cycle analysis of ideal engines allows us to look at the characteristics of aircraft engines in the simplest possible ways so that they can be compared. Following this, realistic assumptions as to component losses will be introduced in Chapter 6 and the parametric cycle analysis repeated for the different aircraft engines in Chapter 7. Performance trends of these engines with losses (real engines) will also be analyzed in Chapter 8. In the last chapter on engine cycle analysis, models will be developed for the performance characteristics of the engine components. The aerothermodynamic relationships between the engine components will be analyzed for several types of aircraft engines. Then the performance of specific engines at all flight conditions and throttle settings will be predicted.
5.2
Notation
The total or stagnation temperature is defined as that temperature reached when a steadily flowing fluid is brought to rest (stagnated) adiabatically. If T, denotes the total temperature, T the static (thermodynamic) temperature, and V the flow velocity, then application of the first law of thermodynamics to a calorically perfect gas gives Tt = T + V2/(2g,.cp). However, the Mach number M = V/a = V/~/y&.RT can be introduced into the preceding equation to give
The total or stagnation pressure Pt is defined as the pressure reached when a steady flowing stream is brought to rest adiabatically and reversibly (i.e., isentropically). Since Pt/P = (Tt/T) v/(~ 1~ then P,=P(I+Y12
M2) ~/(~1)
(5.2)
Ratios of total temperatures and pressures will be used extensively in this text, and a special notation is adopted for them. We denote a ratio of total pressures across a component by ~, with a subscript indicating the component: d for diffuser (inlet), c for compressor, b for burner, t for turbine, n for nozzle,
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
263
and f for fan: total pressure leaving component a total pressure entering component a
,Ba=
(5.3)
Similarly, the ratio of total temperatures is denoted by r, and %=
total temperature leaving component a total temperature entering component a
(5.4)
There are the following exceptions: 1) We define the total/static temperature and pressure ratios of the freestream (rr and 1rr) by
T~ rr
"JTr
To  
,',o Po

1
+ ~@M02 1+ y
1
(5.5) 2\ ~/(~1)
(5.6)

Thus the total temperature and pressure of the freestream can be written as
Tto = To'rr Pto = Po rrr 2) Also, rA is defined as the ratio of the burner exit enthalpy cs,Tt to the ambient enthalpy cpTo: ra 
ht burner exit (CpTt)burnerexit ho (cpT)o
(5.7)
Figure 5.1 shows the cross section and station numbering of a turbofan engine with both afterburning and duct burning. This station numbering is in accordance with Aerospace Recommended Practice (ARP) 755A (see Ref. 30). Note that
18
Fr{ v
urner
Primary nozzle
789 0
2
2.5
Fig. 5.1
3
4 4.5
5
Station numbering for gas turbine engines.
264
ELEMENTS OF PROPULSION T a b l e 5.1
T e m p e r a t u r e a n d p r e s s u r e r e l a t i o n s h i p s for all r a n d 7r
Freestream rr =
1 + ~ M
2
1rr =
Core stream
cptrt4
• A 
cp,,To
rd
Tt 2 7",o
Tc
Tt3 Tt2 =
%
Tt 4
Bypass stream
Zt TAB
1",, =
= 
Tt5
Tt9 T,~
taD B

c pc To
Pt2 Pro
rrd
Pt3
~c = 
Pt2 qTb
=
rrt=
Tf 
Ttl3 Tt2


Cpc To Pt¿3 Tf = Pt2
T,17
TDB
= 
TtI3
TtL9
Pt4 Pt3
Tt 4 Tt 7
CpDBTtl7
CpABTt7
raAB  
Tr3
T,5
y  1 tl/12"~T/(T I) (1   ~   " * 0 !
Tfn  Ttl7
Pm 3TDB Pd3
Ptl9 ~'Jh = P,I~
Pt5 P t4 Pt7
qTAB = Pt5
Y r r , = Pt9 Pry
the station numbers 1 3  1 9 are used for the bypass stream and decimal numbers such as station number 4.5 are used to indicate an intermediate station. Table 5.1 contains most of the shortform notation temperature ratios r and pressure ratios 1r that w e will use in our analysis. (Note that the za are expressed for calorically perfect gases.) These ratios are shown in terms of the standard station numbering. 3°
5.3
Design Inputs
The total temperature ratios, total pressure ratios, etc., can be classified into one of four categories: 1) 2) 3) 4)
Flight conditions: Po, To, Mo, cp, rr, ~'r Design limits: (CpTt)bumerexi t Component performance: ~'d, ~b, ~'n, etc. Design choices: ~r,, ~'f, etc.
5.4 Steps of Engine Parametric Cycle Analysis The steps of engine parametric cycle analysis listed next are based on a jet engine with a single inlet and single exhaust. Thus these steps will use only the station numbers for the core engine flow (from 0 to 9) shown in Fig. 5.1. W e will use these steps in this chapter and Chapter 7. When more than one
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
265
exhaust stream is present (e.g., highbypassratio turbofan engine), the steps will be modified. Parametric cycle analysis desires to determine how the engine performance (specific thrust and fuel consumption) varies with changes in the flight conditions (e.g., Mach number), design limits (e.g., main burner exit temperature), component performance (e.g., turbine efficiency), and design choices (e.g., compressor pressure ratio). 1) Starting with an equation for uninstalled engine thrust, we rewrite this equation in terms of the total pressure and total temperature ratios: the ambient pressure Po, temperature To, and speed of sound ao, and the flight Mach number Mo as follows: F =
1
 (l~9Ve  111oVo)~A9(P9  P0) go
)
(,0)
F
a o ( l h 9 V 9 M 0 q_Ag.P9 1__~9
rho
gc \rho ao
mo
2) Next express the velocity ratio(s) V9/ao in terms of Mach numbers, temperatures, and gas properties of states 0 and 9: (V9~2 ao/
a2M2 a2
__ ~ 9 R 9 g c T 9 M
ToRogcT°
2 9
3) Find the exit Mach number M9. Since
Pt9 =P9( 1  }  ~  M ~ ) ~/(3' 1) then
2I(Pt9~(Yl)/Y M~  31 1 L \ P99/
] 1
where
Pt9 P9
Po Pro Pt2 Pt3 Pt4 Pt5 Pt7 Pt9
P9 Po Pro Pt2 Pt3 Pt4 Pt5 Pt7 z
P0
p9~r~d~c~bTrt~AB~n
m
4) Find the temperature ratio T9/To:
T9 To
rt9 / ro rt9 / ro Tt9/T9 (Pt9/P9) (yl)~7
266
ELEMENTS OF PROPULSION
where
Tt9
To
Tto Tt2 Tt3 Tt4 Tt5 Tt7 Tt9 = TrTdTcTbTtTABTn TO Tto rt2 rt3 rt4 Tt5 Tt7
5) Apply the first law of thermodynamics to the burner (combustor), and find an expression for the fuel/air ratio f i n terms of z's, etc.: /noc pTt3 + / n f h e R = 1;noc pTt4
6) When applicable, find an expression for the total temperature ratio across the turbine ~ by relating the turbine power output to the compressor, fan, and/or propeller power requirements. This allows us to find ~t in terms of other variables. 7) Evaluate the specific thrust, using the preceding results. 8) Evaluate the thrust specific fuel consumption S, using the results for specific thrust and fuel/air ratio: s _
f
(5.8)
F / /no
9) Develop expressions for the thermal and propulsive efficiencies.
5.5 Assumptions of Ideal Cycle Analysis For analysis of ideal cycles, we assume the following: 1) There are isentropic (reversible and adiabatic) compression and expansion processes in the inlet (diffuser), compressor, fan, turbine, and nozzle. Thus we have the following relationships: "rj = "r,~ = 1
7rj = 7rn = 1
"c~. = ~'~!Y I)/y
Tt = 7T¢Yl)/y
2) Constantpressure combustion (Trb = 1) is idealized as a heat interaction into the combustor. The fuel flow rate is much less than the airflow rate through the combustor such that  
Combustion zone
I I.. Fig. 5.4
C o m b u s t o r model.
(4)
i
ELEMENTS OF PROPULSION
270 or
•
/T,4
)
thfhpR = thocp(Tt4  Tt2) = mocpTt2 t~t2  1 The fuel/air ratio f is defined as
l~lf
cpTt2
(rt4 )
Sm~ ~ t~l
(5.13a)
For the ideal ramjet, Tt0 = Tt2 = ToTr and r t 4 / T t 2 = Tb. Thus Eq• (5.13a) becomes
cpTo~r , ~ trb1)
f
(5.13b)
However, TAx
Cp4Tt4  Tt4
rt2 Tt4
cporo
To Tt2
To
for the ramjet, and Eq. (5.13b) can be written as
Cpro.
f = h~R tra  ~'r)
(5.13c)
Step 6: This is not applicable for the ramjet engine• Step 7: Since
M9 = M0 and
Tg/To = rb, then
v9] ao/I
2
r9
 To
9
,,,u0
(5.14)
and the expression for thrust can be rewritten as
F/n°a°M~°(~fb1)lh°a°M°(7~rl)g~
gc
(5.15a)
or
=F _ aoMo (
mo
gc
~ 
1)
=
aoMo(~TAr gc
1
)
(5.15b)
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
271
Step 8:
s
f F//no
S =
cpTogc(ZA  7"r) aoMohpR( w / ~ / 7"r  1)
(5.16a)
or S =
cpTog,,'rr('Cb  1) aoMoheR(v/~1)
(5.16b)
Step 9: Development of the following efficiency expressions is left to the
reader. Thermal efficiency: 1
T/r = i  ~'r
(5.17a)
Propulsive efficiency: 2 ~/P  ~
+ 1
(5.17b)
Overall efficiency: 2("rr  1)
7/0 = Y/TT/p  T%/~r + "/'r
(5.17c)
5.6.2 Summary of Equations~tdeal Ramjet INPUTS:
e
kJ
Btu
M0, r0(K, OR), % P(kggK' lbm°l~J '
h
Btu) PR(~~,l~m]' Tt4(K, °R)
OUTPUTS: F ( N lbf "~ s ( m g / s lbm/la_'~ k ~ s ' I ~ S / / ' f ' \ N ' lbf //' ~/r, ~/e, 7/o
ELEMENTS OF PROPULSION
272 EQUATIONS:
y1
R = Cp
(5.18a)
Y
a0 = ( T k g c T 0 ~r = 1 +
(5.18b) M~
(5.18c)
Tt4 To
zA =  
V9
(5.18d)
(5.18e)
Mo,[~~
ao 
V "rr
F _ aO(~o_Mo )
(5.18f)
cpTo f = hp~t%  Tr)
(5.188)
rno
gc
S 
f
(5.18h)
F/&o
1 ~T = 1    rr
(5.18i)
2 ~Tp  V ~ / r ~ + 1
(5.18j)
2(rr  1) r/o = ~/r 7/,,
~
+ r~
(5.18k)
Example 5.1 The performance of ideal ramjets is plotted in Figs. 5 . 5 a  5 . 5 d vs flight Mach number Mo for different values of the total temperature leaving the combustor. Calculations were performed for the following input data: To = 216.7 K,
y = 1.4,
Tt4
c),  1.004 kJ/(kg  K), =
hpR = 42,800 kJ/kg
1600, 1900, and 2 2 0 0 K
5.6.3 Optimum Mach Number The plot of specific thrust vs Mach number (see Fig. 5.5a) shows that the m a x i m u m value of specific thrust is exhibited at a certain Mach number for each value of Tt4. An analytical expression for this optimum Mach number can be found by taking the partial derivative of the equation for specific thrust with respect to flight Mach number, setting this equal to zero, and solving as follows.
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
273
1000
2200 K 800
600
400
200
I 1
I
I 3
2
4
5
I 7
6
M0 Fig. 5.5a
Ideal ramjet
performance
v s M a c h n u m b e r : specific thrust.
100
90
80
I
70
60 = 2200 K 50
K K I 1
I 2
I 3
I 4
I 5
I 6
I 7
M0 Fig. 5.5b Ideal consumption.
ramjet
p e r f o r m a n c e vs M a c h
number:
thrustspecific f u e l
ELEMENTS OF PROPULSION
274
0.05
o
oo
1
2
3
4
5
6
7
Mo Fig.
5.5c
Ideal ramjet performance vs Mach number: fuel/air ratio.
100 1600 K t ~" ~ 80
/ 1 ////
r/p
rip
/,,,/
6O r/T
1900 K
/
~"'~2200 K / / /
40 
1600 K !
.o
2200 K
./~Z/ /~Z'x,
////~.o
20  /
0~"
I
1
I
2
I
I
3
4
I
5
I
6
Mo
Fig. 5.5d
Ideal ramjet performance vs Mach number: efficiencies.
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
275
Combining Eqs. (5.18e) and (5.18f) and differentiating gives
0 ( F )   a ° 0 IMo(,//~AI)]:O 8Mo ~oo gc 0114ok \ V Yr g~rAr_
8 1 =o l+Mo4V;() OMo\~rJ
Now 0 (~rrr)
OMo
1 10Tr 2 Tr3/2 OMo (T 1)mo
2r3r/2 Thus
V•,A.
_ 1 = Mo~/~(T 1)Mo ,, 3/2 Z~'r
or
~_1:
/~a(T_ 1)M2 V Tr
27r
However, y1 2 M2 = "rr  1 Then ~   l = ~ / ~ T r = l ~ ~ / ~ ° r T@/2r Thus F//no is maximum when (5.19)
"/'r max F#ho = or
M o max F/th o ~
3 (~A

1)
(5.20)
276 5.6.4
ELEMENTS OF PROPULSION
Mass Ingested by an Ideal Ramjet
Because the specific thrust of a ramjet has a m a x i m u m at the flight Mach number given by Eq. (5.20) and decreases at higher Mach numbers, one might question how the thrust of a given ramjet will vary with the Mach number. Does the thrust of a ramjet vary as its specific thrust? Because the thrust of a given ramjet will depend on its physical size (flow areas), the variation in thrust per unit area with Mach number will give the trend we seek. For a ramjet, the diffuser exit Mach number (station 2) is essentially constant over the flight Mach number operating range (M2 = 0.5). Using this fact, we can find the engine mass flow rate in terms of A2, M0, M2, and the ambient pressure and temperature. With this flow rate, we can then find the thrust per unit area at station 2 from F
F rho
A2  &oA2 As we shall see, the mass flow rate is a strong function of flight Mach number and altitude. For our case,
/~0

A2
/~/2  m2A~ /~/2 {~*'~ ~ ] A2
A~ A 2
A~
A
(i)
M2
However, mass flow parameter (MFP),
mi,/gT.
MFP(Mg)   
AiPtg
and
MFP* = M F F ( @ M = 1) 
A*Pt
Then rh2 = MFP* Pt2
,/g2'
Tt2 =
Tt0 ,
and
Pt2 = ¢rjP,o
where
Pt2
7rjPto
¢raPo Pto/Po
7rjPo(Tto/To) 7/(71)
(ii)
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
277
Pt2 __ "rrdPo (Tto'~ 7/(y1)1/2_ "rrdPO(Tr)(7+l)/[2(7_l) ]
(iii)
or
,/rs
,/~ kTo)
,/~
For air, 7 = 1.4 and (TF 1 ) / [ 2 ( y  1)] = 3. Thus, combining Eqs. (i), (ii), and (iii), we get
rh2°= MFP* A2
A*) 7rdPo~r
(5.21)
AM~
The variations of mass flow per unit area [Eq. (5.21)], specific thrust, and thrust per unit area at station 2 with flight Mach number are plotted in Fig. 5.6 for M2 of 0.5, altitude of 12 km, Tt4, of 1900 K, and 7ra of 1.0. Although the specific thrust variation with Mach number reaches a m a x i m u m at Mach 2.30 and then falls off, the thrust of an ideal ramjet continues to increase with flight Mach number until about Mach 5.25 due to the very rapid increase in mass flow per unit area.
750 
2.5
600
2.0
450 % ~~ 300
1.o
%
,
150
Fig. 5.6
0.5
1
2
3
4
Mo
5
6
0.0 7
Ideal ramjet thrust per unit area vs M a c h number.
278
5.7
ELEMENTS OF PROPULSION
Ideal Turbojet
The thrust of a ramjet tends to zero as the Mach number goes to zero. This poor performance can be overcome by the addition of a compressorturbine unit to the basic Brayton cycle, as shown in Fig. 5.7a. The thermal efficiency of this ideal cycle is now
"Or =
1
T° = T,3
1
1
(5.22)
rr'rc
Whereas an ideal ramjet's thermal efficiency is zero at Mach 0, a compressor having a pressure ratio of 10 will give a thermal efficiency of about 50% for the ideal turbojet at Mach 0. For the ideal turbojet
Wt = Wc,
rt4 > Tt3,
and
Pt4 = Pt3
Thus the compressorburnerturbine combination generates a higher pressure and temperature at its exit and is called, therefore, a gas generator. The gas leaving the gas generator may be expanded through a nozzle to form a turbojet, as depicted in Figs. 5.7a and 5.7b, or the gases may be expanded through a turbine to drive a fan (turbofan), a propeller (turboprop), a generator (gas turbine), an automobile (gas turbine), or a helicopter rotor (gas turbine). We will analyze the ideal turbojet cycle as we did the ideal ramjet cycle and will determine the trends in the variation of thrust, thrust specific fuel consumption, and fuel/air ratio with compressor pressure ratio and flight Mach number.
Freestream
Compressor
Combustor
Turbine Nozzle
V
M "E>
,./.,.
2
Fig. 5.7a
3
4
Station numbering of ideal turbojet engine.
9
u2z Lc q2z.~.~p.g 2.gOd =
~d ~d ~d :;d °~d °d 0,4 =
6;d
6;d ~;d ~ d CJd ~;d ° ; d
pug
:g dv~s~
6 o z _ ox,~o~o/~  ov _ ( o r ) ~z 6 ~ z ~ / ~ ~6 ~ ~a
° / ~  7Aj ~ = (°A  6A) I
j :[ dal~
"e'g uo!looS u! pols![ JopJo oql u ! IXOLIpOlUOS oad s[ o u ~ u o la.foq~nl leap! oql ol s!sJ(ieue o i a ~ j o sdols osoql j o uope~!ldd V
s!sdleuv alodO
•a u ! ~ u a l a . f o q a n l l u a p ! u u j o m u a ~ u ! p ~ /  H p u u m u a ~ u ! p s Z aqj~
O'f
""t°d'%:"A;~' /z = >1 £~JaU3 ~!lou!51ssaIuo!suotu!(I ~'~; 0"~ ~'[ 0'/ ~'0
X
6/~
0"0
qL'~
t'z'g
"~!zl
~'
/
SHNIONH 7VHQI 4 0 SISA7VNV 9 7 0 1 0 OlEI/HIAIVEIVcl
280
ELEMENTS OF PROPULSION
However, 7rd = % = % = 1; thus Pt9 = Pocrr~cTrt, and so
2 [(et9"~ ( y l ) / y ] M~  3 '  1 L \ P 9 }  1 where
Pt9
Pt9 Po
Po D
P9   P o P9


z
"n'r"a'c~ t P9
7rr ~'c'at
Then M92 _
2 [(,nrTrcTrt) ( ~ '  l ) / v y1
11
However, ~,7  l v ~ = ~'r and for an ideal turbojet ~ 7  0 / ~ = ~,.and v'¢t~I)/~ = Thus M92 = 2  ~ ( ~ ' r % r t y1
 1)
(5.23)
Step 4: __ Tto Tt2 Tt3 Tt4 Tt5 Tt9 Tt9 = 10   __   = TOTrTdTcTbTtT n ~T()Tr~'cTbTt To Tto Tt2 Tt3 Tt4 Tt5 Then To
To
Tt9 / To Tt9/T9
Tr Tc Tb Tt (Pt9/P9)(y 1)/3,
Tr Tc Tb Tt (,lrr,.B.cTrt)(y l)/y
Tr Tc Tb Tt TrTcTt
Thus T9 = To
%
(5.24)
Step 5: Application of the steady flow energy equation to the burner gives thoht 3 Jr/nfhpR = (fit() q/rtf)ht4 For an ideal cycle, rh0 + rhf ~ rno and Cp3 = cp4 = cp. Thus
~loCpTt3 Jr lJlfhpR = t;nocpTt4
it,4
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
281
or
_ epro (r,4
r,q
her \ To
To ]
f  mo
However, ra = 
Tt4
and
To
r~rc = 
Tt3
To
Then f __ inf
cpTo,
mo   ~
~
(5.25)
 ~)
or
f _ lhf
cpTorrrc,
mo
~
trb 
(5.26)
1)
Step 6: The power out of the turbine is TcVt = (1;frO + ~lf )(ht4  ht5) ~ ~ t o C p ( T t 4
•
= mocpTt4
(1

Tt5)
Tt4] = lhocpTt4(1  7))
The power required to drive the compressor is lJgc = r h o ( h t 3

ht2) = mocp(Tt3  Tt2)
= mocpTt2(~t21)=mocpTt2('rc1)
Since W,. = Wt for the ideal turbojet, then rrtocprt2(T c  1) = r h o c p T t 4 ( 1

yt)
or Tt2
rt = 1  ~ t r c 
1)
Tt = 1  r r ( r c 
1)
Thus
TA
(5.27)
282
ELEMENTS OF PROPULSION
Step 7:
(V9~ 2_ T9 M2 ao /  To
2
~A 1)
(5.28)
(~'r'rc'rt  1)  M0]
(5.29)
 
9
(~rzc~,
" y  1 zr'rc
However,
F aO(~oo_MO) gc
/no
Thus
/no
gc
1 z~r~.
Step 8: According to Eq. (5.8),
S 
f
F//no
The thrust specific fuel consumption S can be calculated by first calculating the fuel/air ratio f and the thrust per unit of airflow F//no, using Eqs. (5.25) and (5.29), respectively, and then substituting these values into the preceding equation. An analytical expression for S can be obtained by substituting Eqs. (5.25) and (5.29) into Eq. (5.8) to get the following: S =
(5.30)
cpTogc('G  "rr%) aohp~
[j
y
1
1 Zr'rc
Step 9: Again the development of these expressions is left to the reader.
Thermal efficiency: ~r=l
1
(5.31a)
Propulsive efficiency: rt~ 
2Mo Vg/ao + Mo
(5.31 b)
Overall efficiency: 7o = riper
(5.31c)
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
283
5.7.2 Summary of Equations~ldeal Turbojet INPUTS:
(kJ,
Mo, To(K, °R), y, cp k g  K
Btu )
Ibm. °R7' hpR
(k~' lb~m]' Btu) TMK, °R), ~
OUTPUTS: F (N
lbf \
s(mg/s
lbm/h'~
k~ls' l~~s) 'f' t ~ ' i ~ ),TT, ,7,, 770
EQUATIONS: 3'1 Cp 3'
(5.32a)
ao = g/~gcTo
(5.32b)
R =
Tr = 1  P "  ~  M
2
(5.32c)
Tt4 To
~;~ =  
(5.32d)
Tc = ( ~rc)(~1)/~
(5.32e)
~'r ~, = 1    ( 7 c  1) TA
(5.32f)
W9__ ~ 2_ "cA (TrTcT t ao y 1 TrTc /no f
I)
(5.32g)
gc cpTo, he~t'r;t V"rc)
S
f F//no
rlr = 1   
(5.32i) (5.32j)
1
TrTc
(5.32k)
2Mo T#p   V 9 / a o IMo ')70 = 7~p~r
(5.321) (5.32m)
284
ELEMENTS OF PROPULSION E x a m p l e 5.2
In Figs. 5.8a5.8d, the performance of ideal turbojets is plotted vs compressor pressure ratio 7rc for different values of flight Mach number Mo. Figures 5.9a5.9d plot the performance vs flight Mach number Mo for different values of the compressor pressure ratio ~rc. Calculations were performed for the following input data: To = 390°R,
Cp = 0.24Btu/(lbm • °R)
Y = 1.4,
hpR = 18,400 Btu/lbm,
T,4 = 3000°R
,3) Figures 5.8a5.8d. Figure 5.8a shows that for a fixed Mach number, there is a compressor pressure ratio that gives maximum specific thrust. The loci of the compressor pressure ratios that give maximum specific thrust are indicated by the dashed line in Fig. 5.8a. One can also see from Fig. 5.8a that a lower compressor pressure ratio is desired at high Mach numbers to obtain reasonable specific thrust. This helps explain why the compressor pressure ratio of a turbojet for a subsonic flight may be 24 and that for supersonic flight may be 10. Figure 5.8b shows the general trend that increasing the compressor pressure ratio will decrease the thrust specific fuel consumption. The decrease in fuel/ air ratio with increasing compressor pressure ratio and Mach number is shown
120 V
~
1°° F
///
/
~

