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Air enters a diffuser witha velocity of 400 m/s, a pressure of 1 bar and temperature of 25 C. It exits with a temperature of 100 C. What is the exit velocity of the air? Assume there are no heat losses or change in potential energy Data:= 0.718 kJ/kg.°C. MW = 28.9 g/mol

User Henny
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1 Answer

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Answer:

Exit velocity of air is 96.43 m/s.

Step-by-step explanation:

Given that


V_1=400\ m/s


T_1=25C


T_2=100C

For air


C_p=1.005\ KJ/kg.K

Now from first law of thermodynamics for open system at steady state


h_1+(V_1^2)/(2000)+Q=h_2+(V_2^2)/(2000)+w

For diffuser


h_1+(V_1^2)/(2000)=h_2+(V_2^2)/(2000)

We know that


h=C_pT


h_1+(V_1^2)/(2)=h_2+(V_2^2)/(2000)


1.005* 25+(400^2)/(2000)=1.005* 100+(V_2^2)/(2)


V_2=96.43\ m/s

So the exit velocity of air is 96.43 m/s.

User Haseeb A
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