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UCK 421E PROPULSION CLASS NOTES
Prof. Dr. Ali Kodal, I.T.U. AERONAUTICAL ENGINEERING 2012
CONTENTS
I. CLASSIFICATION AND GENERAL PERFORMANCE
A. INTORDUCTION
- Air Breathing Engines
- Injection of stored mass engines
- Propulsive system classifications
- Typical propulsive system example: Turbofan engines
- Basic propulsive system problems
- Separation of engine
- Control volume momentum conservation
- External flow effects
- Acceleration effects
B. PERFORMANCE PARAMETERS
- Jet engines performance parameters
- Reciprocating engines (piston) performance parameters
- Rocket engines performance parameters
- Other engine performance parameters
C. PROPULSIVE, THERMAL AND TOTAL EFFICIENCIES
- Definitions
- Efficiency derivations for stationery and moving axis
- Takeoff propulsion
II. THERMODYNAMICS CYCLES A. INTRODUCTION B. REVERSIBLE CYCLES C. IRREVERSIBLE CYCLES AND LOSSES D. IMPORTANT CYCLES
- Carnot cycle
- Rankine cycle
- Spark ignition engines and Otto cycle
- Compression ignition engines and Diesel cycle
- Gas Turbine cycle ( Brayton )
III. COMBUSTION
A. INTRODUCTION
B. SIMPLE APPROACH
C. CHEMICAL TRANSFORMATIONS
D. OXIDATION COMBUSTION REACTIONS
- Mixture ratio
- Fuel equivalence ratio
- Combustion with air
E. GAS MIXTURES
- Dalton law
- Mole fractions
- Mixture total mass
- Mixture molecular mass
- Mass fractions
H. ENERGY BALANCE FOR CHEMICAL REACTIONS
- Energy conservation, 1st law of Thermodynamics
- Standard state
- Constant pressure reaction enthalpy
- Constant volume reaction
- Lower and higher heating values
- Adiabatic flame temperature
- Chemical equilibrium
- Combustion products composition problem
IV. JET POPULSION SYTEM ANALYSIS
A. RAMJET
- Ideal ramjet
- Ramjet with losses
- Entry performance
B. TURBOJET
- Component efficiencies
- Static Turbojet performance
- Flight Turbojet Performance
C. TURBOFAN
- Static ideal Turbofan engine
- Optimum by-pass ratio
I. CLASSIFICATION AND GENERAL PERFORMANCE
A. INTRODUCTION
Propulsive devices produce thrust to accelerate a vehicle and overcome drag force. For aerospace vehicles, thrust is produced by acceleration of fluid stream entering and leaving propulsive device.
- Air breathing engines a) Propeller b) Turboprop c) Turbojet d) Turbofan e) Ramjet
- Injection of stored mass a) Rockets b) Ion thrusters c) Plasma thrusters (plasma = electric conduction gas)
These require addition of thermal or electrical energy to the propellant which is usually a fluid analysis of the addition of work and heat to compressible fluid dynamics.
Assume one dimensional compressible flow and apply fundamental principles to study of propulsive system.
Fig. 1.1 Typical turboprop engine (Hill and Peterson, 1992).
- Basic interactive components of propulsive sytems
Fig. 1.2 Afterburner turbojet.
- Typical example of propulsive system : Turbofan engines
Combustors Nozzles Propellers
Compressors, Fans, Pumps
Turbines Ion Accelerators
Plasma Accelerators
Inlets
IC
Engines
Propulsion System
- Basic Questions in Propulsion System Design
- Automotive propulsion i) Range : 1000 km, ii) Max. Speed : 160 km/h, iii) Passenger : 5 iv) Acceleration: to 160 km/h at 30 s
Questions: Engine Power, How much fuel, Type of engine, Cost, Maintenance considerations, Transmission, Fuel consumption Low or high speed performance.
- Rocket propulsion i) 50000 kg payload ii) 7500 m/s max. speed
Questions: Engine thrust, How much fuel, engine design
- Combustor Design:
H 2 -O 2 ratio, Combustion temperature, rocket nozzle cooling.
