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 316: 6-55

Answer

a) $COP=2.64$ b) $\dot{Q}_L=1.96\ kW$

Work Step by Step

From tables A-11 to A-13: Inlet ($P_1=0.8\ MPa, T_1=35°C$): $h_1=271.24\ kJ/kg$ Outlet ($P_2=0.8\ MPa, x_2=0$): $h_2=95.48\ kJ/kg$ The heat rejected (with $\dot{m}=0.018\ kg/s$): $\dot{Q}_H=\dot{m}(h_1-h_2)$ $\dot{Q}_H=3.164\ kW$ $COP_{HP}=\dot{Q}_H/\dot{W}_i$ Given $\dot{W}_i=1.2\ kW$ $COP=2.64$ $\dot{Q}_H=\dot{W}_i+\dot{Q}_L$ $\dot{Q}_L=1.96\ kW$
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