Aircraft Electrical Systems Engine Performance Coursework - 1375 Words

Aircraft Electrical Systems Engine Performance (Coursework Sample) Content: AIRCRAFT ELECTRICAL AND MECHANICAL SYSTEMS: CASE STUDY AIRCRAFT ELECTRICAL SYSTEMS ENGINE PERFORMANCE(Student Name)(Course No.)(Lecturer)(University)(City State)(Date) Task 1Name of SectionExit Duct Area(Se)Exit Pressure (pe)Exit Air Velocity (Ve)Air Mass Flow ( QUOTE

) Aircraft Velocity (V0)Compressor200100350275200Diffuser220120300275200Combustion Chamber500110250275200Turbine40025900275200Exhaust60025700275200Propelling Nozzle3246.51080275200Average64.42596.67275200The thrust equation for a turbojet engine can be derived from Newtons 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 = 129,632.33Where; 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

 = bhpÆÅ ¾p= = 47,139Where; T = Thrust (lb)Vo = Velocity (ft/s)bhp = Engine break horsepower550 = conversion factor from ft-lbs to horsepowerÆÅ ¾p = propeller efficiencyPropulsive efficiency (ÆÅ ¾p) for the turbojet will be given by the equation;Èp = = = = = 0.5025Where; = Propulsive power output (thrust power given as FTV0) = Mechanical power input = Flight Velocity = Exit VelocityThe solution of Èp 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 (ÆÅ ¾th) for the turbojet engine will be given by the equation;ÆÅ ¾th = Where; m = Mechanical power i = Input energyIn order to find we use the formulae Èp =  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 Èth = = 0.4006Overall efficiency (È0) for the turbojet engine will be given by the equation;È0 = = = 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;È0 = Èth x Èp È0 = 0. 4006 x 0.5025 = 0.2013Task 2Name of SectionExit Duct AreaExit Pressure (pe)Exit Air Velocity (Ve)Air Mass Flow ( QUOTE

) Aircraft Velocity (V0)Compressor20010035055200Diffuser22012030055200Combustion Chamber50011025055200Turbine4002590055200Exhaust6002570055200Propelling Nozzle3246.5108055200Fan90030500220900The 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 = 114,100Where; 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 speedThrust Shaft Horsepower for the turbofan engine will be given by the equation; thp =  QUOTE

 = bhpÆÅ ¾p= = 186,70 9Where; T = Thrust (lb)Vo = Velocity (ft/s)bhp = Engine break horsepower550 = conversion factor from ft-lbs to horsepowerÆÅ ¾p = propeller efficiencyPropulsive efficiency (ÆÅ ¾p) for the turbofan engine will be given by the equation;Èp = = = = 1.285Where; = Flight Velocity = Exit VelocityThermal efficiency (ÆÅ ¾th) for the turbofan engine will be given by the equation;ÆÅ ¾th = Where; m = Mechanical power i = Input energyIn order to find we use the formulae Èp = 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

 Èth = HYPER13 QUOTE HYPER14 = 0.7757Overall efficiency (È0) for the turbofan engine will be given by the equation;È0 = ÃƒË †th x Èp È0 = 0.7757 x 1.285 = 0.9968Task 3Name of SectionExit Duct AreaExit Pressure (pe)Exit Air Velocity (Ve)Air Mass Flow ( QUOTE

) Aircraft Velocity (V0)Compressor200100350275200Diffuser220120300275200Combustion Chamber500110250275200Turbine400251000275200Exhaust60025700275200Propelling Nozzle3246.51500275200Average64.42683.33275200The 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 = 150547.83Thrust Shaft Horsepower for the modified turbojet engine will be given by the equation; thp =  QUOTE

