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 319: 6-100

Answer

$T_L=-3°C$

Work Step by Step

$\dot{Q}_H=\dot{W}_i+\dot{Q}_L$ Given $\dot{Q}_H=2000\ kW,\ \dot{W}_i=200\ kW$: $\dot{Q}_L=1800\ kW$ $COP_R=\dot{Q}_L/\dot{W}_i$ $COP_R=9$ $COP_{R,max}=\dfrac{1}{1-\frac{T_L}{T_H}}$ Given $T_H=300\ K$: $T_L=270\ K=-3°C$
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