/ i"
0.5
.o
40if
 about 3 (due to combined ram and mechanical compression), the difference between PDE and Brayton performance is significantly reduced. Readers are now empowered to freely investigate comparisons of their own choice.
344
ELEMENTS OF PROPULSION 200
.::::_: ::::::
150
F/rhO [lbfs/Ibm] 100
PDE
.
.
.
.
Brayton 5O
0j
3
~,
=
4
7"3/7"0
Fig. 5.36 Specific thrust F#ho of ideal PDE and Brayton cycles as functions of ~, for ~ = 5 and 10, 3/= 1.36, and vehicle speed Vo = 0.
2
PDE . . I Brayt°n ~
1
. .
S [lbnffhlb~
I ,
......
~5
4 ~, = T3/To
Fig. 5.37 Specific fuel consumption S of ideal PDE and Brayton cycles as functions of ~, for ~ = 5 and 10, 3~ = 1.36, her = 19,000Btu/lbm, and Vo = 0.
Problems 5.1
Show that the thermal and propulsive efficiencies for an ideal ramjet engine are given by Eqs. (5.17a) and (5.17b), respectively.
5.2
Calculate the variation with Tt4 of exit Mach number, exit velocity, specific thrust, fuel/air ratio, and thrust specific fuel consumption of an ideal
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
345
turbojet engine for compressor pressure ratios of 10 and 20 at a flight Mach number of 2 and To = 390°R. Perform calculations at Tt4 values of 4400, 4000, 3500, and 3000°R. Use h e r = 18,400 Btu/lbm, cp = 0.24 Btu/ (Ibm. °R), and y = 1.4. Compare your results with the output of the PARA computer program.
5.3
Calculate the variation with Tt4 of exit Mach number, exit velocity, specific thrust, fuel/air ratio, and thrust specific fuel consumption of an ideal turbojet engine for compressor pressure ratios of 10 and 20 at a flight Mach number of 2 and To = 217 K. Perform calculations at Tt4 values of 2400, 2200, 2000, and 1800 K. Use hpR = 42,800 kJ/kg, Cp = 1.004 kJ/ (kg. K), and y = 1.4. Compare your results with the output of the PARA computer program.
5.4
Show that the thermal efficiency for an ideal turbojet engine is given by Eq. (5.22).
5.5
Show that the thermal efficiency for an ideal afterburning turbojet is given by Eq. (5.42).
5.6
A major shortcoming of the ramjet engine is the lack of static thrust. To overcome this, it is proposed to add a compressor driven by an electric motor, as shown in the model engine in Fig. P5.1. For this new ideal engine configuration, show the following: (a) The specific thrust is given by
trio
gc
Y
l
1 "rrTc
(b) The compressor power requirement is given by ~Vc = thocpTo'rr('r c  1)
(c) Determine the static thrust of this engine at the following conditions: To = 518.7°R,
Tt4 = 3200°R,
Cp = 0.24 Btu/(lbm • °R),
[ 0
[ 2
\~..
/
3
4 Fig. PS.1
i
7rc = 4 y = 1.4
_
I
9 Model engine.
/ °
346 5.7
ELEMENTS OF PROPULSION Determine the optimum compressor pressure ratio, specific thrust, and thrust specific fuel consumption for an ideal turbojet engine giving the maximum specific thrust at the following conditions: M0 = 2.1, To = 220 K,
Tt4
1700 K,
=
Cp = 1.004 kJ/(kg • K),
her = 42,800 kJ/kg y = 1.4
5.8
Show that the thermal efficiency for an ideal turbojet engine with optimum compressor pressure ratio is given by ~/T = 1  1 / ~ / ~ .
5.9
Show that the thermal efficiency for an ideal turbofan engine is given by Eq. (5.22).
5.10
Compare the performance of three ideal turbofan engines with an ideal turbojet engine at two flight conditions by completing Table P5.1. The first flight condition (case 1) is at a flight Mach number of 0.9 at an altitude of 40,000 ft, and the second flight condition (case 2) is at a flight Mach number of 2.6 and an altitude of 60,000 ft. Note that the fuel/air ratio need be calculated only once for each case since it is not a function of a or ,TTf. The following design information is given:
Tt4
 20,
T: 5.11
Cp = 0.24 Btu/(lbm • °R)
3000°R,
=
1.4,
her  18,400Btu/lbm
Repeat Problem 5.10 with the first flight condition (case 1) at a flight Mach number of 0.9 and altitude of 12 km and the second flight condition (case 2) at a flight Mach number of 2.6 and an altitude of 18 km. Use the following design information:
Tt4
7rc = 20,
Cp = 1.004 kJ/(kg • K)
1670 K,
=
3' = 1.4,
her = 42,800 kJ/kg
Table P5.1
Engine Turbofan (a) Case 1 Case 2 (b) Case 1, a* Case 2, a* (c) Case 1, err* Case 2, ~rf* Turbojet (a) Case 1 Case 2
a
zrf
1 1
4 4 4 4
V9/ao
V19/ao
1 1 0 0
n/a n/a
n/a n/a
F/mo
f
S
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
347
5.12
Show that the propulsive efficiency for an ideal turbofan engine with optimum bypass ratio is given by Eq. (5.62).
5.13
Show that the propulsive efficiency for an ideal turbofan engine with optimum fan pressure ratio is given by Eq. (5.68).
5.14
For an ideal turbofan engine, the maximum value of the bypass ratio corresponds to V9 = Vo. (a) Starting with Eq. (5.53), show that this maximum bypass ratio is given by T,~ + 1  T~T~ T;t/(r~%) ~max ~'rO'f 
(b)
Show that the propulsive efficiency for this maximum bypass ratio is given by 2
~Tp ~/O'rTf
5.15
l)

1)/(~'r  1) + 1
Under certain conditions, it is desirable to obtain power from the freestream. Consider the ideal airpowered turbine shown in Fig. P5.2. This cycle extracts power from the incoming airstream. The incoming air is slowed down in the inlet and then heated in the combustor before going through the turbine. The cycle is designed to produce no thrust (F = 0), so that V9 = Vo and P9 Po. (a) Starting with Eq. (5.29), show that % =  1+    1 Y  l M ~ "rr "r~, 2 (b)
Then show that the turbine output power is given by
w
\rr
Generator
t3 0
2
3
/
4
Fig. P5.2
I
5
"
"l
0~.. 
9
Ideal airpowered turbine.
9
348
ELEMENTS OF PROPULSION Comoressor rnF
•
r~ C
)
Highpressure 13
19
Lre
4.5 Fig. P5.3 5.16
Seal b e t w e e n fan s t r e a m and l o w  p r e s s u r e turbine s t r e a m
Turbofan engine with aft fan.
In the early development of the turbofan engine, General Electric developed a turbofan engine with an aft fan, as shown in Fig. P5.3. The compressor is driven by the highpressure turbine, and the aft fan is directly connected to the lowpressure turbine. Consider an ideal turbofan engine with an aft fan. (a) Show that the highpressure turbine temperature ratio is given by TtH  
Tt4.5
7"r
Tt4
(b)
1 (%
 1)
rA
Show that the lowpressureturbine temperature ratio is given by
Tt5 aTr ztL     1 ( Tf Tt4.5 ~'a~'tH 5.17

l)
Starting with Eq. (5.48), show that for known values of Mo, To, T t 4 , 7Tc,and 7rf, the bypass ratio a giving a specified value of specific thrust F/nio is given by the solution to the quadratic equation Ac~2 + B a +
C= 0
where A= ~ D
2
B=(T1)(D2+ CYl( F gc D  mo ao
D Vl9 ~o ) + 7"r('rf 1)
D 2 + 2DVm~ E ao / (V19 _ Mo) \ a~
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
E = "ra  "rr(% 
V19 a0 5.18
I2
:
1)    'rr'rc
,['7(7"r'rf 
(~rr'rf 
349
1)
1)
Using the system of equations listed in Problem 5.17, determine the bypass ratio that gives a specific thrust of 400 N/(kg/s) for the following data: To=216.7K,
Tt4=1670K,
Mo=0.8,
7rc:24,
~f=4
with y = 1.4, cp = 1.004 kJ/(kg • K), h e r = 42,800 kJ/kg 5.19
Using the system of equations listed in Problem 5.17, determine the bypass ratio that gives a specific thrust of 40.0 lbf/(lbm/s) for the following data: To:390°R, with y = 1.4,
Tt4=3000°R,
Mo:2,
Cp = 0.24 Btu/(lbm °R),
arc= 16,
9=5
her = 18,400Btu/lbm
Problems for Supporting Material SM5.1
Considerable research and development effort is going into increasing the maximum Tt4 in gas turbine engines for fighter aircraft. For a mixedflow turbofan engine with 7r~.= 16 at a flight condition ofMo = 2.5 and To = 216.7 K, use the PARA computer program to determine and plot the required engine fan pressure, specific thrust, and specific fuel consumption vs Tt4 over a range of 1600 to 2200 K for bypass ratios of 0.5 and 1. Use h p R = 4 2 , 8 0 0 k J / k g , C p = 1.004kJ/(kgK), and y = 1.4. Comment on your results in general. Why do the plots of specific fuel consumption vs specific thrust for these engines all fall on one line?
SM5.2
Considerable research and development effort is going into increasing the maximum Tt4 in gas turbine engines for fighter aircraft. For a mixedflow turbofan engine with ~c = 20 at a flight condition of Mo = 2 and To = 390°R, use the PARA computer program to determine and plot the required engine bypass ratio a, specific thrust, and specific fuel consumption vs Tt4 over a range of 3000 to 4000°R for fan pressure ratios of 2 and 5. Use h p R = 18,400Btu/lbm, C p = 0 . 2 4 B t u / (Ibm °R), and y = 1.4. Comment on your results.
SM5.3
Show that the thermal efficiency of the afterburning mixedflow turbofan engine given by Eq. (SM5.16) can be rewritten as _
cpTo
1
350
ELEMENTS OF PROPULSION
SM5.4
For an afterburning mixedflow turbofan engine with 7rc = 16 and Tt7 = 2200 K at a flight condition of Mo = 2.5 and To = 216.7 K, use the P A R A computer program to determine and plot the required engine fan pressure, specific thrust, and specific fuel consumption vs Tt4 over a range of 1600 to 2200 K for bypass ratios of 0.5 and 1. Use heR=42,800kJ/kg, Cp= 1 . 0 0 4 k J / ( k g . K ) , and 3,= 1.4. Comment on your results in general. Why do the plots of specific fuel consumption vs specific thrust for these engines all fall on one line? Compare these results to those for Problem SM5.1, and comment on the differences.
SM5.5
For an afterburning mixedflow turbofan engine with 7rc = 20 and Tt7 4000°R at a flight condition of Mo = 2 and To = 390°R, use the PARA computer program to determine and plot the required engine bypass ratio a, specific thrust, and specific fuel consumption vs Tt4 over a range of 3000 to 4000°R for fan pressure ratios of 2 and 5. Use her = 18,400 Btu/lbm, Cp = 0.24 Btu/(lbm. °R), and 3' = 1.4. Comment on your results. Compare these results to those for Problem SM5.2, and comment on the changes. =
SM5.6
Use the P A R A computer program to determine and plot the thrust specific fuel consumption vs specific thrust for the turboprop engine of Example SM5.2 over the same range of compressor pressure ratios for turbine temperature ratios of 0.8, 0.7, 0.6, 0.5, 0.4, and optimum.
SM5.7
Use the PARA computer program to determine and plot the thrust specific fuel consumption vs specific thrust for the turboprop engine over the range of compressor pressure ratios from 2 to 40 at To = 425°R, Mo = 0.65, y = 1.4, Cp = 0.24 Btu/(lbm. °R), hpR = 18,400 Btu/lbm, Tt4 = 2460°R, and "qprop= 0.8 for turbine temperature ratios of 0.8, 0.7, 0.6, 0.5, 0.4, and optimum.
SM5.8
Use the P A R A computer program to determine and plot the specific fuel consumption vs specific power for the turboshaft with regeneration over compressor pressure ratios of 4 to To 290 K, Mo = 0, 3' = 1.4, Cp  1.004 kJ/(kg. K), 42,800 kJ/kg, and x = 1.02 for combustor exit temperatures 1300, 1400, 1500, and 1600 K.
SM5.9
Use the P A R A computer program to determine and plot the power specific fuel consumption vs specific power for the turboshaft engine with regeneration of Example SM5.3 at x = 1.02 over the same range of compressor ratios for combustor exit temperatures Tt4 of 2400, 2600, 2800, and 3000°R.
power engine 18 at her = Tt4 of
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
351
Gas Turbine Design Problems 5.D1
You are to determine the range of compressor pressure ratios and bypass ratios for ideal turbofan engines that best meet the design requirements for the hypothetical passenger aircraft, the HP1.
HandCalculate Ideal Performance (HP1 Aircraft). Using the parametric cycle analysis equations for an ideal turbofan engine with Tt4 = 1560 K, handcalculate the specific thrust and thrust specific fuel consumption for an ideal turbofan engine with a compressor pressure ratio of 36, fan pressure ratio of 1.8, and bypass ratio of 10 at the 0.83 Mach, 11kmaltitude cruise condition. Assume y = 1.4, Cp = 1.004 kJ/ (kg. K), and hpR = 42,800 kJ/kg. Compare your answers to results from the parametric cycle analysis program PARA. ComputerCalculated Ideal Performance (HP1 Aircraft). For the 0.83 Mach, 11kmaltitude cruise condition, determine the performance available from turbofan engines. This part of the analysis is accomplished by using the PARA computer program with Tt4 = 1560 K. Specifically, you are to vary the compressor pressure ratio from 20 to 40 in increments of 2. Fix the fan pressure ratio at your assigned value o f . Evaluate bypass ratios of 4, 6, 8, 10, 12, and the optimum value. Assume y = 1.4, cp = 1.004 kJ/(kg. K), and hpze = 42,800 kJ/kg. Calculate Minimum Specific Thrust at Cruise (HP1 Aircraft). You can calculate the minimum uninstalled specific thrust at cruise based on the following information: 1) The thrust of the two engines must be able to offset drag at 0.83 Mach and 11km altitude and have enough excess thrust for Ps of 1.5 m/s. Determine the required installed thrust to attain the cruise condition using Eq. (1.28). Assuming (~inlet "~ ~noz z 0.02, determine the required uninstalled thrust. 2) Determine the maximum mass flow into the 2.2mdiam inlet for the 0.83 Mach, 11kmaltitude flight condition, using the equation given in the background section for this design problem in Chapter 1. 3) Using the results of steps 1 and 2, calculate the minimum uninstalled specific thrust at cruise. 4) Perform steps 2 and 3 for inlet diameters of 2.5, 2.75, 3.0, 3.25, and 3.5 m.
Select Promising Engine Cycles (HP1 Aircraft). Plot thrust specific fuel consumption vs specific thrust (thrust per unit mass flow) for the engines analyzed in the preceding. Plot a curve for each bypass ratio, and crossplot the values of the compressor pressure ratio (see Fig. P5.D 1). The result is a carpet plot (a multivariable plot) for the cruise condition. Now draw a dashed horizontal line on the carpet plot corresponding to the maximum allowable uninstalled thrust specific fuel consumption Smax for the cruise condition (determined in the Chapter 1 portion of this design problem). Draw a dashed vertical line for each minimum uninstalled specific thrust
ELEMENTS OF PROPULSION
352
3.5 m
2.75 m
2.2m diameter
Zc = ~
"
a 4
Smax
a~....~
40
F/dto Fig. P5.D1
Example carpet plot for HP1 aircraft engine.
determined in the preceding. Your carpet plots will look similar to the example shown in Fig. P5.D1. What ranges of bypass ratio and compressor pressure ratio look most promising? 5.D2
You are to determine the ranges of compressor pressure ratio and bypass ratio for ideal mixedflow turbofan engines that best meet the design requirements for the hypothetical fighter aircraft, the HF1.
HandCalculate Ideal Performance (HF1 Aircraft). Using the parametric cycle analysis equations for an ideal mixedflow turbofan engine with Tt4 = 3250°R, handcalculate the specific thrust and thrustspecific fuel consumption for an ideal turbofan engine with a compressor pressure ratio of 25 and bypass ratio of 0.5 at the 1.6 Mach, 40kftaltitude supercruise condition. Assume T = 1.4, cp = 0.24 Btu/(lbm.°R), and hpR = 18,400 Btu/lbm. Compare your answers to the results from the parametric cycle analysis program PARA. ComputerCalculated Ideal Performance (HF1 Aircraft). For the 1.6 Mach, 40kftaltitude supercruise condition, determine the performance available from turbofan engines. This part of the analysis is accomplished by using the PARA computer program with T~4 = 3250°R. Specifically, you are to vary the bypass ratio from 0.1 to 1.0 in increments of 0.05. Evaluate compressor pressure ratios of 16, 18, 20, 22, 24, and 28. Assume ~/= 1.4, Cp = 0.24 Btu/(lbm. °R), and hpR = 18,400 Btu/lbm. Calculate Minimum Specific Thrust at Cruise (HF1 Aircraft). You can calculate the minimum uninstalled specific thrust at supercruise based on the following information:
PARAMETRIC CYCLE ANALYSIS OF IDEAL ENGINES
Minimumspecificthrust
S~
.... ~
353
/ ~
.... ~= 0.1
S~a~
o=1.o 2"
",,,,
YX/X
F/~ o
Fig. P5.D2
Example carpet plot for HF1 aircraft engine.
1) The thrust of the two engines must be able to offset drag at 1.6 Mach, 40kft altitude, and 92% of takeoff weight. Assuming (])inlet "q~bnoz = 0.05, determine the required uninstalled thrust for each engine. 2) The maximum mass flow into a 5ft 2 inlet for the 1.6 Mach, 40kftaltitude flight condition is rh = pAV = O'PrefAMa = (0.2471 0.07647)(5) (1.6 × 0.8671 × 1116) = 146.3 lbm/s. 3) Using the results of steps 1 and 2, calculate the minimum uninstalled specific thrust at supercruise.
Select Promising Engine Cycles (HF1 Aircraft). Plot the thrust specific fuel consumption vs specific thrust (thrust per unit mass flow) for the engines analyzed in the preceding. Plot a curve for each bypass ratio, and crossplot the values of compressor pressure ratio (see Fig. P5.D2). The result is a carpet plot (a multivariable plot) for the supercruise condition. Now draw a dashed horizontal line on the carpet plot corresponding to the maximum allowable uninstalled thrust specific fuel consumption Smax for the cruise condition (determined in the Chapter 1 portion of this design problem). Draw a dashed vertical line for the minimum uninstalled specific thrust determined in the preceding section. Your carpet plots will look similar to the example shown in Fig. P5.D2. What ranges of bypass ratio and compressor pressure ratio look most promising?
6 Component Performance
6.1
Introduction
In Chapter 5, we idealized the engine components and assumed that the working fluid behaved as a perfect gas with constant specific heats. These idealizations and assumptions permitted the basic cycle analysis of several types of engines and the analysis of engine performance trends. In this chapter, we will develop the analytical tools that allow us to use realistic assumptions as to component losses and to include the variation of specific heats.
6.2
Variation in Gas Properties
The enthalpy h and specific heat at constant pressure Cp for air (modeled as a perfect gas) are functions of temperature. Also, the enthalpy h and specific heat at constant pressure Cp for a typical hydrocarbon fuel JP8 and air combustion products (modeled as a perfect gas) are functions of temperature and the fuel/air ratio f. The variations of properties h and Cp for fuel/air combustion products vs temperature are presented in Figs. 6.1a and 6.1b, respectively. The ratio of specific heats y for fuel/air combustion products is also a function of temperature and of fuel/air ratio. A plot of 7 is shown in Fig. 6.2. These figures are based on Eq. (2.64) and the coefficients of Table 2.2. Note that both h and Cp increase and y decreases with temperature and the fuel/air ratio. Our models of gas properties in the engines need to include changes in both Cp and y across components where the changes are significant. In Chapter 7, we will include the variation in Cp and y through the engine. To simplify the algebra, we will consider Cp and y to have constant representative values through all engine components except the burner (combustor). The values of Cp and y will be allowed to change across the burner. Thus we will approximate cp as Cpo (a constant for the engine upstream of the burner) and cp as Cpt (a constant average value for the gases downstream of the burner). Likewise, y will be % upstream of the burner and Yt downstream of the burner. The release of thermal energy in the combustion process affects the values of
355
356
ELEMENTS OF PROPULSION Fuel/air ratio 0.0676 0.030 0.0
1200 
1000 
800

E 600
400 
200 
I
I
I
I
I
I
I
I
500
1000
1500
2000 T (°R)
2500
3000
3500
4000
Fig. 6.1a
E n t h a l p y vs t e m p e r a t u r e f o r J P  8 a n d a i r c o m b u s t i o n p r o d u c t s .
0.36
Fuel/air ratio 0.0676 / / O . O 6 / / / 0 . 0 5 0.04 0.03 0.02
0.34 0.32
=
oo,
o~
0.30
~
0.28
0.26 0.24
022 Fig. 6.1b
o'
5 0
'o
10 0
15
'oo
o'
2 00 T (°R)
5'
2 00
o'
3 00
'o
35 0
40 0
Specific h e a t cp vs t e m p e r a t u r e f o r J P  8 a n d a i r c o m b u s t i o n p r o d u c t s .
COMPONENT PERFORMANCE 1.44
357

1.40
1.36
1.32

Fuel/air ratio

1.28
1.24 0
I 500
I 1000
I 1500
I 2000
I 2500
I 3000
I 3500
[0.0676 4000
T(°R)
Fig. 6.2 Ratio of specific heats ~, vs temperature for JP8 and air combustion products.
Cpt
and %, but these two are related by Cpt    %1
Rt   %ldg
(6.1)
where ~ , = universal gas constant J / = molecular weight Thus if the chemical reaction causes the vibrational modes to be excited but does not cause appreciable dissociation, then the molecular weight J / w i l l be approximately constant. In this case, a reduction in y is directly related to an increase in Cp by the formula cpt Cpc
%
%
m
%

