MAC 2282.002

22.6 ANALYSIS OF A TURBOJET ENGINE

Student: Abdel Ouhib

ADVISORS

Mathematics Advisor: Fernando BurgosAssociate Professor in Engineering Calculus

University of South Florida, Tampa, Fl.Subject Area Advisor: Scott Campbell

Associate Professor in Chemical EngineeringUniversity of South Florda, Tampa, Fl.

Problem Suggested by: Scott Campbell

Associate Professor in Chemical Engineering

University of South Florda, Tampa, Fl.

Fall 2011

PROBLEM STATEMENT Given certain airspeed and properties of a turbojet engine, the goal is to estimate the total power

delivered by the engine.

Contents

1. Abstract.......................................................................................................................................1

2. Motivation...................................................................................................................................1

3. Mathematical Description and Solution......................................................................................1

4. Discussion...................................................................................................................................6

5. Conclusions and Recommendations...........................................................................................7

6. Nomenclature..............................................................................................................................7

7. References...................................................................................................................................8

Abstract

Given certain air speed and air properties, it is possible to compute the total output of work done by a turbojet engine. There are five different major components of a turbojet engine, which include the diffuser, compressor, burner, turbine, and nozzle. To estimate the work done, it is necessary to work through all five parts of the engine. Air properties are only given at the diffuser, and the burner section of the engine. To compute the air properties for each section of the turbojet engine, several calculus and thermodynamics formulas can be used.

Motivation

When working with aeronautical engineering, it is important to understand how to compute the work done by the engine. Computing the work done by an engine can help determine how much thrust is necessary to fly at a certain speed. Scientists and engineers can also use the power of a turbojet engine to determine how much fuel would be needed to fly a certain altitude, airspeed, and length of time.

Mathematical Description

For this project, there are several relevant equations that can be used. The first equation to look at is the following.

Cp

R= 3.3+ 0.006T

This equation can be simplified to the following.

Cp = R(3.3+ 0.0006T )

For all intents and purposes, this will be used as Equation 1.The diffuser, compressor, turbine, and nozzle all operate isentropically, so Sout=Sin in the following equation.

Sout − Sin =

Cp

TdT − R ln Pout

PinTin

Tout

∫Since Sout=Sin=0, the following equation can be substituted.

R ln Pout

Pin=

Cp

TdT

Tin

Tout

∫Using equation 1, the integral can be solved and the equation can be simplified.

R 3.3+ 0.0006T

TdT = R ln Pout

PinTin

Tout

∫

1

This equation can be simplified as well.

3.3lnTout

Tin+ 0.0006 Tout −Tin( ) = ln Pout

PinThis equation will be known as Equation 2.Another equation is given to help solve for entropy given certain air velocities.

hin +

Vin2

s= hout +

Vout2

sThis equation can be simplified as well.

Vin2 −Vout

2

2= hin − hout

h2-h1 are not given in this problem. However, there is a formula that can help solve for hin-hout:

hout − hin = Cp dT

Tin

Tout

∫Equation 1 can be used to solve for the integral.

hin − hout = R(3.3T + 0.0003T 2 )]TinTout

This formula can be substituted back into the equation

Vin2 −Vout

2

2= hin − hout

.....

Vin2 −Vout

2

2= R[(3.3T + 0.0003T 2 )]Tin

Tout

This will be used as Equation 3. Since the project claims that the velocity drops to a negligible value, Vout2 can be ignored.

In a turbojet engine, the turbine powers the compressor. An equation to solve for the power of the compresor and turbine is given in the project. The equation is:

W = m(hin − hout )

M is 50 kg/s. We have already simplified hin-hout to be R[(3.3T + 0.0003T 2 )]TinTout

, so this formula can be simplified to:

W = (m)R[(3.3T + 0.0003T 2 )]TinTout

This will be known as Equation 4. The last equation that is used in this project is the equation that computes the work done by the engine. The equation is as follows:

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Wengine = m(V6 −V1)V1To solve for the total work, the variable needed is V6 because m and V1 are given in the project.

The final simplified equations that were used to solve the project are the following:

Equation 1: Cp = R(3.3+ 0.0006T ) Equation 2: 3.3lnTout

Tin+ 0.0006 Tout −Tin( ) = ln Pout

Pin

Equation 3:

Vin2 −Vout

2

2= R[(3.3T + 0.0003T 2 )]Tin

Tout

Equation 4: W = (m)R[(3.3T + 0.0003T 2 )]TinTout

Equation 5: Wengine = m(V6 −V1)V1

The problem gives the following givens:1. m=50 kg/s2. V1=250 m/s3. P1=0.3 bars4. T1=230 K5. R=0.287 kJ/kgK

In addition, the problem states that the velocity drops to a negligible value in the diffuser and remains negligible until the outlet of the nozzle. The compression ratio for Pout/Pin in the compressor is 10.0. The temperature of the air leaving the burner is 1400 K, as well as no pressure change within the burner. The last piece of information we are given is that the turbine produces only enough power to drive the compressor.

