Jet Engine Ideal Analysis
Engine Efficiency
β’ Propulsive Efficiencyβ’ Thermal Efficiencyβ’ Overall Efficiency
Propulsive Efficiency
β’ The propulsive efficiency compares how much work is done on the aircraft, by supplying kinetic energy to the air.
Propulsive Efficiency
CombustorTurbine=10
ExhaustCompressor=10
Combustor
Va
m=100kg/s m=100kg/s
VJ
hπ ππ’π π‘=ππhπ ππ’π π‘=π(π π½βπ π)
ππππ=πΉπππππ₯ π·ππ π‘πππππ=π(π π½βπ π)π π
KE=KE=
KE =
ππ=π (π π½βππ)π π
π( (π π½β)2β (π π
β )2 )2
ππ=2(π π½βπ π)π π
( (π π½β)2β (π π
β )2 ) ππ=2(π π½βπ π)π π
(π π½βπ π ) (π π½+π π ) ππ=2π π
(π π½+π π )
Thermal Efficiency
β’ The thermal efficiency of an engine is the efficiency of the conversion of the heat energy released by the fuel into kinetic energy in the jet stream.
KE =
π=πΉπ’ππ πΈπππππ¦ππ=
π( (π π½β )2β (π π
β)2)2π
Overall Efficiency
β’ The overall Efficiency compares the work done on the aircraft to the energy given by the fuel.
πππ£πππππ=π .π π
πππ=
π( (π π½β )2β (π π
β)2)2π
ππ=2π π
(π π½+π π )
ππππ=2π π
(π π½+π π )
π ( (π π½β)2β (π π
β )2 )2π ππππ=
π π
(π π½+π π )π( (π π½
β)2β (π πβ)2 )
π
ππππ=π π
(π π½+π π )π (π π½+π π) (π π½βπ π )
π ππππ=π(π π½βπ π )π π
πππππ=
π π π
π=πππ£πππππ
Combustor
Arrangement of Engine
Turbine=10
ExhaustCompressor=10
Combustor
T3=1112KT1=288KP1=101kPa
m=100kg/s
Turbine Entry Temperature 1112Β°K Inlet Air Temperature 288Β°K
Compressor compression ratio = Turbine expansion ratio.
10 Outside air pressure 101 kPa
Specific Heat Capacity of Air at constant Pressure (Cp)
1 kJ/kg Β°K Mass flow of air 100kg/s
Ratio of Specific Heat Capacities for air ()
1.4 Calorific value of fuel is 43,000 kJ/kg
Universal Gas Constant (R) 287 kJ/kg Β°K
Combustor
Compressor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KT1=288KP1=101kPa
m=100kg/s
Calculate the compressor outlet temperature. (T2)
T2β= 556Β°K
T2=556KP2=1010kPa
Combustor
Compressor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KT1=288KP1=101kPa
m=100kg/s
T2=556KP2=1010kPa
Calculate the work done by the compressor. (Wc)
Wc
Combustor
Turbine
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2=556KP2=1010kPa
Calculate the turbine exhaust temperature. (T4)
T4β=576Β°K
T4=576KP4=101kA
Combustor
Turbine
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2=556KP2=1010kPa
T4=576KP4=101kA
Calculate the work done on the turbine. (WT)
WT
Combustor
Useful Work
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2=556KP2=1010kPa
T4=576KP4=101kA
Subtract the Work done by the compressor (WC)from the work done on the turbine (WT) to determine the useful work done by the engine on the aircraft. Useful Work = WT β WC. Useful Work = (53,600-26,800)=26,800 kJ
26,800kJ
Combustor
Combustor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2=556KP2=1010kPa
T4=576KP4=101kA
Calculate the amount of heat energy (Q) required to heat the compressed gases from T2 to T3.
Q
Combustor
Fuel
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2=556KP2=1010kPa
T4=576KP4=101kA
Q
Calculate the amount of fuel required to heat the gases from T2 to T3. The calorific value of fuel is 43000kJ/kg. Therefore to produce 55600kJ we will require:
Combustor
Efficiency
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KP2=1010kPa
T4β=576KP4=101kA
Q
Determine the efficiency of the engine by comparing the amount of useful work done, to the amount of heat energy input to the system.
Combustor
Part 2
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KP2=1010kPa
T4β=576KP4=101kA
Q
Repeat the analysis but with the compressor and turbine efficiencies at 85%.
Combustor
Compressor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KP4=101kA
QCalculate the compressor outlet temperature. (T2) T2β= 556Β°K
Combustor
Compressor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KP4=101kA
Q
Calculate the work done by the compressor. (Wc)
Combustor
Turbine
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 656KP4=101kA
Q
Calculate the turbine exhaust temperature. (T4)
T4=(1112-455.6)=656Β°K
Combustor
Turbine
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 656KP4=101kA
Q
Calculate the work done on the turbine. (WT)
Combustor
Turbine
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 656KP4=101kA
Q
Subtract the Work done by the compressor (WC)from the work done on the turbine (WT) to determine the useful work done by the engine on the aircraft. Useful Work = WT β WC. Useful Work = (45,600-31,500)=14,100 kJ
14,100kJ
Combustor
Combustor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 656KP4=101kA
Q=50900kJ
14,100kJ
Calculate the amount of heat energy (Q) required to heat the compressed gases from T2 to T3.