1 %

1
(6.2)
6.3 Component Performance In this chapter, each of the engine components will be characterized b y f i g u r e s that model the component's performance and facilitate cycle analysis of real airbreathing engines. The total temperature ratio ~, the total pressure ratio 7r, and the interrelationship between ~and 7r will be used as much as possible in a component's figure of merit. of merit
358
6.4
ELEMENTS OF PROPULSION
Inlet and Diffuser Pressure Recovery
Inlet losses arise because of the presence of wall friction and shock waves (in a supersonic inlet). Both wall friction and shock losses result in a reduction in total pressure so that wd < 1. Inlets are adiabatic to a very high degree of approximation, and so we have ~'a = 1. The inlet's figure of merit is defined simply as ~'a. The i s e n t r o p i c e f f i c i e n c y ria of the diffuser is defined as (refer to Fig. 6.3) ht2s  h o rid   hto  ho
~
Tt2s 
To
Tto  To
(6.3)
This efficiency can be related to % and ~a to give ~.(v1)/v _ 1 rid  °r(~rd) ~'r1
(6.4)
Figure 6.4 gives typical values of ~'d for a subsonic inlet. The diffuser efficiency rid was calculated from 7rd by using Eq. (6.4). In supersonic flight, the flow deceleration in inlets is accompanied by shock waves that can produce a total pressure loss much greater than, and in addition to, the wall friction loss. The inlet's overall pressure ratio is the product of the ram pressure ratio and the diffuser pressure ratio. Because of shocks, only a portion of the ram total pressure can be recovered. W e now define 7Tdmax as that portion of ~'a that is due to wall friction and define rir as that portion of ~'d due to ram recovery. Thus "rid =
(6.5)
"/Tdmaxrir
T~
Pt2~P2
2gcCp
1
Fig. 6.3
I, / p t
PO
Definition of inlet states.
COMPONENT PERFORMANCE
359
1.00 0.98 0.96 0.94 0.92 0.90 0.88 0.86 0.1
t
I
I
t
i
t
I
t
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Mo Fig. 6.4
Typical subsonic inlet ~ra and ~/a
For subsonic and supersonic flow, a useful reference for the ram recovery ~/~ is Military Specification 5008B, 36 which is expressed as follows:
~Tr =
1 10.075(Mo1) 800 M 4 + 935
]'3s
Mo~l 1 <Mo 1
(7.20h)
7"i"d ~ 7"£dmaxT]r
(7.20i)
" G   cptTt4
(7.20j)
cpcTo
(7.20k)
~'c = ~r(z' 1
(7.52i)
7Tdmax 97r Cpt Tt4
(7.52j)
Ci,cTo
(7.52k)
7"c = 77~ff,"l)/(Z'ec)
~3"(cYc1)/'Yc  1 "Oc
(7.521)
tc1 (?~l)/('y~ef)
~s =
~s
"qf
7r7:1)/7c  1 ~S1 "r,~ 
f=
(7.52m)
(7.52n)
%%
1
~" [~'c 
r/m(1 + f ) ~ ' ~
1~t  1 

P9
Po
(7.52 p)
rt
(7.52r)
7rr 7"rd ,nc ,nb 7rt 7rn
(7.52s)
P9
(7.52t) ";;t "rt
Cpc
(7.52u)
( P t 9 / P 9 ) q'l)/'r' cpt
V9
(7.52v)
ao
P19
1)]
T]/e'
m 9 =
Ptl9
1 + ~(~f 
(7.52q)
"7"l't = T~t'/[(%l)e']
T9 To
(7.52o)
nbhpR/(cp~To)  ":A
%= 1
Pt9
(7.52h)
V',/~ c o Po P19
~'r~a~f~fn
(7.52w)
PARAMETRIC CYCLE ANALYSIS OF REAL ENGINES
M19  ~
L\ P ' 9 /
1
411
(7.52x)
T19 Tr~'f ~0 = (Ptl9/P19) (~''1)/3'~
(7.52y)
V19 T~ ao  M 9V
(7.52z)
F _ 1 ao[(l+f)V9 rho 1 + a gc ao
Mo+(l+f)
RtT9/To 1  Po/P9 ] + _ _a ao × RcV9/ao Yc J 1 + lagc × (V19 _ Mo + T19/T   0 1  Po/PI9~ \ ao Vl9/ao Tc /
S 
f
(7.55ab)
(1 + a ) F / m o
V9
. M R t r 9 / r o 1  Po/P9
(1 + f )    Mo + (1 +y)~:::, K,. v9 / ao Yc a0 Thrust ratio (FR) = VI9 T19/T 0 1  Po/PI9    M o +
ao

V19/ao
ao2[(1 + f)(V9/ao) 2 + 
Tc
ol(V19/ao) 2 
2gcfhpR

(7.52ac)

2M0[(1 + f ) V 9 / a o + oz(V19/ao)  (1 + ot)Mo] r/e = (1 + f ) ( V 9 / a o ) 2 + o z ( V 1 9 / a o ) 2  (1 + 0 0 M 2
~1T
(7.52aa)
(1 + a)M~]
(7.52ad)
(7.52ae) (7.52af)
7.4.3
Exit Pressure Conditions
Separatestream turbofan engines are generally used with subsonic aircraft, and the pressure ratio across both primary and secondary nozzles is not very large. As a result, often convergentonly nozzles are utilized. In this case, if the nozzles are choked, we have
Ptl9
P19  
3'c/('y 1) and
P99 =
(7.53)
412
ELEMENTS OF PROPULSION
Thus
Po
Ptl9/P19 P19/Po
P19
__
[(% +
1)/2] v'/~y~l)
"WrTTdTTf'lTfn
(7.54)
and
Po P9
m
Pt9/P9 P9/Po
[(Tt + 1)/2ff '/(v'l)
~r~d~c~bTTt~n
(7.55)
Note that these two expressions are valid o n l y w h e n both P9 and P19 are greater than Po. If these expressions predict P9 and P19 less than Po, the nozzles will not be choked. In this case, we take PI9 = Po a n d / o r P9 = Po
Example 7.6 As our first e x a m p l e for the turbofan with losses, we calculate the performance of a turbofan e n g i n e cycle with the following input data. INPUTS:
M0 = 0.8, To = 390°R, Yc = 1.4, Cp¢ = 0.240 B t u / ( l b m . °R) Yt = 1.33, cpt ~'dmax = 0.99,
=
0.276 B t u / ( l b m • °R), hpn = 18,400 B t u / l b m
7rb = 0.96,
% = 0.99,
7Tfn = 0.99, ec = 0.90, ef = 0.89
 0.89, ~b = 0.99, r/m 0.99, Po/P9 = 0.9, Po/P19 = 0.9 Zt4 =
3000°R,
7re = 36, Try = 1.7,
a 8
EQUATIONS:
Rc  Yc%,  1 Cpc
0.4  i ~ ( 0 . 2 4 x 778.16)
53.36 f t . l b f / ( I b m . °R)
Rt  "Yt0.33 Yt 1 Cpt = 1.~(0.276 X 778.16) = 53.29 f t ' l b f / ( l b m . °R) a0 = ~/1.4 x 53.36 x 32.174 x 390 968.2 ft/s
Vo = aoMo = 968.2 x 0.8 = 774.6 ft/s ~'r = 1 + Y ~   ~ M 2 = 1 + 0 . 2 x 0.82 = 1.128
7rr = ~'VrC/(z'c 1) = 1.1283.5 = 1.5243 ~Tr=l 7Td =
sinceM0
00break and 7/c < ~cmax, the specific fuel consumption is more than its inherent thermal efficiency would make possible. The designer would therefore strongly prefer to have the engine always operate at or very near 0o = 00bre,k, but this is impossible because every aircraft has a flight envelope with a range of 0o [see Fig. 8.11]. The best available compromise is to choose a 00bre,k that provides the best balance of engine performance over the expected range of flight conditions. It is interesting to note that, since early commercial and military aircraft primarily flew at or near 0o = 1, they were successfully designed with 00break = 1. Consequently, several generations of propulsion engineers took it for granted that aircraft engines always operated at ~rcmax and Tt4max under standard sealevel static conditions. However, the special requirements of more recent aircraft such as [the F22 Raptor (0o > 1.2 at supercruise)] have forced designers to select theta breaks different from 1.0. These engines may operate either at ~'cmax or Tt4max at standard sealevel static conditions, but never both.
Example 8.1 W e n o w c o n s i d e r c o m p r e s s o r operation at different Tt4 and 0o, specifically a c o m p r e s s o r that has a pressure ratio o f 15 and c o r r e c t e d mass flow rate of 100 l b m / s for Tt2 o f 518.7°R (sealevel standard) and Tt4 o f 3200°R. A t these
ENGINE PERFORMANCE ANALYSIS
459
conditions, 0o is 1, and constants K1 and K2 in Eqs. (8.17) and (8.18) are 3.649 × 10 4 and 3771, respectively. In addition, we assume that an engine control system limits 7U to 15 and Tt4 to 3200°R. By using Eqs. (8.17) and (8.18), the compressor pressure ratio and corrected mass flow rate are calculated for various values of Tt4 and 0o. Figures 8.12 and 8.13 show the resulting variation of compressor pressure ratio and corrected mass flow rate, respectively, with flight condition 0o and throttle setting T,4. Note that at 0o above 1.0, the compressor pressure ratio and corrected mass flow rate are limited by the maximum combustor exit temperature Tt4 of 3200°R. The compressor pressure ratio limits performance at 00 below 1.0. Thus 0 0 b r e a k = 1.0.
8.2.6
Variation
in Engine
Sp~cl
As will be shown in Chapter 9, the change in total enthalpy across a fan or compressor is proportional to the rotational speed N squared. For a calorically perfect gas, we can write Tt3  Tt2 = K 1 N 2 or
rc  1 = K 1 Af2
(i)
T r e f • ' c2
where Nc2 is the compressor corrected speed. The compressor temperature ratio is related to the compressor pressure ratio through the efficiency, or %  1 = ('iT('Yc1)/'Yc  1)/77 ~
Combining this equation with Eq. (i), rewriting the resulting equation in terms of pressure ratio and corrected speed, rearranging into variable and constant terms, and equating the constant to reference values give for constant compressor efficiency 71cK 1 _ "n"(~"~)/~ cR  1
7r(c3'C1)/3'~ 1 N c2

Tref
Solving Eq. (ii) for the corrected speed ratio N c 2 / N c 2 R ,
No2
/'rr (~'l)/'rc  1
(ii)
2 Nc~2R we
have
(8.19a)
v Tr~R This equation can also be used to estimate the variation in engine speed N with flight condition. Equation (8.19a) is plotted in Fig. 8.14 for a reference compressor pressure ratio of 16. Note that a reduction in compressor pressure ratio from 16 to 11 requires only a 10% reduction in corrected speed Arc. Equation (8.19a)
460
ELEMENTS OF PROPULSION 1.10
1.00
0.90
Nc 0.80
0.70
0.6£ 4
I
6
I
8
I
I
I
10
12
14
I
16
18
nc Fig. 8.14
Variation in corrected speed with compressor pressure ratio.
can be written in terms of Tt4/0o by using Eq. (8.17), yielding
/
Nc2 Tt4/00 Nc2R V(Tt4/ Oo)R i
(8.19b)
Because the compressor and turbine are connected to the same shaft, they have the same rotational speed N, and we can write the following relationship between their corrected speeds:
Nc2  ~
V ~o
c4
(8.20)
Comparison of Eqs. (8.19b) and (8.20) gives the result that the corrected turbine speed is constant, or
Nc4 = const
(8.21)
This result may surprise one at first. However, given that the turbine's temperature ratio rt, pressure ratio 7rt, and efficiency r/t are considered constant in this analysis, the turbine's corrected speed must be constant (see Fig. 8.5c).
ENGINE PERFORMANCE ANALYSIS
461
8.2.7 Gas Generator Equations The pumping characteristics of a simple gas generator can be represented by the variation of the gas generator's parameter ratios with corrected compressor speed. The equations for the gas generator's pressure and temperature ratios, corrected air mass flow and fuel flow rates, compressor pressure ratio, and corrected compressor speed can be written in terms of Tt4/Tt2, reference values (subscript R), and other variables. The gas generator's pressure and temperature ratios are given simply by
Pt6 = Pt2
~r~rb~t
(8.22)
Tt6 Tt4 rt Tt2 Tt2
(8.23)
From Eq. (8.10) and referencing, the corrected mass flow rate can be written as I
rnc2 _ rrc / ( T t 4 / T t 2 ) R l;rtc2R 77"cRV Tt4/Tt2
(8.24)
where the compressor pressure ratio is given by Eq. (8.17), rewritten in terms of Tt4/Tt2, or
~rc =
Tt4/Tt2 . (7c1)/7,.
1 q ( T , 4 ~  , 2 ) R t~'cR

1)
(8.25)
Equation (8.19b) for the corrected speed can be rewritten in terms of Tt4/Tt2 as
Nc2
/ Tt4/Tt2
N~2R  V1 (Tt4/Ttz)R
(8.26)
An expression for the corrected fuel flow rate results from Eqs. (7.9), (8.2), and (8.7) as follows. Solving Eq. (7.9) for the fuel flow rate gives rnf z in 0 cptTt4  cpcTt3 rlbht, l¢  c p t T t 4
From Eqs. (8.2) and (8.7), this equation becomes
l~lc2 cptTt4 cpcTt3 mfc = 02 rlbhpR  cptTt4 •


or &fc = the2
Tt4/rt2  'Tc(Cpc/Cpt) "rlbhpR/ ( CptTref )  Zt4 / Tref
(8.27)
462
ELEMENTS OF PROPULSION
where by using Eq. (8.13) and referencing, ~c is given by %
1 + (~'cR =
1" T t 4 / T t 2 
(8.28)
) (Tt4/Tt2)R
Equations (8.228.28) constitute a set of equations for the pumping characteristics of a simple gas generator in terms of Tt4/Tt2 and reference values. Only Eq. (8.27) for the corrected fuel flow rate has the term Tt4/Tref that is not strictly a function of Tt4/Tt2. The first term in the denominator of Eq. (8.27) has a magnitude of about 130, and Tt4/Tref has a value of about 6 or smaller. Thus the denominator of Eq. (8.28) does not vary appreciably, and the corrected fuel flow rate is a function of Tt4/Tt2 and reference values. In summary, the pumping characteristics of the gas generator are a function of only the temperature ratio T t 4 / T t 2 .
Example 8.2 We want to determine the characteristics of a gas generator with a maxim u m compressor pressure ratio of 15, a compressor corrected mass flow rate of 100 lbm/s at 7~,2 of 518.7°R (sealevel standard), and a maximum Tt4 of 3200°R. This is the same compressor we considered in Example 8.1 (see Figs. 8.12 and 8.13). We assume the compressor has an efficiency r/¢ of 0.8572 (ec = 0.9), and the burner has an efficiency ~?b of 0.995 and a pressure ratio 7rb of 0.96. In addition, we assume the following gas constants: Yc = 1.4, cm. = 0.24 Btu/(lbm. °R), Yt = 1.33, and Cpt 0.276 Btu/ (lbm • °R). By using Eq. (7.10), the reference fuel/air ratio fn is 0.03381 for h p R = 18,400 Btu/lbm, and the corrected fuel flow rate is 12,170 lb/h. From Eq. (7.12), the turbine temperature ratio ~ is 0.8124. Assuming et = 0.9, Eqs. (7.13) and (7.14) give the turbine pressure ratio 7rt as 0.3943 and the turbine efficiency ~/t, as 0.910. The reference compressor temperature ratio ~'cR is 2.3624. Calculations were done over a range of Tt4 with Tt: = 518.7°R and using Eqs. (8.228.28). The resulting gas generator pumping characteristics are plotted in Fig. 8.15. We can see that the compressor pressure ratio and corrected fuel flow rate decrease more rapidly with decreasing corrected speed than corrected airflow rate. As discussed previously, the gas generator's pumping characteristics are a function of only Tt4/Ttz, and Fig. 8.15 shows this most important relationship in graphical form. Because the maximum Tt4 is 3200°R and the maximum pressure ratio is 15, the operation of the gas generator at different inlet conditions (Tta, Pte) and/or different throttle setting (7",4) can be obtained from Fig. 8.15. For example, consider a 100°F day (Tt2) at sea level with maximum power. Here Tt2 = 560°R, P t e = 14.7 psia, and Tt4 = 3200°R; thus Tt4/Tta 5.71, and Fig. 8.15 gives the following data: Nc/N~R = 0.96, mc/lhcR = 0.88, ~'c/~'~R = 0.84, &fc/&fcR = 0.78, Tt6/Tte = 4 . 6 , and Pt6/Pta = 4 . 8 . With these data, the pressures,
ENGINE PERFORMANCE ANALYSIS
463
].0
7.0
0.8
6.0
?h C mcR 0.6

5.0
ml h c
rncR TCc
~cR 0.4
~cR
4.0
mfc
~R ~R
0.2
0 0.75
3.0
0.80
0.85
0.90
2.0 1.00
0.95
Nc/NcR Fig. 8.15
Gas generator pumping characteristics.
temperatures, and flow rates can be calculated as follows:
(Pt2T~)(th~). ~RcRmcR =
1x
t~l = \Pt2R V Tt2 ,] Pt2
Tt/~2mfcl: n
m f = P~VT~2Rm)~~R ~fcR = 1 X
T,~
Tt6 = ~
Tt2 =
1,t2
Pt6
Pt6n = Pt~ rt2 =
~c
4.6(560) =
5 ~ V~
5/~&.7 (0.88)(100) 84.7 lbm/s V 560
(0.78)(12,170)
=
=
9860 lbm/h
2576°R
4.8(14.7) 70.6 psia =
~c =   ~'¢R = 0.84(15) = q'gcR
12.6
As another example, consider flight at Mach 0.6 and 4 0 k f t ( 0 = 0.7519, 6 = 0.1858) with m a x i m u m throttle. Since Tt2 ( = 4 1 8 . 1 ° R ) is less than Tt2R,
464
ELEMENTS OF PROPULSION
the maximum value for Tt4 is 2579.4°R (= 3200 x 418.1/518.7), and the compressor has a pressure ratio of 15 and corrected mass flow rate of 100 lbm/s. The air mass flow rate is reduced to 20.7 lbm/s and the mass fuel flow rate is reduced to 2030 lbm/h.
8.3 Turbojet Engine In this section, the performance equations of the singlespool turbojet engine, shown in Fig. 8.16, are developed and the results are studied. We assume choked flow at stations 4 and 8. In addition, the throttle (Tt4), flight conditions (Mo, To, and Po), and the ambient pressure/exhaust pressure ratio Po/P9 can be independently varied for this engine. The performance equations for this turbojet can be obtained easily by adding inlet and exhaust nozzle losses to the singlespool gas generator studied in the previous section. This engine has five independent variables (T,4, Mo, To, Po, and Po/P9). The performance analysis develops analytical expressions for component performance in terms of these independent variables. We have six dependent variables for the singlespool turbojet engine: engine mass flow rate, compressor pressure ratio, compressor temperature ratio, burner fuel/air ratio, exit temperature ratio T9/To, and exit Mach number. A summary of the independent variables, dependent variables, and constants or knowns for this engine is given in Table 8.4. The thrust for this engine is given by
F /no
ao[(l+f)V9_Mo+(l+f)Rtr9/TolPo/P9] gc ao Rc V9/ao %. J
(i)
where
(ii)
T9 rt4 ~ TO0 = ( P t 9 / P 9 ) (y'I)/yt
Pt9 P9
2

Po P9
(iii)
7rrTTd"B'c3Tb TTt"B"n
3
4
5
Fig. 8.16 Singlespool turbojet engine. (Courtesy of Pratt & Whitney.)
8
ENGINE PERFORMANCE ANALYSIS Table 8.4
465
Performance analysis variables for singlespool turbojet engine
Variables Component
Independent
Engine Diffuser Compressor Burner Turbine Nozzle Total number
M0, To, P0
Constant or known
rho 7rd =
Zt4
f(Mo) "qc
"nc, %
"B'b, ~b
f
~'t, % P9 / Po
M9, Tg/To
"n'n
5
M9 =
Dependent
6
~
L\ ~ g j
j
(iv)
and V9 = M 9 ~ tRtT9 ao V 3'cRcTo
(v)
The thrustspecific fuel consumption for this engine is given by
s
f
F/Fno
(vi)
"cA  "rr'Cc hpR ~?j(cpcTo)  "cA
(vii)
where f =
Equations (ivii) can be solved for given Tt4, Mo, To, Po, Po/P9, gas properties with expressions for ~'A, 7rr, % ~'a, 7rc, %, and engine mass flow rate in terms of the five independent variables and other dependent variables. In the previous section, we developed Eq. (8.28), repeated here, for the compressor's temperature ratios in terms of Tta/Tt2 and reference values:
rc =
1 + (T~R 
1" T,4/Ta ~( ~ R
The compressor pressure ratio is related to its temperature ratio by its efficiency. An equation for the engine mass flow rate follows from the mass flow parameter (MFP) written for station 4 with choked flow and the definitions of
466
ELEMENTS OF PROPULSION
component ~ values. We write
A4_MFP(1)P°~rlra~c[~.bA4MFP(1)] ~/Tt4 1 +J c ~ L1 +j j
rho %
Because the terms within the square brackets are considered constant, we move the variable terms to the left side of the equation, and, using referencing, equate the constant to reference values:
&ov'T~ ffbA4MFP(1) = ( Po"rrr'n'd'n'c
rho~ .~ \ Po ~r'trd ~rc/ R
1 t J
Solving for the engine mass flow rate, we get •
PoTrr'lrdqrc . T ~
mo = moR ( ~ ) ~
V T,4
(8.29)
Relationships for ~'~, 7rr, ~'r, and ~'d follow from their equations in Chapter 7. The throat area of the exhaust nozzle is assumed to be constant. With Po/P9 an independent variable, the exit area of the exhaust nozzle A 9 must correspond to the nozzle pressure ratio Pt9/P9. An expression for the exhaust nozzle exit area follows from the mass flow parameter and other compressible flow properties. The subscript t is used in the following equations for the gas properties (7, R, and F) at stations 8 and 9. Using Eq. (8.11) for choked flow at station 8 gives
Pt9A8
Ft
(i)
~h8 4 ~ R4~7~,/g~ From the equation for the mass flow parameter [Eq. (2.76)], the mass flow rate at station 9 is
Pt9A9 ~ t
M{1
,.9  Wg,9
7t 
1
2"~(%+1)/2(7t1)1
t +sm)
(ii)
Using the nozzle relationships Tt8 = Z,9 and % = Pt9/Pt8 and equating the mass flow rate at station 8 [Eq. (i)] to that at station 9 [Eq. (ii)] give A9
A8
Ft
1 1
,/g, ~'.M9
1
"}st  l
+~M~)
2"~ (%+1)/[2(%1)]
ENGINE PERFORMANCE ANALYSIS
467
Replacing the Mach number at station 9 by using
M9 ~ T t  ~ [(Pt9/P9)(~"l)/3't 
l]
gives A9 = Ft
2 1 1 .
A8
.
.
(Pt9/P9) (z',+1)/(23',) .
.
"Yt 77"nv/(Pt9/P9) (%l)/y'  1
(8.30)
Because the throat area A8 is constant, Eq. (8.30) can be used to obtain the ratio of the exit a r e a A 9 to a reference exit a r e a A9R that can be written as
A9 A9R
_
_
rPtg/e9 .] ('Yt+I)/(2'yt) /(Pt9/P9)(R7'I)/~ I [(Pt9/P9)RJ V(et9/P9) (y,1)/y, i
(8.31)
8.3.1 Summary of Performance Equations~SingleSpool Turbojet Without Afterburner INPUTS: Choices Flight parameters: Throttle setting: Exhaust nozzle setting: Design constants '71": 'T:
ri: Gas properties: Fuel: Reference conditions Flight parameters: Throttle setting: Component behavior:
Mo, To (K, °R), Po (kPa, psia) Tt4 (K, °R) Po/P9 "/Tdmax , 7rb, 3Tt, "B"n
zt ric, rib, rim
%, Yt, Cpc, Cpt [kJ/(Kg. K), Btu/(lbm. °R)]
hpR, (kJ/kg, Btu/lbm) MoR, ToR (K, °R), PoR (kPa, psia), ~'rR, "fi'rR Tt4R (K, °R) TdR, 7rcR, %R
OUTPUTS: Overall performance:
F(N, lbf), rno (kg/s, lbm/s),f,
,(rag, lbm . lbf ] ' rip, rir, rio Component behavior:
"n'd, 7rc, "rc,f, M9, N/NR
468
ELEMENTS OF PROPULSION
EQUATIONS: Rc
Tc  1 Cpc Yc
(8.32a)
Rt  Yt  1 Cpt 3',
(8.32b)
ao = v / ~ R c g c To
(8.32e)
V 0 ~ aoM 0
(8.32d)


