Answer to Question #210627 in Mechanical Engineering for Mohamed sadique

Question

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.

Answer 1

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

Table of steam properties

The exit area is then;

$A_2 =
\dfrac{
mv_2}{
V_2}
 
 =
 
 \dfrac{(2 kg)(0.1628 m³ /kg)}{
 665 m/s}
 = 4.9×10
^{-4}\ m
^2$