Answer to Question #232543 in Mechanical Engineering for Mubarra

Question #232543

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)


1
Expert's answer
2021-09-04T01:00:56-0400

Solution;

Given;

"m=70kg\/s"

"v_1=20m\/s"

"v_2=10m\/s"

But;

"K.E=\\frac12\u00d7m\u00d7v^2"

Hence;

"\\Delta K.E\/s=\\frac12\u00d7m\u00d7(v_2^2-v_1^2)"

"\\Delta K.E\/s=\\frac12\u00d770\u00d7(10^2-20^2)"

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


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