Refrigeration and heat pumps

Everything in the second law told you that a heat engine cannot convert all of its heat input to work, and that moving heat from cold to hot costs work you must supply. This topic is about the machine built on the second half of that statement. It is not a new device requiring new physics. It is the same four-process loop as a power cycle, traversed the other way, and the entire design problem is how far the real hardware is allowed to drift from the reversible ideal before the drift starts costing real money.

Reversed cycles and COP

Take a Carnot engine and run it backwards. Every arrow on the diagram reverses: heat that left the cold reservoir now enters it, heat that entered the hot reservoir now leaves it, and the net work that used to come out now has to go in. The device still has four reversible steps and it still moves between the same two temperatures, so the bookkeeping is unchanged in every respect except direction. What it buys you is the ability to pump heat uphill, from a cold space to a warm one, provided you are willing to pay for it in work.

Performance for this machine is measured the way it was defined in the second law: what you want, divided by what you pay. For a refrigerator or an air conditioner, what you want is the heat removed from the cold space,

and for the reversed Carnot cycle specifically, where \(Q_H/Q_L = T_H/T_L\), this reduces to a ceiling set by temperature alone:

Take a domestic freezer holding \(253\) K inside a kitchen at \(303\) K, and the best any machine could possibly do between those two temperatures is \(253/50 = 5.06\): over five units of cooling for every unit of work. That number is not a target to design towards. It is a bound that no amount of cleverness will cross, the same way \(1 - T_L/T_H\) bounds a power cycle, and it is worth calculating first in any refrigeration problem precisely because it tells you, before a single component is chosen, whether a claimed performance figure is physically possible.

Here is the complication that this section exists to raise and the next one exists to resolve. A reversed Carnot cycle is drawn as a rectangle on a T–s diagram: two isotherms for the heat transfers, two isentropes joining them. Nothing forbids that shape. What is wrong with it is not the thermodynamics but the hardware needed to trace it. The isothermal heat transfers are easy to get for free, because a fluid boiling or condensing at fixed pressure does them automatically. That is the entire content of the saturation dome in properties of pure substances. The two isentropic steps are the problem, and they are a problem in different ways at each end of the cycle.

Isentropic compression of a saturated or two-phase mixture is wet compression: liquid droplets are present at the compressor inlet, and a reciprocating or rotary compressor built to handle vapour is not built to handle droplets, which erode valves and blades and can hydraulic-lock a cylinder outright. Isentropic expansion at the other end has the opposite problem. It is perfectly buildable (a small turbine would do it), but the work it recovers is a few per cent of the compressor's work at best, and no one has found it worth the capital cost of a second rotating machine to reclaim that little. Both objections are engineering objections, not thermodynamic ones, and both are removed by replacing one process each, which is exactly the redesign the next section makes.

The vapour-compression cycle

Keep the two isothermal heat transfers (they were free) and replace the two troublesome isentropic steps with ones that real hardware handles well. Compress only vapour, never a two-phase mixture, by arranging for the fluid to be fully vaporised before it reaches the compressor. And replace the expansion turbine with a throttle: a valve, or even a length of narrow capillary tube, that drops the pressure with no moving parts and no work recovered. The result is the ideal vapour-compression cycle, the pattern behind essentially every refrigerator, air conditioner and chiller in service.

Four states, four processes, numbered around the loop. Saturated vapour at the evaporator pressure enters the compressor at state 1 and leaves at state 2, compressed isentropically to the condenser pressure. This step is exactly as good as it can be, because nothing has been given up here; wet compression was the problem, and there is none of it. At constant high pressure the vapour gives up heat to the surroundings, desuperheating and then condensing to saturated (or slightly subcooled) liquid at state 3. The throttle drops it to the evaporator pressure at state 4, a mixture of liquid and vapour at low quality, without changing its enthalpy. No work crosses a throttle's boundary and it exchanges no heat, so the steady-flow energy balance collapses to \(h_4 = h_3\). Finally the cold mixture absorbs heat from the space being refrigerated, evaporating at constant low pressure back to saturated vapour at state 1, and the loop closes.

