Question 14.13: Heat Pump Carnot COP with Heat Exchangers Find the COP of a ......
Heat Pump Carnot COP with Heat Exchangers
Find the COP of a Carnot heat pump for the conditions of Example 14.12 if \dot{C}_{i} = \dot{C}_{o} = 0.5 kW/K, and if the heat exchangers are of identical counterflow design, with areas A_{HXi} = A_{HXo} = 4 m², and heat transfer coefficients U_{HXi} = U_{HXo} = 200 W/(m² ⋅ K)
Given: T_{sup} = 30°C, T_{o} = 0°C, (K_{tot})_{HXi} = (K_{tot})_{HXo} = 0.80 kW/K
\dot{C}_{i} = \dot{C}_{o} = 0.5 kW/K on indoor and outdoor sides
Find: COP
Lookup values: Heat exchanger effectiveness from C_{min}/C_{max} = 0 curve of Figure 12.17 for NTU = (K_{tot})_{HX}/\dot{C} = 1.6 is \varepsilon = 0.80 = \varepsilon_{i} = \varepsilon_{o}
The heating load of Example 14.12 is
\dot{Q}_{h} = (20°C – 0°C)(250 W/K) = 5 kW
Solving Equation 14.45 for T_{cond}, we obtain
\dot{Q}_{h} = \varepsilon_{i} \dot{C}_{i} (T_{cond} – T_{sup}) (14.45)
T_{cond} = T_{sup} + \frac{\dot{Q}_{h}}{\varepsilon_{i} \dot{C}_{i}} = 30°C + \frac{5 kW}{0.80 \times 0.5 kW/K}= 42.50°C = 273.15 + 42.50 = 315.65 K
From Equation 14.48, noting that \varepsilon_{i} \dot{C}_{i} = \varepsilon_{o} \dot{C}_{o} and that the temperature ratio must be evaluated in unit of Kelvins, we obtain
T_{evap} = \left\lgroup \frac{1}{1 + 1 – T_{sup}/T_{cond}} \right\rgroup T_{o} = \left\lgroup \frac{1}{1 + 1 – 303.15/315.65} \right\rgroup \times T_{o}= 0.962 \times T_{o} = 262.75 K = -10.40 °C
Finally, the COP is
COP = \frac{T_{cond}}{T_{cond} – T_{evap}} = \frac{315.65 K}{315.65 K – 262.75 K} = 5.97Comments
This is only about 40% of the value 14.7 with innite flow rates and without heat exchanger penalties.