Problem — Real air conditioner (SEER)

Problem

An air conditioner consumes $1{,}5\;\text{kW}$ and delivers $4{,}5\;\text{kW}$ of “cooling” to the indoor space (SEER rating). Estimate the real COP and compare it with the Carnot limit, assuming $T_f = 295\;\text{K}$, $T_c = 308\;\text{K}$.

Solution

Real COP. The coefficient of performance is the ratio between the cooling power delivered and the electrical power absorbed:

$\text{COP}_\text{real} = \frac{P_\text{cooling}}{P_\text{electrical}} = \frac{4{,}5\;\text{kW}}{1{,}5\;\text{kW}} = \ev{3}$

Carnot limit. The theoretical maximum for a refrigerator between $T_f$ and $T_c$ is

$\text{COP}_\text{Carnot} = \frac{T_f}{T_c - T_f} = \frac{295}{308 - 295} = \frac{295}{13} \approx \ev{22{,}7}$

Comparison. The ratio between real performance and the ideal limit is

$\frac{\text{COP}_\text{real}}{\text{COP}_\text{Carnot}} = \frac{3}{22{,}7} \approx 0{,}13 = \ev{13\%}$

The real air conditioner only achieves $\sim 13\%$ of the Carnot limit. It’s not the “fault” of the second law (which would allow up to $22{,}7$): the small temperature gap ($13\;\text{K}$) makes the ideal ceiling very high, but friction, imperfect heat exchangers, and real compression cut down performance. The room for technological improvement is therefore huge.

Collegamenti

Argomenti: Macchine termiche Concetti: Frigorifero e pompa di calore · Ciclo di Carnot Competenze: Stime di Fermi Oggetti: Frigorifero