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Now that all simplified equations have been found, computations can start.

DIFFUSERThe beginning is the diffuser. Using the givens, and equation 3, it is possible to find the temperature out.

2502 − 02

= 287[3.3Tout − 3.3(230)+ 0.0003Tout2 − 0.0003(230)2 ]

Simplifying this equation, a quadratic equation is found.

0.0003Tout2 + 3.3T − 883.8

Using the quadratic formula, Tout=261.6 K. Now that Tout, Tin and Pin are known, Pout can be computed using equation 2.

3.3ln 261.6

230+ 0.0006 261− 230( ) = ln Pout

0.3Solving for Pout, the result comes to 0.47 bars. Energy balances, and the four requirements for the next component of the engine are known.

COMPRESSOR

For the compressor, Tin, Pin, and Pout are known, since the compression ratio of 10.0 is given. Using equation 2, Tout can be solved.

4

3.3ln Tout261.6

+ 0.0006(Tout − 261.6) = ln10

Tout is the only unknown, so solving for Tout, the result comes to 503 K. The compression ration is 10, so Pin(10)=4.7 bars.

The compressor is also driven by the turbine. The project claims that the compressor consumes all of the energy of the turbine. Now that all four requirements are given, the work can be computed using equation 4.

W = 50(287)[3.3(241.4)+ 0.0003(184,574.4)]Solving for work, the result comes to 12,226.1 kW. Every requirement has been solved for the compressor at this point.

BURNER

From the compressor, we are given Tin as well as Pin. The project tells us that Tout=1400 K. The project also tells us there is no pressure change within the burner, so Pin=Pout. All four requirements of the burner have been solved.

TURBINE

For this component, Tin=1400 K, and Pin=4.7 bars. We also know what the work done is, as it was computed in the compressor. Using equation 4, we can solve for Tout. Looking back at the work equation, the following equation can be written and solved.

−12,226.1= 50(.287)[3.3Tout − 4620 + 0.0003Tout2 − 588

When simplifying this equation, it becomes quadratic.0 = 0.0003Tout

2 + 3.3Tout − 4356When the quadratic equation is solved, Tout=1191 K. Now that three out of the four requirements have been solved for, equation 2 can be used.

3.3ln 11911400

+ 0.0006(1191−1400) = ln Pout4.7

Solving for Pout, the result comes to 2.45 bars. All four requirements have been solved for in this component of the engine.

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NOZZLE

Air is being expelled out of the engine, and back into the atmosphere. Knowing this, Pout can be assumed to drop back to atmospheric pressure at 0.3 bars. Pin=2.45, and Tin is 1191 K. Using this information, equation 2.

3.3ln Tout1191

+ 0.0006 Tout −1191( ) = ln 0.32.45

Solving for Tout, the result comes to 690 K. Using all of the given information, it is now possible to compute the final air speed by using equation 3.

0 −Vout2

2= 1.18(637 −1197)

This solves for the final airspeed, which comes out to 1150 m/s.

TOTAL WORK DONE BY ENGINE

Using the mass flow rate, the final air velocity, and the initial air velocity, we can now solve for the total work done by the engine.

Wengine = 50(1150 − 250)(250) = 11250kW

Discussion As the final result, the work done by the engine came out to be 11,250 kW. The objective of the project was to estimate, for a given air speed and air properties, the power delivered by the engine. The objective of this project was met by working through each component of the engine. Applying this problem directly to the field, it is shown that if a scientist or engineer knows the pressure, mass flow rate, temperature, and velocity of air as it enters the diffuser, then it is possible to determine the total work done by the engine by utilizing calculus and thermodynamics.

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Conclusions and Recommendations In this project, I was able to compute the total work done by an engine by using several calculus and thermodynamics equations, as well as several air properties. The project required me to work through each component of the engine and balance the energy of each component. After doing the project, I would recommend to anyone attempting the project to derive the equations necessary to do the problem from the calculus equations given. After working the problem the first few times by using only thermodynamics equations and assuming constants, my answer was close, but not exact. When using the given calculus equations, a rigorous calculation and much more accurate answer can be computed.

NomenclatureCp Heat capcity kJ/kgKR Universal gas constant kJ/kgKT Temperature KV Velocity m/sM Mass flow rate kg/sW Work kW

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References

Bouter, Zouhir. "Thermodynamics/Functions of a Jet Engine." Personal interview. 27 Nov. 2011.

Campbell, Scott. Personal interview. 6 Dec. 2011.

How A Jet Engine Works. How A Jet Engine Works. 12 Jan. 2007. Web. 1 Dec. 2011. ! <http://!www.youtube.com/watch?v=p1TqwAKwMuM>.

Stewart, James. Essential Calculus: Early Transcendentals. Belmont: Brooks/Cole, 2011. Print.

Tatum, J.B. "Physics - Thermodynamics." University of Victoria --- Astronomy -- A Window of ! the World Beyond. Freebookcentre. Web. 26 Nov. 2011. <http://!! www.astro.uvic.ca/~tatum/thermod.html>.

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