MAE 4261: AIR-BREATHING ENGINES

Exam 2 Review

Exam 2: November 18th, 2008

Mechanical and Aerospace Engineering Department

Florida Institute of Technology

D. R. Kirk

EXAM 2 TOPICS

• Turbofans

– Section 5.5

– Figure 5.23, 5.29, 5.30, 5.31, 5.32, 5.33, 5.34

• Non-rotating components

– Inlets: Section 6.1- (first half of) 6.3

– Nozzles: Section 6.7

– Combustors (burners): Section 6.4-6.5

• Figures 6.21, 6.22, 6.23, 6.24, 6.26

• Energy exchange with moving blade rows

– Section 7.1-7.2

• Axial compressors

– Section 7.3-7.7

– Figures 7.7, 7.8, 7.11, 7.12, 7.27, 7.32, 7.33

BYPASS RATIO: TURBOFAN ENGINES

Bypass Air

Core Air

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

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

hm

TU

f

ooverall

hm

UmUm

f

ooee

thermal

22

22

propulsivethermaloverall

o

eooee

opropulsive

UUUmUm

TU

1

2

22

22

PROPULSIVE EFFICIENCY AND SPECIFIC THRUST AS A FUNCTION OF EXHAUST VELOCITY

o

epropulsive

U

U

1

2

1o

e

o U

U

Um

T

Conflict

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

CRUISE FUEL CONSUMPTION vs. BYPASS RATIO

SUMMARY OF IN-CLASS EXAMPLES

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

1000

bMam

T hm

TU

foverall

0

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

TURBOJET SUMMARY

oco

ttco

oo

Mam

T

11

2

occo

too

Mam

T

1

11

1

20

0

000 1

ctoverall

am

TM

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 TRENDS: IN-CLASS EXAMPLE

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

Max

imu

m S

pec

ific

Th

rust

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

Sp

ecif

ic I

mp

uls

e, M

axim

um

T

hru

st,

s

Max Non-Dim Thrust: Theta_t=6Max Non-Dim Thrust: Theta_t=9Max Thrust Isp: Theta_t=6Max Thrust Isp: Theta_t=9

TURBOJET TRENDS: IN-CLASS EXAMPLE

Plot of Thrust Normalized by Compressor Inlet Area and Ambient Pressurevs. Flight Mach Number for Compressor Inlet Mach Number, M2=0.5

0

5

10

15

20

25

30

0 0.5 1 1.5 2 2.5 3

Flight Mach Number

Th

rust

No

rmal

ized

by

A2

and

P0

Theta_t=6

Theta_t=9

TURBOJET TRENDS: HOMEWORK #3, PART 1Tt4 = 1600 K, c = 25, T0 = 220 K

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

4.50

5.00

0 0.5 1 1.5 2 2.5 3

Mach Number

Sp

ec

ific

Th

rus

t

0%

20%

40%

60%

80%

100%

120%

Eff

icie

nc

y

Specific ThrustPropulsive EfficiencyThermal EfficiencyOverall Efficiency

TURBOJET TRENDS: HOMEWORK #3, PART 2a Tt4 = 1400 K, T0 = 220 K, M0 = 0.85 and 1.2

0.00

0.50

1.00

1.50

2.00

2.50

3.00

0 10 20 30 40 50

Compressor Pressure Ratio

Sp

ec

ific

Th

rus

t

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

Eff

icie

nc

y

Specific Thrust, M=0.85Specific Thrust, M=1.2Propulsive Efficiency, M=0.85Thermal Efficiency, M=0.85Overall Efficiency, M=0.85Propulsive Efficiency, M=1.2Thermal Efficiency, M=1.2Overall Efficiency, M=1.2

