Problem — Refrigerator as a reverse cycle

Problem

A domestic refrigerator has COP $\beta = 4$. Explain the physical meaning and the relation to the efficiency $\eta$ of the reverse Carnot cycle.

Solution

Meaning of the COP. A refrigerator is a heat engine run backwards: it consumes work $W$ (from the compressor) to extract heat $Q_f$ from the cold compartment and dump $Q_c = Q_f + W$ into the warm environment. The coefficient of performance measures how much heat is extracted per unit of work spent:

$\beta = \frac{Q_f}{W}$

With $\beta = 4$: for every joule of electrical energy absorbed, the refrigerator removes $4$ J of heat from its interior. The COP can be greater than $1$ because it does not “create” energy: it only moves it, spending work to push it up the thermal gradient.

Carnot limit. A reverse Carnot cycle, the best possible, has a COP that depends only on the temperatures:

$\beta_{\text{Carnot}} = \frac{T_f}{T_c - T_f}$

Relation to efficiency. The direct cycle has $\eta = W/Q_c$ and the reverse cycle $\beta = Q_f/W$. Using $Q_c = Q_f + W$ in the reversible case gives the link:

$\beta = \frac{Q_f}{W} = \frac{Q_c - W}{W} = \frac{1}{\eta} - 1 = \frac{1-\eta}{\eta}$

$\ev{\beta = \frac{Q_f}{W}, \qquad \beta_{\text{Carnot}} = \frac{T_f}{T_c - T_f} = \frac{1-\eta}{\eta}}$

A refrigerator is therefore a Carnot engine run in reverse: where the engine produces work from the thermal gap, the refrigerator spends work to pump heat uphill.

Collegamenti

Argomenti: Heat engines Concetti: Refrigerator and heat pump · Carnot cycle Competenze: Symbolic setup Oggetti: Refrigerator