Problem — Ideal heat pump for a house

Problem

An ideal heat pump heats a house at $T_H = 295\;\text{K}$, drawing heat from outside air at $T_C = 275\;\text{K}$. Calculate the maximum theoretical coefficient of performance (COP). If the house needs a thermal power of $3\;\text{kW}$ to stay at constant temperature, what minimum electrical power does the pump need? Comment on why real heat pumps perform much worse.

Solution

Maximum COP (Carnot model). For a heat pump $\text{COP}_\text{hp} = Q_H/L$. In the reversible Carnot limit, using $|Q_C|/Q_H = T_C/T_H$, we get

$\text{COP}_\text{hp}^\text{max} = \frac{T_H}{T_H - T_C}$

Substituting:

$\text{COP}_\text{hp}^\text{max} = \frac{295}{295 - 275} = \frac{295}{20} = \ev{14{.}75}$

Minimum electrical power. The required thermal power is $P_\text{th} = 3\;\text{kW} = Q_H$ per second. Since $\text{COP}_\text{hp} = P_\text{th}/P_\text{el}$:

$P_\text{el,min} = \frac{P_\text{th}}{\text{COP}_\text{hp}^\text{max}} = \frac{3\;\text{kW}}{14{.}75} = \ev{0{.}20\;\text{kW}}$

Comment. Theoretically, only $\approx 200\;\text{W}$ of electrical power is enough to supply $3\;\text{kW}$ of thermal power: the pump does not “create” energy, it moves it from outside to inside (the missing $2{.}8\;\text{kW}$ comes from the outside air). Real pumps are irreversible — friction, finite temperature differences in the heat exchangers, compressor losses — so their COP is well below this Carnot ceiling, typically $3$–$5$.

Links

Topics: Heat engines Concepts: Refrigerator and heat pump · Carnot cycle Skills: Symbolic set-up Objects: Refrigerator