Embed Size (px)
DESCRIPTION
A deck of slides that goes through a simple analysis of an Ideal Jet Engine
Citation preview
Jet Engine Ideal Analysis
Engine Efficiency
• Propulsive Efficiency• Thermal Efficiency• Overall Efficiency
Propulsive Efficiency
• 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
h𝑇 𝑟𝑢𝑠𝑡=𝑚𝑎h𝑇 𝑟𝑢𝑠𝑡=𝑚(𝑉 𝐽−𝑉 𝑎)
𝑊𝑜𝑟𝑘=𝐹𝑜𝑟𝑐𝑒𝑥 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑊=𝑚(𝑉 𝐽−𝑉 𝑎)𝑉 𝑎
KE=KE=
KE =
𝜂𝑝=𝑚 (𝑉 𝐽−𝑉𝑎)𝑉 𝑎
𝑚( (𝑉 𝐽❑)2− (𝑉 𝑎
❑ )2 )2
𝜂𝑝=2(𝑉 𝐽−𝑉 𝑎)𝑉 𝑎
( (𝑉 𝐽❑)2− (𝑉 𝑎
❑ )2 ) 𝜂𝑝=2(𝑉 𝐽−𝑉 𝑎)𝑉 𝑎
(𝑉 𝐽−𝑉 𝑎 ) (𝑉 𝐽+𝑉 𝑎 ) 𝜂𝑝=2𝑉 𝑎
(𝑉 𝐽+𝑉 𝑎 )
Thermal Efficiency
• 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 =
𝑄=𝐹𝑢𝑒𝑙 𝐸𝑛𝑒𝑟𝑔𝑦𝜂𝑇=
𝑚( (𝑉 𝐽❑ )2− (𝑉 𝑎
❑)2)2𝑄
Overall Efficiency
• The overall Efficiency compares the work done on the aircraft to the energy given by the fuel.
𝜂𝑜𝑣𝑒𝑟𝑎𝑙𝑙=𝑇 .𝑉 𝑎
𝑄𝜂𝑇=
𝑚( (𝑉 𝐽❑ )2− (𝑉 𝑎
❑)2)2𝑄
𝜂𝑝=2𝑉 𝑎
(𝑉 𝐽+𝑉 𝑎 )
𝜂𝑝𝜂𝑇=2𝑉 𝑎
(𝑉 𝐽+𝑉 𝑎 )
𝑚 ( (𝑉 𝐽❑)2− (𝑉 𝑎
❑ )2 )2𝑄 𝜂𝑝𝜂𝑇=
𝑉 𝑎
(𝑉 𝐽+𝑉 𝑎 )𝑚( (𝑉 𝐽
❑)2− (𝑉 𝑎❑)2 )
𝑄
𝜂𝑝𝜂𝑇=𝑉 𝑎
(𝑉 𝐽+𝑉 𝑎 )𝑚 (𝑉 𝐽+𝑉 𝑎) (𝑉 𝐽−𝑉 𝑎 )
𝑄 𝜂𝑝𝜂𝑇=𝑚(𝑉 𝐽−𝑉 𝑎 )𝑉 𝑎
𝑄𝜂𝑝𝜂𝑇=
𝑇 𝑉 𝑎
𝑄=𝜂𝑜𝑣𝑒𝑟𝑎𝑙𝑙
Combustor
Arrangement of Engine
Turbine=10
ExhaustCompressor=10
Combustor
T3=1112KT1=288KP1=101kPa
m=100kg/s
Turbine Entry Temperature 1112°K Inlet Air Temperature 288°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 °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 °K
Combustor
Compressor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KT1=288KP1=101kPa
m=100kg/s
Calculate the compressor outlet temperature. (T2)
T2’= 556°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)
T4’=576°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 – 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
T2’=556KP2=1010kPa
T4’=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
T2’=556KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=576KP4=101kA
QCalculate the compressor outlet temperature. (T2) T2’= 556°K
Combustor
Compressor
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=576KT4 = 656KP4=101kA
Q
Calculate the turbine exhaust temperature. (T4)
T4=(1112-455.6)=656°K
Combustor
Turbine
Turbine=10
Exhaust
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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 – 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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=576KT4 = 797KP4=?kA
Q=50900kJ
Therefore:
Calculate T4.
T4 = 797°K
Nozzle
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2’=556KT2=603 KP2=1010kPa
T4’=741KT4 = 797KP4=?kAQ=50900kJ
Having calculated T4, and knowing the efficiency of the turbine is 85%, use the value to calculate T4
’.
T4’=(1112-370)=741°K
Nozzle
Part 3
CombustorTurbine=10
Compressor=10
Combustor
T3=1112KP3=1010kPa
T1=288KP1=101kPa
m=100kg/s
T2’=556KT2=603 KP2=1010kPa
T4’=741KT4 = 797KP4=244kPaQ=50900kJ
Use the value of T4’ 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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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 T2’=556K
T2=603 KP2=1010kPa
T4’=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 T2’=556K
T2=603 KP2=1010kPa
T4’=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 T2’=556K
T2=603 KP2=1010kPa
T4’=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 T2’=556K
T2=603 KP2=1010kPa
T4’=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 T2’=556K
T2=603 KP2=1010kPa
T4’=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 T2’=556K
T2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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
T2’=556KT2=603 KP2=1010kPa
T4’=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:
