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: 6-101

Answer

a) $\dot{m}_r=0.0498\ kg/s$ b) $\dot{Q}_L=5.07\ kW$ c) $COP_R=1.54$ d) $\dot{W}_{i,min}=1.13\ kW$

Work Step by Step

From tables A-11 to A-13: Inlet ($P_1=1.2\ MPa, T_1=50°C$): $T_{sat}=46.3°C,\ h_1=278.28\ kJ/kg$ Outlet ($P_2=1.2\ MPa, T_2=T_{sat}-5°C$): $h_2=110.19\ kJ/kg$ From tables A-4 to A-6: Inlet ($x_3=0, T_3=18°C$): $h_3=75.54\ kJ/kg$ Outlet ($x_4=0, T_4=26°C$): $h_4=109.01\ kJ/kg$ For the water $\dot{m}_w=0.25\ kg/s$: $\dot{Q}_H=\dot{m}_w(h_4-h_3)$ $\dot{Q}_H=8.367\ kW$ For the refrigerant: $\dot{Q}_H=\dot{m}_r(h_1-h_2)$ $\dot{m}_r=0.0498\ kg/s$ $\dot{Q}_H=\dot{W}_i+\dot{Q}_L$ Given $\dot{W}_i=3.30\ kW$: $\dot{Q}_L=5.07\ kW$ $COP_R=\dot{Q}_L/\dot{W}_i$ $COP_R=1.54$ $COP_{R,max}=\dfrac{1}{1-\frac{T_L}{T_H}}=\dot{Q}_L/\dot{W}_{i,min}$ Given $T_L=-35°C,\ T_H=T_3=18°C$: $\dot{W}_{i,min}=1.13\ kW$
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