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The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and 30°C enters it with a velocity of 357 m/s and the exit state is 200 kPa and 90°C. The specific heat of air at the average temperature of 60°C = 333 K is cp = 1.007 kJ/kg·K.

Respuesta :

anniwuanyanwu1

Answer:

Exit velocity = 498.3m/s

Explanation:

The energy balance for the system is given by:

Ein - Eout = â—‡Esystem

Ein = Eout

Therefore

m' (h1 + V1^2/2) = m' (h2 + V2^2/2)

h1 + V1^2/2 = h2 + V2^2/2

The exit velocity will be:

V2 = [V1^2 + 2(h1 - h2)^0.5 = [V1^2 + 2(cp(T1 - T2)]^0.5

But cp = 1.007kJ/kg

V2 = [(357)^2 + 2(1.007)(90-30)(1000m^2/s^2)]^0.5

V2 = [127449 + 120840]^0.5

V2 = [248289]^0.5

V2 = 498.3m/s

Rau7star

Answer:

81.29 m/s

Explanation:

Given:

60°C = 333 K is cp = 1.007 kJ/kg·K.

at air inlet:

Pressure [tex]P_{1}[/tex]= 100 kPa

Temperature [tex]T_{1}[/tex]= 30°C

Velocity [tex]V_{1}[/tex]= 357 m/s

at air outlet:

Pressure [tex]P_{2}[/tex]= 200kPa

Temperature [tex]T_{2}[/tex] = 90°C

We can define the energy balance as:

[tex]E _{in}[/tex] - [tex]E_{out}[/tex]=  Δ[tex]E_{system}[/tex] = 0

therefore,      [tex]E _{in}[/tex]  =   [tex]E_{out}[/tex]

m[tex]h_{1}[/tex] + m[tex]\frac{V_{1}^{2} }{2}[/tex] =m[tex]h_{2}[/tex] + [tex]\frac{V_{2}^{2} }{2}[/tex]  (taking m common and cancelling it)

4[tex]h_{1}[/tex] + [tex]\frac{V_{1}^{2} }{2}[/tex] =  [tex]h_{2}[/tex]  +  [tex]\frac{V_{2}^{2} }{2}[/tex]

The velocity at the exit of a diffuser:

[tex]V_{2}[/tex] = [ [tex]V_{1}^{2}[/tex] + 2 [tex]c_p{}[/tex] ([tex]T_{1}[/tex] - [tex]T_{2}[/tex]) [tex]]^{0.5}[/tex]

[tex]V_{2}[/tex]= [ [tex]357^{2}[/tex] +  2  x   1000  x  1.007  x  (30-90) [tex]]^{0.5}[/tex]

[tex]V_{2}[/tex]= 81.29 m/s

therefore, the velocity at the exit of a diffuser is 81.29 m/s

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