MAE 4261: AIR-BREATHING ENGINES Review for Exam 1 Exam 1: October 21, 2008 Mechanical and Aerospace Engineering Department Florida Institute of Technology.

MAE 4261: AIR-BREATHING ENGINES

Review for Exam 1

Exam 1: October 21, 2008

Mechanical and Aerospace Engineering Department

Florida Institute of Technology

D. R. Kirk

READING: HILL AND PETERSON• Chapter 1

– Can a jet or rocket engine exert thrust while discharging into a vacuum (with no atmosphere to “push against”)? YES

– Could a rocket vehicle be propelled to a speed much higher than the speed at which the jet leaves the rocket nozzle? YES, See Homework 1, Problem 2

• Chapter 2: 2.1-2.3 and Chapter 3: 3.1-3.4

– If you need a review of mass, momentum and energy equations

– Detailed review of thermodynamics is located online (Lecture 6)

• Chapter 5:

– 5.1 and 5.2: Review of overall concepts, thrust and efficiency definitions (See Lectures 4 and 5)

– 5.3: Ramjet Engines (See Lectures 7 and 8)

• Be familiar with trends shown in Figures 5.9 and 5.10

– 5.4: Turbojet Engines (See Lectures 9 and 10)

• Be familiar with trends shown in Figures 5.19-5.22

– 5.5: Turbofan Engines (See Lecture 11)

• Be familiar with trends shown in Figures 5.29-5.34

CROSS-SECTIONAL EXAMPLE: GE 90-115B

• Why does this engine look the way that it does?

• How does this engine push an airplane forward, i.e. how does it generate thrust?

• What are major components and design parameters?

• How can we characterize performance and compare with other engines?

CONSERVATION OF MASS

• This is a single scalar equation

– Velocity doted with normal unit vector results in a scalar

• 1st Term: Rate of change of mass inside CV

– If steady d/dt( ) = 0

– Velocity, density, etc. at any point in space do not change with time, but may vary from point to point

• 2nd Term: Rate of convection of mass into and out of CV through bounding surface, S

• 3rd Term (=0): Production or source terms

0ˆ dSnUdVdt

d

CV S

0ˆ dSnUUdVdt

d

CV S

CS

MOMENTUM EQUATION: NEWTONS 2nd LAW

FdSnUUdVUdt

d

CV S

ˆ

FdSnUUUdVUdt

d

CV S

CS

ˆ

• This is a vector equation in 3 directions

• 1st Term: Rate of change of momentum inside CV or Total (vector sum) of the momentum of all parts of the CV at any one instant of time

– If steady d/dt( ) = 0

– Velocity, density, etc. at any point in space do not change with time, but may vary from point to point

• 2nd Term: Rate of convection of momentum into and out of CV through bounding surface, S or Net rate of flow of momentum out of the control surface (outflow minus inflow)

• 3rd Term:

– Notice that sign on pressure, pressure always acts inward

– Shear stress tensor, , drag

– Body forces, gravity, are volumetric phenomena

– External forces, for example reaction force on an engine test stand

• Application of a set of forces to a control volume has two possible consequences

1. Changing the total momentum instantaneously contained within the control volume, and/or

2. Changing the net flow rate of momentum leaving the control volume

ext

CVSS

FdVgdSdSnpF

ˆ

HOW AN AIRCRAFT ENGINE WORKS

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VVmT

APPVmVmT

• Flow through engine is conventionally called THRUST

– Composed of net change in momentum of inlet and exit air

• Fluid that passes around engine is conventionally called DRAG

ChemicalEnergy

ThermalEnergy

KineticEnergy

NON-DIMENSIONAL THRUST EQUATION

1

1

00

0

00

0

00

00

0

0

00

V

VM

am

T

RTa

a

VM

V

VV

m

T

VVm

T

VVmT

APPVmVmT

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Result from control volumeanalysis employing conservationof mass and momentum equation