- Separation of engine
Thrust = push or pull exerted by engine and its enclose
Want to compare different engine + power plants. Separate engine performance from the effect of enclose of airplane. Maybe difficult.
Consider Rocket Engine:
For Jet Engines:
F = Thrust
(Produced by the thrust generator. If the external flow were isentropic, i.e external losses are neglected, can determine F in terms of engine parameters by momentum consideration for control volume. )
- Momentum Conservation for Control Volume
Propulsive flow = flow of working fluid
External flow = flow of fluid
Payload
Fuel tanks
Engine
Rocket nozzle:
Control volume momentum:
Newton 2nd^ law: ∑ F = dmvdt
For control volume:
d dt ∭ ρ VdV + ∬(ρVreln) CV CS
Vds = − ∬ pnds + ∑ Fext CS Where: V : Velocity measure w.r.t. virtual coordinates Vrel : Velocity relative control surface n : Outward normal unit vector
Stationary (control volume) systems:
Over A∞ :
V = Vrel =i V∞
P = P∞
ρ = ρ∞
Over Az—Ae :
V = Vrel =i V∞
P = P∞
ρ = ρ∞
S and S’ are stream tubes that fluid does not cross.
Ae
Ve
Mass conservation between S and S’:
ρ∞(Az − Ae)V∞ = ρ∞(A∞ − Ac)V∞
(Az − Ae) = (A∞ − Ac) (2.1)
(Az − A∞) = (Ae − Ac) (2.2)
Over section Ae :
V = Vrel =i Ve
P = Pe ρ = ρe
Steady flow, thus
d dt ∭( ) = 0
∬CS (ρV rel ∙ n)Vds = ρ∞(Az − Ae)V∞^2 + ρeAeVe^2 − ρ∞(A∞ − Ac)V∞^2 − ρ∞AcV∞^2 (2.3)
ṁe = ṁ i + ṁf = ṁ i (1 + f)
f = mf
̇ ṁ i = fuel air ratio
Using (2.1) :
∬CS (ρV rel ∙ n)Vds = ṁe Ve − ṁ i V∞ = ṁ i [(1 + f)Ve − V∞]^ (2.4)
Pressure force = − ∬CS pnds
On S :
− ∬ pnds = PS ∞(Az − A∞) = P∞(Ae − Ac) (2.5)
On A∞ :
F = ṁe Ve + (Pe − P∞)Ae (2.10)
F ~ independent of external flow—depends only processes within engine Ac = capture area varies with flight speed Ve = depend on detailed processes within propulsive system---engine. Performance calculation involve around finding Ve.
- Influence of External Flow
External flow exerts no force on the propulsive flow. Applying momentum conservation to the
volume between S and S’ :
V =i V∞
From mass conservation (A∞ − Ac) = (Az − Ac)
→ (^) ∬CS (ρV rel ∙ n)Vds = 0
Steady state : (^) dtd ∭( ) = 0
No external force : ∑ Fext = 0
− ∬CS pnds = 0
P∞(Az − A∞) − ∫ A Ace PEdA+ P∞(A∞ − Ac) − P∞(Az − Ae) = 0 (^) ∫ A Ace (PE − P∞)dA= 0
For real life situation: ∫ (PE − P∞)dA
Ae Ac
+ FT = D ≠ 0
FA = available thrust = F-D
Frictional force
Drag force on the engine
D = depends on details of external flow.
