Date post: | 26-Dec-2015 |
Category: |
Documents |
Upload: | margery-ryan |
View: | 219 times |
Download: | 4 times |
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
oe
eaeooee
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
e
e
e
e
eaeee
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
am
T
ee
ee
ee
ee
o
eo
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
22
11or
2
2
2
11
2
1
)2(
21
2
2
2
2
MTtT
a
u
TtT
RTa
Rpc
Tpc
u
TtT
pc
uT
tT
Tpch
uh
th
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"
22
1
get to theorembinomial theusing expand ,12For
12
2
11
velocity)zero ally toisentropic ddecelerate is stream if pressure is (
pressure stagnation thedefines 1
constant)1/(
find we using
constant
uptp
M
Mptp
tp
T
T
ptp
T
pRTp
P
t
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
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
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
1000
bMam
T hm
TU
foverall
0
Energy (1st Law) balance across burnerCycle analysis employing general form of mass, momentum and energy
RAMJET RESULTS
RAMJET RESULTS
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 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
Sp
ec
ific
Im
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
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 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
pu
lsiv
e E
ffic
ien
cy
Bypass Ratio = 1
Bypass Ratio = 5
Bypass Ratio = 10
Bypass Ratio = 20