Entropy in the Real World: Refrigerators — Question 10

Question 10

A heat-driven refrigerator is bounded by three reservoir temperatures

An idealized cyclic refrigerator receives driving heat $Q_g$ from 450 K and removes heat $Q_c$ from 270 K. It rejects $Q_r$ to 300 K and has no external work or matter exchange. Define its thermal cooling coefficient as $K_{\mathrm{heat}}=Q_c/Q_g$.

1. Derive the maximum thermal coefficient from energy and entropy balances.

2. For a 900 J cooling load, find minimum driving heat and corresponding rejected heat. Why is this coefficient different from $Q_c/W$?

Question 10 - Solution

A heat-driven refrigerator is bounded by three reservoir temperatures

(a) Method and calculation

$$Q_r=Q_g+Q_c,\quad S_{\mathrm{gen}}=\frac{Q_g+Q_c}{300}-\frac{Q_g}{450}-\frac{Q_c}{270}\ge0.$$$$\frac{Q_c}{Q_g}\le\frac{1/300-1/450}{1/270-1/300}=3.00.$$

(b) Interpretation and further calculation

$$Q_{g,\min}=900/3=300\,\mathrm J,\quad Q_r=1200\,\mathrm J.$$$$1200/300-300/450-900/270=0.$$Driving heat is the resource in the denominator. There is no external work input, so a work-based coefficient would not be the useful performance definition here.

Physical check

Equality is a reversible bound for the stipulated three-reservoir model, not a claim about an actual absorption machine.