Question 14.12: Heat Pump Carnot COP with Finite Air Flow Rate Find the COP ......
Heat Pump Carnot COP with Finite Air Flow Rate
Find the COP of a heat pump for a house with a typical value K_{tot} = 250 W/K [474 Btu/(h· °F)] if T_{i} = 20°C (68°F), T_{o} = 0°C (32°F), and the airflow rate of the heating system is \dot{V} = 0.42 m³/s.
Given: T_{i}, T_{o}, K_{tot}, \dot{V}
Find: COP
Lookup values: \rho C_{p} = 1.2 kJ/(m^{3} \cdot K)
Step-by-Step
The quantity \dot{m} c_{p} = \dot{V}_{\rho} C_{p} = 0.42 m^{3}/s \times 1.2 kJ/(m^{3} \cdot K) = 0.5 kW/K . Inserting this into Equation 14.44, the supply temperature to the house is
T_{sup} – T_{i} = (T_{i} – T_{o}) \frac{K_{tot}}{\dot{V} \rho c_{p}} (14.44)
T_{sup} = 20°C + (20°C – 0°C) \frac{250 W/K}{0.5 kWK} = 30 °C= 303.15 K (86°F)
The corresponding COP is
COP = \frac{T_{sup}}{T_{sup} – T_{o}} = \frac{303.15}{303.15 – 273.15} = 10.11which is much less than the value of 14.7 with infinite flow rate.
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