Refrigerators and heat pumps

A heat engine absorbs heat at a high temperature, produces work, and rejects the remainder at a low temperature. Drive the same cycle round the other way — compress the working substance where it used to expand, expand it where it used to be compressed — and every energy arrow reverses. Heat now enters at the low temperature and leaves at the high temperature, and the machine consumes work instead of producing it. That work is not optional. Left to itself energy always drains from hot to cold and never the other way, so any device that carries heat up a temperature gradient has to be paid for. Draw the diagram with each arrow pointing the way the energy actually travels and label every quantity by its magnitude: Q_C entering from the cold reservoir, W entering from the motor, Q_H leaving into the hot reservoir. Sign errors in this topic almost always come from carrying signed first-law symbols into a picture whose directions are already drawn in.

The working substance ends every cycle in the state it began, so its internal energy change over the cycle is exactly zero: ΔU = 0. Whatever enters must leave. There are two inputs and one output, and the accounting collapses to a single addition, Q_H = Q_C + W, with all three symbols standing for magnitudes. A cooling unit that removes 420 J from its cold box while the compressor does 120 J of work on the refrigerant must therefore hand 420 + 120 = 540 J to the room behind it. Divide every term by the cycle time and the same relation holds between rates, which is the form the data usually arrives in: that same unit lifting heat at 2.8 kW on 0.80 kW of electrical input warms the room at 3.6 kW. Use whichever version matches your numbers, then audit it by asking whether the hot-side figure is the largest of the three. In a reversed cycle it always is.

Nothing inside the machine says whether it is a refrigerator or a heat pump; the name records which reservoir you are paying attention to. Care about the cold side and you have a refrigerator, rated by the cooling coefficient of performance K_cool = Q_C/W. Care about the hot side and the same box is a heat pump, rated by Kₕₑₐₜ = Q_H/W. A reversible air conditioner switches between the two roles with a valve, cooling a room in July and heating it in January with the same compressor. Because Q_H = Q_C + W, the two ratings are locked together: Kₕₑₐₜ = K_cool + 1, always, for any machine whatsoever. The cooling unit of the previous section scores K_cool = 420/120 = 3.5 as a refrigerator and Kₕₑₐₜ = 540/120 = 4.5 as a heat pump. Note what the definitions do not contain: neither is work out over heat in, so neither is an efficiency and neither is capped at 1.

Run a Carnot cycle backwards and it becomes the best possible refrigerator between the same two reservoirs, because any machine that beat it could drive it in reverse and shuttle heat from cold to hot with no net work at all. Its two heat exchanges happen isothermally at T_C and T_H, and for the reversible cycle the heats stand in the ratio of the absolute temperatures, Q_C/Q_H = T_C/T_H. Put that into the definitions. Divide Q_H = Q_C + W by Q_C to get W/Q_C = Q_H/Q_C − 1 = T_H/T_C − 1 = (T_H − T_C)/T_C, so K_cool ≤ T_C/(T_H − T_C). The same step from the hot side gives Kₕₑₐₜ ≤ T_H/(T_H − T_C), and the two ceilings differ by 1, exactly as the general relation demands. For a cold box at 250 K in a 300 K room the ceilings are 250/50 = 5.0 and 300/50 = 6.0. Both temperatures must be in kelvin: the formulae divide by a difference, and Celsius values put the wrong number on top.

Look at what the ceiling depends on. Not the size of the cabinet, not how much heat you want moved, only the two reservoir temperatures — and mostly on their difference. Take a heat pump holding a house at 290 K. With the outside air at 280 K the lift is 10 K and the heating ceiling is 290/10 = 29.0; let the air fall to 270 K and the lift is 20 K and the ceiling drops to 290/20 = 14.5; at 260 K it is 290/30 = 9.67. The cheapest heat is always the heat you lift the shortest distance, which is why ground-source pumps drawing on soil near 283 K beat air-source pumps on a January night, and why underfloor heating at 308 K outperforms radiators at 333 K. The same reading explains a refrigerator: push the kitchen 5 K warmer, or block the condenser coils so they must run hotter still, and the ceiling falls while the compressor works longer for the same cold box.

A domestic fridge is a vapour-compression cycle. Liquid refrigerant boils in the evaporator inside the cold box, absorbing Q_C; the compressor raises its pressure and temperature, doing W; the vapour condenses in the coils outside, releasing Q_H; a throttling valve drops it back to low pressure. Three things cost you. Heat only flows down a gradient at a finite rate, so the evaporator must sit several kelvin below the cold space and the condenser several kelvin above the room: the refrigerant works across a wider lift than the reservoirs suggest. The throttle is violently irreversible, replacing the reversible expansion a Carnot cycle would use. And the compressor is neither frictionless nor adiabatic. A chest freezer holding 240 K inside a 290 K garage faces a ceiling of 240/50 = 4.8, and a measured K_cool of 1.6 is one third of that, which is ordinary for an appliance of that kind.