Aircraft Electrical Systems Engine Performance (Coursework Sample)

AIRCRAFT ELECTRICAL AND MECHANICAL SYSTEMS: CASE STUDY AIRCRAFT ELECTRICAL SYSTEMS ENGINE PERFORMANCE
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(Lecturer)
(University)
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Task 1
Name of SectionExit Duct Area
(Se)Exit Pressure (pe)Exit Air Velocity (Ve)Air Mass Flow
( QUOTE

) Aircraft Velocity (V0)Compressor200100350275200Diffuser220120300275200Combustion Chamber500110250275200Turbine40025900275200Exhaust60025700275200Propelling Nozzle3246.51080275200Average64.42596.67275200
The thrust equation for a turbojet engine can be derived from Newton’s Second law of motion, which implies that force equals the time rate of change of momentum.
F =  QUOTE


For a Turbojet Engine, the Net thrust (FN) is normally given by the difference between the Gross thrust (FG) and the Momentum drug (Vinh, 1995). This is given as;
FN = FG – momentum drag
= momentum thrust + pressure thrust – momentum drug
=  QUOTE

Ve + (pe – po) Se -  QUOTE

Vo
=  QUOTE

 (Ve –Vo) + (pe – po) Se
= 275(596.67 – 200) + (64.42 – 1)324
= 109084.25 + 20548.08
= 129,632.33
Where; W = is weight flow rate of the air passing through the engine. Ve = jet stream velocity pe = static pressure across propelling nozzle po = atmospheric pressure (given as 1 atm)Se = propelling nozzle area Vo = aircraft speed


Thrust Shaft Horsepower for the turbojet engine will be given by the equation;
thp =  QUOTE

 = bhpÆžp
= = 47,139
Where; T = Thrust (lb)
Vo = Velocity (ft/s)
bhp = Engine break horsepower
550 = conversion factor from ft-lbs to horsepower
Æžp = propeller efficiency
Propulsive efficiency (Æžp) for the turbojet will be given by the equation;
È p = =
= =
= 0.5025
Where; = Propulsive power output (thrust power given as FTV0)
= Mechanical power input
= Flight Velocity
= Exit Velocity
The solution of È p < 1; this shows that not the entire mechanical power () goes into propelling the turbojet. Some of it is left as excess kinetic energy.
Thermal efficiency (Æžth) for the turbojet engine will be given by the equation;
Æžth =
Where; m = Mechanical power
i = Input energy
In order to find we use the formulae È p =  QUOTE


But we know that  QUOTE

 = FTV0
Therefore;  QUOTE

 = 129,632.33 x 200 = 25,926,466
Making  QUOTE

 the subject we have;  QUOTE

 =  QUOTE


=  QUOTE

 = 51,594,957
Input energy  QUOTE

is given by; (fuel mass flow rate) x (fuel energy density)
Therefore;  QUOTE

 = 275 x 2500 = 687,500lbs/hr
In to BTU = 687,500 x 18,730 = 128768750
È th = = 0.4006
Overall efficiency (È 0) for the turbojet engine will be given by the equation;
È 0 = = =
Where; = the rate of energy input needed to generate the thrust
= Propulsive power output (thrust power given as FTV0)
Therefore; from the definitions of thermal and propulsive efficiency, the overall efficiency can also be given as;
È 0 = È th x È p
È 0 = 0.4006 x 0.5025 = 0.2013
Task 2
Name of SectionExit Duct AreaExit Pressure (pe)Exit Air Velocity (Ve)Air Mass Flow
( QUOTE

) Aircraft Velocity (V0)Compressor20010035055200Diffuser22012030055200Combustion Chamber50011025055200Turbine4002590055200Exhaust6002570055200Propelling Nozzle3246.5108055200Fan90030500220900
The thrust produced by turbofan engine is given by the equation;
FN = FG – momentum drag
= momentum thrust + pressure thrust – momentum drug
=  QUOTE

Ve + (pe – po) Se -  QUOTE

Vo
=  QUOTE

 (Ve –Vo) + (pe – po) Se
= 220(500-900) + (30-1)900
= 88000 + 26100
= 114,100
Where; W = is weight flow rate of the air passing through the engine. Ve = jet stream velocity pe = static pressure across propelling nozzle po = atmospheric pressure Se = propelling fan area Vo = aircraft speed
Thrust Shaft Horsepower for the turbofan engine will be given by the equation;
thp =  QUOTE

 = bhpÆžp
= = 186,709
Where; T = Thrust (lb)
Vo = Velocity (ft/s)
bhp = Engine break horsepower
550 = conversion factor from ft-lbs to horsepower
Æžp = propeller efficiency
Propulsive efficiency (Æžp) for the turbofan engine will be given by the equation;
È p =
= = = 1.285
Where; = Flight Velocity
= Exit Velocity
Thermal efficiency (Æžth) for the turbofan engine will be given by the equation;
Æžth =
Where; m = Mechanical power
i = Input energy
In order to find we use the formulae È p =
But we know that = FTV0
Therefore;
 = 114,100 x 900 = 102,690,000
Making  QUOTE

 the subject we have;  QUOTE

 =  QUOTE


=  QUOTE

 = 79,914,396
Input energy  QUOTE

is given by; (fuel mass flow rate) x (fuel energy density)
Therefore;  QUOTE

 = 220 x 2500 = 550,000lbs/hr
In to BTU = 550,000 x 18,730 = 103,015,000
 QUOTE

 È th = HYPER13 QUOTE HYPER14 = 0.7757
Overall efficiency (È 0) for the turbofan engine will be given by the equation;
È 0 = È th x È p
È 0 = 0.7757 x 1.285
= 0.9968
Task 3
Name of Section
Exit Duct Area
Exit Pressure (pe)
Exit Air Velocity (Ve)Air Mass Flow
( QUOTE

) Aircraft Velocity (V0)Compressor200100350275200Diffuser220120300275200Combustion Chamber500110250275200Turbine400251000275200Exhaust60025700275200Propelling Nozzle3246.51500275200Average64.42683.33275200
The thrust produced by a modified turbojet engine is given by the equation;
FN = FG – momentum drag
= momentum thrust + pressure thrust – momentum drug
=  QUOTE

Ve + (pe – po) Se -  QUOTE

Vo
=  QUOTE

 (Ve –Vo) + (pe – po) Se
= 275(683.33-200) + (64.42 – 1)324
= 132,915.75 + 17,632.08
= 150547.83
Thrust Shaft Horsepower for the modified turbojet engine will be given by the equation;
thp =  QUOTE


=  QUOTE

 = 54,744.67
Propulsive efficiency (Æžp) for the modified turbojet engine will be given by the equation;
È p =
= = = 0.4502
Thermal efficiency (Æžth) for the modified turbojet engine will be given by the equation;
Æžth =
In order to find we use the formulae È p =
But we know that = FTV0
Therefore; = 15,0547.83 x 200 = 301,094,400
Making the subject we have; =
=  QUOTE

 = 66,880,053
Input energy  QUOTE

is given by; (fuel mass flow rate) x (fuel energy density)
Therefore;  QUOTE

 = 275 x 2500 = 687,500lbs/hr
In to BTU = 687,50...