rr= l}~M "/Jr =
2
(8.32e) (8.32f)
T%'/(TcI)
T~r = 1
for Mo < 1
r b = 1  0.075(Mo  1) 135 ,/1"d ~
(8.32g) for Mo > 1
(8.32h) (8.32i)
7Tdmax T~r
(8.32j)
Tt2 = ToT r
Tt4/T,2 (T,4/Ta)R
(8.32k)
~'c = [1 + ~c(rc  1)] ~/(~'1)
(8.321)
%=l+(rcR1)
Thf=
cptTt4
(8.32m)
cpcTo ra Trrc
Porrrrrdrrc mo = moR (
Pt9
Po
P9  P9
T9
(8.320) (8.32p)
Tl'rTi'dTTc71"bTTt3Tn
Pt9 (Ytl)/%
Tt4TtlTo
To  (Pt9/Po)(%l)/~/, V9 = M9
. T~ V
2
ao
(8.32n)
hpR rlb/(CpTO)  ,ra
~f~RI T9
V %1% TO
.q
(8.32q)
(8.32r)
(8.32s)
ENGINE PERFORMANCE ANALYSIS
F
_
[
V9 
ao (1 + f )  
ino
gc
F=
ao
469
Rt T 9 / T o l  Po/P9] V9/ ao Yc ..1
Mo + (1 +f)~~
rho ( ~ o )
S
(8.32t)
(8.32u)
f
F/ino
(8.32v)
a°2[(1 + f)(V9/a°)2  M2]
*IT =
(8.32w)
2gcfhpR 2gcVo(F/tho)
*1e = ao2[(1 + f)(Vo/ao) 2  M~]
(8.32x)
'1o = *1P*1r
(8.32y)
N
(8.32z)
/ To'rr Or(z,.1)/'/c _ 1
~R = VTORTrR1)/'''~c Or(c~  1 A9 _ F Pt9/e9 ] (yt+I,/(2yt, / (Pt9/P9)(Ryt1)/yt 1 A9R L ( ~ R I V~ ~
(8.32aa)
Example 8.3 We consider the performance of the turbojet engine of Example 7.1 sized for a mass flow rate o f S 0 k g / s at the reference condition and altitude of 12 kin. We are to determine this engine's performance at an altitude of 9 km, Mach number of 1.5, reduced throttle setting (Tt4 = 1670°R), and exit to ambient pressure ratio (Po/P9) of 0.955. REFERENCE: To=216.7K,
%=1.4,
Cpt = 1 . 2 3 9 kJ/(kg. K), % = 2.0771, "/Tdmax = *1b =
0.98,
.1c = 0.8641,
ort = 0.8155,
3Td = 0.8788,
0.95,
*1m = 0.99,
f = 0.03567,
Cpc=1.004kJ/(kg.K), Ttg = 1 8 0 0 K, M o = 2 ,
Pt9/P9 = 11.62,
S = 44.21 (mg/s)/N,
~rc=10
ort = 0.3746
Orb = 0.94,
Po/P9 = 0.5, heR
yt=l.3
Orn = 0.96
= 42,800 k J / k g
F/rho = 806.9 N / ( k g / s )
Po = 19.40 kPa (12 km),
rho = 50 k g / s
F = rho × (F/rho) = 50 x 806.9 = 40,345 N OFFDESIGN CONDITION: To = 229.8 K,
Po 30.8 kPa (9 km),
Po/P9 =
0.955,
Tt4 =
1670 K
Mo  1.5
470
ELEMENTS OF PROPULSION
EQUATIONS:
Rc  %  1 Cpc = ? . 4 ( 1 . 0 0 4 ) = 0 . 2 8 6 9 k J / ( k g • K ) yc 1.4 %1 ~
Rt
c et =
~_~ . (1.239) = 0.2859kJ/(kg.K)
ao = v/TcRcgcTo = ~ / 1 . 4 x 2 8 6 . 9 x 1 x 2 2 9 . 8 = 3 0 3 . 8 m / s Vo = aoMo = 3 0 3 . 8 x 1.5 = 4 5 5 . 7 m / s ~'r =
1 + ~ @  ~ M 2  1 + 0.2 x 1.52  1.45
~'r = Tr?C/(Yc1) = 'Or ,B'd~
TA=
1  0 . 0 7 5 ( M o  1) 135 = 1  0 . 0 7 5 ( 0 . 5 ) 135 0 . 9 7 0 6
Zramax'or = 0.95 x 0 . 9 7 0 6 = 0 . 9 2 2 0 cptTt4 = 1 . 2 3 2 9 x 1670 = 8 . 9 6 8 2
cpcTo
1.004 x 229.8
T,2= To~'r = TrR
1.453.5  3 . 6 7 1
1
2 2 9 . 8 × 1.45 = 3 3 3 . 2 K
+ ~~M2R
1 + 0 . 2 x 2 2 = 1.80
=
Tt2 R
TORTrR = 2 1 6 . 7 x 1.8 = 3 9 0 . 1 K
7TrR
=.y,./(.y,.I) rrrR
Tc=
=
1 + (~'cR  
1.83.5
1)
~
7.824
Tt4/Tt2 (Tt4/Tt2)R
1 + ( 2 . 0 7 7 1  1)
1670/333.2 1800/390.1
 2.170
7rc= [1 + "oc(~'c  1)] z'/(r'l) = [1 + 0 . 8 6 4 1 ( 2 . 1 7 0 
1 ) ] 3.5 =
11.53
TA   Tr"t"c
f=
heR'oJ(cpcTo) ~'~ 8.9682
1.45 x 2 . 1 7 0
42,800 x 0.98/(1.004 Pt9
Po
P9
P9
= 0.03368
x 2 2 9 . 8 )  8 . 9 6 8 2
7rr T'd.'B'c,'lrb T't,'lr n
0 . 9 5 5 x 3.671 × 0 . 9 2 2 0 x 11.53 × 0 . 9 4 x 0 . 3 7 4 6 × 0 . 9 6 = 12.60
M9 = ~ y t ~ [ ( P t 9 / P 9 ) ( z "  l ) / ~ "  l ] = } / O ~ ( 1 2 . 6 0 ° ' 3 / l " 3  1 ) = 2 . 3 0 1
ENGINE PERFORMANCE ANALYSIS
471
T9 ~'xrt Cpc 8.9682 × 0.8155 1.004   3.303 Too = (Pt9/P9) (y'I)/y' Cpt 12.60 °.3/13 1.239 V9_M9 ~ ao 
~/~o
4.023
+f)V9M 0 I(1 kf)~cc T9/T° 1P°/P9]
F _ a o [(1 mo gc = 303.8
= 2.301~/1~.'] x 285"9 . ~~(3.303)=
ao (
V9/ao
Yc
_1
285.93.3030.045"] 1.03368 × 4.023  1.5 + 1.033682ff~.94.b~ ~  ]
= 303.8(2.6585 + 0.0272) = 8 1 5 . 9 N / ( k g / s ) S =
f
0.03368 × 106
F/;no
815.9
= 41.28 ( m g / s ) / N
Po qT"r'n'd'n"c . T ~ ;no = ;noR (Po'rrr~c)R V T,4
;no=50
30.8 × 3.671 × 0.9220 × 11.53 1~/]~0/____ 46.78kg/s 19.4x7.824×0.8788× 10 V1670
F F = ;no7 = 46.78 × 815.9 = 3 8 , 1 7 0 N
mo
a2[(1 + f ) ( V 9 / a o )
riT" = =
2 
M 2]
2gcfhpR 303.82[(1.03368)(4.0232)  1.52] 2 × 1 × 0.03368 × 42,800 × 1,000
= 46.36%
2g~Vo(F/;no) rip  a2[(1 + f)(V9/ao) 2  M~] =
2 x 1 x 455.7 × 815.9 = 55.64% 303.82[(1.03368)(4.0232)  1.52]
rio = "qeri~r = 0.4635 x 0.5564 = 25.79%
N = / Tol"r __TT (%I)/ye  1 NR VToRI"rR~(c~I)/Y~ 1 J 2 2 9 . 8 x_ 1.45 11.53°'4/1"451 = V 216.7 × 1.8 10 0"4/1"4  1 = 0.9278
;no2 ~ /(Tt4/T,z)R 11.53/1800/390.1 ;nc2~  ~cRV Tt4/T,2  ~ W ~ 
1.106
472
ELEMENTS OF PROPULSION
A9  F et9/e91 A9R L(~)RJ
/ (et9/P9)(R%l)/yt 1
V
~
i
{12.60"~2"3/1"3J11.62°'3/1"31
A9 A9R  ~11'~/]
V
~
"1= 1.052
Example 8.4 Consider a turbojet engine composed of the gas generator of Example 8.2, an inlet with 7rdmax 0.99, and an exhaust nozzle with .wn = 0.99 and Po/P9  1. The reference engine has the following values. REFERENCE: To 518.7°R,
Yc = 1.4,
Cpc 0.24 B t u / ( l b m . °R),
Cpt = 0 . 2 7 6 B t u / ( l b m . °R), ~'c  15,
r/c 0.8572,
7rdmax 0.99, r/m 0.99,
Pt9/P9 5.5653,
Tt4 = 3200°R, ~ 0.8124,
~'b = 0.96,
Po/P9  1,
7r, = 0.99,
Yt = 1.33
M0 = 0
7rt 0.3943 ~Tb 0.995
P0 = 14.696 psia (sea level)
rn0 = 100Ibm/s,
F/&o = 113.421bf/(lbm/s)
F &o x (F/&o) 100 x 113.42 = 11,3421bf This engine has a control system that limits the compressor pressure ratio 7rc to 15 and the combustor exit total temperature Zt4 to 3200°R. Calculation of engine performance using Eqs. (8.32a8.32aa) with full throttle at altitudes of sea level, 20 kft, and 40 kft over a range of flight Mach numbers gives the results shown in Figs. 8.178.22. Note the breaks in the plots of thrust, engine mass flow rate, compressor pressure ratio, and station 2 corrected mass flow rate at a M a c h / altitude combination of about 0.9/20 kft and 1.3/40 kft. To the left of these breaks, the combustor exit temperature Tt4 is below its m a x i m u m of 3200°R, and the compressor pressure ratio ~'c is at its m a x i m u m of 15. To the right of these breaks, the combustor exit temperature Tt4 is at its m a x i m u m of 3200°R, and the compressor pressure ratio ~'c is below its m a x i m u m of 15. At the break, both the compressor pressure ratio and combustor exit temperature are at their m a x i m u m values. This break corresponds to the engine's theta break of 1.0. The designer o f a gas generator's turbomachinery needs to know the maximum power requirements o f the compressor and turbine. Because the turbine drives the compressor, the m a x i m u m requirements of both occur at the same conditions. Consider the following power balance between the compressor and turbine:
ENGINE PERFORMANCE ANALYSIS
473
14,000 12,000 10,000 20 k f t ~ 8,000 6,000
40 k f t ~
4,000 2,000 0
I
I
I
I
I
0.4
0.8
1.2
1.6
2.0
Mo Fig. 8.17
Maximum thrust F of a turbojet vs M0.
1.6 
1.5
1.4
~
1.3
1.2
1.1
1.0 0
I
I
I
I
I
0.4
0.8
1.2
1.6
2.0
MO
Fig. 8.18 Thrustspecific fuel consumption S of a turbojet vs Mo.
474
ELEMENTS OF PROPULSION 200
160
120
.~, 80 :
40
0 0
i
i
I
I
J
0.4
0.8
1.2
1.6
2.0
MO F i g . 8.19
E n g i n e m a s s f l o w r a t e o f a t u r b o j e t vs Mo.
16
15
\
14 
SL
!
13 12 11 10 9 8 0
i
i
I
I
I
0.4
0.8
1.2
1.6
2.0
mo Fig. 8.20
C o m p r e s s o r p r e s s u r e r a t i o o f a t u r b o j e t vs M o .
ENGINE PERFORMANCE ANALYSIS
475
105 100
\
95 90 85 80 75 70 65 0
I
I
I
I
I
0.4
0.8
1.2
1.6
2.0
MO
Fig. 8.21
C o m p r e s s o r corrected m a s s flow rate of a turbojet vs Mo.
0.8 0.7 0.6
20 kfl 40 kit.  ~" ~ ",~'Okft
SL
0.5 T/T SL
rl r
0.4 ~, " t *" 0.3
40 kit
¢..~
Oe
SL
/
0.2 rl o
0.1 0 0
0.4
I 0.8
I 1.2
I 1.6
MO
Fig. 8.22
T u r b o j e t efficiencies vs M0.
I 2.0
476
ELEMENTS OF PROPULSION
Rewriting turbine power in terms of its mass flow rate, total temperatures, etc., gives Wc  "0mrh0(1 + f ) C p t ( T t 4  Tt5) or (Vc = t:rtoTt4['rlm(1 + f ) c p t ( 1  zt)]
Because the terms within the square braces of the preceding equation are considered constant, the maximum compressor power will be at the flight condition having maximum engine mass flow rate at maximum Tt4. From Fig. 8.19, the maximum compressor or turbine power corresponds to the maximum engine mass flow rate at sea level and Mach 1.4. At an altitude of 20 kft and a Mach number of 0.8, engine performance calculations at reduced throttle (Tt4) using Eqs. (8.32a8.32aa) were performed, and some of these results are given in Fig. 8.23. The typical variation in thrust specific fuel consumption S with thrust F is shown in this figure. As the throttle is reduced, the thrust specific fuel consumption first reduces before increasing. This plot of thrust specific fuel consumption S vs thrust F is commonly called the t h r o t t l e h o o k because of its shape. We stated at the beginning of this chapter that the principal efficiencies that affect engine performance are the thermal efficiency and the propulsive
1.20
0.7
1.15
0.6
1.10
0.5
Oe 0.4
1.05
or
7" 0.3
1.00
0.2
0.95
f 0.90 0
no I 0.2
I 0.4
qo
I 0.6
I 0.8
F /F R
Fig. 8.23
Turbojet performance at partial throttle.
0.1 1.0
ENGINE PERFORMANCE ANALYSIS
477
efficiency. Figure 8.23 shows the very large changes in both propulsive and thermal efficiency with engine thrust. Note that as thrust is reduced from its maximum, the increase in propulsive efficiency more than offsets the decrease in thermal efficiency such that the overall efficiency increases and the thrust specific fuel consumption decreases until about 40% of maximum thrust. Below 40% thrust, the decrease in thermal efficiency dominates the increase in propulsive efficiency and the overall efficiency decreases, and the thrust specific fuel consumption increases with reduced thrust.
8.3.2 Corrected Engine Performance The changes in maximum thrust of a simple turbojet engine can be presented in a corrected format that essentially collapses the thrust data. Consider the thrust equation for the turbojet engine as given by F = m°[(1 +f)V9  V0] g¢ where V9 = ~/2gccptTt4"rt[1  ('Trr'rrdTTcTTb TTt'rrn) ( yt1)/3"t] and Vo = Moao = Mov/%cRcTo Note that the engine mass flow rate is related to the compressor corrected mass flow rate by mo = m~o = ~
~0
= ,hc~ ~
~2
~d~0
= mc~ ~ o o
The engine thrust can now be written as F =
thc2 "B'd(% I'" .,/~n ttl qf)V9  Vo] gc ~/ tlo
Dividing the thrust by the dimensionless total pressure at station 0 gives F_/nc2"n'd[(l+f)~ooo ° 60 g~
Vo] ~o
(8.33a)
where
V9 ,/~

rff2., J ~?~x/2gccpt Tref'rt[1  7"rrTrdT"rc'lrb~t Trn)( ~' l )/ z' ] ¥1t2
(8.33b)
ELEMENTS OF PROPULSION
478 and
V0
4N
M0 ,/'rr
= ~aSL
(8.33c)
The maximum thrust for the turbojet engine of Example 8.4 can be determined by using the preceding equations. Figures 8.17, 8.20, and 8.21 show the variation of the maximum thrust F, compressor pressure ratio, and corrected mass flow rate from this turbojet engine at full throttle vs flight Mach number Mo. The corrected thrust F/6o of this engine is plotted vs flight condition 00 in Fig. 8.24. The variation of Tt4/Tt2, compressor pressure ratio, corrected mass flow rate, and corrected fuel flow rate are plotted vs 00 in Fig. 8.25. The representation of the engine thrust, as corrected thrust vs 00, essentially collapses the thrust data into one line for 00 greater than 1.0. The discussion that follows helps one see why the plot in Fig. 8.24 behaves as shown. When 00 is less than 1.0, we observe the following: 1) The compressor pressure ratio is constant at its maximum value of 15 (see Fig. 8.25). 2) The compressor corrected mass flow rate is constant at its maximum value of 100 lbm/s (see Fig. 8.25). 3) The value of Tt4/Tt2 is constant at its maximum value of 6.17 (see Fig. 8.25). 4) The corrected exit velocity given by Eq. (8.33b) is essentially constant. 12,000 11,000 10,000
9,000 t~
8,000 7,000 6,000 5,000 4,000 0.8
Fig. 8.24
I
0.9
1.0
1.1
M a x i m u m corrected thrust
1.2
1.3
(F/8o) of a turbojet
1.4 vs Oo.
ENGINE PERFORMANCE ANALYSIS
479
1.10
1.0
0.9
•
0.8
~~t2/(Tt41Tt2)R ~ ~cR
0.7
0.6
0.5
0.4 0.8
I 0.9
I
I
I
I
1.0
1.1
1.2
1.3
~rc ~:cR I 1.4
0o Fig. 8.25
Maximum throttle characteristics of a turbojet vs 0o.
5) The corrected flight velocity [Eq. (8.33c)] increases in a nearly linear manner with Mo. 6) The corrected thrust [Eq. (8.33a)] decreases slightly with increasing 0o. When 00 is greater than 1.0, we observe the following: 1) The compressor pressure ratio decreases with increasing 00. 2) The compressor corrected mass flow rate decreases with increasing 00. 3) The value of Tt4 is constant at its maximum value of 3200°R. 4) The corrected exit velocity given by Eq. (8.33b) decreases with increasing 0o. 5) The corrected flight velocity [Eq. (8.33c)] increases in a nearly linear manner with Mo. 6) The corrected thrust [Eq. (8.33a)] decreases substantially with increasing 00. As shown in Fig. 8.24, the trend in maximum corrected thrust F/6o of this turbojet dramatically changes at the 00break value of 1.0. Both the compressor pressure ratio 7rc and combustor exit temperature Tt4 are at their maximum values when 00 is 1.0. The engine control system varies the fuel flow to the combustor to keep ~'c and Tt4 under control. The control system maintains 7re at its maximum for 00 values less than 1.0, and Tt4 at its maximum for 00 values greater than 1.0. These same kinds of trends are observed for many other gas turbine aircraft engines.
480
ELEMENTS OF PROPULSION 1.6 0 kfi
1.5
1.4 " 1.3
1.2
1.1
1.0 0.0
I
I
I
[
I
0.4
0.8
1.2
1.6
2.0
M0
Fig. 8.26
S/V~of a turbojet
vs Mo.
The thrustspecific fuel consumption S of this turbojet at maximum thrust is plotted vs Mach number in Fig. 8.18. If the values of S are divided by the square root of the corrected ambient temperature, then the curves for higher altitudes are shifted up and we get Fig. 8.26. Note that these curves could be estimated by a straight line. Equations (1.36a1.36f) are based on this nearly linear relationship with flight Mach number M0. When the corrected thrustspecific fuel consumption [Sc, see Eq. (8.8)] is plotted vs 00, the spread in fuel consumption data is substantially reduced, as shown in Fig. 8.27. One could estimate that the corrected thrust specific fuel consumption has a value of about 1.24 for most flight conditions.
8.3.3
Throttle Ratio
The throttle ratio (TR) is defined as the ratio of the maximum value of Tt4 to the value of Tt4 at sealevel static (SLS) conditions. In equation form, the throttle ratio is
TR

(Zt4)max (T~4)SLS
(8.34a)
The throttle ratio for the simple turbojet engine and compressor of Figs. 8.178.27 has a value of 1.0. Both the compressor performance and engine performance curves change shape at a 00 value of 1.0. This change in shape of the performance curves occurs at the simultaneous maximum of ~'c and Tt4. The fact that both the throttle ratio and dimensionless total temperature
ENGINE PERFORMANCE ANALYSIS ].4

1.3
~
481
40 kft
1.2
1.1
1.0
0.8
I
I
I
I
I
I
0.9
1.0
1.1
1.2
1.3
1.4
0o Fig. 8.27
S/~o
of a turbojet vs 0o.
00 have a value of 1.0 at the simultaneous maximum is not a coincidence but is a direct result of compressorturbine power balance given by Eq. (8.17). At the simultaneous maximum of ~'c and Tt4, the throttle ratio equals 00: TR = 0o
at max 7re and max Tt4
or
TR = 00break
(8.34b)
Highperformance fighters want gas turbine engines whose thrust does not drop off as fast with increasing 00 as that of Fig. 8.24. The value of 0o, where the corrected maximum thrust F/~o curves change slope, can be increased by increasing the maximum Tt4 of the preceding example turbojet engine.
Example 8,5 Again, we consider the example turbojet engine with a compressor that has a compressor pressure ratio of 15 and corrected mass flow rate of 100 lbm/s for Tt2 of 518.7°R and Tt4 of 3200°R. The maximum 7rc is maintained at 15, and the maximum Tt4 is increased from 3200 to 3360°R (TR = 1.05). The variation in thrust, thrustspecific fuel consumption, compressor pressure ratio, and corrected mass flow rate of this turbojet engine at full throttle are plotted vs flight Mach number M0 in Figs. 8.28, 8.29, 8.30, and 8.31, respectively. Figure 8.32 shows the corrected thrust F / ~ plotted vs 00. Comparing Figs. 8.17 and 8.28, we note that the thrust of both engines are the same at sealevel static, and the engine with a throttle ratio of 1.05 has higher thrust at high Mach numbers. Figures 8.24 and 8.32 show that changing the throttle ratio from 1.0 to 1.05 changes the 00 value at which the curves change shape 00break and increases the corrected thrust at 00 values greater than 1.0.
482
ELEMENTS OF PROPULSION 16,000 14,000 12,000 10,000 8,000 6,000 4,000 2,000 0 0
I
I
I
I
I
0.4
0.8
1.2
1.6
2.0
MO Fig. 8.28
M a x i m u m thrust F of i m p r o v e d turbojet vs M0.
1.6 S~"/
1.5
20 kfi,,,
1.4
:,.
1
1 ) 1'35
(8.45i)
Tt4/T°
"CcL= 1 +

(TMTo)R
(Tr)~R (TcL 
1)R
(8.45j)
rr
~'cL = [1 + "OcC(~'cL  1)ff 1.0. The decreases in engine thrust, fuel consumption, and air mass flow rate with altitude are shown in Fig. 8.55 for a flight Mach number of 0.5. The decrease in engine thrust with altitude for the highbypassratio turbofan engine is much greater than that of the dry turbojet (see Fig. 8.42). If both a highbypassratio turbofan engine and a dry turbojet engine were sized to produce the same thrust at 9 km and 0.8 Mach, the highbypassratio turbofan engine would have much greater thrust at sealevel static conditions. This helps explain the decrease in takeoff length between the
ENGINE PERFORMANCE ANALYSIS
513
Alt (km) 22
0 .   1.5
.T~.5 3 6  7.6
,n
20
z ~ z ~
9
z