Combustor
Combustor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 656KP4=101kA
Q=50900kJ
14,100kJ
Calculate the amount of fuel required to heat the gases from T2 to T3. The calorific value of fuel is 43000kJ/kg. Therefore to produce 55600kJ we will require:
Combustor
Efficiency
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 656KP4=101kA
Q=50900kJ
14,100kJ
Determine the efficiency of the engine by comparing the amount of useful work done, to the amount of heat energy input to the system.
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = ?KP4=?kA
Q=50900kJ
14,100kJ
The work done by the compressor. (Wc)
Therefore the work done by the turbine is also 31,500kJ
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=?kA
Q=50900kJ
Therefore:
Calculate T4.
T4 = 797Β°K
Nozzle
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=?kAQ=50900kJ
Having calculated T4, and knowing the efficiency of the turbine is 85%, use the value to calculate T4
β.
T4β=(1112-370)=741Β°K
Nozzle
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kPaQ=50900kJ
Use the value of T4β to determine P4 (Remember assume P3=P2).
p4=244.55kPa
Nozzle
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kAQ=50900kJ
Nozzle
When the nozzle is choked the velocity of the gas at the throat of the nozzle = Mach 1. Let the station at the throat of the nozzle be station 5. The pressure at the throat is the critical pressure and the critical pressure equation is:
P5
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kAQ=50900kJ
Nozzle
Determine P5.
P5=129kPa
As P5 is > P1 the nozzle is choked.
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kAQ=50900kJ
Nozzle
P5=129kPa
At the throat of the nozzle, the critical temperature is equal to:
Determine T5.
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kAQ=50900kJ
Nozzle
P5=129kPa
We will assume that the exhaust is at the throat of the nozzle. Using the calculated value of T5 calculate the exhaust jet velocity using the following equation:
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kAQ=50900kJ
Nozzle
P5=129kPaT5=664K
Using the universal gas law calculate the density of the air at the exhaust:
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kAQ=50900kJ
Nozzle
P5=129kPaT5=664K
From the continuity equation calculate the cross sectional area of the exhaust.
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kAQ=50900kJ
Nozzle
P5=129kPaT5=664K
At this stage you are in a position to calculate the static thrust of the engine.
Thrust = 59.7kN
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kAQ=50900kJ
Nozzle
P5=129kPaT5=664K
Calculate the Specific Fuel Consumption of the engine. The burning of the fuel heats the air from T2 to T3
Heat Energy required is: Q=m.cp(T3- T2 ) Q = 100 (1)(1112-603) = 50871kJ
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=741KT4 = 797KP4=244kPaQ=50900kJ
Nozzle
P5=129kPaT5=664K
Specific fuel consumption is:
What happens when we install an afterburner?
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa T2β=556K
T2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPa
Q=50900kJ
Afterburner
Nozzle
The exhaust gas is reheated to 2000K. the calculations are the same as that the dry turbojet, but now the nozzle inlet temperature is 2000K.
T5 = ?KP5=?
m=100kg/s
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa T2β=556K
T2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPaQ=50900kJ
Afterburner
Noz
zle
When the nozzle is choked the velocity of the gas at the throat of the nozzle = Mach 1. Let the station at the throat of the nozzle be station 5. The pressure at the throat is the critical pressure and the critical pressure equation is:
Determine P5.
T5 = ?KP5=129kPa
m=100kg/s
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa T2β=556K
T2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPaQ=50900kJ
Afterburner
Noz
zle
T5 = 1667KP5=129kPa
At the throat of the nozzle, the critical temperature is equal to:
Determine T5.
m=100kg/s
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa T2β=556K
T2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPaQ=50900kJ
Afterburner
Noz
zle
T5 = 1667KP5=129kPa
At the throat of the nozzle, the air is travelling at the speed of sound. Determine the velocity of the jet.
m=100kg/s
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa T2β=556K
T2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPaQ=50900kJ
Afterburner
Noz
zle
T5 = 1667KP5=129kPa
Using the universal gas law calculate the density of the air at the exhaust:
m=100kg/s
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa T2β=556K
T2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPaQ=50900kJ
Afterburner
Noz
zle
T5 = 1667KP5=129kPa
From the continuity equation calculate the cross sectional area of the exhaust.
m=100kg/s
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPaQ=50900kJ
Afterburner
Noz
zle
T5 = 1667KP5=129kPa
The thrust of the core can be calculated from:
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPaQ=50900kJ
Afterburner
Noz
zle
T5 = 1667KP5=129kPa
The fuel required to heat the air in the afterburner from 797 to 2000K is: Heat Energy required is: Q=m.cp(T4.5- T4 ) Q = 100 (1)(2000-797) = 120300kJ
Combustor
Afterburner
Turbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2β=556KT2=603 KP2=1010kPa
T4β=576KT4 = 797KP4=244kPa
Q=50900kJm=1.18kg
Afterburner
Noz
zle
T5 = 1667KP5=129kPa
The total amount of fuel used was: 2.797 (Afterburner) + 1.18 for the Engine Specific fuel consumption is:
Q=120300kJM=2.797kg
Comparison between Afterburner and Jet Engine
With only the Engine
Thrust = 59.7kN
Specific fuel consumption is:
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