Heat will not flow from a cold larder into a warm kitchen on its own, and the whole of refrigeration is the business of paying it to.
Reversing the arrows
Every cycle so far has run in one direction: take heat in at high temperature, deliver work, reject heat at low temperature. Run the same loop backwards and the signs all flip. Work goes in, heat is absorbed at low temperature, and a larger quantity of heat is rejected at high temperature. Nothing in the Second Law forbids it, because the work supplied pays for the entropy bookkeeping. What is forbidden is doing it for free.
The same hardware serves two purposes depending on which end you care about. If the point is the heat absorbed at low temperature, the machine is a refrigerator and the cold space is a larder, a cold store or a cryostat. If the point is the heat delivered at high temperature, it is a heat pump and the warm space is a building. The thermodynamics is identical; only the invoice differs.
Coefficient of performance
Thermal efficiency is the wrong score here, and not because these machines are inefficient. Efficiency is defined as what you want over what you pay for, and here what you want is heat moved while what you pay for is work. The ratio is routinely greater than one, so calling it an efficiency invites the wrong instinct entirely. It is called a coefficient of performance instead:
Since the cycle conserves energy, , and dividing through gives a relation worth remembering:
A heat pump is always better at heating than the same machine is at cooling, by exactly one, because the work itself ends up as heat in the warm space and is not wasted.
The Carnot bound applies here too, in reversed form. Between reservoirs at and , no machine beats . The important feature of that expression is the denominator: performance collapses as the temperature lift grows. A domestic fridge lifting heat from to degrees Celsius has a bound of . A freezer lifting from to has a bound of . Cold is expensive, and very cold is very expensive.
Why the reversed Carnot cycle is not built either
The best conceivable cycle between two temperatures is the reversed Carnot cycle, and its practical objections mirror the ones that killed the forward version.
Two of its four processes are fine. Evaporating a refrigerant at constant low pressure absorbs heat at constant temperature, and condensing it at constant high pressure rejects heat at constant temperature, both automatically isothermal because they happen inside the dome. The trouble is the other two.
The compression would have to start inside the dome, taking a wet mixture and compressing it to saturated vapour. Liquid droplets entering a compressor erode blades and, in a reciprocating machine, can cause hydraulic lock that wrecks it in one stroke. Real machines are designed to compress dry vapour and nothing else.
The expansion would have to be an isentropic turbine handling a nearly-liquid mixture at a few kilowatts of output. Such machines exist and are used at large scale in industrial plant, but for anything domestic the machine costs more than the work it recovers.
So the practical cycle makes two changes: evaporate all the way to saturated vapour so the compressor sees dry gas, and replace the turbine with a valve. That is the ideal vapour-compression refrigeration cycle, and it is what is inside essentially every refrigerator, freezer, air conditioner and heat pump on earth.
The cycle, worked
Four states. At the refrigerant leaves the evaporator as saturated vapour at the low pressure. The compressor raises it isentropically to the condenser pressure at , where it is superheated. The condenser rejects heat at constant pressure, leaving saturated liquid at . The expansion valve throttles it back to the low pressure at , isenthalpically, producing the wet mixture that goes to the evaporator.
Take R-134a between an evaporator at degrees Celsius, where the saturation pressure is kPa, , kJ kg⁻¹ and kJ kg⁻¹ K⁻¹, and a condenser at kPa, where the saturation temperature is degrees Celsius, , kJ kg⁻¹ and kJ kg⁻¹ K⁻¹. The superheated table at kPa and degrees Celsius reads kJ kg⁻¹ and kJ kg⁻¹ K⁻¹.
Example. Find the isentropic compressor work, the refrigeration effect and the coefficient of performance.
The compressor leaves state at , which at kPa lies between saturated vapour () and the degree entry (). The fraction across is , so kJ kg⁻¹ at about degrees Celsius, and kJ kg⁻¹.
The valve gives kJ kg⁻¹, so the refrigeration effect is kJ kg⁻¹ and
Every joule of electricity moves four of heat. The condenser rejects kJ kg⁻¹, which is the sum of the other two as it must be.
Now you. The compressor has an isentropic efficiency of . Find the actual work, the actual discharge enthalpy and the coefficient of performance.
Answer
kJ kg⁻¹, so kJ kg⁻¹. The refrigeration effect is unchanged at kJ kg⁻¹, so .
Set that beside the two bounds. Between the cold space at and the kitchen at degrees Celsius, Carnot allows . Driving the heat transfers requires the refrigerant to run colder than the space and hotter than the room, at and , and Carnot between those allows only . The real compressor takes it to . The three numbers separate the loss due to heat exchanger size from the loss due to machine quality, and they point at different fixes: bigger coils for the first, a better compressor for the second.
Example. A cold room needs kW of cooling on the cycle above with the compressor. Find the refrigerant flow, the compressor power and the condenser duty.
The flow is kg s⁻¹, or kg per minute. The compressor power is kW, and the condenser must reject kW.
