MAE 6530 - Propulsion Systems II Homework 6.2 29 Consider the TurboFan Engine whose Block Diagram is Shown Below Diffuser Compressor core 3 fan 3 Bypass Turbine Combustor Core Exhaust Fan Bypass Exhaust fan 3 Fan Bypass Exhaust Bypass core 4 core 5 core exit fan exit fan exit ∞ 1 2 With Following Conditions Aircraft Mach Number: 0.70 Aircraft Altitude: 6 km Bypass Ratio: 2 Fan Pressure Ratio: 2 Compressor Ratio: 6 Burner Outlet Temperature: 1700 K Fuel: JP4 à h f = 42.68 MJ/kg
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
MAE 6530 - Propulsion Systems II
Homework 6.2
29
Consider the TurboFan Engine whose Block Diagram is Shown Below
Diffuser Compressorcore3
fan3
Bypass
TurbineCombustorCore Exhaust
Fan
Bypass Exhaust
fan3
Fan
Bypass ExhaustBypass
core4
core5
coreexit
fanexit
fanexit
∞
1
2
With Following Conditions Aircraft Mach Number: 0.70Aircraft Altitude: 6 kmBypass Ratio: 2Fan Pressure Ratio: 2Compressor Ratio: 6Burner Outlet Temperature: 1700 K Fuel: JP4 à hf = 42.68 MJ/kg
Stephen Whitmore
MAE 6530 - Propulsion Systems II
Homework 6.2 (2)
30
Assume the Following Component Propertiesi) Diffuser, Compressor, Fan, Turbine, Nozzle ~ Isentropicii) Nozzle exit flow is NOT mixediii) Combustor is 35% efficient àiv) Fuel massflow is NOT negligiblev) Mean specific heats, gamma are constant across engine constant àvi) Fan, Core Nozzle Exits Optimized for Altitude
Diffuser Compressorcore3
fan3
Bypass
TurbineCombustorCore Exhaust
Fan
Bypass Exhaust
fan3
Fan
Bypass ExhaustBypass
core4
core5
coreexit
fanexit
fanexit
∞
1
2
ηb =!Q3−4!mfuel ⋅hf
= τ f ⋅Cp ⋅T∞hf γ≈1.4
Cp ≈1004.96J /kg−K
MAE 6530 - Propulsion Systems II
Homework 6.2 (3)
31
Calculatei) Normalized Thrustii) % of Thrust delivered by Core Flowiii) % of Thrust delivered by Bypass Flowii) Ratio of Bypass Thrust to Core Thrustiii) Normalized Specific Impulseiv) TSFC lbm/lbf-hrv) Bypass Ratio for Optimal Ispvi) Optimal TSFC