Answer
$W=11.47hp$
Work Step by Step
$W=Q\Delta P=0.8\frac{ft^3}{s}*(70psia-15psia)*(\frac{1Btu}{5.40psiaft^3})*(\frac{1hp}{0.71\frac{Btu}{s}})=11.47hp$
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 100: 2-42E
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