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, a:
Ratio of bypass air flow rate to core flow rate
Example: 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
Similar to PWJT4A: T=17,000 lbf, a ~ 1
1995: Boeing 777, FAA Certified
PW4000-112: T=100,000 lbf , a ~ 6
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
TU o
overall 
m f h
 m eU e2 m oU o2 



2
2 

thermal 
m f h
 propulsive 
TU o
2

 m eU e2 m oU o2  1  U e



Uo
2 
 2
overall  thermal propulsive
PROPULSIVE EFFICIENCY AND SPECIFIC THRUST AS A
FUNCTION OF EXHAUST VELOCITY
Ue
T

1
m U o U o
Conflict
2
 propulsive 
Ue
1
Uo
COMMERCIAL AND MILITARY ENGINES
(APPROX. SAME THRUST, APPROX. CORRECT RELATIVE SIZES)
GE CFM56 for Boeing 737 T~30,000 lbf, a ~ 5
•
•
•
•
•
•
•
•
•
•
Demand high T/W
Fly at high speed
Engine has small inlet area
(low drag, low radar crosssection)
Engine has high specific
thrust
Ue/Uo ↑ and prop ↓
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 ↑
P&W 119 for F- 22, T~35,000 lbf, a ~ 0.3
CRUISE FUEL CONSUMPTION vs. BYPASS RATIO
SUMMARY OF IN-CLASS EXAMPLES
RAMJETS


T
 M 0  b 1
m 0 a0
Cycle analysis employing general form
of mass, momentum and energy
 overall
TU 0

m f h
Energy (1st Law) balance across burner
• 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
TURBOJET SUMMARY
 t
T
2
 o c t  1

m o ao
 1
  o c
T
2  
1

 t 1 
m o a o
  1    o c
 overall

  M o



   0  c  1  M o


 T
M 0   1
 m 0 a 0

 t   c 0 



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
7
Heating Value of Fuel = 4.3x10 J/kg, Specific Heat Ratio = 1.4, T0=200K
10000
5
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
Maximum Specific Thrust
4
9000
8000
3.5
7000
3
6000
2.5
5000
2
4000
1.5
3000
1
2000
0.5
1000
0
0
0
0.5
1
1.5
Flight Mach Number
2
2.5
3
Specific Impulse, Maximum
Thrust, s
4.5
TURBOJET TRENDS: IN-CLASS EXAMPLE
Plot of Thrust Normalized by Compressor Inlet Area and Ambient Pressure
vs. Flight Mach Number for Compressor Inlet Mach Number, M 2=0.5
30
Thrust Normalized by A2 and P0
Theta_t=6
Theta_t=9
25
20
15
10
5
0
0
0.5
1
1.5
Flight Mach Number
2
2.5
3
TURBOJET TRENDS: HOMEWORK #3, PART 1
Tt4 = 1600 K, pc = 25, T0 = 220 K
5.00
120%
4.50
100%
4.00
80%
3.00
2.50
60%
2.00
40%
1.50
Specific Thrust
Propulsive Efficiency
Thermal Efficiency
Overall Efficiency
1.00
0.50
20%
0.00
0%
0
0.5
1
1.5
Mach Number
2
2.5
3
Efficiency
Specific Thrust
3.50
TURBOJET TRENDS: HOMEWORK #3, PART 2a
Tt4 = 1400 K, T0 = 220 K, M0 = 0.85 and 1.2
3.00
90%
80%
2.50
2.00
60%
50%
1.50
40%
1.00
30%
Specific Thrust, M=0.85
Specific Thrust, M=1.2
Propulsive Efficiency, M=0.85
Thermal Efficiency, M=0.85
Overall Efficiency, M=0.85
Propulsive Efficiency, M=1.2
Thermal Efficiency, M=1.2
Overall Efficiency, M=1.2
0.50
20%
10%
0.00
0%
0
10
20
30
Compressor Pressure Ratio
40
50
Efficiency
Specific Thrust
70%
Specific Thrust
TURBOJET TRENDS: HOMEWORK #3, PART 2b
Tt4 = 1400 K and 1800 K, T0 = 220 K, M0 = 0.85
4
80%
3.5
70%
3
60%
2.5
50%
2
40%
1.5
30%
1
20%
Specific Thrust, Tt4=1400K
Specific Thrust, Tt4=1800 K
Propulsive Efficiency, Tt4=1400 K
Thermal Efficiency, Tt4=1400 K
Overall Efficiency, Tt4=1400 K
Propulsive Efficiency, Tt4=1800 K
Thermal Efficiency, Tt4=1800 K
Overall Efficiency, Tt4=1800 K
0.5
10%
0
0%
0
10
20
30
Compressor Pressure Ratio
40
50
TURBOFAN SUMMARY

T
2
 o c t  1  t

m ao
 1
  o c



2

  M o   





1

M
0
  1 0 f



 2

T
 0 f  1  M 0 
 1   

ma o
  1

 T

 m ao


Two streams:
Core and Fan Flow
Turbine power = compressor + fan power
Exhaust streams have same velocity: U6=U8