Now you. A domestic refrigerator on the same cycle needs kW. What flow and compressor power does it need?
Answer
kg s⁻¹ and the power is kW.
The valve, and what it costs
The throttling valve is the one deliberately irreversible component in the cycle, and it is worth costing rather than excusing.
Replace it with an isentropic turbine and follow the numbers. Saturated liquid at kPa has kJ kg⁻¹ K⁻¹. Expanded isentropically to kPa, where and kJ kg⁻¹ K⁻¹, the quality would be , giving kJ kg⁻¹. The turbine would recover kJ kg⁻¹.
That is per cent of the compressor work, which is not nothing, and it appears twice over, because the lower exit enthalpy also increases the refrigeration effect. In a domestic machine, recovering it would require a two-phase turbine producing about half a watt, and the answer is obviously no. In a large ammonia plant or an industrial chiller, the answer is sometimes yes, and expanders are fitted.
The valve's irreversibility can be quantified directly. The state after throttling has quality , hence entropy kJ kg⁻¹ K⁻¹, against going in. So kJ kg⁻¹ K⁻¹ every time a kilogram passes through, and the next lesson will convert that into kJ kg⁻¹ of destroyed work potential at ordinary surroundings, which is close to the kJ kg⁻¹ the turbine would have recovered.
What real machines do differently
Three departures from the ideal cycle are universal, and two of them are deliberate.
The refrigerant leaving the evaporator is deliberately superheated by a few degrees, because a compressor that occasionally swallows liquid does not last. Superheating raises the compressor inlet enthalpy and its work, slightly, and buys reliability.
The liquid leaving the condenser is deliberately subcooled by a few degrees, so that the valve is fed liquid rather than a mixture of liquid and flash vapour. Subcooling is nearly free performance: cooling the liquid K below saturation drops by about kJ kg⁻¹, since the liquid specific heat is around kJ kg⁻¹ K⁻¹, and every one of those joules adds directly to the refrigeration effect. The effect is instead of kJ kg⁻¹, a gain of five per cent for no extra work.
Pressure drops through the evaporator and condenser tubes are not deliberate. They widen the pressure ratio the compressor must work against and cost a few per cent.
Heat pumps
Run the same machine for its condenser output and the economics change completely. The heat pump above delivers kJ kg⁻¹ for kJ kg⁻¹ of work, so , which is as promised.
Against direct electric heating, which delivers exactly one joule of heat per joule of electricity, that is a factor of four. A house needing kW of heat draws kW instead of kW.
The catch is in the denominator of the Carnot expression. The colder it is outside, the larger the lift and the worse the coefficient, so a heat pump is least effective exactly when the house needs it most. Air-source machines in real service manage a seasonal average around in a temperate climate and drop below in a hard frost, when supplementary heating cuts in. Ground-source machines see a source that stays near degrees Celsius all winter and hold a higher coefficient, at the cost of digging.
Example. The heat pump above, with , serves a house needing kW. What electrical power does it draw, and what would resistance heating draw?
kW against kW, a saving of kW while the conditions hold.
Now you. In a cold snap the coefficient falls to and the house needs kW. What does the machine draw?
Answer
kW, more than two and a half times the mild-weather figure for only a quarter more heat.
Where this cycle stops working
Vapour compression has a range, and outside it other machines take over.
A single stage cannot span a large temperature lift, because the pressure ratio becomes impractical and the flash losses at the valve grow until most of the flow is doing nothing useful. The standard answer is a cascade: two or more separate loops with different refrigerants, the condenser of the colder one acting as the evaporator of the warmer. Liquefied natural gas plants use three cascaded loops, on propane, ethylene and methane, to reach K.
Where cheap heat is available and electricity is not, an absorption system replaces the compressor with a pump, an absorber and a generator, exploiting the fact that pumping a liquid solution costs almost nothing while driving the refrigerant back out of solution can be done with waste heat at degrees Celsius. The coefficient of performance is under , but it is being paid in a different currency.
Below about K no refrigerant is a liquid, so the working fluid stays a gas throughout and the machine becomes a reversed Brayton cycle, with a turbine as the expansion device and no phase change anywhere. Aircraft cabin air conditioning uses this, because the compressed air is already available from the engine.
Finally, the refrigerant itself is a choice with consequences beyond thermodynamics. R-134a was adopted to replace R-12 after the Montreal Protocol of , because chlorinated refrigerants destroy stratospheric ozone. R-134a contains no chlorine and is a potent greenhouse gas instead, with a global warming potential around times that of carbon dioxide, and it is being phased down in turn under the Kigali Amendment of in favour of R-1234yf, propane, ammonia and carbon dioxide itself. Each replacement trades some combination of efficiency, flammability, toxicity and operating pressure, and none is free.
One question is now unavoidable. This lesson scored machines with a coefficient of performance above , earlier lessons scored power plants with efficiencies around , and the cogeneration lesson produced a plant whose output was mostly hot water. None of those numbers can be compared with any other, and none of them says which component in a plant to fix. Fixing that is the last lesson.