1l+
18
% 16
z / //
,._.,
14
.,'///// 12
0
I
0.0
0.2
I
I
0.4
I
0.6
I
0.8
1.0
M0
Fig. 8.48 Thrustspecific fuel consumption of highbypass.ratio turbofan at maximum thrust.
1200
1000
800
....
600
~
SL 1.5 km 3 km 4.5 km 6 km 7.Skin
9km
400
~
200
I
0.0
~ l l k m 12km
0.2
I
I
0.4
0.6
I
0.8
I
1.0
MO
Fig. 8.49
Mass flow rate of highbypassratio turbofan at maximum thrust.
514
ELEMENTS OF PROPULSION 850
1.5 km 800 SL
" .. "..
~~750
\
~'x
\6+km
\X N\
\
700 \\
600
I
0.0
Fig. 8.50 thrust.
0.2
I
0.4
MO
\ \ \ 1 . 5 km
I
I
0.6
0.8
SL
I
1.0
C o r r e c t e d mass flow rate of highbypassratio turbofan at maximum
10.0
9.5
SL
9.0
.l.5km
/7
//
// 8.5
~
//
3km
~
8.0
6+km
7.5
7.0 0.0
Fig. 8.51
i 0.2
t 0.4
MO
i 0.6
i 0.8
t 1.0
Bypass ratio of highbypassratio turbofan at maximum thrust.
ENGINE PERFORMANCE ANALYSIS
515
1.90
1.85
1.80
~"
1.75
""
6+km
r9 1.70 1.65 
"""\
3km
1.60 .5 km 1.55 ~ 1.50
Fig. 8.52
0.0
SL
I
I
I
I
0.2
0.4
0.6
0.8
Mo
I
1.0
Fan pressure ratio of highbypassratio turbofan at maximum thrust.
23 22
f
1.5 km _
21 ~
SL
" ... .
.
~
6+ km
20 ncZe19
SL ~
~
~
\
4.5 km
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"(6Z "Jail aa]ju) aSgis au!qanl gOBl~!XV
g9"6 "~hl
~n
;n
V l~OJ13
sn[nuu v £
8£
JOlO~
I
8~
:uo!lelS
JOll!lS
6~9
,,kl:l =1N I H O V l t l O 8 I::1FI_L
630
ELEMENTS OF PROPULSION
By using the stage loading parameter ~b, Eq. (ii) can be written as a~r /~3  cos 0/2 tan 0/3 sin 0/2 = ~bV22 u2
(9.103)
The velocity at station 2 can be found from
•I
2gccpTt2
~ ~.~21 I rz/tt ~'  ~X~)~v*21
V2 = M2a2 =
,
(9.104)
If 0/3 = 0, then Eq. (9.103) simplifies to (or
sin 0/2 = ~ ' ~ 2
(9.105)
If 0/3 is not zero, Eq. (9.103) can be solved by substituting x/1  s i n 2 0/2 for cos 0/2, squaring both sides of the equation, and solving the resulting quadratic equation for sin 0/2. The solution is
(o~r)
)~1
0~22
~7 tan 0/3
sin 0/2 =
+
(U3tan \2 [ °9r\2 0/3)  ~ 0 ~ 2 )
\u2 1 + (u3 tan 0/3] 2 \u2 /
(9.106)
The velocity at station 3 can be written in terms of that at station 2 and the two flow angles 0/2 and c~3: V3  u3 cos a_.____V2 _~2
(9.107)
U2 COS 0/3
The degree of reaction can be written in terms of the given data as follows:
°Rt  
T2 
T3
Tt2 
Tt3
=1
Tt2 
Zt3 
Tt2 
V~V~ =1 2gccp(Ta  T~3) 1
= 1
°Rt =
(Tt2  Z2) [ Tt3  T3
2qc(~or) 2 ~ c0S2 0/2 1
V2 2 .
Tt3
V~V~ 2~or) 2
A C0S2 0/3}
1  ~~ (~rr) [1 2 \(u3 ~] c°s ~  j a2] .]
(9.108)
TURBOMACHINERY
631
The Mach number at station 3 can be found from
. v3 r72 M3 =/I//2 ~22 ~/~33
(9.109)
where T3 1 ~22 = 
Rt~t2
1+
M2
(9.110)
An equation for the Mach number at station 2R can be developed as follows: V2R MZR = M 2  V2 where V2R = ~/U 2 "Or (V2 I 09r) 2 = V2
os 2
a2 +
(
sin a2  V2,/
Thus
M2R = M2
os2 c~2 +
sin oe2
V2//
(9.111)
Likewise, an equation for the Mach number at station 3R is developed as follows: V3R
M3R = M3  
V3
where
V3R=~/u~+(v3+o~r) 2=V3
os e a 3 +
sina3+V3j
Thus
M3R = M3
Jc
os 2 a3 +
t
sin a3 + V3,]
9112,
An equation for the rotor relative total temperature (TaR = Tt3n) can be developed by noting that
T3 = Tt3
V2  T,3R 2gcCp
Then Tt3R = T,3 ~
2gcCp
V2R 2gcCp
ELEMENTS OF PROPULSION
632 or
Tt3n=Tt3+
[
V~ cos 2013+ 2gccp
sin 0 1 3 + ~
1
1
(9.113)
9.5.3 Summary of EquationsAxialFlow Turbine Stage INPUTS:
Ttl, Tt3, oJr, etl, M1, M2, 011, 013, Cp, y, u3/u2, and et
o r ~tstator
and ~btrotor
OUTPUTS:
012,V2, u2, v2, T2, et2, P2, M2R, V3, "3, I)3, T3, Pt3, P3, M3, M3R, ~b,VR, °gt, Zscx/S, Zrcx/S, 7rs, and ~s EQUATIONS: T1 =
Ttl
1 + [(~,  1)/2]M~
2gccpTtl 1 + 2 / [ ( y  1)M~]
VI = U1 ~
V 1 COS 01l
Tt2 = Ttl T2
Tt2 1 + [ ( y  1)/2]M22 ~l
Vz =
2gccpTt2 + 2/[(y  1)M221
qJ = g ~ c p ( T .

r~3)
(mr) 2 W R w
m
u3 tan wr _ ~2
O/2 = s i n  1
q'~
013 1 +
U3tan013 \.2 :
1 + (u3 tan a3"~ 2 \u2 / u2 = V 2 c o s 012
v2 = V2 sin 012 V3 _ " 3 c o s ot 2 V2 " 2 c o s 013
~b~22
TURBOMACHINERY
633
t/3 ~ V 3 COS ot 3
v3 = V3sin ce3
1(V2"]2[
°R,= 1~\wrr/
(u3 cosc~2~2] 1,,.u2cosc~3/ _J
T3  7"2  °Rt(Ttl  Tt3)
V2VT3 M2R=342 7c os2a2+ ( sin~2 °gr)
V2} 2
M3R z
M37cos2a3 + (sin o~3+ ~) 2
Tt3 R z
Tt3+~
cos 2oO+ sinc~3+V3]
Tt2 R ~ Tt3 R
P2 Ptl ik~tl.] 7,3 ,Ts~T,1 Zscx _ (2cos2 0(2) tanc~l +U2tancz2 S
U1
\U2fl
ffi2 ~ t a n  1 F2  (.or /,/2
/33 = tan_ 1v3 + oJr U3
Zrcx
~ 3 ) ( t a n ~ 2 + U 3 t2a n ~ 3 )
I. (~rstator a n d ~btrotor
\U3/
given:
Pt2 =
1 'i' ~tstator[1

Ptl (T2/Tt2) y/(yI)] 
[ T2 \ ~,/(z, I)
P2 = Pt2t~t2 )
1
634
ELEMENTS OF PROPULSION
['Tt2R"~ 7/(71) Pt2R = P 2 ~ ) Pt3R 
Pt2R 1 + ~btrotor[1 (T3/Tt3R) 7/(yl)] 
( T 3 ) y/(y1) P3 "=et3R ~t3R (Tt3"~ 'y/(y1) e t3 "="P3 \'T33.] Pt3 ,irs~Ptl "rlt
1   rs 1  ~.~y1)/y
II. et given:
//Tt 3"~3'/[(7 l)etl Pt3=Ptl~Ttl ) "ITs~ 
Pt3 Ptl
r/~  1

1  "rs "IT~yI)/v
["T3"~y/(y1) I'Tt3R'~ Y/(Y1) Pt3R=Pa~w ) With polytropic efficiency specified, P2, Pt2, and Pt2R cannot be calculated without an additional relationship for either P2 o r Pt2. For estimation of Pt2, the program T U R B N has as the user input a value of tht stator.
Example 9.10 Consider meanradius stage calculationflow with losses. Given: Trl
~
1850 K,
Tt3 = 1560 K,
Ptl = 1700 kPa, M2 = 1.1,
u3/u2 = 0.9, 3' = 1.3,
tor=
(])tstator =
M1 = 0.4,
450 m/s,
0.06,
al=0deg o13 = 1 0 d e g
thtrotor= 0.15
R = 0.2873 k J / ( k g . K) [% = 1.245 k J / ( k g . K)]
TURBOMACHINERY
635
Solution: Ttl 1850 K = 1806.6 K 1 + [ ( y   1)/2]Ml2  1 + 0.15 x 0.4 2
T1
V~.
gI
!
2gccpT,1
/2
1 + 2 /  ~     1 ) M 2] = V
x 1 x 1245 × 1850 1 + 2 / ( 0 . 3 × 0 . 4 2)
328.6m/s
=
VI cos 0/1 = 328.6 m / s
U1
vl = Vj sin 0/1 = 0 Tt2 = Ttl ~ 1 8 5 0 K
T2=
V2 =
=
Tt2
1 +[(3'
1850K
= 1565.8 K
1)/2]M22 = 1 + 0 . 1 5 × 1.12
2gccpTt2 1 + 2 / [ ( 7  1)M 2]
2 × 1 x 1 2 4 5 × 18 i7.15) = 8 4 1 . 2 m / s
~¥~Ao77
q, = &cp(Tn 
Tt3) = 1245(1850  1560)
(a~r) 2
1 VR = v~
=
= 1.78296
1  0.5296 ~/2 x 1.78296
() wr
0/2 ~sinI
4502
I/tg22
_ (u3 t a n 0 / 3 ) ¢ l + \u 2
(U3tan0/3~  ( ~ 0 ~ 7 ~ 'ku2
,/
\
l + (U3 tana3] 2 \U 2 /I (.or
U3   t a n 0/3 u2 0/2 ~
1 . 7 8 2 9 6 ( ~ ) 0 . 9 5=3 7 9 . 0.9 tan 10deg = 0.15869 sin_ 1 0.95379  0.15869x/1 + 0.158692  0.953792 1.0158692 sin l 0.87776 = 61.37 deg
U2 = g 2 c o s 0/2 = 8 4 1 . 2 c o s 6 1 . 3 7 deg = 4 0 3 . 1 m / s
v2J
636
ELEMENTS OF PROPULSION
V2 =
V2
_ u2 wr V3
sin a 2
~
841.2 sin 61.37 deg = 738.3 m/s
403.1___ 0.8958 450 {cos 61.37 COS__a2 V2 ~_ 0.9 ~ ~ deg ) (841.2) = 368.4 m/s
g3   U2 COS 0/3
~OS
u3 = 113cos 0/3 = 368.4cos lOdeg = 362.8 m/s v3 = V3 sina3 = 368.4 sin lOdeg = 64.0m/s
,_±cv=fF,_ 2tp \ w r ] [ = 1
l
\u= cos a3/ J
1 (841.2~2 [ ( cos 61.37deg) 2] 2 x 1.78296\4~6] 1  0.9 c o  ~ =0.2080
T3 = T2  °Rt(Ttl  Tt3) = 1565.8  0.2080(1850  1560) = 1505.5 K
V3 ./r~ 1 1 (368"4~, 15fl ff6~'8 M3 = M2~V~33 = • \84i2]V 15ff~.5  0.4913 M2R = Mz~c°s2 a2 + ( sin0/2  °gr~
=1.1
cos261.37deg+
(
sin 61.37 deg
450) 2 84i~.2/ = 0.6481
2
M3R = M3
0s2 0/3 Jr (\sin 43 + V3]
=0.4913 T,3R
~c
os210deg+
(
450'2 sinlOdeg+3~.4 ) = 0.8390
V2 [ cOs2 a3b(sin0/3+~3) ogr\ 2 l J ] T,3 + 2~cCp = 1560 ~
450\ 2 ] 368.42 cos2 10 deg + (sin 10 deg + 3~.4)  l J 2 x 1 × 1245
= 1664.4 K Tt2R = Tt3R = 1664.4 K Tt3 Ttl
zs . . . .
1560 1850
0.8432
TURBOMACHINERY
Zscx=(2c°s2°z2)(
637
+U2tana2](Ul)2ul / \u2/
 (2 cos2 61.371 deg) ( t a n 0 deg + 403"1 61.37 d e g ) (328"6"] 2 328.6.6 tan \40~_1.1] = 0.6857 /32  tan1
v2  mr U2
t a n  1 738.3  450 403.1  35.57 deg
/33 = tan_l v3 + t o r _ tanI 64.0 + 450 u3 362.8  54.78 deg
Zrcx  (2 c°s2 fl3)( tan
d U3 2 tan fl3)./\l~13/ (u2)
= (2 cos 2 54.78 deg)( tan 35.573 deg + 0.9 tan 54.78 deg) = 1.6330
[TI\ z'/(y1)
P1 = Ptl ~  t l )
=
1700 (1806.6"~ 1"3/°3 \ ~ ] = 1533.8 kPa
Ptl
Pt2
1 t t~tstator[1  (T2/Zt2) y/(y1)] 1700 = 1 + 0.0611  (1565.8/1850) 1'3/°3]
(T2"~ 3'/('~1)
P2 = Pt21,~t2]
( 1 5 6 5 . 8 ) 1"3/0"3 = 1649.1
[TtzR'Xz,/(~l) "t2R = P2~Tf) = et3R =
1649.1 kPa
= 800.5kPa
[ 1 6 6 4 . 4 \ 1.3/o.3 800.5 ~15f6~.8) = 1043.0kPa
Pt2R
1 I t~trotor[1  (T3/Tt3R) "r/(r1)] 1043.0 = 990.6 kPa 1 + 0.1511  (1505.5/1664.4) L3/°3]
T3 ) y/(y1)
P3=Pt3R ~
Pt3 = P3 ~,~3,]
/'1505.5~ 1.3/o.3 = 990.6 ~ 16~'~.4)
= 641.3 kPa
3(156° 13/o3= 748.1 kPa
= 641. \ ~ ]
638
ELEMENTS OF PROPULSION
Pt3 "ITs Ptl
748.1
1700
0.441
1  ~~ 1  0.8432 = 90.87% 'r/s  1  "n'~~'l)/v  1  0.4401 °'3/1"3
9.5.4
Flow Path Dimensions
9.5.4.1
Annulus area. The annulus area (see Fig. 9.26) at any station of a turbine stage is based on the flow properties (Tt, Pt, Mach number, and flow angle) at the mean radius and the total mass flow rate. Equation (9.8) is the easiest equation to use to calculate the flow area at any station i:
Ai = Pti(COSai)MFP(Mi) By using the relationships of Fig. 9.26, the radii at any station i can be determined, given the flow annulus area [Eq. (9.8)] and either the mean radius rm or the h u b / t i p ratio rh/r t.
9.5.4.2 Axial dimensions and number of blades. Figure 9.69 shows the cross section o f a typical turbine stage that can be used to estimate its axial length. The chord/height ratio c/h of turbine blades varies from about 0.3 to 1.0. Assuming constant chord length and circular arc chamber line, the program T U R B N calculates the axial blade widths Ws and Wr of a stage, the blade spacings W~/4 and W~/4, and the number of blades based on user inputs of the tangential force coefficient Z and chord/height ratio c/h for both the stator and rotor blades. A minimum width o f i in. (0.6 cm) and spacing of 81in. (0.3 cm) are used in t h e plot of a turbine cross section and calculation of axial length.
1
2
3
/ Stator ws
~ Rotor
ws
.
Wr
q
4 1
rt
h = rt rh rh
ws=
cosos,
h2 + h3 ~h ~c]r c o s
Wr = ~
Orh
Centerline
Fig. 9.69
Typical axial dimensions of a turbine stage.
TURBOMACHINERY
639
The stagger angle 0 of a blade depends on the shape of the chamber line and blade angles 3'/and % (see Fig. 9.62). For a circular arc chamber line, the stagger angle 0 is simply given by 0 = (%  yi)/2. For constantchord blades, the axial chord (and axial blade width) is greatest where the stagger angle is closest to zero. This normally occurs at the tip of the stator and hub of the rotor blades. For estimation purposes, a turbine blade's incidence angle is normally small and can be considered to be zero. Thus Yi = 0/i. The blade's exit angle Ye can be obtained using Eq. (9.89) for the exit deviation. However, Eq. (9.89) requires that the solidity (~r = c/s) be known. For known flow conditions (0/1, 0/2, U2/bll, 012, 0/3, and u3/u2) and given tangential force coefficients (Zs and Zr), Eqs. (9.98a) and (9.98b) will give the required axial chord/spacing ratio Cx/S for the stator and rotor, respectively. An initial guess for the blade's solidity ~ is needed to obtain the stagger angle 0 from cx/s. After the solidities are determined at the hub, mean, and tip that give the desired tangential force coefficient Z, the number of required blades follows directly from the chord/height ratio c/h, the circumference, and the blade spacing at each radius. The following example shows the calculations needed to find the axial blade width and number of blades.
Example 9.11 Here we consider the turbine stator of Example 9.10 with a mass flow rate of 60 kg/s, a mean radius of 0.3 m, a tangential force coefficient Zs of 0.9, and a chord/height ratio c/h of 1.0. The flow annulus areas and radii at stations 1 and 2 are as follows. Station 1: MFP(M,) = 0.024569 A1 =
rhx/~
60x/q8~ z
PtlMFP(M1)(cos 0/1)
1,700,00 x 0.024569 x 1
= 0.0617788 m 2 A1
hi  27rrm
0.061788 0.6~
rtl = 0.3164m l)lh ~
Vim ~
l)lt ~
0.03278 m
rhl = 0 . 2 8 3 6 m 0
Station 2: MFP(M2) = 0.039042 A2 =
m4~S Pt2MFP(M2)(COS 0/2)
60~vq8K6 1,649,100 x 0.039042 x cos61.37 deg
= 0.083654 m 2
640
ELEMENTS OF PROPULSION h2 .
A2 0.083654 . . . 27rrm 0.67r
rtz = 0.3222m ro
/)2h = / 3 2 m 
r2h
0.04438 m
rh2 = 0 . 2 7 7 8 m
= 738 3
(0.3
~
"\0.2778]
= 797.3 m / s
O~2h= tan 1 v2h = tan_l 797.3 _ 63.18 deg u2 403.1 rm 7 3 8 3 ( 0.3 ~ = 6 8 7 . 4 m / s vet = VZmr2t = " \0.3222] o/2t = tan_l v2t = tan_l 687.4 _ 59.61 deg u2 403.1 The chord of the stator is c 
c hi + h2 1.0 0.03278 + 0.04438 = 0.03858 m h 2 2
For the specified tangential force coefficient Zs, we calculate the stagger angle, solidity, and spacing of the stator at the mean line, hub, and tip. Mean line: Zs(CX)\s/m =
(2c°s2 a 2 m )
122
(
= (2cosZ61.37deg)
×
m
\4o3.1/ 0.9
0
b/1
tan O¢lm q ~~1tan O/2m tanOdeg+
403~
tan 61.37 deg
o.6857
6,9
Vim = O~lm ~ 0
Initially, we assume a solidity o of 1.0. Then, 0 + 84q61.37
")/lm "~ 8 ~ F~mOf2m 
Y2m =
84'~m 1

8X/1 1
Om   "Y2m  Tim __ __70" 14 _ 35.07 deg
2 Oem __ (Cx/S)m __
COS Om
2 0.7619  0.9309 COS35.07 deg
 70.14 deg
)
TURBOMACHINERY
641
After several iterations, the results are Y2m = 70.49 deg, Om= 35.25 deg, and O"m = 0.9329. The blade spacing s is 0.04135 m, and the axial chord is 0.03150 m. Hub:
Zs( cx] =(2cosZa2h)(tanalh +U2tana2h~(Ul~ 2 \ S/h
Ul
I k,U2]
= (2 cos2 63.18 d e e ) ( t a n 0 deg + 403"1~ tan 63.18 d e g )
× (328.q2= 0.6565 \403.1/ (?) h
_ 0.6565__ _ 0.7294 0.9
Tlh ~ ° / l h ~ 0
Initially, we assume a solidity ~ of 1.0. Then, Ylh ~ 8 ~ O / 2 h
Y2h=
8x/'~
8~/163.18 = 72.21 deg 8x/i 1
  0 ~
1