  1 2


2
t

  1   
  0  1  M 0 
  1  1 


 max




Maximum power, c selected
to maximize f
TURBOFAN TRENDS: IN-CLASS EXAMPLE
Non-Dimensional Thrust vs. Flight Mach Number
t=6, To=200 K (PW4000 Series,  ~ 5-6)
Higher  of interest in range of Mo < 1 and lower  of interest for supersonic transport
16
Bypass Ratio = 1
Bypass Ratio = 5
Bypass Ratio = 10
Bypass Ratio = 20
Non-Dimensional Thrust
14
12
10
8
6
4
2
0
0
0.5
1
1.5
2
Flight Mach Number, M0
2.5
3
TURBOFAN TRENDS: IN-CLASS EXAMPLE
Propulsive Efficiency vs. Flight Mach Number
t=6, To=200 K
1
0.9
Propulsive Efficiency
0.8
0.7
0.6
0.5
0.4
0.3
Bypass Ratio = 1
Bypass Ratio = 5
Bypass Ratio = 10
Bypass Ratio = 20
0.2
0.1
0
0
0.5
1
1.5
2
Flight Mach Number, M0
2.5
3
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, pd = 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
m 2   2U 2 A2
•
m 2
P

  2U 2  2 M 2 RT2  P2
M2
A2
RT2
RT2
•
m 2

 Pt 2
A2
RT2
m 0
P0

A2
RT0
M2
  1 2 
M2 
1 
2


  1 2 
M2 
1 
2


 T  m 0 
T

a0
 

A2 P0  m0 a0  A2 P0 
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
 1
2  1
 1
2  1
M2
  1 2 
M2 
1 
2


 1
2  1
NON-DIMENSIONAL THRUST FOR A2 AND P0
Plot of Thrust Normalized by Compressor Inlet Area and Ambient Pressure
vs. Flight Mach Number for Compressor Inlet Mach Number, M 2=0.5
30
Theta_t=6
Theta_t=9
Thrust / (A2 P0)
25
20
15
10
5
0
0
0.5
1
1.5
Flight Mach Number
•
•
Thrust at fixed altitude is nearly constant up to Mach 1
Thrust then increases rapidly, need A2 to get smaller
2
2.5
3
OPERATIONAL OVERVIEW
High Thrust
Low Speed, M0 ~ 0
High Mass Flow
Stream Tube Accelerates
Low Thrust
High Speed, M0 ~ 0.8
Low Mass Flow
Stream Tube Decelerates
2
 Ui 
T

 1 
1 U 2 A  U 0 
0 i
Aerodynamic force is always favorable for thrust production
2
NORMAL SHOCK TOTAL PRESSURE LOSSES
1
0.9
0.8
M2, P02/P01
0.7
0.6
0.5
0.4
0.3
0.2
Downstream Mach Number, M2
Total Pressure Ratio, P02/P01
0.1
0
1
1.5
2
2.5
3
3.5
Upstream Mach Number, M1
4
4.5
5
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
• 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
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 conditions
2. Efficiently expand high pressure, high temperature gases to atmospheric pressure
(convert thermal energy → kinetic energy)
KEY EQUATIONS FOR NOZZLE DESIGN
Nozzle area ratio as a function of engine parameters
Once nozzle area is set, operating point of engine depends only on t
A7 Athroat Pt 2


A2
A2
Pt 7



Tt 7
 1 

M2
Tt 2
 1  1 M 2 

2 
2


 1
2  1
Compare with Section 6.7 H&P
A7 is the throat area, how do we find the exit area of the nozzle?
Found from compressible channel flow relations, recall that M7=1
Set by jet stagnation pressure and ambient
A8
A
 exit
A7 Athroat
  1 2 
1
M8 
1 
2



 1 
M8 


2


 1
2  1
Compare with Equation 3.15
COMBUSTORS
MAJOR COMBUSTOR COMPONENTS
• Key Questions:
– Why is combustor configured this way?
– What sets overall length, volume and geometry of device?
Turbine
Compressor
Fuel
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 f > 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 f < 0.5
•
What do we do?
– Burn (keep combustion going) near f=1 with
some of ingested air
– Then mix very hot gases with remaining air to
lower temperature for turbine
Compressor
Air
Turbine
SOLUTION: BURNING REGIONS
Primary
Zone
f~0.3
f ~ 1.0
T>2000 K
RELATIVE LENGTH OF AFTERBURNER
J79 (F4, F104, B58)
Combustor
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)
AXIAL COMPRESSORS
WHERE IN THE ENGINE? PW2000
Fan
Compressor
2 SPOOL DEVICE: PW2000
Low Pressure Compressor (wlow)
High Pressure Compressor (whigh)
High and Low Pressure Turbines
REVIEW: PRESSURE DISTRIBUTION
•
•
Rotor
– Adds swirl to flow
– Adds kinetic energy to flow
with ½v2
– 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
w
Centerline
•
•
Rotor
– Adds swirl to flow
– Adds kinetic energy to flow
with ½v2
– 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)
EXAMPLES OF BLADE TWIST