Answer
c) $76\%$
Work Step by Step
The energy give it by the pump is:
$P_{p}=Q\Delta P=(18\frac{L}{s}*\frac{1m^3}{1000L})*211kPa=3.798kW$
Then the efficiency of the pump is:
$\eta_{p}=\frac{3.798kW}{5kW}=0.7596$
$\eta_{p}=75.96\%$
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 108: 2-133
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