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 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 106: 2-114E

Answer

$W=0.016Btu$

Work Step by Step

We have: $F=\frac{Constant}{x^k}$ Integrating on both sides by $dx$ we obtain: $W=Fx=\frac{Constant(x_{2}^{1-k}-x_{1}^{1-k})}{1-k}$ Substituting: $W=\frac{200lbf*in^{1.4}*(7in^{1-1.4}-2in^{1-1.4})}{1-1.4}=149.35lbf*in$ $W=149.35lbf*in*\frac{1ft}{12in}*\frac{1Btu}{778.169lbf*ft}=0.016Btu$
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