Answer to Question #210445 in Mechanical Engineering for Mohamed sadique

Question

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.

Answer 1

$h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}$

But $V_1=0$

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

$V_2= \sqrt{2c_p(T_1-T_2)}$

$V_2= \sqrt{2*1000(1100-300)}$

$V_2= 1265 m/s$