Question 16.3: A Carnot refrigerator Now let’s see what happens if the Carn...
A Carnot refrigerator
Now let’s see what happens if the Carnot engine described in Example 16.2 is run backward as a refrigerator. What is its performance coefficient?
SET UP AND SOLVE We use the values from Example 16.2, with all the signs reversed because the cycle is run backward:
\mathrm{Q_C = +1400 J, Q_H = -2000 J, W = -600 J}.
From Equation 16.6, the performance coefficient K is
\mathrm{K=\frac{Q_C}{\left|W\right| }=\frac{\left|Q_C\right| }{\left|Q_H\right| -\left|Q_C\right| } } (16.6)
\mathrm{K=\frac{Q_C}{\left|W\right| }=\frac{1400 J}{600 J}=2.33. }
REFLECT For a Carnot cycle, e and K depend only on the temperatures, as shown by Equations 16.8 and 16.9, and we don’t need to calculate Q and W. For cycles containing irreversible processes, however, these equations are not valid, and more detailed calculations are necessary.
\mathrm{e_{Carnot}=1-\frac{T_C}{T_H}=\frac{T_H-T_C}{T_H} } . (16.8)
\mathrm{K_{Carnot}=\frac{T_C}{T_H-T_C} }. (16.9)
Practice Problem: Suppose you want to increase the performance coefficient K in Example 16.3. Would it be better to decrease \mathrm{T_H} by 50 K or to increase \mathrm{T_C} by 50 K? Answer: Increase \mathrm{T_C} .