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: 1-47E

Answer

a) $31330 lbf/ft^2$ b) $217.6 psia$

Work Step by Step

a) $P = (1500 kPa)(\dfrac{20.886 lbf/ft^2}{1kPa}) = 31330 lbf/ft^2$ b) $P = (1500 kPa)(\dfrac{20.886 lbf/ft^2}{1kPa})(\dfrac{1 ft^2}{144 in^2})(\dfrac{1 psia}{1 lbf/in^2}) = 217.6 psia$
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