Answer
a) $248W$
Work Step by Step
Here we are working with kinetic energy:
The kinetic work is: $W_{k}=\frac{1}{2}mV^2$
Where $m=\rho Q$
Then:
$W_{k}=\frac{1}{2}\rho QV^2=\frac{1}{2}*1.15\frac{kg}{m^3}*3\frac{m^3}{min}*(12\frac{m}{s})^2=248.4W$
Thermodynamics: An Engineering Approach 8th Edition
by Cengel, Yunus; Boles, Michael
Published by
McGraw-Hill Education
ISBN 10:
0-07339-817-9
ISBN 13:
978-0-07339-817-4
Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 107: 2-126
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