TURBOJET TRENDS: HOMEWORK #3, PART 2b Tt4 = 1400 K and 1800 K, T0 = 220 K, M0 = 0.85

0

0.5

1

1.5

2

2.5

3

3.5

4

0 10 20 30 40 50

Compressor Pressure Ratio

Sp

ec

ific

Th

rus

t

0%

10%

20%

30%

40%

50%

60%

70%

80%

Specific Thrust, Tt4=1400KSpecific Thrust, Tt4=1800 KPropulsive Efficiency, Tt4=1400 KThermal Efficiency, Tt4=1400 KOverall Efficiency, Tt4=1400 KPropulsive Efficiency, Tt4=1800 KThermal Efficiency, Tt4=1800 KOverall Efficiency, Tt4=1800 K

TURBOFAN SUMMARY

00 1

1

21

1

2MM

am

Tfo

co

ttco

o

00 1

1

21 M

am

Tf

o

00

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 TRENDS: IN-CLASS EXAMPLE

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 3Flight Mach Number, M0

No

n-D

ime

ns

ion

al T

hru

st

Bypass Ratio = 1Bypass Ratio = 5Bypass Ratio = 10Bypass Ratio = 20

TURBOFAN TRENDS: IN-CLASS EXAMPLE

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, M0

Pro

pu

lsiv

e E

ffic

ien

cy

Bypass Ratio = 1Bypass Ratio = 5Bypass Ratio = 10Bypass Ratio = 20

INLETS

OVERVIEW: INLETS AND DIFFUSERS

• Purpose:

1. Capture incoming stream tube (mass flow)

2. Condition flow for entrance into compressor (and/or fan) over full flight range

• At cruise, slow down flow to 0.4 < M2 < 0.7

• At take-off, accelerate flow to 0.4 < M2 < 0.7

• Remain as insensitive as possible to angle of attack, cross-flow, etc.

• Requirements

1. Bring inlet flow to engine with high possible stagnation pressure

• Measured by inlet pressure recovery, d = Pt2/Pt1

2. Provide required engine mass flow

• May be limited by choking of inlet

3. Provide compressor (and/or fan) with uniform flow

EFFECT OF MASS FLOW ON THRUST VARIATION

• Mass flow into compressor = mass flow entering engine

• Re-write to eliminate density and velocity

• Connect to stagnation conditions at station 2

• Connect to ambient conditions

• Resulting expression for thrust

– Shows dependence on atmospheric pressure and cross-sectional area at compressor or fan entrance

– Valid for any gas turbine

002

0

0002

12

1

22

212

1

22

0

0

2

0

12

1

22

2

22

2

2

22

2222

222

2

2

2222

21

12

11

21

1

aPA

m

am

T

PA

T

M

MM

RT

P

A

m

M

M

RTP

A

m

MRT

PRTMRT

PU

A

m

AUm

t

NON-DIMENSIONAL THRUST FOR A2 AND P0

• Thrust at fixed altitude is nearly constant up to Mach 1

• Thrust then increases rapidly, need A2 to get smaller

Plot of Thrust Normalized by Compressor Inlet Area and Ambient Pressurevs. Flight Mach Number for Compressor Inlet Mach Number, M2=0.5

0

5

10

15

20

25

30

0 0.5 1 1.5 2 2.5 3

Flight Mach Number

Th

rus

t / (

A2

P0

)

Theta_t=6

Theta_t=9

OPERATIONAL OVERVIEW

High ThrustLow Speed, M0 ~ 0High Mass FlowStream Tube Accelerates

Low ThrustHigh Speed, M0 ~ 0.8Low Mass FlowStream Tube Decelerates

2

020

1

21

U

U

AU

T i

iAerodynamic force is always favorable for thrust production

NORMAL SHOCK TOTAL PRESSURE LOSSES

• As M1 ↑ p02/p01 ↓ very rapidly• Total pressure is indicator of how much useful work can be done by a flow

– Higher p0 → more useful work extracted from flow• Loss of total pressure are measure of efficiency of flow process

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 1.5 2 2.5 3 3.5 4 4.5 5

Upstream Mach Number, M1

M2,

P02

/P01

Downstream Mach Number, M2

Total Pressure Ratio, P02/P01

Example: Supersonic Propulsion System

• Engine thrust increases with higher incoming total pressure which enables higher pressure increase across compressor

