Answer to Question #232543 in Mechanical Engineering for Mubarra

Question

A steady flow system loses 3 kW of heat also loses 20 kW of work. The fluid flows through the system at a steady rate of 70 kg/s. The velocity at inlet is 20 m/s and at outlet it is 10 m/s. The inlet is 20 m above the outlet. Calculate the following.

i. The change in K.E./s (-10.5 kW)

Answer 1

Solution;

Given;

$m=70kg/s$

$v_1=20m/s$

$v_2=10m/s$

But;

$K.E=\frac12×m×v^2$

Hence;

$\Delta K.E/s=\frac12×m×(v_2^2-v_1^2)$

$\Delta K.E/s=\frac12×70×(10^2-20^2)$

$\Delta K.E/s=-10.5kW$