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 325: 6-149

Answer

a) $\dot{m}_R=0.0338\ kg/s$ b) $\dot{Q}_H=7.04\ kW$ c) $COP=4.40$ d) $\dot{W}_{i,min}=0.740\ kW$

Work Step by Step

From tables A-11 to A-13, for R-134a: Inlet ($x_1=0.15, T_1=12°C$): $P_1=443.3\ kPa\ h_1=96.54\ kJ/kg$ Outlet ($P_2=P_1, x_2=1$): $h_2=257.33\ kJ/kg$ From tables A-4 to A-6, water: Inlet ($x_3=0, T_3=60°C$): $h_3=251.18\ kJ/kg$ Outlet ($T_4=40°C, x_4=0$): $h_4=167.53\ kJ/kg$ a) $\dot{Q}_L=\dot{m}_w(h_3-h_4),\ \dot{m}_w=0.065\ kg/s$ $\dot{Q}_L=5.437\ kW=\dot{m}_R(h_2-h_1)$ $\dot{m}_R=0.0338\ kg/s$ b) $\dot{Q}_H=\dot{W}_i+\dot{Q}_L,\ \dot{W}_i=1.6\ kW$ $\dot{Q}_H=7.04\ kW$ c) $COP_{HP}=\dot{Q}_H/\dot{W}_i$ $COP=4.40$ d) $COP_{HP, max}=\dfrac{1}{1-\frac{T_L}{T_H}},\quad T_L=25°C,\ T_H=T_3$ $COP_{HP, max}=9.51=\dot{Q}_H/\dot{W}_{i,min}$ $\dot{W}_{i,min}=0.740\ kW$
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