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 320: 6-108

Answer

a) $COP_R=4.33$ b) $COP_{R,max}=26.91$ c) $\dot{V}_{1,min}=12.9\ L/min$

Work Step by Step

From tables A-11 to A-13, at the compressor: Inlet ($P_1=0.4\ MPa, x_1=1$): $v_1=0.05127\ m³/kg,\ h_1=255.61\ kJ/kg$ Outlet ($P_2=1.2\ MPa, T_2=70°C$): $h_2=300.63\ kJ/kg$ $\dot{m}_r=\dot{V}_1/v_1$ Given $\dot{V}_1=80\ L/s$ $\dot{m}_r=0.02601\ kg/s$ $\dot{W}_i=\dot{m}_r(h_2-h_1)$ $\dot{W}_i=1.171\ kW$ $\dot{Q}_L=250\ kJ/min+900W$ $\dot{Q}_L=5.067\ kW$ $COP_R=\dot{Q}_L/\dot{W}_i$ $COP_R=4.33$ $COP_{R,max}=\dfrac{1}{1-\frac{T_L}{T_H}}$ With $T_L=23°C,\ T_H=34°C$: $COP_{R,max}=26.91$ $COP_{R,max}=\dot{Q}_L/\dot{W}_{i,min}$ $\dot{W}_{i,min}=0.1883\ kW$ $\dot{W}_{i,min}=\dot{m}_{r,min}(h_2-h_1)$ $\dot{m}_{r,min}=0.004182\ kg/s$ $\dot{m}_{r,min}=\dot{V}_{1,min}/v_1$ $\dot{V}_{1,min}=0.0002144\ m³/s=12.9\ L/min$
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