Consider (^) ∫ A Ace (PE − P∞)dA= 0
Break S’ into two parts:
∫ A Ace (PE − P∞)dA= ∫ A Aci (PE − P∞)dA+ ∫ A Aie (PE − P∞)dA=^ Pre-entry drag + Pressure drag
- Influence of Acceleration
Consider accelerating rocket:
F = ṁe Ve + (Pe − P∞)Ae (2.14)
Mass conservation:
ṁ = ρeAeVe =
dM dt
∭ ρVdV = M(t)V(t) CV
V(t) M(t)
Mg
D
Ve
Pe
Ac (^) Ai Ae
dV = −Ve
dM M
Suppose Ve is constant V(t) = −Ve lnM + const
Initial conditions: V(0)=0, M(0)=M 1
V(t) = −Ve ln
M(t) M 1 If M, ṁ and F were constants, then MV=F∙t
Example Jet engine
f <<1 V∞=300 m/s, F=3.56.10^5 N Te=1200K, ℳ=25, =1.3, Pe=P∞ Find 𝑚̇ =?, Ac=? Solution: F = ṁ i [(1 + f)Ve − V∞] + (Pe − P∞)Ae
f <<1 and Pe=P∞ :
F = ṁ i (Ve − V∞)
ṁ i = ṁ =
F
(Ve − V∞)
V∞
∞≈1.0 kg∕m^3
Ve Me=
ṁ =
3.56 ∙ 10^5
(720 − 300) = 847.6 kg/s
ṁ = ρ∞AcV∞ → 𝐴𝐶 = (^) ρ𝑚̇ ∞V∞
= (^) 1.0∙300874.6 = 2.82 𝑚^2 or Dc=1.895 m
B. PERFORMANCE PARAMETERS
- Aircraft Range
Consider level flight, constant velocity, altitude and lift to drag ratio (L/D)
M(t) = aircraft mass
F : required thrust
L=Mg
F=D
Mass flow rate: 𝑚̇𝑓 = − 𝑑𝑀𝑑𝑡
Problem = How much fuel is required to fly a given distance, or how far can the aircraft fly with a given amount of fuel. For a given range what fraction of gross weight to payload.
𝐹 = (^) 𝐿 𝑀𝑔 ⁄𝐷 → (^) 𝑚̇𝐹 𝑓
= 𝑀𝑔𝐿 ⁄𝐷 ∙ 1 (−𝑑𝑀𝑑𝑡 )
→ (^) 𝑚̇𝐹 𝑓
= − (^) 𝐿 𝑀𝑔 ⁄𝐷 ∙ (^) 𝑑𝑀𝑑𝑡
V x
Mg
Maximum takeoff thrust
Example:
Boeing 747, M 1 = 750 000 lbm, L/D =15, V = 550 mil/h, R= 3000 mil,
s = 0.9 lbm/h∙lbf
Estimate fuel required in steady constant velocity flight?
Solution:
𝑙𝑛
𝑀 1 𝑀 2
=
𝑅𝑔𝑠
𝑉 (𝐷𝐿)
=
3000𝑚𝑖𝑙 32.2 𝑓𝑡𝑠 2 0.9 𝑙𝑏𝑚/ℎ𝑙𝑏𝑓
550 𝑚𝑖𝑙ℎ 15
1 slug= 32.2 lbm lb=slug ft/s^2
𝑙𝑛
𝑀 1 𝑀 2 = 0.3273^ →^
𝑀 1 𝑀 2 = 1.387^ →^ Mfuel^ = M^1 -M^2 = 2.093∙
(^5) lbm
Using SI units
R = 4828 km, g = 9.807 m/s^2 , s = 2.549∙10-5^ kg/Ns = 9.176∙10-2^ kg/Nh,
V=885 km/h = 245.9 m/s
- Performance Parameters for Reciprocating Engines with a Propeller
Reciprocating (piston) engines are not good for high speed flights. However, recent researches have been modified their status.
BHP = Engine Brake Horsepower
𝑚̇𝑓 𝐵𝐻𝑃 = 𝐵𝑆𝐹𝐶^ = Brake Specific Fuel Consumption
Useful Power = F∙V 1 hp = 550 ft lb/s SI : 1 hp = 745.71 W
Propeller Efficiency:
𝑢𝑠𝑒𝑓𝑢𝑙 𝑝𝑜𝑤𝑒𝑟
𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 =^ propulsive efficiency for any propulsion system
Range problem:
L = Mg, 𝑚̇𝑓 = − 𝑑𝑀𝑑𝑡 , 𝐹 = 𝑀𝑔𝐿
⁄𝐷
, dx = V dt
⁄𝐷
𝑃
𝑉 𝐵𝑆𝐹𝐶)^ = −^
𝑑𝑀 𝑑𝑥
𝑃 𝐵𝑆𝐹𝐶
Integrate →
P ↑
BSFC ↓
} R increased range
P might range from 0.6~0.
BSFC ~ 0.5 lbm/BHP h → gasoline engine
