$$U$$ Internal energy
$$W$$ Work done on the fluid
$$Q_h$$ Energy leaving the refrigerant at the heat sink
$$Q_c$$ Energy entering the refrigerant on the cool side
dots $$\dot{Q_c}\;\dot{Q_h}\;\dot{W}$$ Energy movement per time
$$T_h$$ Temperature at the hot side
$$T_c$$ Temperature on the cool side
$$\eta$$ Coefficient of refrigerator performance $$\frac{|\dot{Q_c}|}{\dot{W}}$$
$$R$$ Ideal gas constant
$$V$$ Specific volume (volume per mole)
$$P$$ Pressure

# How efficient could a refrigerator possibly be?

Photo credit to Dev Benjamin

Air conditioners and refrigerators are the same machine in the way that chainsaws and automobiles are the same machine. The operating conditions for which they have been optimized are different but the underlying physics is the same. Air conditioners (in places where they extensively used) and refrigerators are the largest drivers of electrical usage that have room to be substantially improved at the engineering level. In this article, we will examine the theoretical limits of the efficiency of refrigeration systems from thermodynamics.

## Why is it so hard to cool things?

To illustrate the difficulty of refrigeration, let us examine the limiting case. The most energy efficient refrigerator would run on an ideal gas in a completely reversible way. It is called the Carnot refrigeration cycle*. Here's how it works [1,2].

All efficient cooling is accomplished by taking a fluid, compressing it until hot and high pressure, allowing it to cool in a warm place (removing heat from the fluid), allowing it to expand (which causes cooling) and allowing it to warm back up in a cool place (adding heat to the fluid). The fluid (known as "refrigerant") recirculates through this loop continuously while the unit is running. The refrigerant picks up heat from the cold place (inside the refrigerator) and drops it off in a hot place (outside the refrigerator).

There are other ways are to cool things such as solid state electrical systems (Peltier element) but none of them are close to the efficiency of vapor compression, a technology which is several hundred years old [3].

[Caption] To understand a refrigerator, we must consider the energy balances around two control volumes. The refrigerator control volume is the space we desire to cool. If the refrigerator is colder than its surroundings, $$T_c < T_h$$, then heat is always entering the control volume through imperfect insulation, opening of doors, improper seals, etc. We call these various leaks the rate of heat entry, $$\dot{Q}_{in}$$. Heat is also being removed by the refrigerant, blue arrow, $$\dot{Q_c}$$. Next we have to consider the energy balance on the refrigerant, the heat pump control volume. Here we have three sources of energy transfer - the rate of heat removed from the refrigerator by the heat pump, $$\dot{Q_c}$$, the rate of heat ejected into the external surroundings, $$\dot{Q_h}$$, and the rate of work done on the refrigerant by the compressor, $$\dot{W}$$. Refrigerant shown in purple.

## Considerations for the refrigerator control volume

For the change in internal energy of the refrigerator volume, recall the first law of thermodynamics and the equation for internal energy, $$\Delta \dot{U}$$. The internal energy of the refrigerator is only affected by the colored arrows indicating work and heat that cross the dashed line of the refrigerator control volume: $$\dot{Q}_{in}$$ and $$\dot{Q_c}$$.

\begin{align} \Delta U &= Q+W \quad\quad \text{or}\\ \Delta\dot{ U} &= \dot{Q}+\dot{W}\\ \Delta\dot{U} &= \dot{Q_c}+\dot{Q}_{in}\\ \end{align}

Because one arrow is entering the control volume and one is exiting, $$\dot{Q_c}$$ and $$\dot{Q}_{in}$$ have opposite signs ($$\dot{Q_c}<0$$). This equations leaves us with two choices for lowering $$U$$. We can reduce $$\dot{Q}_{in}$$ by either fixing leaks in the refrigerator seals, improving the refrigerator insulation, or opening the door less. Alternatively, we can use more electrical power to increase the magnitude of $$\dot{Q_c}$$. This analysis does not give us anything that we could not achieve through common sense. The analysis of the heat pump control volume, however, contains some surprising conclusions.