Writing right side as a velocity ratio

Introduce non-dimensional Mach number, M0

Speed of sound, a0

Non-dimensional or Specific ThrustEquation is only conservation of mass and momentumStarting point for all analyses (ramjet, turbojet, turbofan)

NOW INTRODUCE THERMODYNAMICS

1

1

1

1

00

00

00

00

00

000

0

0

TM

TMM

am

T

RTM

RTMM

am

T

RTa

RTa

aM

aMM

am

T

V

VM

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Non-Dimensional result from control volumeanalysis employing conservation of mass and momentum equation

Goal is to tie this equation in with behavior of the engine, which is characterized thermodynamically

Introduce V=Ma, which introduces Mach number and speed of sound, which depends on temperature

For the ideal cycle analysis, assume that the specific heat ratio, , and the gas constant R are remain unchanged throughout the engine

Non-dimensional or Specific ThrustEquation now ties in mass, momentum and energyStarting point for all analyses (ramjet, turbojet, turbofan)Find Me and Te by accounting Tt and Pt through engine

MAJOR COMPONENTS: TURBOJET(LOW BYPASS RATIO TURBOFAN)

EXAMPLE OF COMMERCIAL ENGINE: HIGH BYPASS RATIO TURBOFAN

MAJOR GAS TURBINE ENGINE COMPONENTS

1. Inlet:

– Continuously draw air into engine through inlet

– Slows, or diffuses, to compressor

2. Compressor / Fan:

– Compresses air

– Generally two, or three, compressors in series

– Raises stagnation temperature and pressure (enthalpy) of flow

– Work is done on the air

3. Combustor:

– Combustion or burning processes

– Adds fuel to compressed air and burns it

– Converts chemical to thermal energy

– Process takes place at relatively constant pressure

MAJOR GAS TURBINE ENGINE COMPONENTS

4. Turbine:

– Generally two or three turbines in series

– Turbine powers, or drives, the compressor

– Air is expanded through turbine (P & T ↓)

– Work is done by the air on the blades

– Use some of that work to drive compressor

– Next:

• Expand in a nozzle

– Convert thermal to kinetic energy (turbojet)

– Burning may occur in duct downstream of turbine (afterburner)

• Expand through another turbine

– Use this extracted work to drive a fan (turbofan)

5. Nozzle:

– Flow is ejected back into the atmosphere, but with increased momentum

– Raises velocity of exiting mass flow

ENGINE STATION NUMBERING CONVENTION

ENGINE STATION NUMBERING CONVENTION

2.0-2.5: Low Pressure Compressor

2.5+: High PressureCompressor

3: Combustor

4: Turbine

0: Far Upstream

1: Inlet

5-6: Nozzle

One of most important parameters is TT4: Turbine Inlet TemperaturePerformance of gas turbine engine ↑ with increasing TT4 ↑

8

7

TYPICAL PRESSURE DISTRIBUTION THROUGH ENGINE

AIRCRAFT ENGINE BASICS

• All aircraft engines are HEAT ENGINES

– Utilize thermal energy derived from combustion of fossil fuels to produce mechanical energy in the form of kinetic energy of an exhaust jet

– Momentum excess of exhaust jet over incoming airflow produces thrust

– Remember: Thrust is a Force and Force = Time Rate Change of Momentum

• In studying these devices we will employ two types of modeling

1. Thermodynamic (Cycle Analysis)

• Thermal → mechanical energy from thermal is studied using thermodynamics

• Change in thermodynamic state of air as it passes through engine is studied

• Physical configuration (geometry) of engine NOT important, but rather processes are important

2. Fluid Mechanic

• Relate changes in pressure, temperature and velocity of air to physical characteristics of engine

INTRODUCTION TO CYCLE ANALYSIS

• Cycle Analysis → What determines engine characteristics?

