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 6 - The Second Law of Thermodynamics - Problems - Page 313: 6-25E

Answer

$\dot{Q}_H=2.05\times10^7\ Btu/h$

Work Step by Step

$\eta = \dot{W}_e/\dot{Q}_H$ Given $\eta = 0.03,\ \dot{W}_e=180\ kW$: $\dot{Q}_H=6000\ kW=2.05\times10^7\ Btu/h$
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