Suppose we were to operate a Brayton heat engine backward, going ccw rather than cw in the pV diagram. FIGURE 21.16a (which you should compare to Figure 21.14a) shows a device for doing this. FIGURE 21.16b is its pV diagram, and FIGURE 21.16c is the energy-transfer diagram. Starting from point 4, the gas is adiabatically compressed to increase its temperature and pressure. It then flows through a high-temperature heat exchanger where the gas cools at constant pressure from temperature $T_3$ to $T_2$. The gas then expands adiabatically, leaving it significantly colder at $T_1$ than it started at $T_4$. It completes the cycle by flowing through a low-temperature heat exchanger, where it warms back to its starting temperature.

FIGURE 21.16 A refrigerator that extracts heat from the cold reservoir and exhausts heat to the hot reservoir.

Full Alternative Text

Full Alternative Text

Suppose that the low-temperature heat exchanger is a closed container of air surrounding a pipe through which the engine’s cold gas is flowing. The heat-exchange process $1\to 4$ cools the air in the container as it warms the gas flowing through the pipe. If you were to place eggs and milk inside this closed container, you would call it a refrigerator!

Going around a closed pV cycle in a ccw direction reverses the sign of W for each process in the cycle. Consequently, the area inside the curve of Figure 21.16b is $W_{\text{in}}$, the work done on the system. Here work is used to extract heat $Q_{\text{C}}$ from the cold reservoir and exhaust a larger amount of heat $Q_{\text{H}}=Q_{\text{C}}+W_{\text{in}}$ to the hot reservoir. But where, in this situation, are the energy reservoirs?

Understanding a refrigerator is a little harder than understanding a heat engine. The key is to remember that heat is always transferred from a hotter object to a colder object. In particular,

• The gas in a refrigerator can extract heat $Q_{\text{C}}$ from the cold reservoir only if the gas temperature is lower than the cold-reservoir temperature $T_{\text{C}}$. Heat energy is then transferred from the cold reservoir into the colder gas.

• The gas in a refrigerator can exhaust heat $Q_{\text{H}}$ to the hot reservoir only if the gas temperature is higher than the hot-reservoir temperature $T_{\text{H}}$. Heat energy is then transferred from the warmer gas into the hot reservoir.

These two requirements place severe constraints on the thermodynamics of a refrigerator. Because there is no reservoir colder than $T_{\text{C}}$, the gas cannot reach a temperature lower than $T_{\text{C}}$ by heat exchange. The gas in a refrigerator must use an adiabatic expansion $(Q=0)$ to lower the temperature below $T_{\text{C}}$. Likewise, a gas refrigerator requires an adiabatic compression to raise the gas temperature above $T_{\text{H}}$.

EXAMPLE 21.3 Analyzing a refrigerator

A refrigerator using helium gas operates on a reversed Brayton cycle with a pressure ratio of 5.0. Prior to compression, the gas occupies $100{\text{cm}}^3$ at a pressure of 150 kPa and a temperature of $-23\text{°}\text{C}$. Its volume at the end of the expansion is $80{\text{cm}}^3$. What are the refrigerator’s coefficient of performance and its power input if it operates at 60 cycles per second?

MODEL The Brayton cycle has two adiabatic processes and two isobaric processes. The work per cycle needed to run the refrigerator is $W_{\text{in}}=Q_{\text{H}}-Q_{\text{C}}$; hence we can determine both the coefficient of performance and the power requirements from $Q_{\text{H}}$ and $Q_{\text{C}}$. Heat energy is transferred only during the two isobaric processes.

VISUALIZE FIGURE 21.17 shows the pV cycle. We know from the pressure ratio of 5.0 that the maximum pressure is 750 kPa. Neither $V_2$ nor $V_3$ is known.

FIGURE 21.17 A Brayton-cycle refrigerator.