## Considerations for the heat pump control volume

The heat pump control volume shows a device that takes some input energy, $$\dot{W}$$, and uses it to remove heat from a cold place, $$\dot{Q_c}$$, and dump it in a hot a place, $$\dot{Q_h}$$. The natural way to consider the efficiency is to consider the ratio of heat removed to energy used.

$$\eta = \frac{\text{heat removed from the refrigerator}}{\text{electrical power used by the refrigerator}}=\frac{|\dot{Q_c}|}{\dot{W}}$$

In the rest of this section, we are going to show ideal heat pumps have an upper efficency limit of $$\eta = \frac{T_c}{T_h-T_c}$$.

In a Carnot refrigerator, the fluid is an ideal gas; in a real refrigerator, the fluid spends some time as a gas and some as a liquid as it circulates. This increases the rate at which heat can be moved but it requires more energy than it would to use an ideal gas. Since our refrigerant is recirculating, the change in internal energy in the fluid after one cycle, $$\Delta U$$, must be 0. After one cycle, some heat entered the fluid from the cold side ($$Q_c$$), some heat left the fluid on the hot side ($$Q_h$$), and some work was done on the fluid by the compressor ($$W$$). Because our refrigerator is operating at a certain rate, it will be more helpful to think of these quantities per time. Instead of $$Q_c$$ for "heat transfered from the cold side per refrigerant cycle," we will use $$\dot{Q_c}$$ to indicate "heat transfered from the cold side per unit time." Recall the first law of thermodynamics.

\begin{align} \Delta\dot{ U} &= \dot{Q}+\dot{W}\\ 0 &= \dot{Q_c}+\dot{Q_h}+\dot{W} \end{align}

Quick note about sign conventions. Some people say that work on the fluid is positive, some people say work on the surroundings the system is positive. There are groups of chemists, engineers, and physicists who argue about this. It does not matter (and they all agree on that), it only matters that you stay consistent. To make life easier we will shift to absolute values. Since we know all the work put into the system and all the heat leaving the cold side must go to the hot side, we can write:

\begin{align} \dot{W} +|\dot{Q_c}| &= |\dot{Q_h}| \\ \dot{W} &= |\dot{Q_h}| -|\dot{Q_c}|\\ \frac{\dot{W}}{|\dot{Q_c}|} &= \frac{|\dot{Q_h}|}{|\dot{Q_c}|}-1\\ \frac{|\dot{Q_c}|}{\dot{W}} &= \frac{|\dot{Q_c}|}{|\dot{Q_h}|-|\dot{Q_c}|} \label{eq:balance} \end{align}

### Finding a better way

All we are going to do now is find a better way to write $$\frac{|\dot{Q_c}|}{|\dot{Q_h}|-|\dot{Q_c}|}$$. This may become a little technical for those without some familiarity with thermodynamics.

[Caption] In the Carnot refrigeration cycle shown above in a PV diagram, the adiabates (no heat transfer) and isotherms (constant temperature) are blue and red, respectively. The refrigerant moves from $$a$$ to $$b$$ to $$c$$ to $$d$$ to $$a$$.

The Carnot refrigerator contains two adiabatic processes (no heat transfer) and two isothermal processes per cycle. For an isothermal process, $$\partial U = C_v\partial T =$$ 0. Recall the ideal gas law (volume shown as volume per mole, eliminating the need for $$n$$) $$PV=RT$$.

\begin{align} \partial U &= C_v\partial T = 0= \partial Q+\partial W\\ \int_0^Q \partial Q &= RT \int_{V_0}^{V_1} \frac{\partial V}{V}\\ Q &= RT \log\left( \frac{V_1}{V_0}\right) \end{align}

In an adiabatic process, $$\partial T \neq$$ 0 but $$Q =$$ 0. Our equation changes a bit:

\begin{align} \partial U = C_v\partial T &= \partial Q+\partial W\\ C_v\partial T &= \partial W = -P\partial V = RT \frac{\partial V}{V} \\ C_v \int_{T_0}^{T_1} \partial T &= -R\int_{V_0}^{V_1} \frac{\partial V}{V}\\ C_v\log\left(\frac{T_1}{T_0} \right) &= R\log\left( \frac{V_0}{V_1}\right) \end{align}

Next, apply these formulas to the Carnot PV diagram pictured above.

