Answer to Question #210627 in Mechanical Engineering for Mohamed sadique

Question #210627

Steam enters a converging-diverging nozzle operating at steady state with P1= 40 bar. T1= 400°C, and a velocity of 10 m/s. The steam flows through the nozzle with negligible heat transfer and no significant change in potential energy. At the exit, P2= 15 bar, and the velocity is 665 m/s. The mass now rate is 2 kg/s. Determine the exit area of the nozzle in m2.


1
Expert's answer
2021-06-28T08:52:54-0400


Apply the steady state energy balance between (1) and (2) gives




Table of steam properties



The exit area is then;

"A_2 =\r\\dfrac{\rmv_2}{\rV_2}\r\n\r\n=\r\n\r\n\\dfrac{(2 kg)(0.1628 m\u00b3 \/kg)}{\r\n665 m\/s}\r\n= 4.9\u00d710\r^{-4}\\ m\r^2"

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