• Cycle analysis is study of thermodynamic behavior of air as it flows through engine without regard for mechanical means used to affect its motion

• Characterize components by effects they produce

• Actual engine behavior is determined by geometry; cycle analysis is sometimes characterized as representing a “rubber engine”

• Main purpose is to determine which characteristics to choose for components of an engine to best satisfy a particular need

– Express T, , Isp as function of design parameters

STAGNATION QUANTITIES DEFINED• Quantities used in describing engine performance are the stagnation pressure,

enthalpy and temperature

• Stagnation enthalpy, ht , enthalpy state if stream is decelerated adiabatically to zero velocity

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pc

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Ideal gas

Stagnation temperature

Speed of sound

Total to static temperature ratioin terms of Mach number

FOR REVERSIBLE + ADIABATIC = ISENTROPIC PROCESS

flow speed lowfor Equation" Bernouli"

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get to theorembinomial theusing expand ,12For

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velocity)zero ally toisentropic ddecelerate is stream if pressure is (

pressure stagnation thedefines 1

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find we using

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THERMODYNAMIC PROCESSES IN THE ENGINE

• How should we represent thermodynamic process in engine?

• It is cyclic

– Air starts at atmospheric pressure and temperature and ends up at atmospheric pressure and temperature

– Definition of ‘Open’ vs. ‘Closed’ Cycles

• Consider a parcel of air taken round a cycle with heat addition and rejection

• Need to consider thermodynamics of propulsion cycle

• To do this we make use of First and Second Laws of Thermodynamics

THERMODYANMICS: BRAYTON CYCLE MODEL

• 1-2: Inlet, Compressor and/or Fan: Adiabatic compression with spinning blade rows

• 2-3: Combustor: Constant pressure heat addition

• 3-4: Turbine and Nozzle: Adiabatic expansion

– Take work out of flow to drive compressor

– Remaining work to accelerate fluid for jet propulsion

• Thermal efficiency of Brayton Cycle, th=1-T1/T2

– Function of temperature or pressure ratio across inlet and compressor

P-V DIAGRAM REPRESENTATION

• Thermal efficiency of Brayton Cycle, th=1-T1/T3

– Function of temperature or pressure ratio across inlet and compressor

BYPASS RATIO: TURBOFAN ENGINES

Bypass Air

Core Air

Bypass Ratio, B, :Ratio of bypass air mass flow rate to core mass flow rateExample: Bypass ratio of 6:1 means that air mass flow through fan and bypassing core engine is six times air mass flow flowing through core

TRENDS TO HIGHER BYPASS RATIO

1958: Boeing 707, United States' first commercial jet airliner 1995: Boeing 777, FAA Certified

PW4000-112: T=100,000 lbf , ~ 6Similar to PWJT4A: T=17,000 lbf, ~ 1

COMMERCIAL AND MILITARY ENGINES(APPROX. SAME THRUST, APPROX. CORRECT RELATIVE SIZES)

• Demand high T/W• Fly at high speed• Engine has small inlet area

(low drag, low radar cross-section)

• Engine has high specific thrust

• Ue/Uo ↑ and prop ↓ P&W 119 for F- 22, T~35,000 lbf, ~ 0.3

• Demand higher efficiency • Fly at lower speed (subsonic, M∞ ~ 0.85)• Engine has large inlet area• Engine has lower specific thrust• Ue/Uo → 1 and prop ↑

GE CFM56 for Boeing 737 T~30,000 lbf, ~ 5

EFFICIENCY SUMMARY• Overall Efficiency

– What you get / What you pay for

– Propulsive Power / Fuel Power

– Propulsive Power = TUo

– Fuel Power = (fuel mass flow rate) x (fuel energy per unit mass)

• Thermal Efficiency

– Rate of production of propulsive kinetic energy / fuel power

– This is cycle efficiency

• Propulsive Efficiency

– Propulsive Power / Rate of production of propulsive kinetic energy, or

– Power to airplane / Power in Jet

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PROPULSIVE EFFICIENCY AND SPECIFIC THRUST AS A FUNCTION OF EXHAUST VELOCITY

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ENGINE AND OVERALL AIRPLANE PERFORMANCE

• Most MAE 4261 lectures focused on characterizing propulsion system

• Also look at behavior of entire airplane

– Which parameters from engine performance feed directly into overall airplane performance (Thrust, Isp, TSFC, etc.)