 =  QUOTE

 = 54,744.67Propulsive efficiency (ÆÅ ¾p) for the modified turbojet engine will be given by the equation;Èp = = = = 0.4502Thermal efficiency (ÆÅ ¾th) for the modified turbojet engine will be given by the equation;ÆÅ ¾th = In order to find we use the formulae Èp = But we know that = FTV0 Therefore; = 15,0547.83 x 200 = 301,094,400Making 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... Aircraft Electrical Systems Engine Performance Coursework - 1375 Words Aircraft Electrical Systems Engine Performance (Coursework Sample) Content: AIRCRAFT ELECTRICAL AND MECHANICAL SYSTEMS: CASE STUDY AIRCRAFT ELECTRICAL SYSTEMS ENGINE PERFORMANCE(Student Name)(Course No.)(Lecturer)(University)(City State)(Date) Task 1Name of SectionExit Duct Area(Se)Exit Pressure (pe)Exit Air Velocity (Ve)Air Mass Flow ( QUOTE

) Aircraft Velocity (V0)Compressor200100350275200Diffuser220120300275200Combustion Chamber500110250275200Turbine40025900275200Exhaust60025700275200Propelling Nozzle3246.51080275200Average64.42596.67275200The thrust equation for a turbojet engine can be derived from Newtons 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 = 129,632.33Where; 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

 = bhpÆÅ ¾p= = 47,139Where; T = Thrust (lb)Vo = Velocity (ft/s)bhp = Engine break horsepower550 = conversion factor from ft-lbs to horsepowerÆÅ ¾p = propeller efficiencyPropulsive efficiency (ÆÅ ¾p) for the turbojet will be given by the equation;Èp = = = = = 0.5025Where; = Propulsive power output (thrust power given as FTV0) = Mechanical power input = Flight Velocity = Exit VelocityThe solution of Èp 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 (ÆÅ ¾th) for the turbojet engine will be given by the equation;ÆÅ ¾th = Where; m = Mechanical power i = Input energyIn order to find we use the formulae Èp =  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 Èth = = 0.4006Overall efficiency (È0) for the turbojet engine will be given by the equation;È0 = = = 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;È0 = Èth x Èp È0 = 0. 4006 x 0.5025 = 0.2013Task 2Name of SectionExit Duct AreaExit Pressure (pe)Exit Air Velocity (Ve)Air Mass Flow ( QUOTE

) Aircraft Velocity (V0)Compressor20010035055200Diffuser22012030055200Combustion Chamber50011025055200Turbine4002590055200Exhaust6002570055200Propelling Nozzle3246.5108055200Fan90030500220900The 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 = 114,100Where; 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 speedThrust Shaft Horsepower for the turbofan engine will be given by the equation; thp =  QUOTE

 = bhpÆÅ ¾p= = 186,70 9Where; T = Thrust (lb)Vo = Velocity (ft/s)bhp = Engine break horsepower550 = conversion factor from ft-lbs to horsepowerÆÅ ¾p = propeller efficiencyPropulsive efficiency (ÆÅ ¾p) for the turbofan engine will be given by the equation;Èp = = = = 1.285Where; = Flight Velocity = Exit VelocityThermal efficiency (ÆÅ ¾th) for the turbofan engine will be given by the equation;ÆÅ ¾th = Where; m = Mechanical power i = Input energyIn order to find we use the formulae Èp = 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

 Èth = HYPER13 QUOTE HYPER14 = 0.7757Overall efficiency (È0) for the turbofan engine will be given by the equation;È0 = ÃƒË †th x Èp È0 = 0.7757 x 1.285 = 0.9968Task 3Name of SectionExit Duct AreaExit Pressure (pe)Exit Air Velocity (Ve)Air Mass Flow ( QUOTE

) Aircraft Velocity (V0)Compressor200100350275200Diffuser220120300275200Combustion Chamber500110250275200Turbine400251000275200Exhaust60025700275200Propelling Nozzle3246.51500275200Average64.42683.33275200The 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 = 150547.83Thrust Shaft Horsepower for the modified turbojet engine will be given by the equation; thp =  QUOTE

 =  QUOTE

 = 54,744.67Propulsive efficiency (ÆÅ ¾p) for the modified turbojet engine will be given by the equation;Èp = = = = 0.4502Thermal efficiency (ÆÅ ¾th) for the modified turbojet engine will be given by the equation;ÆÅ ¾th = In order to find we use the formulae Èp = But we know that = FTV0 Therefore; = 15,0547.83 x 200 = 301,094,400Making 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...