Entropy in the Real World: Refrigerators — Question 6

Question 6

Refrigerators in a reversible cascade have the same overall work limit

Two reversible refrigerator stages lift heat from 240 K to an intermediate 270 K and then to 300 K. The lower stage removes 1200 J from 240 K; all its rejected heat enters the upper stage. There is no intermediate heat leak or accumulation.

1. Find each work input, intermediate heat transfer, and total coefficient of performance.

2. Derive total work for arbitrary intermediate $T_m$ and show that it is independent of $T_m$ for reversible stages.

Question 6 - Solution

Refrigerators in a reversible cascade have the same overall work limit

(a) Method and calculation

$$W_1=1200(270/240-1)=150\,\mathrm J,\quad Q_m=1350\,\mathrm J.$$$$W_2=1350(300/270-1)=150\,\mathrm J,\quad K_{R,\mathrm{total}}=1200/(150+150)=4.00.$$

(b) Interpretation and further calculation

$$Q_m=Q_cT_m/T_c,\quad Q_h=Q_mT_h/T_m=Q_cT_h/T_c.$$$$W_{\mathrm{total}}=Q_h-Q_c=Q_c(T_h/T_c-1)=300\,\mathrm J.$$An ideal cascade reallocates work between stages; it cannot outperform the reversible single-stage limit.

Physical check

The upper stage must remove the lower stage work input as well as its original cooling load.