FIGURE 21.17 Full Alternative Text

SOLVE To calculate heat we’re going to need the temperatures at the four corners of the cycle. First, we can use the conditions of state 4 to find the number of moles of helium:

$$n={\displaystyle \frac{p_4{V}_4}{R{T}_4}}=0.00722\text{mol}$$

Process $1\to 4$ is isobaric; hence temperature $T_1$ is

$$T_1={\displaystyle \frac{V_1}{V_4}}{T}_4=\left(0.80\right)\left(250\text{K}\right)=200\text{K}=-73\text{°}\text{C}$$

With Equation 21.14 we found that the quantity $p^{(1-\gamma )\text{/}\gamma }T$ remains constant during an adiabatic process. Helium is a monatomic gas with $\gamma =\frac{5}{3}$, so $(1-\gamma )\text{/}\gamma =-\frac{2}{5}=-0.40$. For the adiabatic compression $4\to 3$,

$$p_3^{}{}^{-0.40}{T}_3=p_4^{}{}^{-0.40}{T}_4$$

Solving for $T_3$ gives

$$T_3={\left({\displaystyle \frac{p_4}{p_3}}\right)}^{-0.40}{T}_4={\left({\displaystyle \frac{1}{5}}\right)}^{-0.40}(250\text{K})=476\text{K}=203\text{°}\text{C}$$

The same analysis applied to the $2\to 1$ adiabatic expansion gives

$$T_2={\left({\displaystyle \frac{p_1}{p_2}}\right)}^{-0.40}{T}_1={\left({\displaystyle \frac{1}{5}}\right)}^{-0.40}(200\text{K})=381\text{K}=108\text{°}\text{C}$$

Now we can use $C_{\text{P}}=\frac{5}{2}R=20.8\text{J/mol K}$ for a monatomic gas to compute the heat transfers:

$$\begin{array}{ccc}\hfill Q_{\text{H}}& \hfill =\hfill & \left|Q_{32}\right|=n{C}_{\text{P}}(T_3-T_2)\hfill \\ \hfill & \hfill =\hfill & \left(0.00722\text{mol}\right)\left(20.8\text{J/mol K}\right)\left(95\text{K}\right)=14.3\text{J}\hfill \\ \hfill Q_{\text{C}}& \hfill =\hfill & \left|Q_{14}\right|=n{C}_{\text{P}}(T_4-T_1)\hfill \\ \hfill & \hfill =\hfill & \left(0.00722\text{mol}\right)\left(20.8\text{J/mol K}\right)\left(50\text{K}\right)=7.5\text{J}\hfill \end{array}$$

Thus the work input to the refrigerator is $W_{\text{in}}=Q_{\text{H}}-Q_{\text{C}}=6.8\text{J}$. During each cycle, 6.8 J of work are done on the gas to extract 7.5 J of heat from the cold reservoir. Then 14.3 J of heat are exhausted into the hot reservoir.

The refrigerator’s coefficient of performance is

$$K={\displaystyle \frac{Q_{\text{C}}}{W_{\text{in}}}}={\displaystyle \frac{7.5\text{J}}{6.8\text{J}}}=1.1$$

The power input needed to run the refrigerator is

$$P_{\text{in}}=6.8{\displaystyle \frac{\text{J}}{\text{cycle}}}\times 60{\displaystyle \frac{\text{cycles}}{\text{s}}}=410{\displaystyle \frac{\text{J}}{\text{s}}}=410\text{W}$$

REVIEW These are fairly realistic values for a kitchen refrigerator. You pay your electric company for providing the work $W_{\text{in}}$ that operates the refrigerator. The cold reservoir is the freezer compartment. The cold temperature $T_{\text{C}}$ must be higher than $T_4(T_{\text{C}}>-23\text{°}\text{C})$ in order for heat to be transferred from the cold reservoir to the gas. A typical freezer temperature is $-15\text{°}\text{C,}$ so this condition is satisfied. The hot reservoir is the air in the room. The back and underside of a refrigerator have heat-exchanger coils where the hot gas, after compression, transfers heat to the air. The hot temperature $T_{\text{H}}$ must be less than $T_2(T_{\text{H}}<108\text{°}\text{C})$ in order for heat to be transferred from the gas to the air. An air temperature $\approx 25\text{°}\text{C}$ under a refrigerator satisfies this condition.