– How fast can airplane fly?

– How far can airplane fly on a single tank of fuel (range)?

– How long can airplane stay in air on a single tank of fuel (endurance)?

• Tie in MAE 4261 with aerodynamics and structures

3 TYPES OF AIR-BREATHING ENGINES

• Apply cycle analysis to control volume result for conservation of mass, momentum and energy

• Consider 3 engine types

1. Ramjets

2. Turbojets

3. Turbofans

Symbol Physical Description Ratio of stagnation (total) pressures across component

(d: diffuser (inlet), c: compressor, b: burner (combustor), t: turbine, a: afterburner, n: nozzle)

Ratio of stagnation (total) temperatures across component (d: diffuser (inlet), c: compressor, b: burner (combustor), t: turbine, a: afterburner, n: nozzle)

Ratio of stagnation (total) pressure to ambient static pressure, p0 Ratio of stagnation (total) temperature to ambient static temperature, T0

RAMJETS

• Thrust performance depends solely on total temperature rise across burner

• Relies completely on “ram” compression of air (slowing down high speed flow)

• Ramjet develops no static thrust

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bMam

T hm

TU

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0

Energy (1st Law) balance across burnerCycle analysis employing general form of mass, momentum and energy

RAMJET RESULTS

RAMJET RESULTS

TURBOJET SUMMARY

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Mam

T

1

11

1

20

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Cycle analysis employing general form of mass, momentum and energy

Turbine power = compressor power

How do we tie in fuel flow, fuel energy?Energy (1st Law) balance across burner

TURBOJET RESULTS

Plot of Non-Dimensional Thrust and Specific Impulse for Maximum Thrust Condition

Heating Value of Fuel = 4.3x107 J/kg, Specific Heat Ratio = 1.4, T0=200K

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.5 1 1.5 2 2.5 3

Flight Mach Number

No

n-D

ime

ns

ion

al

Th

rus

t, M

ax

imu

m T

hru

st

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

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ific

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pu

lse

, M

ax

imu

m T

hru

st,

s

Max Non-Dim Thrust: Theta_t=6

Max Non-Dim thrust: Theta_t=9

Max Thrust Isp: Theta_t=6

Max Thrust Isp: Theta_t=9

TURBOFAN SUMMARY

00 1

1

21

1

2MM

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Tfo

co

ttco

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00 1

1

21 M

am

Tf

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2

max

11

1

1

21 M

am

T t

o

Two streams:Core and Fan Flow

Turbine power = compressor + fan powerExhaust streams have same velocity: U6=U8

Maximum power, c selectedto maximize f

TURBOFAN RESULTS

Non-Dimensional Thrust vs. Flight Mach Numbert=6, To=200 K (PW4000 Series, ~ 5-6)

Higher of interest in range of Mo < 1 and lower of interest for supersonic transport

0

2

4

6

8

10

12

14

16

0 0.5 1 1.5 2 2.5 3

Flight Mach Number, Mo

No

n-D

ime

ns

ion

al

Th

rus

t

Bypass Ratio = 1

Bypass Ratio = 5

Bypass Ratio = 10

Bypass Ratio = 20

TURBOFAN RESULTS

Propulsive Efficiency vs. Flight Mach Numbert=6, To=200 K

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5 3

Flight Mach Number, Mo

Pro

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ien

cy

Bypass Ratio = 1

Bypass Ratio = 5

Bypass Ratio = 10

Bypass Ratio = 20