• Modern compressors desire entrance Mach numbers of around 0.5 to 0.8, so flow must be decelerated from supersonic flight speed

• Process is accomplished much more efficiently (less total pressure loss) by using series of multiple oblique shocks, rather than a single normal shock wave

NOZZLES

OVERVIEW: NOZZLES• Subsonic Aircraft: Usually a fixed area convergent nozzle is adequate

– Can be more complex for noise suppression• Supersonic Aircraft: More complex, variable-area, convergent-divergent device• Two Primary Functions:

1. Provide required throat area to match mass flow and exit conditions2. Efficiently expand high pressure, high temperature gases to atmospheric pressure

(convert thermal energy → kinetic energy)

KEY EQUATIONS FOR NOZZLE DESIGN

12

1

28

87

8

12

1

22

22

7

7

2

22

7

21

21

11

21

1

1

M

MA

A

A

A

MM

T

T

P

P

A

A

A

A

throat

exit

t

t

t

tthroat

Nozzle area ratio as a function of engine parameters

Once nozzle area is set, operating point of engine depends only on t

A7 is the throat area, how do we find the exit area of the nozzle?Found from compressible channel flow relations, recall that M7=1Set by jet stagnation pressure and ambient

Compare with Equation 3.15

Compare with Section 6.7 H&P

COMBUSTORS

MAJOR COMBUSTOR COMPONENTS

• Key Questions:

– Why is combustor configured this way?

– What sets overall length, volume and geometry of device?

Com

pres

sor

Tur

bine

Air

Fuel

Combustion Products

WHY IS THIS RELEVANT?• Most mixtures will NOT burn so far away from

stoichiometric– Often called Flammability Limit– Highly pressure dependent

• Increased pressure, increased flammability limit

– Requirements for combustion, roughly > 0.8

• Gas turbine can NOT operate at (or even near) stoichiometric levels– Temperatures (adiabatic flame temperatures)

associated with stoichiometric combustion are way too hot for turbine

– Fixed Tt4 implies roughly < 0.5

• What do we do?– Burn (keep combustion going) near =1 with

some of ingested air– Then mix very hot gases with remaining air to

lower temperature for turbine

SOLUTION: BURNING REGIONS

Air

Com

pres

sor

Tur

bine

~ 1.0T>2000 K

~0.3

PrimaryZone

RELATIVE LENGTH OF AFTERBURNER

• Why is AB so much longer than primary combustor?

– Pressure is so low in AB that they need to be very long (and heavy)

– Reaction rate ~ pn (n~2 for mixed gas collision rate)

J79 (F4, F104, B58)

Combustor Afterburner

AXIAL COMPRESSORS

WHERE IN THE ENGINE? PW2000

FanCompressor

2 SPOOL DEVICE: PW2000

High Pressure Compressor (high)

Low Pressure Compressor (low)

High and Low Pressure Turbines

REVIEW: PRESSURE DISTRIBUTION• Rotor

– Adds swirl to flow

– Adds kinetic energy to flow with ½v2

– Increases total energy carried in flow by increasing angular momentum

• Stator

– Removes swirl from flow

– Not a moving blade → cannot add any net energy to flow

– Converts kinetic energy associated with swirl to internal energy by raising static pressure of flow

– NGV adds no energy. Adds swirl in direction of rotor motion to lower Mach number of flow relative to rotor blades (improves aerodynamics)

AXIAL COMPRESSOR ENERGY EXCHANGE• Rotor

– Adds swirl to flow

– Adds kinetic energy to flow with ½v2

– Increases total energy carried in flow by increasing angular momentum

• Stator

– Removes swirl from flow

– Not a moving blade → cannot add any net energy to flow

– Converts kinetic energy associated with swirl to internal energy by raising static pressure of flow

– NGV adds no energy. Adds swirl in direction of rotor motion to lower Mach number of flow relative to rotor blades (improves aerodynamics)

Centerline

EXAMPLES OF BLADE TWIST