Oh  Y2h  Yl~ _ 72.21__ _ 36.10deg 2
2
crh  (Cx/S)~ _ 0.7294  0.9027 cos Oh cos 361.0 deg After several iterations, the results are 72h = 72.73 deg, Oh = 36.37 deg, and ~rh = 0.9058. The blade spacing s is 0.04259 m, and the axial chord is 0.03107 m. Tip:
Ul
] \g2.]
= (2 cos 2 59.61 deg) _(tan 0 deg + ~403 tan1
(328.q 2
x \4~[.1J = 0.7115 (_~) _ 0.71__15 _ 0.7905 t 0.9 "Ylt ~
OLlt ~ 0
59.61 d e g )
ELEMENTS OF PROPULSION
642
Table 9.13
Location Tip Mean Hub
Summary of Example 9.11 results
Average radius, m
Solidity
Spacing, m
Number of blades
Axial chord, m
0.3193 0.3000 0.2807
0.9555 0.9329 0.9058
0.04038 0.04135 0.04259
49.7 45.6 41.4
0.03192 0.03150 0.03107
Initially, we assume a solidity ~ of 1.0:
Y2t=
Ot
Tit + 8 V c ~ a 2 t 0 + 8,,/]59.61  68.13 deg 8~/~1 8~q1 Y2t  Tit _ 68.13__ _ 34.06 deg 2 2
(cx/s)t 0.7905 ot = cos Ot  cos 34.06 deg  0.9542 After several iterations, the results are Y 2 t = 6 8 . 3 5 deg, Ot = 34.18 deg, and o't = 0.9555. The blade spacing s is 0.04038 m, and the axial chord is 0.03192 m. For this stator blade with a chord of 0.03858 m, we require the information in Table 9.13. Thus the number of required stator blades is 50, which will have a meanradius spacing s of 0.03770 m and solidity o of 1.023 ( = 0.03858/ 0.03770) on the mean radius. The blade has an axial width Ws of 0.03192 m (see Fig. 9.69).
9.5.4.3 Blade profile. The shapes of turbine stator and rotor blades are based on airfoil shapes developed specifically for turbine applications. Two airfoil shapes are included in the program T U R B N to sketch the blade shapes for a stage: the C4 and T6 British profiles. The base profile of the C4 airfoil is listed in Table 9.14 and shown in Fig. 9.70 for a 10% thickness. Table 9.15 and Fig. 9.71 give the base profile of the T6 airfoil for a 10% thickness. The program T U R B N assumes a circular arc mean line for sketching the blade shapes.
_101
I
I
10
t
L
I
20
Fig. 9.70
I
I
30
I
40
I
I
50
I
I
60
I
I
70
I
I
80
The C4 turbine airfoil base profile.
I
t
90
I
I
100
TURBOMACHINERY C4 airfoil profile (t/c
T a b l e 9.14
xl~, ~
643
0,10) a'b
=
y/c, ~
x/c, %
y/c, %
0.0 1.65 2.27 3.08 3.62 4.02 4.55 4.83 5.00
40 50 60 70 80 90 95 100
4.89 4.57 4.05 3.37 2.54 1.60 1.06 0.0
0.0 1.25 2.5 5 7.5 10 15 20 30
aLeadingedge radius = 0.12t. bTrailingedge radius = 0.06t.
T 6 airfoil profile (t/c = 0.10) a'b
T a b l e 9.15
x/c, ~
y/c, ~
x/c, %
y/c, %
0.0 1.17 1.54 1.99 2.37 2.74 3.4 3.95 4.72
40 50 60 70 80 90 95 100
5.00 4.67 3.70 2.51 1.42 0.85 0.72 0.0
0.0 1.25 2.5 5 7.5 10 15 20 30
aLeadingedge radius = 0.12t. bTrailingedge radius = 0.06t.
_101
i
I 10
I
I
I
20
Fig. 9.71
I
I
30
I
I
40
I
I
50
I
I
I
60
I
I
70
I
I
I
80
The T6 t u r b i n e airfoil base profile.
I
90
100
644
ELEMENTS OF PROPULSION
Example 9.12 Consider a singlestage turbine using the computer program TURBN. Given: rh = 200 l b m / s ,
Tt3 = 2860°R,
M2=l.1,
Ptl=250psia,
wr = 1 5 0 0 ft/s,
al=0deg,
R = 53.40ft.lbf/(lbm.°R),
Ttl = 3400°R
Ml = 0.4,
a3=0deg,
u3/u2 = 0.90,
rm = 1 2 in. y=
tPtstator= 0.06,
1.3 ~btrotor= 0.15
Solution: The program T U R B N is run with az as the unknown. The results are given in Table 9.16 with the hub and tip tangential velocities based on freevortex swirl distribution. The cross section of the stage sketched by the computer program is shown in Fig. 9.72 for stator c/h = 0.8 and rotor c/h = 0.6. The very high A N 2 of 4.18 × 10 ~° in. 2. rpm 2 at a relative total temperature of 3052°R is not possible with current materials unless the blades are cooled. Also, the high rim speed of about 1206 f t / s would require existing materials to be cooled.
9.5.5 AxialFlow Turbine Stagc
Ol2
Known
The preceding method of calculation assumed that a2 was unknown. The program T U R B N will handle the case when any one o f the following four variables is unknown: c~2, Tt3, ce3, or Ma
When Tt3 is unknown, Eqs. (9.20) and (9.103) are solved for Tt3, giving
_(
u3
)
wrV2 sin ol2 +   cos c~2 tan c~3 gcCp u2
Tt3 = Ttl
(9.114)
W h e n a3 is unknown, Eqs. (9.20) and (9.103) are solved for ce3, giving tana3  ua/u2
co
2 V2
tanaz
(9.115)
W h e n Mz is unknown, the velocity at station 2 (v2) is obtained from Eq. (9.103), giving V2 =
~bwr . sin oL2 q (u3/u2) cos oL2 tan a3
(9.116)
Then M2 is obtained from Eq. (9.104) rewritten as M2 =
V2
(9.117)
~ / ( y  1)gccpTt2  [ ( ' y  1)/2IV 2
9.5.6 AxialFlow Turbine Stage AnalysisNo Exit Swirl Consider the flow through a singlestage turbine as shown in Fig. 9.73 with zero exit swirl. W e will consider the case where there is no exit swirl (v3 = O,
Z
=
m
.=. L

.= i
i
o=
TURBOMACHINERY
~
~.~~F=~ ~~.
ht3
~.
I II
II
o II
II
II
II
II
H
c~
fl
II
II
c~
× 0o
H
645
646
ELEMENTS OF PROPULSION
F~ ~h= 11.04 it= 12.96 Back rh= 10.20 tt 13.80 L=4.45m
_l ................. C e ~ Line T u ~ e C,ossSec6~
fic ~ i r , g = 6.0 in
Fig. 9.72
Sketch of cross section for turbine stage of Example 9.12 from TURBN.
1
2
3
Annulus area A
Rotor
Stator 
Station:
2R
1
3R
rt3~1~
~ V 3 ~
3 for
u2 192
192R
v3R/ v3R ~ u3
Fig. 9.73
Generalized turbine stage, zero exit swirl.
K~). .3
TURBOMACHINERY
647
a3 = 0) and the axial velocities at stations 2 to 3 are the same (u2 = u3). The flows through the nozzle (stator) and rotor are assumed to be adiabatic. For solution, we assume the following data are known:
M2, Ttl, Tt3, tor, cp, and 7 The equations for solution of a zero exit swirl, axialflow turbine based on these known data are developed in this section. At station 2,/)2 is given by
~1 V2 =
2gccpTt2 + 2 / [ ( 7  1)M 2] 2 ¢ , / ( 1  ~,)
=~or
1 + 2 / [ ( 7  1 ) M 2]
(9.118)
and 0/2 by
sin0/2 = ~0~2 =
(1  ~'s)
1 ~ ( y   ] ) M
(9.119)
Because the axial velocities at stations 2 and 3 are equal, the velocity at station 3 is given by 1/3 = V2 cos 0/2
(9.120)
The degree o f reaction is given by Eq. (9.87):
°Rt = 1
tp 2
The stage exit Mach number/143 is derived as follows. Given that
M3 M2
u3/a3  V2/a2
_
!u a___22 /L2 V2 a 3   c o s 0/2
Vv3
Then from Eq. (9.81b), we write
T2T3
°Rt   Ttl  Tt3
q, 1  
2
which allows the temperature ratio T2/T3 to be written as T2 
T3
1 =
1  (Tt2/T2)(1  rD(1  q,/2)
(9.121)
Thus the Mach number M3 can be written as M 2 cos a 2 M3 =
(9.122)
x/1  (1  rs)(1  ~9/2){ 1 + ( 3 /  1)/2]M 2 }
648
ELEMENTS OF PROPULSION
A compact equation for the rotor exit relative Mach number M3R is developed as follows. Since
M3R M2
V3R a2 _ V3R a3 V2 V2 V r3
the velocity ratio is obtained by first writing V2R = u 2 + (wr) 2 = V22  v2 + (o)r) 2 Thus
V~R I V~

=
v~4 (wr)2 
1 
V2"
\v2]
~
1
('#

\ V=,I L\oor/
]
1)
Since
M3R M2
V3R T~ 172 Vr3
then by using Eq. (9.121), the Mach number ratio is found by g
M3R
/
1  (oor/V2)2(qt 2 
1)
M2 = V1  (T~/T~    z ~ i   ~ O / 2 )
(9.123)
which with Eq. (9.119) becomes (and is actually used as)
(9.124)
This equation contains an interesting and unexpected piece of guidance for the design of turbine stages. To ensure that stator cascade choking controls the turbine mass flow rate, M2 should be greater than unity and M3R should be less than unity. Equation (9.124) reveals, however, that when the stage loading coefficient is unity (degree of reaction is 0.5), the opposite must be true. Therefore, even though it would appear preferable to aim for a degree of reaction near 0.5 to balance the difficulty of designing the stator and rotor airfoils, the requirement to reduce M3R translates to lower allowable values of the degree of reaction and correspondingly higher stage loadings. In actual practice, the degree of reaction is usually found in the range of 0.2 to 0.4, so that a substantial, but minority, fraction of the overall static enthalpy (and static presssure) drop still takes place across the rotor and is available to prevent the separation of the suction surface boundary layer. It is important to bear in mind that even turbine airfoil boundary layers can separate, and when they do, the effect on efficiency (and heat transfer) is usually disastrous.
TURBOMACHINERY
649
The rotor relative total temperature (Tt2R = Tt3R), which is useful for heat transfer and structural analyses, is given by Eq. (9.113). For our case of zero exit swirl and constant axial velocity (u3 = u2), this equation reduces to Tt3R
7,2
= "rs +
1  ~'s

(9.125)
2q,
Example 9.13 To illustrate the application of this method, a singlestage turbine with zero exit swirl will be designed for the following conditions: M1 = 0.4 w r = U = 300 m / s R = 0.2872 kJ/(kg • K) Tt3 = 1260 K ~s 
T,3  0.900 T,1
Me = 1.10 Tt2 = Ttl = 1400 K gcCp = 1.158 KJ/(kg • K) y = 1.33 ~b= 1.8006 Eq. (9.20)
If one chooses to assume et = 0.90, the results are °Rt = ce2 = M3 = ~'s =
0.0997 Eq. (9.87) 47.35 deg Eq. (9.119) 0.7498 Eq. (9.122) 0.6239 Eq. (9.91)
M3R= 0.8756 Tt3R = 1299 K ~b = 1 . 6 5 8 6 ~Tt= 0.9053
Eq. Eq. Eq. Eq.
(9.124) (9.125) (9.78a) (9.76)
These results provide the basis for stepbystep calculations leading to the summary of flow properties given in Table 9.17.
Table 9.17
Results for Example 9.13 axialflow turbine state calculation with zero exit swirl
Station Property
T, T Pt Ptl
P
1
2
2R
3R
3
1400.0 1364.0
1400.0 1167.0
1298.9 1167.0
1298.9 2012.7
1260.0 2012.7
1.0000
?
?
0.7053
0.6239
0.9003
?
?
0.4364
0.4364
0.400 288.7 288.7 0 0
1.100 734.4 497.6 540.2 47.35
0.8586 552.5 497.6 240.2
0.8756 581.0 497.6 300.0
0.7498 497.6 497.6 0 0
25.76
31.09
Ptl
M V bl l) O~
/3
m/s m/s m/s deg deg
650
ELEMENTS OF PROPULSION
9.5.6. 1 General solution. When 3' and e t are fixed, closer examination of the turbine stage design equation set reveals that the results depend only on the dimensionless quantities M2, zs, and ~0. Under these conditions, it is therefore possible to generate graphical representations that reveal the general tendencies of such turbine stages and serve the important purpose of defining the limits of reasonable stage designs. This has been done for typical ranges of the prevailing dimensionless parameters and 3' values of 1.33 and 1.3; the results are presented in Figs. 9.74 and 9.75, respectively. These charts may be used to obtain initial ballparkstage design estimates and also reveal some important trends. If values of M3R less than 1 and stage loading coefficients ~bless than or equal to 2.0 are taken as reasonable limits, it is clear that better (i.e., lower) values of ~Oare available for the same ATt as zs decreases (lower inlet temperatures) and/or mr increases (higher wheel speeds). By noting that ~0 does not depend on M2 [Eq. (9.87)], it is also clear that M2 determines only M3R. Larger values of M2 are desirable because they reduce the annulus flow area A and the rotating airfoil centrifugal stresses, and ensure choking of the stator airfoil passages over a wider turbine operating range; but Figs. 9.74 and 9.75 show that increasing M2 reduces the number of acceptable solutions. Finally, all other things being equal, stages having lower % (i.e., more energy extraction) suffer the dual disadvantages of increased stage loading coefficient ~b and increased annulus flow area A. Figures 9.74 and 9.75 exhibit an extremely interesting mathematical behavior in the region where all the curves for a value of M2 appear to, and indeed do, pass through a single point. This fact may be verified by equating the numerator and denominator on the right side of the equals sign in Eq. (9.124), and this reveals that M 3 R = M2 is independent of either Me or ~s provided only that
~1
= y 21
(2~~)M2 2
1.3 1.2 1.1 1.0
~R
(deg)
0.9 ,~ 0
~"
~ " ' ~ 50
0.8 0.825
0.7
~
0.6
~
1.0
r 1.2
I 1.4
I 1.6
I 1.8
I 2.0
0.5
0.4
0.3
0.2
0.1
0.0
0.5
0.9
60
~70 I 2.2 0.1
°R t
Fig. 9.74 Generalized turbine stage behavior, zero exit swirl (y = 1.33).
TURBOMACHINERY
651
1.3 1.2 1.1 1.0
M3R
0.9
¥2 = l.O~,~....~
~
0.8
~
~
o
~
(deg)
.:~5 ~
~~.5o 875
0.8
o
0.7 ~
~
~ 5 0
0.6 0.5 1.0
I 1.2
[ 1.4
I 1.6
I 1.8
0.875
I 2.0
~
~
~ 7 0 2.2
v 0.5
Fig. 9.75
0.4
0.3
0.2 °Rt
0.1
0.0
0.1
Generalized turbine stage behavior, zero exit swirl (Y = 1.3).
Hence, for each y, there are two values of ~p that satisfy this condition. When M3R = M2 = 1.0 and 3' =1.3, they are 1.072 and 0.82, the former obviously being the one that appears in Fig. 9.75 and the latter having no practical use. For M3R = M2 = 1.2 and 3' = 1.3, they are 1.102 and 0.746. The physical meaning of this convenient convergence is clear enough, namely, that near M3R M2, where the stator and rotor airfoil exit conditions are similar, the stage loading parameter t/, must be near unity regardless of the other stage parameters.
9.5.6.2 Multistage turbine design. When the required stage loading coefficient q, for a design is greater than 2.0, a singlestage design would require a hopeless negative reaction [Eq. (9.87)] and would be impossible to design with high aerodynamic efficiency. A desirable multistage design would have the total temperature difference distributed evenly among the stages: (mzt)turbin e =
(number of stages)
x (mTt)stag e
This would result in stages with the same stage loading coefficients [Eq. (9.20)] and same degree of reaction [Eq. (9.87)] for the same rotor speed U. For a threestage design, we get Iltsl = ~Js2 = @s3
and
(°Rt)sl = (°Rt)s2 = (°Rt)s3
To obtain the choked flow in the firststage stator (nozzle), the Mach number entering the rotor M2 is slightly supersonic. The Mach numbers in the remaining stages are kept subsonic. The net result is that the stage loading of the first stage is larger than the loading of any of the other stages. For a threestage design, the
652
ELEMENTS OF PROPULSION
stage loading coefficient and degree of reaction of the second and third stages are nearly equal.
9.5.7 Shaft Speed The design rotational speed of a spool (shaft) having stages of compression driven by a turbine is initially determined by that component that limits the speed because of high stresses. For a lowpressure spool, the first stage of compression, since it has the greatest A N 2, normally dictates the rotational speed. The first stage of turbine on the highpressure spool normally determines that spool's rotational speed because of its high AN 2 or high disk rim speed at elevated temperature.
9.5.8
Design Process
The design process requires both engineering judgment and knowledge of typical design values. Table 9.18a gives the range of design parameters for axialflow turbines that can be used for guidance. The comparison of turbines for Pratt & Whitney engines in Table 9.18b shows typical turbine design values and the leading trends in turbine technology. Note the increases over the years in inlet temperature, mass flow rate, and output power. From Table 9.18b, comparison of the JT3D and JT9D highpressure turbines shows that the stage loading coefficient did not appreciably change between the designs. However, the turbine inlet temperature increased to a value above the working temperature of available materials, requiring extensive use of cooling air. The stage loading coefficient for the lowpressure turbine increased dramatically, reducing the number of stages to four. If the stage loading coefficient of the lowpressure turbine of the JT3D were not increased significantly in the design of the JT9D, about six or seven stages of lowpressure turbine would have been requiredincreasing both cost and weight.
Table 9.18a Range of axialflow turbine design parameters
Parameter Highpressure turbine Maximum A N 2 Stage loading coefficient Exit Mach number Exit swirl angle, deg Lowpressure turbine Inlet corrected mass flow rate Hub/tip ratio at inlet Maximum stage loading at hub Exit Mach number Exit swirl angle, deg
Design range 45 × 101° in. 2.rpm 2 1.42.0 0.400.50 020 4044 lbm/(s •fie) 0.350.50 2.4 0.400.50 020
TURBOMACHINERY Table9.18b
653
ComparisonofPratt& Whi~eyen~nes
Parame~r
JT3D
JT9D
Year of introduction Engine bypass ratio Engine overall pressure ratio Core engine flow, lb/s Highpressure turbine Inlet temperature, °F Power output, hp Number of stages Average stage loading coefficient Coolant plus leakage flow, % Lowpressure turbine inlet temperature, °F Power output, hp Number of stages Average stage loading coefficient Coolant plus leakage flow, %
1961 1.45 13.6 187.7
1970 4.86 24.5 272.0
1745 24,100 1 1.72 2.5
2500 71,700 2 1.76 16.1
1410 31,800 3 1.44 0.7
1600 61,050 4 2.47 1.4
9.5.8.1 Stepsof design. The material presented in previous sections can now be applied to the design of an axialflow turbine. The complete design process for a turbine will include the following items: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10)
Selection of rotational speed and annulus dimensions Selection of the number of stages Calculation of airflow angles for each stage at the mean radius Calculation of airflow angle variations from the hub to tip for each stage Selection of blade material Selection of blading using experimental cascade data Selection of turbine cooling, if needed Verification of turbine efficiency based on cascade loss data Prediction of offdesign performance Rig testing of design
Items 1  5 will be covered in this section. The other steps are covered in Refs. 42, 43, 22, and 29. The design process is inherently iterative, often requiring the return to an earlier step when prior assumptions are found to be invalid. Many technical specialities are interwoven in a design, e.g., an axialflow turbine involves at least thermodynamics, aerodynamics, structures, materials, heat transfer, and manufacturing processes. Design requires the active participation and disciplined communication by many technical specialists.
Example 9.14 We will consider the design of a turbine suitable to power the eightstage, axialflow compressor designed earlier in this chapter for a simple turbojet gas turbine
654
ELEMENTS OF PROPULSION
engine (see Example 9.7). From engine cycle and compressor design calculations, a suitable design point for the turbine of such an engine at sealevel, standardday conditions (P = 14.696 psia and T = 518.7°R) may emerge as follows: Compressor Compressor Compressor Compressor Compressor Compressor
pressure ratio: 10.41 flow rate: 150 lbm/s efficiency: 86.3% exit Tt: 1086°R y: 1.4 R: 53.34 ft. lbf/(lbm. °R)
Rotor speed to: 800 rad/s Turbine flow rate: 156 lbm/s Tt entering turbine: 3200°R Pt entering turbine: 143.1 psia Turbine y: 1.3 Turbine R: 53.40 ft. lbf/(lbm. °R)
From these specified data, we now investigate the aerodynamic design of an axialflow turbine. The compressor input power is Vllc = mcp(Tte  Tti) = (150 × 0.240)(1086  518.7) = 20,423 Btu/s
 21.55 MW Assuming that the compressor input power is 0.98 of the turbine output power (the other 2% of turbine power goes to shaft takeoff power and bearing losses), the required output power of the turbine is 22.0 MW (21.55 MW/0.98). The total temperature leaving the turbine is
Ttc= Tti
Wt
thCp

