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 1 - Introduction and Basic Concepts - Problems - Page 43: 1-55E

Answer

$P=92.81psia$

Work Step by Step

$\rho=1.03*62.4\frac{lbm}{ft^3}=64.27\frac{lbm}{ft^3}$ $P=P_{atm}+\rho*g*h$ $P=14.7psia+(64.27\frac{lbm}{ft^3})*(32.2\frac{ft}{s^2})*(175ft)*(\frac{1lbf}{32.2\frac{lbm*ft}{s^2}})*(\frac{ft^2}{144in^2})$ $P=14.7psia+78.11psia=92.81psia$
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