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A turbofan operates at 25,000 ft and moves at 815 ft/s. It ingests 1.2 times the amount of air into the fan than into the core, which all exits through the fan exhaust. The fuel-flow-to-core airflow ratio is 0.0255. The exit densities of the fan and core are 0.00154 and 0.000578 slugs/ft3, respe~tively. The exit pressures from the fan and core are 10.07 and 10.26 psia, respectively. The developed thrust is 10,580 !bf, and the exhaust velocities from the fan and core are 1147 and 1852 ft/s, respectively. (a) Find the ingested air mass flow rate for the core and TSFC. (b) What are the exit areas of the fan and core nozzles

Respuesta :

Answer:

a)

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = 3.301 x [tex]10^{-5}[/tex]

b)

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Explanation:

Data Given:

Height = 25000 ft

Vehicle velocity = [tex]u_{a}[/tex] = 815 ft/s = 248.41 m/s

[tex]m_{s} = 1.2m_{e}[/tex]

[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]  

Where,

[tex]m_{s}[/tex] = Mass flow rate of fan

[tex]m_{e}[/tex] = Mass flow rate of core

F = Thrust

Density of core = [tex]D_{e}[/tex] = 0.000578 slugs/[tex]ft^{3}[/tex] = 0.2979 kg/[tex]m^{3}[/tex]

Density of fan = [tex]D_{s}[/tex] = 0.00154 slugs/[tex]ft^{2}[/tex] = 0.7937 kg/[tex]m^{3}[/tex]

Ambient Pressure of Fan = [tex]P_{s}[/tex] = 10.07 Psi = 69430.21 Pa

Ambient Pressure of core = [tex]P_{e}[/tex] = 10.26 Psi = 70740.2 Pa

Thrust = F = 10580 lbf = 47062.2 N

Velocity of fan = [tex]u_{s}[/tex] = 1147 ft/s = 349.6 m/s

Velocity of core = [tex]u_{e}[/tex] = 1852 ft/s = 564.5 m/s

At the height of 25000 ft, P = 37600 [tex]P_{a}[/tex]

Now,

we have:

[tex]m_{e}[/tex] = [tex]u_{e}[/tex] x  [tex]D_{e}[/tex]  x [tex]A_{e}[/tex]

Plugging in the values, we get:

[tex]m_{e}[/tex]  = 168.16 [tex]A_{e}[/tex]   Equation 1

And,

[tex]m_{s}[/tex]  = [tex]D_{s}[/tex]  x [tex]A_{s}[/tex] x  [tex]u_{s}[/tex]

[tex]m_{s}[/tex]  = 277.5 [tex]A_{s}[/tex]  Equation 2

As, we know,

[tex]m_{s} = 1.2m_{e}[/tex]  

[tex]m_{s}[/tex]  = 277.5 [tex]A_{s}[/tex]

And now for Thrust, we have:

F = [tex]A_{e}[/tex] x ([tex]P_{e}[/tex]  - [tex]P_{a}[/tex] ) + [tex]A_{s}[/tex] x ([tex]P_{s}[/tex] - [tex]P_{a}[/tex] ) + [tex]m_{e}[/tex]x ([tex]u_{e}[/tex]  - [tex]u_{a}[/tex] ) + [tex]m_{s}[/tex] x ([tex]u_{s}[/tex]  -  [tex]u_{a}[/tex] ) Equation 3

Now, substitute equation 1 and 2 in equation 3, we get:

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = Thrust Specific Fuel Consumption = fuel mass flow rate  / Thrust

TSFC = [tex]m_{f}[/tex]/F

And,

[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]  

[tex]m_{e}[/tex]   =  60.94

[tex]m_{f}[/tex]  = 0.0255 x 60.94

[tex]m_{f}[/tex]  = 1.55397

TSFC = [tex]m_{f}[/tex]/F

TSFC = 1.55397/47062.2

TSFC = 3.301 x [tex]10^{-5}[/tex]

Low TSFC = High efficiency

High TSFC = Low efficiency

a)

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = 3.301 x [tex]10^{-5}[/tex]

b)

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

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