Every energy quantity in the cycle is now a difference of two tabulated enthalpies, because every process is either a steady-flow device at constant pressure or a throttle at constant enthalpy. Nothing here calls for an integral. The refrigeration effect, the work input and the heat rejected are

and the coefficient of performance is simply their ratio, \(\mathrm{COP}_R = q_L / w_{\text{in}}\), read straight off a property table with no cycle integration anywhere in the calculation. For R-134a operating between roughly \(-19\) °C and \(31\) °C saturation (evaporator and condenser pressures a first-year problem would hand you directly), typical enthalpies are \(h_1 \approx 239\), \(h_2 \approx 275\) and \(h_3 = h_4 \approx 95\) kJ/kg, giving \(q_L \approx 144\) kJ/kg, \(w_{\text{in}} \approx 36\) kJ/kg and \(\mathrm{COP}_R \approx 3.97\).

Notice what has been given up against the reversed Carnot cycle, and where. The two heat transfers are still at constant pressure and, inside the dome, at constant temperature too, so they remain exactly as good as the ideal allowed. The throttle is the one process in the whole loop that is not reversible. Entropy rises across it for nothing, since it produces no work at all, and it is the price paid for deleting the expansion turbine. That the ideal vapour-compression cycle still comes respectably close to its Carnot ceiling, rather than falling far short, is the entire reason the redesign was worth making.

Refrigerants and the P–h diagram

The fluid doing this work is chosen, not incidental. A refrigerant needs a saturation pressure that sits in a convenient range at the temperatures the cycle actually operates at: high enough at the evaporator that air cannot leak in through a seal, low enough at the condenser that the vessel does not need to be built like a boiler, together with a large latent heat, so that a given mass flow moves a useful amount of energy, and chemical stability, low toxicity and compatibility with compressor lubricants. Historically this settled on chlorofluorocarbons, which fit the pressure and stability requirements almost too well and turned out to destroy stratospheric ozone once released; the industry moved to hydrofluorocarbons such as R-134a, and is moving again, under climate regulation, toward fluids with lower global warming potential: R-1234yf, and a return to older working fluids such as ammonia and carbon dioxide in the applications that can tolerate their higher toxicity or higher operating pressure.

Whichever fluid is chosen, the cycle is read off a P–h diagram rather than the T–s diagram used for power cycles, and the substitution is not a matter of taste. Every energy quantity in the previous section was an enthalpy difference, so putting \(h\) on the horizontal axis turns every one of those differences into a distance you can see. Better still, the shape of each process becomes diagnostic rather than incidental. The condenser and the evaporator run at constant pressure, so they are horizontal lines. The throttle runs at constant enthalpy, so it is vertical. The compressor is the only process that is neither, because it alone holds entropy fixed rather than pressure or enthalpy, and on this diagram that makes it the only sloped line in the cycle. A glance at a P–h plot tells you which leg is which without reading a single label.

Two departures from the idealised cycle are worth being able to place on this diagram, because real equipment is built with both of them on purpose. The vapour leaving the evaporator is usually given a few degrees of superheat (state 1 sits slightly outside the dome rather than exactly on the saturated vapour line) as insurance against liquid slugging the compressor if the load runs low and the mixture does not fully evaporate. The liquid leaving the condenser is often subcooled a few degrees below saturation for the same reason in reverse: it guarantees the throttle sees pure liquid rather than a flash of vapour that would reduce the refrigeration effect. Both moves cost a small amount of heat exchanger area and buy a real margin against a failure mode that is expensive to have happen in the field.

The other habit worth building on this diagram is checking a claimed cycle for physical sense before trusting any number pulled from it. A condenser pressure below the evaporator pressure, an evaporator temperature above the condenser temperature, a compressor exit state that lands inside the dome rather than in the superheated region. Each is an equation-balancing error that a glance at the P–h picture catches immediately and a column of numbers does not.

