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 197: 4-17E

Answer

$W = 180.89 BTU$

Work Step by Step

For the first point: $P = aV+b$ $15\ psia = 5 \frac{psia}{ft³}\times 7\ ft³ + b$ $b = -20\ psia $ For the second point: $P = aV+b$ $100\ psia = 5 \frac{psia}{ft³}\times V_2 -20\ psia$ $V_2 = 24\ ft³$ The work done: $W = \int_{V_1}^{V_2} P dV$ $W = \int_{V_1}^{V_2} (aV+b) dV$ $W = (\frac{a}{2}V^2+bV)|_{V_1}^{V_2}$ $W = 977.5\ psi.ft^3 \times \frac{1\ BTU}{5.40395\ psi.ft³}$ $W = 180.89 BTU$
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