Worked example — Carnot refrigerator and energy balance

Problem

A freezer keeps $T_f = 255\;\text{K}$ in a kitchen at $T_c = 300\;\text{K}$. It extracts $500\;\text{W}$ of heat from the cold compartment. Calculate: (a) ideal COP, (b) minimum electrical power at the compressor, (c) power rejected to the surroundings.

Solution

(a) Ideal COP (Carnot refrigerator). The coefficient of performance of a refrigerator measures the heat extracted per unit of work spent, $\beta = Q_f/L$. In the reversible limit:

$\beta = \frac{T_f}{T_c - T_f} = \frac{255}{300 - 255} = \frac{255}{45} \approx \ev{5{,}67}$

(b) Minimum electrical power. Since $\beta = Q_f/L$, the power at the compressor is $P_W = P_{Q_f}/\beta$ (the powers are in the same ratio as the energies). With $P_{Q_f} = 500\;\text{W}$:

$P_W = \frac{500\;\text{W}}{5{,}67} \approx \ev{88\;\text{W}}$

(c) Power rejected to the surroundings. The first law requires that everything entering the machine (extracted heat + work) is rejected into the kitchen:

$P_{Q_c} = P_{Q_f} + P_W = 500 + 88 \approx \ev{588\;\text{W}}$

A real freezer draws $200$-$300\;\text{W}$ instead of the ideal $88\;\text{W}$: the ratio of real to ideal performance is about $40\%$. Note also the domestic paradox: at steady state the kitchen receives $\approx 588\;\text{W}$, more than the $500\;\text{W}$ extracted from the compartment — a refrigerator with the door open heats the room, it doesn’t cool it.

Collegamenti

Argomenti: Macchine termiche Concetti: Frigorifero e pompa di calore · Ciclo di Carnot · Primo principio della termodinamica Competenze: Risoluzione di un ciclo termodinamico Oggetti: Frigorifero