Carnot Heat Pump
If a Carnot heat pump operates between an outdoor temperature of −10°C and a building interior temperature of 22°C, what is the COP?
Given: T_{l} = 273 + (−10) = 263 K, T_{h} = 273 + 22 = 295 K
Figure: See Figure 12.3.
Assumptions: Reversible, Carnot cycle
Find: PF_{Carnot,hp}
The last part of Equation 12.5 gives the solution directly:
COP_{Carnot,h} = PF_{Carnot,hp} \equiv \frac{Q_{h}}{W_{net}} = \frac{Q_{1} + W_{net}}{W_{net}} = COP_{Carnot,c} + 1 = \frac{T_{h}}{T_{h} – T_{l}} (12.5)
PF_{Carnot,hp} = \frac{295}{295 – 263} = 9.22Comments
If we were able to operate a heat pump of this efciency each kW of electricity used to operate the heat pump (W_{net} in Figure 12.3) would result in 9.22 kWh of heat being provided to the building’s interior. Of course, practical heating equipment cannot achieve PFs of this level for reasons we shall discuss in greater detail in Chapter 14.