Refrigerators and Heat Pumps

Running the engine backwards

A Carnot engine is reversible, so every leg can be run in the opposite direction. The work and the two heats keep their sizes and swap their directions. The reversed engine takes heat \(Q_C\) from the cold reservoir. It receives work \(W\) from outside. And it gives heat \(Q_H\) to the hot reservoir. That is a refrigerator.

Follow Fig 4.1. The gas first expands adiabatically from A to D and cools to \(T_C\). Then, touching the cold body, it expands at \(T_C\) and takes in \(Q_C = 1728\,\text{J}\). An adiabatic compression then warms it to \(T_H\). Finally it is squeezed at \(T_H\) and gives out \(Q_H = 2880\,\text{J}\). The loop now runs anticlockwise, so its area, \(1152\,\text{J}\), is work done on the gas.

The energy balance is the engine's, read the other way: \(Q_H = Q_C + W\). The hot body receives the heat taken from the cold body and the work as well. This answers the warm grille. The back of the refrigerator gives the kitchen both the heat pulled out of the food and the electrical energy used by the motor.

A household refrigerator does this with a liquid called a refrigerant. Inside the cold compartment the refrigerant evaporates in thin pipes at a low pressure, and evaporation takes in latent heat. A compressor then squeezes the vapour, which heats it. In the coils at the back the hot vapour condenses and gives out its heat. Through a narrow valve the liquid returns to the low pressure, and the cycle repeats.

How good is a refrigerator?

For a refrigerator the useful result is the heat taken out of the cold space, \(Q_C\). The cost is the work \(W\). So its measure of merit is their ratio, the coefficient of performance. We write it \(K\); many books write \(\beta\) or \(\omega\), and the workbook uses \(\beta\). Unlike an efficiency, \(K\) is usually larger than \(1\).

For a Carnot refrigerator the heats are again in the ratio of the temperatures, \(Q_C/Q_H = T_C/T_H\). Dividing the top and bottom of \(K = Q_C/(Q_H - Q_C)\) by \(Q_H\) gives \(K = T_C/(T_H - T_C)\). By Carnot's theorem, run backwards, no refrigerator between the same two temperatures does better.

The last form links the refrigerator to the engine it came from. Put \(T_C/T_H = 1 - \eta\) into \(K = (T_C/T_H)/(1 - T_C/T_H)\) and you get \(K = (1 - \eta)/\eta\). A cycle that is a poor engine is a good refrigerator. With \(\eta = 0.4\), \(K = 0.6/0.4 = 1.5\).

Start the sim at a freezer compartment of \(260\,\text{K}\), which is \(-13\,°\text{C}\), in a summer kitchen at \(312\,\text{K}\), which is \(39\,°\text{C}\). The best possible \(K\) is \(260/52 = 5\). So \(200\,\text{J}\) of work moves each \(1000\,\text{J}\) out, and the kitchen gets \(1200\,\text{J}\). Now lower \(T_C\) to \(240\,\text{K}\). The work per \(1000\,\text{J}\) grows to \(300\,\text{J}\). The colder the inside, the more each joule costs.

In the figure above, the refrigerator takes \(300\,\text{J}\) from the air in front of it each second and gives \(400\,\text{J}\) to the air behind it. Both sides are in the same closed room. So the room gains \(100\,\text{J}\) each second, the work of the motor, and it grows warmer. Cooling a room needs the hot side outside the room, as in an air conditioner.

Air conditioners and heat pumps

An air conditioner is a refrigerator whose cold side is the room and whose hot side is the unit outdoors. On a Delhi summer day it may keep a room at \(25\,°\text{C}\), \(298\,\text{K}\), while the air outside is at \(45\,°\text{C}\), \(318\,\text{K}\). The best possible \(K\) is \(298/20 = 14.9\). Real air conditioners reach about \(3.5\).

Why so far below the limit? A real unit needs its coils colder than the room and hotter than the street. Otherwise heat would not flow into and out of them fast enough. So the refrigerant works between perhaps \(280\,\text{K}\) and \(330\,\text{K}\), not \(298\,\text{K}\) and \(318\,\text{K}\). The compressor and the fans add friction and electrical losses.

A "1.5 ton" air conditioner removes heat at about \(5.3\,\text{kW}\). At \(K = 3.5\), it draws \(5.3/3.5 = 1.5\,\text{kW}\) from the mains. The outdoor unit then pours out \(5.3 + 1.5 = 6.8\,\text{kW}\) into the street. A perfect Carnot unit would need only about \(0.36\,\text{kW}\) for the same cooling.

Now turn the same machine round. In a Ladakh winter, a heat pump takes heat from the cold air outside and delivers it into a house. Its useful output is \(Q_H\), the heat delivered, so its heating coefficient is \(Q_H/W\). Since \(Q_H = Q_C + W\), this is \(Q_C/W + 1 = K + 1\), always more than \(1\).

Keep a house in Leh at \(20\,°\text{C}\), \(293\,\text{K}\), while the air outside is at \(-15\,°\text{C}\), \(258\,\text{K}\). The best heating coefficient is \(293/35 = 8.4\). So each joule of electricity could deliver \(8.4\,\text{J}\) of heat. An electric heater delivers only \(1\,\text{J}\) per joule. That is why heat pumps are spreading in cold countries.

Now return to the warm grille. The refrigerator takes heat from the food and adds the work of its motor. It gives all of it to the kitchen from the coils at its back. So the back is warm, and the heat there always exceeds the heat taken from inside by the work, as the Clausius statement requires.