Entropy in the Real World: Refrigerators — Question 2

Question 2

Entropy generation raises the work required for a fixed cooling load

For a cyclic refrigerator, $Q_h=Q_c+W$, where $Q_c$ is heat removed from the cold region and $W$ is work input. Define $K_R=Q_c/W$ and heating coefficient $K_H=Q_h/W$. A refrigerator removes 1350 J per cycle from 270 K and rejects heat to 300 K.

1. Derive the minimum work and maximum $K_R$ using the reservoir entropy balance.

2. If actual work is 225 J, find actual $K_R$, rejected heat and entropy generation.

Question 2 - Solution

Entropy generation raises the work required for a fixed cooling load

(a) Method and calculation

$$S_{\mathrm{gen}}=\frac{Q_c+W}{T_h}-\frac{Q_c}{T_c}\ge0\ \Rightarrow\ W\ge Q_c\left(\frac{T_h}{T_c}-1\right).$$$$W_{\min}=1350(300/270-1)=150\,\mathrm J,\quad K_{R,\max}=\frac{270}{300-270}=9.00.$$

(b) Interpretation and further calculation

$$K_R=1350/225=6.00,\quad Q_h=1575\,\mathrm J.$$$$S_{\mathrm{gen}}=1575/300-1350/270=.250\,\mathrm{J/K}.$$$$W-W_{\min}=75\,\mathrm J=T_hS_{\mathrm{gen}}.$$The extra work ultimately increases heat rejection to the warm reservoir.

Physical check

The cold load is held fixed in the work comparison.