3200
22,000/1.055  3200  450.1 = 2749.9°R 156 x 0.297
The turbine temperature ratio (~'t = Tte/Tti) is 0.8593. If the flow entering the rotor has a Mach number of 1.2 at 60 deg to the centerline of the turbine and a 1% total pressure loss through the turbine stator (nozzle), the annulus area entering the rotor is
A2 = (cos ot2)(Pt2)MFP(M2) 156 3 ~ ' ~ cos 60 deg x 143.1 × 0.99 × 0.502075~/53.34/53.40
= 248.3 in. 2
For a rotor angular speed to of 800 rad/s, A N 2 for the rotor is 1.45 × 101° in. 2. rpm2this blade stress is within the capability of modem cooled turbine materials (about 2 to 3 x 101°in. 2. rpm2). Calculation of the stage loading coefficient for the turbine helps in determining the number of turbine stages. For the turbine mean radius equal to that of the compressor, the stage loading coefficient on the mean line is gccpATt 
7455 × 450.1
(torm) 2 = (866Z
2 = 2.600
Using Fig. 9.73, we see that this value of stage loading coefficient is larger than that possible for a singlestage turbine. Either the mean rotor speed torm must be
TURBOMACHINERY Table 9.19
655
Variation of stage loading, radii, and rim speed with mean radius for Example 9.14 singlestage design
rm, in.
rt, in.
rh, in.
rh/rt
Ur, ft/s
16.00 17.04 18.00 19.00 20.00 21.00
2.949 2.600 2.330 2.091 1.887 1.712
17.23 18.20 19.10 20.04 20.99 21.94
14.77 15.88 16.90 17.96 19.01 20.06
0.857 0.873 0.885 0.896 0.906 0.914
918 992 1060 1131 1201 1271
increased to reduce the stage loading coefficient for a singlestage turbine, or a twostage turbine will be required. Because increasing the rotor angular speed o) will increase the blade stress A N 2 and because only a little margin exists, we will investigate the effect of increasing the mean radius on stage loading coefficient, annulus radii (rt and rh), and rim speed (Urassuming the rim radius is 1 in. smaller than that of the hub). From the results given in Table 9.19, we can see that a singlestage turbine would require a mean radius of 19 to 20 in. to reduce the stage loading coefficient and keep the rim speed below about 1200 ft/s. This would result in a tip radius equal to or larger than the compressor's inlet radius. In addition, the tip radius of 20 to 21 in. is much larger than current turbines for gas turbine engines that range between 10 and 17 in. Although a singlestage turbine is desirable because of the reduced weight, the low rotor angular speed of 800 rad/s makes this size undesirable. For a smaller turbine, the designer might consider increasing the rotor angular speed and redesigning the compressor. A rotor angular speed of 1000 rad/s for a turbine with a 16in. mean radius has a stage loading coefficient for the mean radius of 1.885 and a rim speed of 1148 ft/s, which is possible for a single stage. The designs of both a singlestage turbine and a twostage turbine are performed in the following sections. The computer program TURBN is used to ease the calculational burden in both designs.
9.5.8.2
Singlestage design.
We consider a singlestage turbine with the
following characteristics: Rotor angular speed w: 800 rad/s Tt entering turbine: 3200°R Tt leaving turbine: 2749. I°R Turbine R: 53.40 ft. lbf/(lbm. °R)
Turbine mass flow rate: 156 lbm/s
Pt entering turbine: 143.1 psia Ratio of specific heats: 1.3
To keep the degree of reaction at the hub from being too negative at a reasonable value of the stage loading coefficient ~b, we consider a nonzero exit swirl angle a 3 for a stage with a hub speed of about 1200 ft/s (this corresponds to a rim speed of about 1130 ft/s, which will limit the disk stress). This hub speed corresponds to a hub radius of 18 in. and a tip radius of about 20.1 in. The
656
ELEMENTS OF PROPULSION
computer program TURBN was run with the exit swirl angle a3 unknown, the data just listed, and the following additional input data: Mean rotor speed ~or: 1270 ft/s M2:1.1 M~: 0.4
u3/u2:1.0 Zs: 0.9 Zr: 0.9
oJ: 800 rad/s or2:60 deg cq: 0 deg ~btstato r = 0.06 and (c/h)s: 1.0 (c/h)r: 1.0
~btrotor =
0.15
Computer calculations yield the singlestage turbine summarized in Table 9.20 with hub and tip tangential velocities based on freevortex swirl distribution. This is a viable singlestage design with moderate exit swirl a3, positive reaction, and subsonic M3R at the tip. 10 ~° 2 This design gives a blade AN 2 of 1.44 x in.2.rpm and hub speed of 1201 ft/s. This AN 2 value is well within the limits of cooled turbine materials, and the low hub speed is below the limiting speed of turbine disk materials. A crosssectional sketch of the singlestage turbine just designed and plotted by the computer program TURBN is shown in Fig. 9.76. Note that this sketch does not show the required exit guide vanes that will turn the flow back to axial. The estimated axial length L shown in Fig. 9.76 is based on the input values of Z and c/h for the stator and rotor blades and the scaling relationships of Fig. 9.69. For the input values of Z and c/h, the resulting solidity at the mean radius, number of blades, and chord length for the stator and rotor are as shown in Table 9.21. The selected axial chord and number of blades for the stator or rotor depend on many factors (e.g., flow through the blades, vibration, blade attachment). Figure 9.77 shows the computer sketch of the blades at the mean radius, using C4 base profiles.
9.5.8.3 Twostage design. In a twostage design, the stage loading coefficients ~p are lower and the temperature ratios ~~.are higher than those for a singlestage design. This results in higher reactions, less turning of the flow, and lower loss coefficients. For good flow control of the turbine, the firststage stator (nozzle) should be choked, which requires that M2 for this stage be supersonic. Inspection of Fig. 9.75 shows that at low q, and high %, the value of M3R is a little less than M2thus, we will want to select a low supersonic value of M2 (about 1.05) for the first stage. A balanced design would have about the same a2 values for both stages with the firststage M3Rt below 0.9. The twostage turbine will be designed with a 17.04in. mean radius (same as multistage compressor) at an rpm of 7640 (oJ = 800 rad/s), giving a mean rotor speed Um= O~rmof 1136 ft/s. An initial starting point for the design of this twostage turbine is constant axial velocity through the rotor (u3 = u2), zero exit swirl (a3 = 0), and a secondstage M2 of 0.7. The stage loading coefficients and other flow properties depend on the split in temperature drop between the stages. Calculations were performed by using the computer program TURBN
=
.=. f~
=
=,
i
m =
[...
c~
c~
O
~
~ ~
~ ~ . ~
. ~
, 1 ' ~ " ~
TURBOMACHINERY
~ =
~
~'~~ ~
cq
U'h
cq
:=
...=. r ~
~.
II II II
..
C'q
il
ti
E E"
[i
.~~×
II
657
658
ELEMENTS
OF PROPULSION
S~e ~) Front rh= 18.25 a= 19.85 Back rh= 17.84 rt = 20.26 L = 4.88
in
.I . . . . . . . . . . . .
CenterLine TurbineCross Section
= 6,0 in Fig. 9.76
Sketch of cross section for singlestage turbine design.
with ca2 unknown at different values of the temperature leaving the firststage turbine. The resulting a2 and M3nt values are listed in Table 9.22. An interstage temperature of 2925°R gives a balance design for a2 values with the first stage M3Rt, above 0.9. The value of M3R t c a n be reduced by selecting a value for the axial velocity ratio u3/u2 less than unity. A design with an interstage temperature of 2925°R and firststage u3/u2 of 0.9 is selected to reduce M3m. The losses for the first stage and second stage are estimated by using polytropic efficiencies of 0.9 and 0.92, respectively, and stator loss coefficients of 0.06 and 0.02, respectively. For all blades, a value of 0.9 is used for the tangential force coefficient Z, and a value of 1.0 is used for the chord/height ratio c/h. Results for both stages are presented in Tables 9.23
Table 9.21
Stator Rotor
Blade results for singlestage turbine
Solidity
Number of blades
Chord, in.
0.979 1.936
64 103
1.831 2.250
TURBOMACHINERY
659 Stage: 1 Inlet E~ Thckn~, Pfofil~ Chord(in] Redid Poe.,r,
Nozzle Ro~ OEO 38.4 69,7 57,5 1ZO~ 1ZO~ TG TG 1.826 2.234 50~;
Tip
~
Fig. 9.77
Hub
Sketch of blades for singlestage turbine design.
and 9.24 with hub and tip tangential velocities based on freevortex swirl distribution. A crosssectional sketch of the twostage turbine just designed and plotted by TURBN is shown in Fig. 9.78. The sketch shows the stator and rotor for both stages. Note that the turbine exit stator is not shown in the sketch. For the input values of Z and c/h, the resulting solidity at the mean radius, number of blades, and chord length for the stator and rotor blades of the two stages are as shown in Table 9.25. The selected axial chord and number of blades for the stator and rotor depend on many factors (e.g., flow through the blades, vibration, blade attachment). Figures 9.79 and 9.80 show the computer sketch of the blades at the mean radius, using C4 base profiles for the first and second stages, respectively.
Table 9.22
Variation of stage parameters with interstage temperature for Example 9.14 twostage design Stage 1
Tt3, °R 2875 2900 2925 2950 2975
Stage 2
rs
0/2, deg
1.8750 1.7307 1.5865 1.4423 1.2981
0.8984 0.9063 0.9141 0.9219 0.9297
55.0 49.12 43.88 39.06 34.55
M3m 0.7774 0.8467 0.9068 0.9587 1.0034
~P
rs
a2, deg
1.0034 0.8659 1.0102 1.1544 1.2986
0.9565 0.9482 0.9401 0.9322 0.9243
28.61 34.90 41.65 49.13 57.90
660
=,
m
~
II
II
II
II ~
ELEMENTSOFPROPULSION
er~(~l,...,.~l,,~(.,,i,...~,..~
II
II
~5
II
X
II
@
@
= @
"=
C~
TURBOMACHINERY
C',I
C~
:=
li
II
li
II
II
il
II
X
,
~/r sp~c) at Mach numbers less than 1.7. The inlet size and mass flow characteristics of the inlet and engine are considered.
a) Inlet performance. The total pressure recovery of this inlet ~Tr is tabulated in Table 10.1, is plotted in Fig. 10.35, and meets the requirement for efficient total pressure recovery (~/~ > r/r spec) at Mach numbers less than 1.75. Analysis of the mass flow characteristics of the inlet and the engine is required before this inlet can be sized. As long as the normal shock between stations b and c touches the lip of the inlet, either station b or c of Fig. 10.34 can be equated to station s of Fig. 10.43. From Eq. (10.14b), we have Aot Ptb MFP(Mb) As ProMFP(M0)
Ptc MFP(Mc) ProMFP(M0)
(10.16a)
722
ELEMENTS OF PROPULSION
1.28
Aoi req a~ ref
J ................
.....
l ....
Ao
Aoi spec A0 ref
1.24 1.20
///,. 
1.16
.
*~ 1.12 1.08
1.04 ~

~ / / 4~_______________~

A~ ref
1.00 '
0.96
0.0 Fig. 10.44
I
I
I
I
0.5
1.0
1.5
2.0
Airflow requirement for example inlet at altitude.
and Ptc Pt~ = "Or
Ptb "or Pro  Ptc/Ptb
(10.16b)
The area ratio Aot/A s can be calculated by using the results of Table 10.1, assuming that Ms = Mb at subsonic values of Mb (no local flow acceleration or deceleration). For subsonic Mo, a reasonable approximation is for choked flow at station s; thus, A~i = A~. The results of the area ratio Aoi/A s calculations for this example inlet are plotted in Fig. 10.45. Note that this plot has a minimum value of 1 at Mo = 1. Also note the jumps in value of Aoi/As corresponding to the transition between normal and oblique shocks at each ramp (these also correspond to the jumps in total pressure shown in Fig. 10.35). The selected design point of an external compression inlet sets the value of A~/ A 1. Since Aoi = A1 at an inlet's supersonic design point, then A~/A1 = 1/(Aoi/As) evaluated at this design point. For this example, the inlet design point is the maximum Mo of 2.0. Thus As/A1 = 0.7681 and the resulting values of Aoi/Al, from using Eq. (10.14a), are tabulated in Table 10.2.
b) Inlet size. Now that the mass flow ratio of the inlet has been determined, the inlet mass flow rate requirements must be established before it can be sized. By using these data, the maximum calculated value of the inlet size (A1/A~ ref), from Eq. (10.15), sets the size of this fixedgeometry inlet. Comparison of Figs. 10.44 and 10.45 indicates that the two most demanding operating points will most likely be at Mo = 1.23 and Mo = 1.42. Sizing calculations at the two Mach
INLETS NOZZLES, AND COMBUSTION SYSTEMS
723
1.35 1.3( 1.2' 1.20 1.15 1.10 
f
1.05 1.00 1.95
0.0
I
[
0.4
0.8
1.2
I
I
1.6
2.0
M0 Fig. 10.45
Area ratio of Example
10.4 inlet (Fig. 10.34).
numbers are presented in Table 10.3. The flight condition at Mo = 1.42 determines the inlet size as A1 = 1.425A~ref. The resulting flow ratios of the sized inlet and its required performance are plotted for altitudes between 36 and 60 kft in Fig. 10.46. This plot shows the difference in flow behavior of the inlet between actual and required flows. As the flight Mach number increases above 1.5, the inlet mass flow rate Aoi//Al
Table 10.2
Example
10.4 inlet performance
Mo
Aoi/A1
0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
0.7749 0.7681 0.7681 0.7681 0.8121 0.8259 0.8895 0.9151 0.9378 0.9592 0.9798 1.0
ELEMENTS OF PROPULSION
724
Table 10.3
M0 1.23 1.42
E x a m p l e 10.4 inlet sizing
Ao/A~ref
Aoireq/A~ref
Aoi/At
Alreq/A~ref
1.0290 1.1004
1.0854 1.1803
0.7681 0.8283
1.413 1.425
increases while the required inlet mass flow r a t e (Aoi//A 1)req decreases. The difference between these mass flow rates is airflow that is accepted by the inlet and then bypassed about the engine back to the atmosphere, or spilled about the inlet, or is a combination of bypassed and spilled. The large quantity of excess air for this inlet will correspond to a high inlet installation loss at Mach numbers above 1.5. If a better match of inlet and engine is needed, the inlet will require variable geometry.
10.4.7 Examples of Existing Inlet Designs Three examples of supersonic inlet designs are shown in Figs. 10.47a, 10.47b, 10.48, and 10.49. Figure 10.47a shows the fixed doubleramp (6deg ramp followed by a 6.67deg isentropic ramp) external compression inlet with a throat slot bleed system developed in the J79 engine installation in the F16 aircraft. The total pressure recovery of this inlet is shown in Fig. 10.47b.
1.05 Ao :
II
: AO
Aoi /A ~
1.00 0.95 0.90
"~ 0.85
Aoi req/A 1
0.80
0.75 0.70 0
I
i
I
I
I
0.4
0.8
1.2
1.6
2
M0 Fig. 10.46 Mass flow performance of sized E x a m p l e 10.4 inlet at altitude and full engine throttle.
INLETS NOZZLES, AND COMBUSTION SYSTEMS
/6deg
725
i
..... L 7 +
\"l
Cowl I
ers
(incomingair) Fig. 10.47a The F16/J79inlet: side and isometricviews(Ref.65).
1.00 0.96 ~ ~2

16/79inlet
0.92 /
F16inlet! ~
0.88 I
0.840
I
0.4
I
I
0.8 1.2 Machnumber
I I,
1.6
I
2.0
Fig. 10.47b The F16/J79inlet: inlet pressurerecoverycomparison(Ref.65).
726
ELEMENTS OF PROPULSION
P
F Secondandthirdrampbleedexits ~/~ A3 F Bypassd°°r
l ~l~i P +
, ~
Engine
~:~ "'~kDiffusertrarap\ \~\~Q ~.Throatslotbypass ~I ~Cowlrotationpivot \ \ ~  Thirdramp
\\
~"Firstramp Note:Sideplate bleed notshown Fig. 10.48
The F15 inlet system (Ref. 66).
The variable tripleramp external compression inlet of the F15 aircraft is shown in Fig. 10.48. This side view of the inlet shows the ramps as they would be positioned when operating at the supersonic design point (ramp angles of 7, 8, and 8 deg for the first, second, and third ramps, respectively). The first ramp angle is fixed, and the second and third ramp angles are variable. The capture area of this inlet is variable with movement of the first ramp/top of inlet assembly from  4 to + 11 deg (this assembly is shown at 0 deg). The takeoff, transonic acceleration, and supersonic cruise modes of operation for the Concorde propulsion system are shown in Fig. 10.49. It has a complex variablegeometry inlet to satisfy the engine mass flow rate requirements at many diverse flight conditions. Note that the supersonic cruise dump control (3) is opened as an auxiliary inlet for takeoff.
10.5
Exhaust Nozzles
The purpose of the exhaust nozzle is to increase the velocity of the exhaust gas before discharge from the nozzle and to collect and straighten the gas flow. For large values of specific thrust, the kinetic energy of the exhaust gas must be high, which requires a high exhaust velocity. The pressure ratio across the nozzle controis the expansion process, and the maximum thrust for a given engine is obtained when the exit pressure Pe equals the ambient pressure P0. Nozzles and their operation are discussed in many textbooks. The two basic types of nozzles used in jet engines are the convergent and convergentdivergent (CD) nozzles.
INLETS NOZZLES, AND COMBUSTION SYSTEMS !
727
2
3
4
S
6
7
Engine variable geometry. The takeoff settings: 1ramp sections retracted: 2secondary control valve closed; 3dump control open as auxiliary inlet; 4engine bay cooling air flap open; 5after spill flap closed; 6tertiary doors open; 7secondary nozzle in convergent position !
.2
3
4
5
6
7
Transonic acceleration: lramp section retracted; 2secondary control valve open; 3dump control closed; 4cooling air flap closed; 5spill flap ctosed; 6tertiary doors open; 7secondary nozzle trailing !
2

/~
~mq::b,,,....
3
4
5
6
7
Cruise at Math 2,2: Iramp sections extended; 2secondary control valve open; 3dump control open to dump; 4cooling air flap closed; 5spill flap open; 6terfary doors closed; 7secondary nozzle divergent. Fig. 10.49
Concorde propulsion system, modes of operation.
The functions of an exhaust nozzle may be summarized as follows: 1) Accelerate the flow to a high velocity with minimum total pressure loss. 2) Match exit and atmospheric pressures as closely as desired. 3) Permit afterbumer operation without affecting main engine operationthis function requires a variablearea nozzle.
728 4) 5) 6) 7) 8)
ELEMENTS OF PROPULSION Allow for cooling of walls if necessary. Mix core and bypass streams of turbofan if necessary. Allow for thrust reversing if desired. Suppress jet noise and infrared radiation (IR) if desired. Thrust vector control if desired.
Do all of these with minimal cost, weight, and boattail drag while meeting life and reliability goals.
10.5.1 Nozzle Types 10.5. 1.1 Convergent nozzle. The convergent nozzle is a simple convergent duct, as shown in Fig. 10.50. When the nozzle pressure ratio Pte/Po is low (less than about 4), the convergent nozzle is used. The convergent nozzle has generally been used in engines for subsonic aircraft.
10.5. 1.2 Convergentdivergent (CD) nozzle. The convergentdivergent nozzle is a convergent duct followed by a divergent duct. Where the crosssectional area of the duct is at a minimum, the nozzle is said to have a throat. Most convergentdivergent nozzles used in aircraft are not simple ducts, but incorporate variable geometry and other aerodynamic features. The convergentdivergent nozzle is used if the nozzle pressure ratio Pte/Pois high (greater than about 6). Highperformance engines in supersonic aircraft generally have some form of a convergentdivergent nozzle. If the engine incorporates an afterburner, the nozzle throat is usually scheduled to leave the operating conditions of the engine upstream of the afterburner unchanged (in other words, vary the exit nozzle area so that the engine does not know that the afterburner is operating). Also, the exit area must be varied to match the different flow conditions and to produce the maximum available thrust. Earlier supersonic aircraft used ejector nozzles (Fig. 10.51) with their highperformance turbojets. Use of the ejector nozzle permitted bypassing varying amounts of inlet air around the engine, providing engine cooling, good inlet recovery, and reduced boattail drag. Ejector nozzles can also receive air from outside the nacelle directly into the nozzle for better overall nozzle matchingthese are called twostage ejector nozzles. For the modern highperformance afterburning turbofan engines, simple convergentdivergent nozzles are used without secondary air, as shown in Fig. 10.52 for the FI00 engine.
Nozzle entrance
Fig. 10.50
Nozzle throat
Convergent exhaust nozzle.
INLETS NOZZLES, AND COMBUSTION SYSTEMS 2
3
5
6
729
7
Supersonic nozzle configuration with afterburning: (1) secondary flow; (2) outer case engine; (3) movable primary nozzle shown at maximum area; (4) primary flow, effective throat; (5) movable secondary nozzle shown at maximum exit area; (6) mixing layer between primary and secondary streams; and (7) supersonic primary flow
8
9
10
11
12
Subsonic nozzle configuration with no afierburning: (8) primary nozzle at minimum area; (9) separation point of external flow; (10) secondary nozzle at minimum area; (11) sonic primary stream; and (12) region of separated flow in external flo w
I
1,

13
f
14
15
16
17
Subsonic nozzle configuration, not aflerburning, and blowin door in use: (13) tertiary flow of ambient gas into nozzle; (14) blowin door and inflow configuration; (15) reversible hinge/latch; (16) movable secondary nozzle; and (17) separation point of external flow.
Fig. 10.51
Ejector nozzle configuration (Ref. 67).
730
ELEMENTS
~..._Actuator
Tailpipe'J r hner
~_/__ ~ ~
OF
PROPULSION
t Actuationring ~ Nozzlecavity Outer , ~" fairing / ..L _ ~ ~ ~ ~ ~ ../
/
z..\ nozzle
cam
Open position
Closed position
Primary ~ nozzle ~L___Secondary nozzle
' '   Secondary nozzle link
Fig. 10.52 Convergentdivergent exhaust nozzle sehematie (Ref. 67).
10.5.2 Nozzle Functions One can think of the exhaust nozzle as dividing the power available from the main burner exit gas between the requirements of the turbine and the jet power. 67 Thus the nozzle serves as a backpressure control for the engine and an acceleration device converting gas thermal energy to kinetic energy. A secondary function of the nozzle is to provide required thrust reversing and/or thrust vectoring.
10.5.2. 1 Engine backpressure control
T h e t h r o a t a r e a o f the n o z z l e is
one of the main means available to control the thrust and fuel consumption characteristics of an existing engine. In preliminary engine cycle analysis, selection of specific values for the engine design parameters and the design mass flow rate fixes the throat area of the nozzle. The engine performance methods of Chapter 8 assume that the nozzle throat area and the other internal flow areas of the engine remain constant. This assumption of constant areas establishes the offdesign operating characteristics of the engine and the resulting operating lines for each major component. Changing the nozzle throat area from its original design value will change the engine design and the operating characteristics of the engine at both on and offdesign conditions. At times, it is necessary to change the offdesign operation of an engine in only a few operating regions, and variation of the throat area of the exhaust nozzle may provide the needed change. At reduced engine corrected mass flow rates (normally corresponding to reduced engine throttle settings), the operating line of a multistage compressor moves closer to the stall or surge line (see Fig. 10.53). Steadystate operation close to the stall or surge line is not desirable because transient operation may cause the compressor to stall or surge. The operating line can be moved away from the stall or surge line by increasing the exhaust nozzle throat area, as shown in Fig. 10.53. This increase in nozzle throat area reduces the engine backpressure and increases the corrected mass flow rate through the compressor (see Figs. 8.9 and 10.53). Large changes in the exhaust nozzle throat area are required for afterburning engines to compensate for the large changes in total temperature leaving the
INLETS NOZZLES, AND COMBUSTION SYSTEMS
731
150
Design point  . ~ Original operating line ~ /,'~ Operating line at ~/// increasedA8 ~ ///'~
100 O
Stall line ~
~
//
~./ ' z,t~
50
/0.75 I 110
%Nc2 11"~ i
"
I
I
80
70
65 0
20
I
I
I
I
I
40
60
80
100
120
% design rhc2 Fig. 10.53
Compressor map with exhaust nozzle area change.
afterburner. The variablearea nozzle required for an afterburning engine can also be used for back pressure control at its nonafterburning settings. One advantage of the variablearea exhaust nozzle is that it improves the starting of the engine. Opening the nozzle throat area to its maximum value reduces the backpressure on the turbine and increases its expansion ratio. Thus the necessary turbine power for starting operation may be produced at a lower turbine inlet temperature. Also, because the backpressure on the gas generator is reduced, the compressor may be started at a lower engine speed, which reduces the required size of the engine starter.
10.5.2.2 Exhaust nozzle area ratio. Maximum engine thrust is realized for ideal flow when the exhaust nozzle flow is expanded to ambient pressure (Pe = P0) When the nozzle pressure ratio is above choking, supersonic expansion occurs between aftfacing surfaces. A small amount of underexpansion is less harmful to aircraft and engine performance than overexpansion. Overexpansion can produce regions of separated flow in the nozzle and on the aft end of the aircraft, reducing aircraft performance.
732
ELEMENTS OF PROPULSION
Short convergent
Simple ejector
Iris
Condi iris
Blownindoorejector
Fully variable ejector
o
~ Plug
o
Max A/B, LowM Max A/B, High M
Isenliopic ramp
Fig. 10.54 Typical nozzle concepts for afterburning engines (Ref. 67). The exhaust nozzle pressure ratio Pte/Po is a strong function of flight Mach number. Whereas convergent nozzles are usually used on subsonic aircraft, convergentdivergent nozzles are usually used for supersonic aircraft. When afterburning engine operation is required, complex variablegeometry nozzles must be used (see Fig. 10.52). Most of the nozzles shown in Fig. 10.54 are convergentdivergent nozzles with variable throat and exit areas. The throat area of the nozzle is controlled to satisfy engine backpressure requirements, and the exit area is scheduled with the throat area. The sophisticated nozzles of the F15 and B1 aircraft have two schedules: a lowspeed mode and a highspeed mode. 67
10.5.2.3 Thrust reversing and thrust vectoring. The need for thrust reversing and thrust vectoring is normally determined by the required aircraft and engine system performance. Thrust reversers are used on commercial transports to supplement the brakes. Inflight thrust reversal has been shown to enhance combat effectiveness of fighter aircraft. 67 Two basic types of thrust reversers are used: the cascadeblocker type and the clamshell type (Fig. 10.55). In the cascadeblocker type, the primary nozzle exit is blocked off, and cascades are opened in the upstream portion of the nozzle duct to reverse the flow. In the clamshell type, the exhaust jet is split and reversed by the clamshell. Because both types usually provide a change in effective throat area during deployment or when deployed, most reversers are designed such that the effective nozzle throat area increases during the brief transitory period, thus preventing compressor stall. High bypass turbofan engines use cascadeblocker type thrust reverser in the fan nozzle.
INLETS NOZZLES, AND COMBUSTION SYSTEMS Primary propulsion nozzle
~fl/))
Blocker
Fig. 10.55
Deployed clamshell7 ~.. \ ~,/
J Clamshell reverser
Cascade reverse
733
/
Thrust reversers (Ref. 67).
The exhaust system for the Concorde is shown in Fig. 10.56a. There are two nozzles, a primary nozzle and a secondary nozzle. The secondary nozzle is positioned as a convergent nozzle for takeoff and as a divergent nozzle for supersonic cruise. The modes of operation for this exhaust system are shown in Fig. 10.49 along with the inlet. Development of thrust vectoring nozzles for combat aircraft has increased in the last decade. Vectoring nozzles have been used on vertical takeoff and landing (VTOL) aircraft, such as the AV8 Harrier and F35 Joint Strike Fighter, and are proposed for future fighters to improve maneuvering and augment lift in combat.
3 Details of the exhaust system: (1) tertiary doors; (2) nozzle in supersonicconfiguration; (3) subsonic configuration;(4) thrust reverse buckets; (5) primary nozzle; (6) silencer lobes; (7) .secondarynozzle Fig. 10.56a
Concorde exhaust system.
734
ELEMENTS OF PROPULSION
Fig. 10.56b Pratt & Whitney Fll9PW100 turbofan engine with twodimensional thrust vectoring nozzle. (Courtesy of Pratt & Whitney.)
Thrust vectoring at augmented power settings is being developed for use in future fighters. However, cooling of the nozzle walls in contact with the hot turning or stagnating flows is very difficult and will require increased amounts of nozzlecooling airflow. The operation of the Pratt & Whitney Fll9PW100 augmented turbofan engine's thrust vectoring nozzle is shown in Fig. 10.56b. This twodimensional nozzle was developed for use in the F22 Advanced Tactical Fighter. Figure 10.57 shows the schematic of a twodimensional convergentdivergent nozzle with thrust vectoring of _+ 15 deg and thrust reversing. This is typical of the capabilities sought for use in future fighter aircraft.
10.5.2.4 Infrared signature. The rear of the engine is very hot and can produce a large infrared signature. Considerable effort is being used in modern military aircraft to reduce this signature. Most effective methods involve hiding the exit of the lowpressure turbine from direct view. Shown in Fig. 10.58 is a platypus exhaust duct used on the F117A Nighthawk stealth fighter. This exhaust nozzle is canted up 10 deg to prevent lingofsight to the turbine face. 6°'68 For afterburning engines, the nozzle in concert with the devices (e.g., flameholders) downstream of the turbine can hide the turbine exit and thus reduce high temperature signatures. Generally speaking, reduced engine performance is the cost of stealth.
INLETS NOZZLES, AND COMBUSTION SYSTEMS
735
ForwardThrustMode ~ ~ r ' D i v e r g e n t flapsprovide' ' ~ ~ . ' ~ [ optimumexpansion .
.
.
.
~ _ _
.
~
AglA9=IAearatt°
8
/
A
flap
9
= 1.28
(__ Lowerdivergent'a \
Convergentreversing Idle to intermediate power only Fully modulated
Idle to maximumpower..~. +15° capability
Vectored Fig. 10.57
Vectored
Typical twodimensional t h r u s t vectoring nozzle with t h r u s t r e v e r s i n g
(Ref. 68).
Side View Installation Sketch
(Engine)
Vertical TensionPosts (21 per tailpipe) B Top View (Engine) A
B Fig. 10.58
Section AA
Section BB
E x h a u s t nozzle of F117A ( f r o m Ref. 68).
736
ELEMENTS OF PROPULSION
A9
Station:
8
7
9
A 8 = Primary nozzle throat area A 9 = Secondary nozzle exit area a = Secondary nozzle halfangle 0 = Primary nozzle halfangle Ls = Secondary nozzle length
Fig. 10.59
10.5.3
Nozzle geometric parameters.
Nozzle Coefficients
Nozzle performance is ordinarily evaluated by two dimensionless coefficients: the gross thrust coefficient and the discharge or flow coefficient. Figure 10.59 shows a convergentdivergent exhaust nozzle with the geometric parameters used in the following definitions of nozzle coefficients. Only total pressure losses downstream of station 8 are included in the gross thrust coefficient.
10.5.3. 1 Gross thrust coefficient. The gross thrust coefficient Cfg is the ratio of the actual gross thrust Fg actual to the ideal gross thrust Fg iaea~, or
C f g  Fg  actual
( 10.17 )
F g ideal
Empirically derived coefficients are applied to Eq. (10.17) to account for the losses and directionality of the actual nozzle flow. Each engine organization uses somewhat different coefficients, but each of the following basic losses is accounted for: 1) Thrust loss due to exhaust velocity vector angularity 2) Thrust loss due to the reduction in velocity magnitude caused by friction in the boundary layers 3) Thrust loss due to loss of mass flow between stations 7 and 9 from leakage through the nozzle walls 4) Thrust loss due to flow nonuniformities
10.5.3.2 Discharge or flow coefficient. The ratio of the actual mass flow rh8 to the ideal mass flow rh8/is called the discharge coefficient CD:
m8 m8i
CD~:
(10.18)
INLETS NOZZLES, AND COMBUSTION SYSTEMS
737
This coefficient can be shown to be identically equal to the ratio of the effective onedimensional flow area required to pass the total actual nozzle flow ABe to the nozzle physical throat area A8 as follows:
CD
in8 _ P8 V s A s e _ a 8 e
th8i
P8V8A8
As
The variation of the discharge coefficient with nozzle pressure ratio is shown in Fig. 10.60a for a conic convergent nozzle. When the nozzle is choked, the
CD max
P,8/Po a) Convergent nozzle 1.00
0.98
0.96
0.94
0.92 0
I
10
20 30 Primary nozzle half angle 0 b)
40
CO m a x v s 0
CDmax I
1
2
3
4
Pt8]Po
c) CD nozzle Fig. 10.60
Nozzle discharge coefficient (Ref. 67).
738
ELEMENTS OF PROPULSION
discharge coefficient reaches a m a x i m u m value Co max The value o f Co m a x is a function of the primary nozzle halfangle 0, as shown in Fig. 10.60b. Figure 10.60c shows the variation in discharge coefficient for a convergentdivergent nozzle with nozzle pressure ratio. Note the change in behavior of Co between that of the convergentdivergent nozzle and that of the convergent nozzle as the nozzle pressure ratio drops below choking. This is due to the venturi behavior of the convergentdivergent nozzle. The discharge coefficient is used to size the nozzle throat area to pass the desired mass flow rate. For example, consider a nozzle with
Pt8 = 30psia,
rhs = 2 0 0 I b m / s ,
Tt8 = 2000°R,
R = 53.34 f t . l b f / ( l b m . °R),
3 / = 1.33
0 = 20 deg
At M8 = 1, G A S T A B with 3' = 1.33 and J g = 28.97, then M F P = 0.5224, and thus Ase = 570.7in. 2. Figure 10.60b gives CDmax = 0.96 for 0 = 2 0 d e g and thus the required throat area is 594.5 in. 2.
10.5.3.3 Velocity coefficient. The velocity coefficient Cv is the ratio of the actual exit velocity V9 to the ideal exit velocity ggi corresponding to Pt9 = Pt8, or 1/9 Vgi
Cv =  
(10.19)
and represents the effect of frictional loss in the boundary layer of the nozzle. It is mainly a function o f the nozzle ratio As/A9 and the halfangle a, as shown in Fig. 10.61. 1.000
0.996
Cv 6 deg 0.992
0.988
I
I deg
1.0
eg
I 2.0
[ 3.0
A9/A8
Fig. 10.61
CD nozzle velocity coefficient (Ref. 67).
I 4.0
INLETS NOZZLES, AND COMBUSTION SYSTEMS
739
V9 axial= V9cos aj
Fig. 10.62
L o c a l a n g u l a r i t y coefficient.
10.5.3.4 Angularity coefficient. The angularity coefficient CA represents the axial friction of the nozzle momentum; thus it is proportional to the thrust loss due to the nonaxial exit of the exhaust gas (see Fig. 10.62). For a differential element of flow, this coefficient is the cosine of the local exit flow angle aj. The local flow angle aj varies from zero at the centerline to a at the outer wall; thus, the nozzle angularity coefficient is the integral of aj across the nozzle:
,f
CA ~ ~
cos aj drn
(10.20)
Figure 10.63 presents the correlation of the angularity coefficient with the nozzle area ratio AsIA 9 and halfangle a. This figure is based on analytical evaluations of the inviscid flowfield in convergentdivergent nozzles for a range of practical nozzle geometries.
10.5.4
Nozzle Performance
Many nozzle coefficients simplify to algebraic expressions or become unity for the special case of onedimensional adiabatic flow. This is a useful
1.00
2
\
0.99
6 ...8 . 10
\ /
ca
/
~
12
.k 0.98
14 X Uncertain ~
0.97 1.0
/ /" \
I 1.2
I 1.4
~ \
I 1.6
~ I 1.8
A9[A8 Fig. 10.63
CD nozzle angularity coefficient (Ref. 67).
i 2.0
740
ELEMENTS OF PROPULSION
limit for understanding each coefficient and for preliminary analysis of nozzle performance using engine cycle performance data. For onedimensional adiabatic flow, CA = 1, m8 C o

