Answer to Question #210445 in Mechanical Engineering for Mohamed sadique

Question #210445

Air (1100 K/ 101.32 kPa) is passing through a converging nozzle and leave it at 300 K. Determine the velocity of air at the nozzle outlet. The nozzle is laid horizontal. The inlet velocity of air can be ignored.


1
Expert's answer
2021-06-25T14:41:03-0400

h1+V122=h2+V222h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}

But V1=0V_1=0

So, V2=2(h1h2)V_2 = \sqrt{2(h_1-h_2)}

V2=2cp(T1T2)V_2= \sqrt{2c_p(T_1-T_2)}

V2=21000(1100300)V_2= \sqrt{2*1000(1100-300)}

V2=1265m/sV_2= 1265 m/s


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