Entropy in the Real World: Refrigerators — Question 3

Question 3

A heat pump delivers heating by combining work with environmental heat

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 heat pump supplies 12000 J per cycle to a home at 300 K with measured $K_H=4.00$. It extracts heat from an outdoor reservoir at 270 K.

1. Find work input, extracted outdoor heat and entropy generation.

2. Find the reversible minimum work for the same heating load and compare the pump with an ideal resistance heater converting work entirely to indoor heat.

Question 3 - Solution

A heat pump delivers heating by combining work with environmental heat

(a) Method and calculation

$$W=Q_h/K_H=3000\,\mathrm J,\quad Q_c=12000-3000=9000\,\mathrm J.$$$$S_{\mathrm{gen}}=12000/300-9000/270=20/3\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$K_{H,\max}=T_h/(T_h-T_c)=10,\quad W_{\min}=12000/10=1200\,\mathrm J.$$$$W_{\mathrm{resistance}}=12000\,\mathrm J,\quad W/W_{\mathrm{resistance}}=1/4.$$The pump uses less work because it also transfers outdoor heat; the ideal resistance heater supplies no such transferred heat.

Physical check

The measured coefficient is below the reversible limit for these reservoir temperatures.