•
A8e 
m8i
Pt8 
A8
et7
and the velocity coefficient Cv is given by [Eq. (10.19)]
V9 Vgi
Cv =  
where V9 is the exit velocity corresponding to Tt8 and (A/A*)9 ~ (Pt9/Pts) x [A9/(CDA8)] and Vgi is the ideal exit velocity corresponding to Tt8 and ( a / a * ) 9 i = A9/(CDA8).
The gross thrust for a onedimensional flow can be expressed as
rhsV9
Fgactual    
gc
and the ideal gross thrust (corresponds to
FgideaJ 
~ (P9  P o ) A 9
P9 =/DO)
(10.21)
as
/n8i Vs gc
(10.22)
where Vs is the isentropic exit velocity based on Pts/Po and Tt8. For onedimensional flow of a calorically perfect gas, Eq. (10.21) can be written as
Fgactual •
rh8V9[l_f y  I 1Po/P9 ] gc 2y (Pt9/Pg) (yI)/y  1
(10.23)
The gross thrust coefficient for onedimensional flow of a calorically perfect gas can be obtained by substituting Eqs. (10.22) and (10.23) into Eq. (10.17), giving
1 1  Po/P9 .] 2y (Pt9/P9) (71)/3'  1d
3/
(10.24)
This equation reduces to CA, = CDCv for the ideal expansion (P9 = P9i = Po). For isentropic flow, P 9 = P9i, Pt9 = Pts, Cv = 1, and Co = 1. Equation (10.24) is plotted in Fig. 10.64 for isentropic flow vs the nozzle area ratio A 9 / A s for different nozzle pressure ratios Pts/Po. Note that ideal expansion (P9 = P 0 ) gives a gross thrust coefficient of unity and that both underexpansion
INLETS NOZZLES, AND COMBUSTION SYSTEMS
741
1.00
o9t//11/\ \ \ \ \ C&
Ill/Ill/
090 ll [~ 
0.88 
1.0
Fig. 10.64
\
\
~
~
~
~
12 13 13 "
2
// VFlow separationlimit P9/Po 0.37 i i 2.0 3.0
A9/A8
I 4.0
5.0
Thrust coefficient for onedimensional isentropic flow (Y = 1.3).
(P9 > Po) and overexpansion (P9 < Po) reduces the gross thrust coefficient below unity. The extent of overexpansion in nozzles is limited by flow separation resulting from the interaction of the nozzle boundary layer and the strong oblique shock waves at the exit of the nozzle. In extreme overexpansion, Summerfield et al. 2° noted that the oblique shock waves moved from the exit lip into the nozzle (see Fig. 3.17), the flow downstream of the shock waves was separated in the vicinity of the wall, and as a result, the wall static pressure downstream of the shock waves was nearly equal to the ambient pressure Po. A simple estimate for the ratio of the pressure just preceding the shock waves P, to the ambient pressure Po, suggested by Summerfield et al., 2° is given by
Ps
~ P0
0.37
(10.25)
This flow separation limit can be included in the onedimensional gross thrust coefficient of Eq. (10.24) for isentropic flow by considering the effective exit pressure (P9 = P9i) to be the pressure just preceding the shock wave (Ps). Equations (10.24) and (10.25) were used to obtain the flow separation limit shown in Fig. 10.64. The design area ratio A9/A8 of convergentdivergent nozzles is selected such that the nozzle flow does not separate due to overexpansion for most throttle settings. This is because the increase in gross thrust coefficient associated with flow separation does not normally offset the accompanying increase in installation loss.
742
ELEMENTS OF PROPULSION 1.00
Fg cony
0.95
Fgideal 0.90 0.85 0
10
20
Pt8/Po Fig. 10.65 Ratio of convergent nozzle gross thrust to ideal gross thrust vs pressure ratio ( y = 1.3).
Nozzle pressure ratios are 3 to 5 in the subsonic cruise speed range of turbofan and turbojet engines. Typically, a subsonic engine uses a convergent exhaust nozzle. This is because, in the nozzle pressure range of 3 to 5, the convergent gross thrust (interception of lines with vertical axis, A9/A8 = 1) is 13% below the peak gross thrust (P9 = P0). Consequently, there may be insufficient gross thrust increase available in going to a convergentdivergent nozzle on a subsonic cruise turbofan or turbojet engine to pay for the added drag and weight of such a nozzle. In some applications, this loss in gross thrust coefficient of a convergent nozzle is too much, and a CD nozzle is used. The design pressure ratio across the nozzle increases rapidly with supersonic flight Mach number. At Mach 2, a pressure ratio of about 12 is typical. At this pressure ratio, the convergent nozzle gross thrust penalty is about 9%, as shown in Fig. 10.65. This figure is a plot of the ratio of the gross thrust in Fig. 10.64 of a convergent nozzle (A9/A 8 = 1) to the peak thrust (P9  P0) vs Pts/Po. Substitution of convergentdivergent nozzles for convergent nozzles provides large thrust gains for supersonic aircraft.
Example 10.5 Consider the calculation based on onedimensional flow. Given: th8 = 200 lbm/s A9
=2.0 A8
Pt9 =0.98 Pt8
Pt8 = 30 psia Tts = 2000°R
y = 1.33
C0=0.98
R=53.34ft.lbf/(lbm.°R) P0=5psia
Find the dimensions of an axisymmetric nozzle and the vaues of and Cv.
Cfg, Fg,
INLETS NOZZLES, AND COMBUSTION SYSTEMS
Solution: At M8 = 1 GASTAB MFP = 0.5224, thus
A8e =
with y = 1.33 and J # = 28.97, then
rhs4'~
PtsMFP(M8 = 1)
200 2~,'26~  570.7 in. 2 30 × 0.5224
With CD = 0.98, thus A8 = 582.3 in. 2 and r8 = 13.61 in. Since then A9 = 1165 in9 and r 9 = 19.25 in.:
(~)
A9
743
A9/A8 = 2.0,
2.0
9i. C. DA8. . 0.98
~
2.041
M9i = 2.168
and
Pt9iP9i= 0.0990
ThUS, P9i = (0.0990)(30 psia) = 2.970 psia,
W9i ~ ~
1
_
~Pt8,]
V9i = ~ / ( 1 7 1 6 ) ( 2 0 0 0 ) ~ [ 1 . (~2) Pt9 A9 _ 0 " 9 8 x 2 " 0 _ 2 . 0 9
Pt8 CDA8
0.98
 ( 0 . 0 9 9 0 ) °33/133] = 3475 ft/s + M 9 = 2.146 and p ~ = 0 . 1 0 2 5 t9
Thus, P9 = (0.1025)(0.98)(30 psia) = 3.014 psia, 2(1.33) 33] V9 = ~ / ( 1 7 1 6 ) ( 2 0 0 0 ) ~    d ~ [ 1  ( 0 . 1 0 2 5 ) 0"33/1 = 3456 ft/s
Cv = V9 = 0.9945 Vg,
............
% =
/1  (2.97/30) 0"33/1"33
v [0.33 15.0/3.014] X 1 t 2 X 1.33 (29.4/3.014) °33/j'33  1"
= 0.9593 Figure 10.3b gives 0 = 10 deg for CD = 0 . 9 8 . Likewise, Fig. 10.64 gives a = 6 deg for Cv = 0.9945 and A9/A8 2. Thus Ls = 54 in. and the dimensions of the exhaust nozzle are shown in Fig. 10.66. The gross thrust can be calculated
ELEMENTS OF PROPULSION
744
2
1
~
I~ 13.61 in. J
Centerline Fig. 10.66
~54 in.
~
19.25in. 1
Dimensions of Example 10.5 exhaust nozzle.
in several ways: directly from Eq. (10.23),
Fgactual = 
rhsV9 gc
q (P9  P0)A9
(200)(3456) F (3.014  5.0)(1165) 19,170 lbf 32.174
or from the ideal gross thrust and Cfg with
Vs=v/(1716)(2000)
2(1.33)[
~
1
30]
.]=3151ft/s
then
1718tVs
fgideal  gc
(200/0.98)(3151) 32.174
19,990 lbf
Fg actual = CugFg ideaJ  (0.9593)(19,990) = 19,170 lbf
10.6
Introduction to Combustion Systems
Combustion systems of aircraft gas turbine engines largely encompass the main burners (also called burners or combustors) and afterburners (also called augmenters or reheaters). Both main burners and afterburners are covered in this section because they have many basic principles in common. The basic principles of the combustion process, combustion stability, total pressure ratio, length scaling, diffusers, and fuels are presented in the following sections and provide the means for understanding the design of the main burner and afterburner. The thermal energy of the air/fuel mixture (reactants) flowing through an airbreathing engine is increased by the combustion process. The fuel must be vaporized and mixed with the air before this chemical reaction can occur. Once this is done, the combustion process can occur and thus increase the thermal energy of the mixture (products of combustion). All of this takes time and space.
INLETS NOZZLES, AND COMBUSTION SYSTEMS
745
The design of the main burner and afterburner of an airbreathing engine differs in many ways from that of conventional combustion devices. Space (especially length) is at a premium in aircraft applications, and the length of the combustion chamber is reduced by hastening completion of the combustion process. The combustion intensity (rate of energy released per unit volume) is much higher for the main burner of a turbojet [40,000 Btu/(s.ft 3) or 11,150 M W / m 3] compared to a typical steam power plant [10 Btu/(s. ft 3) or 2.8 MW/m3]. The following properties of the combustion chambers are desired: 1) Complete combustion 2) Low total pressure loss 3) Stability of combustion process 4) Proper temperature distribution at exit with no "hot spots" 5) Short length and small cross section 6) Freedom from flameout 7) Relightability 8) Operation over a wide range of mass flow rates, pressures, and temperatures However, many of these desirable properties are in competition. For example, both complete combustion and low total pressure loss are contrary to small size. Hence the design of a main burner or afterburner is a compromise. We examine many of these desirable combustion properties in order to understand the design and operation of main burners and afterburners, starting with the combustion process.
10.6.1
Combustion Process
The combustion processes occur with the vaporized fuel and air mixed on a molecular scale. The rate of this reaction depends on both the static pressure P and temperature T in a very complex way. For many situations, the reaction rate can be approximated by a form of the Arrhenius equation 67' 69 written for the mass rate of reaction as
Reaction rate cc
E
P~f(T)exp ~ T
(10.26)
where n is an exponent that depends on the number of molecules involved in a reactive collision (for example, n = 2 for two molecules, n = 3 for three molecules); f(T) is a function that relates the reaction rate to the forms of energy (translation, rotation, and vibration) the molecules have; the term exp[E/(~T)] accounts for the number of molecular collisions in which the energy of one molecule relative to another exceeds the activation energy E; and ~ is the universal gas constant. For hydrocarbonair combustion, n = 1.8. At low pressures, the reaction rate becomes slow and can become limiting for aircraft engines at very high altitudes. However, under most operating conditions, the rate of combustion is limited by the rate at which the fuel is vaporized and mixed with air. In most combustors, the fuel is injected as an atomized liquiddroplet spray into the hot reaction zone where it mixes with air and hot combustion gases. The atomized fuel vaporizes,
746
ELEMENTS OF PROPULSION
and then the vapor is mixed with air. If the temperature and pressure in the reaction zone are sufficiently high, the reaction rate will be fast and the fuel vapor will react as it comes in contact with sufficient oxygen. The stoichiometric fuel/air ratio for the typical hydrocarbon fuel can be estimated by assuming octane as a representative hydrocarbon and writing the stoichiometric chemical reaction: 2C8H18+25
O2+~N2
~
16CO2+18H20+25
~  N2
For this reason, the stoichiometric fuel/air ratio is found to be fstoich =
2(96 + 18) = 0.0664 25[32 + (79/21)28]
The equivalence ratio (o is the actual fuel/air ratio divided by the fuel/air ratio required for complete combustion (stoichiometric fuel/air ratio), or 4~ 
f
(10.27)
fstoich
The equivalence ratio ~bis greater than 1.0 for rich fuel/air ratio and less than 1.0 for a lean fuel/air mixture. To prevent excessive temperatures at the exit of the main burner or afterburner and protect its walls, the overall fuel/air ratio must be much less than stoichiometric with 4~ < 1.0. As an example, an engine being flown at Mach 0.9, a 12km altitude, and full throttle with a compressor pressure ratio of 20 and 85% isentropic efficiency will have a compressor pressure outlet temperature of about 653 K. If the turbine inlet temperature is limited to 1670 K, hpR is 42,800 kJ/kg, and Cpc and Cpt are 1.004 and 1.235 kJ/(kg K), then the fuel/air ratio for 100% efficient combustion in the main burner is, from Eq. (7.9), f _
1.235 x 1 6 7 0  1.004 × 653 = 0.0346 4 2 , 8 0 0  1.235 x 1670
which corresponds to an overall equivalence ratio of 0.519 for a main burner using JP8 fuel (fstoich = 0.0667). Figure 10.67 shows the flammability characteristics of a kerosenetype fuel. The 0.52 equivalence ratio of the preceding example is at the lower limit of flammability shown in Fig. 10.67. This presents a design problem at the fullthrottle value of ~b and at the engine's partialthrottle values of ~b. The problem of lean mixtures in a burner can be overcome by mixing and burning a rich fuel/air mixture in a small region where the local equivalence ratio is near unity. By using only a portion of the total air in a region, a locally rich mixture can be efficiently burned and then the products of combustion diluted and cooled to an acceptable turbine inlet temperature by the remaining air (see Fig. 10.77). At usual pressures and temperatures, hydrocarbon/air mixtures will react over only a rather narrow range of q~from approximately 0.5 to 3 and not at all below
INLETS NOZZLES, AND COMBUSTION SYSTEMS
'°F
/
6
4 .o
2
._~
1.0
747
Saturation line
Rich flammabilitylimit
0.6 0.4
0.2
0.1 I 0 Fig. 10.67 atmospheric
I 50
~¢'/
I I 100 150 Temperature (°C)
I 200
I 250
Flammability characteristics for a kerosene type fuel in air at pressure (Refs. 38 and 67).
0.2 atm at standard temperature. Hydrogen has much wider flammability limits than do hydrocarbonsapproximately 0.25 < ~b < 6 at standard temperature and 1 atm. Some special, rather expensive fuels have flammability limits intermediate between hydrogen and hydrocarbons. These special fuels have been used at times for testing and for extending the altitude limits of engines for special applications. A limitation is imposed by the combustion because it is necessary to maintain a stationary flame within a highvelocity airstream. Imagine the flame as propagating through the combustible mixture at the flame speed, while the mixture is carried downstream. To have a stable flame, the velocity of the mixture must be maintained within certain limits: If the velocity is too high, the flame will be "blown out" the exit; if it is too low, the flame will travel upstream and be extinguished. This problem of holding the combustion flame within the combustion system is solved by establishing regions of recirculation at the front of the main burner (see Fig. 10.77) or behind a bluff body, called a flame holder, at the front of an afterburner (see Fig. 10.87). These regions of recirculation create areas of local low velocity that "hold" the flame, and at the same time, the resulting turbulence gently increases the rate of energy transfer from these regions. Provided stable combustion is attained, complete combustion in the case of lean mixtures is virtually ensured since, with excess oxygen, local fuelrich areas are unlikely. On the other hand, combustion of a nearstoichiometric
748
ELEMENTS OF PROPULSION
mixture requires an essentially uniform distribution of constituents to avoid wasting some fuel in local fuelrich (oxygenpoor) regions.
10.6.2 Ignition Ignition of a fuel/air mixture in a turbine engine combustion system requires inlet air and fuel conditions within flammability limits, sufficient residence time of a combustible mixture, and location of an effective ignition source in the vicinity of the combustible mixture. The flammability limits for a kerosenetype fuel are shown in Fig. 10.67. Note that the flammability region is further subdivided into two regions separated by the spontaneous ignition temperature (SIT). The spontaneous ignition temperature is the lowest temperature at which visible or audible evidence of combustion is observed. Typical values of SIT are presented in Table 10.4. When the temperature in the combustion system is below the SIT, an ignition source is required to bring the local temperature above the spontaneous ignition temperature. The minimum amount of energy necessary to achieve ignition is shown in Fig. 10.68. Note that the minimum amount of energy is not always at a stoichiometric mixture ratio. For heavy fuels, such as C7H16, the minimum is nearer ~b = 2. Once the flammability limits and SIT requirements are met, then the ignition delay time becomes the key combustion characteristic. The ignition delay time tig n is related to the initial temperature T by E
tign 0