Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 254: 5-20E

Answer

$w=0.201\ Btu/lbm$

Work Step by Step

The flow work is given by: $w=P_2v_2-P_1v_1$ Approximating compressed liquid by saturated, and constant specific volume: $v_2=v_1=0.01672\ ft³/lbm$ With $P_2=80\ psia,\ P_1=15\ psia$: $w=1.0868\ ft³.psia=0.201\ Btu/lbm$
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