Hot side Cold side
$$\displaystyle Q_h = RT_h \log\left( \frac{V_c}{V_b}\right)$$ $$\displaystyle Q_c = RT_c \log\left( \frac{V_a}{V_d}\right)$$

And then we can apply the equations for the isotherms:

Left Right
$$\displaystyle C_v\log\left(\frac{T_h}{T_c} \right) = R\log\left( \frac{V_a}{V_b}\right)$$ $$\displaystyle C_v\log\left(\frac{T_h}{T_c} \right) = R\log\left( \frac{V_d}{V_c}\right)$$

Take the ratio of the hot and cold sides:

\begin{align} \frac{|Q_h|}{|Q_c|} &= \frac{\left| RT_h \log\left( \frac{V_c}{V_b}\right)\right|}{\left| RT_c \log\left( \frac{V_a}{V_d}\right)\right|}\\ \frac{|Q_h|}{|Q_c|} &= \frac{\left| T_h \log\left( \frac{V_c}{V_b}\right)\right|}{\left| T_c \log\left( \frac{V_a}{V_d}\right)\right|}\\ \end{align}

We seem to be stuck on the ratio of logarithms but we can resolve that with the equation from the isotherms.

\begin{align} C_v\log\left(\frac{T_h}{T_c} \right) = R\log\left( \frac{V_a}{V_b}\right) &= R\log\left( \frac{V_d}{V_c}\right) \\ \log\left( \frac{V_a}{V_b}\right) &= \log\left( \frac{V_d}{V_c}\right) \\ \frac{V_a}{V_b} &= \frac{V_d}{V_c} \\ \end{align}

Because our ratios are in absolute values, the $$\log\left( \frac{V_c}{V_b}\right)$$ factor cancels with the $$\log\left( \frac{V_a}{V_d}\right)$$ factor in the ratio above.

\begin{align} \frac{|Q_h|}{|Q_c|} &= \frac{ \left| T_h \log\left( \frac{V_c}{V_b}\right) \right|}{ \left| T_c \log\left( \frac{V_a}{V_d}\right)\right|}\\ \frac{|Q_h|}{|Q_c|} &= \frac{T_h}{T_c}\\ \end{align}

We can substitute $$\frac{|Q_h|}{|Q_c|}$$ into the equation at the top.

\begin{align} \frac{|\dot{Q_c}|}{\dot{W}} &= \frac{|\dot{Q_c}|}{|\dot{Q_h}|-|\dot{Q_c}|} = \left[\frac{|\dot{Q_h}|}{|\dot{Q_c}|}-1\right]^{-1}\\ \frac{|\dot{Q_c}|}{\dot{W}} &= \left[\frac{T_h}{T_c}-1\right]^{-1} = \frac{T_c}{T_h-T_c}\\ \end{align}

This is $$\eta$$ the coefficient of refrigerator performance by inverting the equation above. This result is important.

$$\eta = \frac{\text{heat removed from the refrigerator}}{\text{electrical power used by the refrigerator}}=\frac{|\dot{Q_c}|}{\dot{W}} = \frac{T_c}{T_h-T_c}$$

This is the maximal theoretical efficiency of a refrigerator, the upper boundary from thermodynamics. All temperatures must be in absolute units for this equation to be valid. For a refrigerator in a 75°F room holding food at 40°F, the coefficient of performance for a Carnot refrigerator is 14. It is not possible to remove more than 14 J of heat per Joule of electricity used. In the real world, every Joule of energy put into the compressor does not translate to 1 Joule of energy applied to the refrigerant, there are non-idealities in how the refrigerant behaves, and the refrigerant must move quickly enough that the whole process is far from isentropic. Although it is hard to find data on coefficients of performance for consumer refrigerators, what is available indicates most consumer refrigerators run with a coefficient of performance between 2 and 4. Even though their efficiency has improved since 1980, there is a lot of room for technical innovation.

*My professors used to joke that the people who make Carnot refrigerators also make frictionless ramps and massless pulleys.

## References

[1] J. M. Smith, H. C. Van Ness, M. W. Abbott, and M. T. Swihart, "The second law of thermodynamics," in Introduction to Chemical Engineering Thermodynamics 8th Ed, pp. 174—180, 2017.

[2] J. M. Smith, H. C. Van Ness, M. W. Abbott, and M. T. Swihart, "Refrigeration and liquification," in Introduction to Chemical Engineering Thermodynamics 8th Ed, pp. 327—334, 2017.

[3] J. M. Calm, "The next generation of refrigerants—Historical review, considerations, and outlook," International Journal of Refrigeration, vol. 31, pp. 1123—1133, 2008.