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 4 - Energy Analysis of Closed Systems - Problems - Page 201: 4-60E

Answer

$V_1=V_2=64.78\ ft³$ $Q=919.4\ Btu$

Work Step by Step

For an ideal gas: $PV=mRT$ Plugging in the values of $P_1=30\ psia, m=10\ lbm, T_1=65°F=524.67°R, R=0.3704\ psia.ft³/lbm.°R$ We get to: $V_1=V_2=64.78\ ft³$ For the final state ($P_2=2P_1=60\ psia)$ We get to $T_2=1049.34°R$ The energy balance for this system: $Q=\Delta U $ Interpolating from table A-17E: $u_1=89.42\ Btu/lbm, u_2=181.36\ Btu/lbm$ Therefore $Q=919.4\ Btu$
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