Heat pumps

Nothing about the hardware in the previous two sections cares which direction the building's wall faces. Put the evaporator outside and the condenser inside, and the same loop that pulled heat out of a cold box now pulls heat out of the outside air (cold air still carries thermal energy, right down to the point where the refrigerant's evaporating temperature is lower still, which is why a heat pump keeps working, if with falling performance, well below the freezing point) and delivers it, together with the compressor work, to the room. It is the same machine. The second law already gave the relation between the two ways of grading it:

What changes between a refrigerator and a heat pump is not the cycle. It is which of the two heat transfers you are being paid for, and that decides which coefficient belongs in the conversation. A refrigerator's job is measured by what it removes from the cold space; a heat pump's job is measured by what it delivers to the warm one. Using \(\mathrm{COP}_R\) to judge a heating installation understates its performance by exactly one, which sounds trivial until the number in question is close to one to begin with.

It is close to one exactly when it matters most. Take the same 253 K to 303 K span used earlier: the reversed Carnot ceiling gives \(\mathrm{COP}_{HP,\text{rev}} = 303/50 = 6.06\), meaning a reversible heat pump would deliver over six units of heat for every unit of electrical work. Compare that against the alternative of heating with a resistance element, which converts work to heat with an efficiency of exactly one by definition. Every joule in becomes a joule of heat, and not one more, because there is no cold reservoir being drawn from at all. A heat pump running at even half its reversible ceiling still beats a resistance heater by a wide margin, and the reason has nothing to do with the heat pump inventing energy. It moves heat that was already there in the outside air, and the work is only the toll for moving it uphill.

The same physics explains why heat pump performance degrades as the outside temperature falls. Widen the gap between \(T_L\) and \(T_H\) and the Carnot ceiling falls with it, and the real machine falls along with the ceiling. Manufacturers quote a balance point, the outside temperature below which the heat pump alone cannot supply the building's full heating load and a backup resistance strip switches in, and choosing that balance point correctly (low enough that the efficient machine carries most of the season, high enough that it is not asked to do the impossible on the coldest night) is a genuine design decision rather than a safety margin tacked on afterwards.

What limits COP in practice

The ideal vapour-compression cycle already gave up some ground to the reversed Carnot ceiling at the throttle. A real machine gives up more, and the two biggest losses sit at exactly the two components the ideal cycle treats as perfect: the compressor, which is not isentropic, and the throttle, whose entropy generation the ideal cycle already counted but whose consequences are worth following through in numbers.

A real compressor needs more work than the isentropic ideal to reach the same exit pressure, because friction and fluid turbulence inside it generate entropy of their own. The shortfall is captured in an isentropic efficiency, the same idea used to grade turbines and nozzles in isentropic processes and efficiency, with the ratio inverted because a compressor is being penalised for consuming too much rather than delivering too little:

where \(h_{2s}\) is the ideal isentropic exit enthalpy and \(h_{2a}\) the actual one, at the same exit pressure. Take the R-134a cycle from earlier, with \(w_s = 36\) kJ/kg, and a compressor running at \(\eta_C = 0.80\), which is a realistic figure for a small reciprocating machine. The actual work rises to \(w_a = 36/0.80 = 45.0\) kJ/kg, a quarter more than the ideal, and since the refrigeration effect \(q_L\) is set entirely by the evaporator and throttle and does not depend on the compressor at all, the whole penalty lands on the denominator:

against 3.97 for the ideal cycle and 5.06 for the reversed Carnot ceiling. Three numbers, each one a real machine's worth of distance from the last, and each gap traceable to a named cause rather than a vague allowance for inefficiency.

The throttle's loss is smaller in this example but worth understanding rather than writing off, because it is structural rather than a matter of better engineering. No throttle, however well made, does work, and none ever will. That is what makes it a throttle rather than a turbine. Entropy rises from state 3 to state 4 with nothing recovered, and two consequences follow together. Some of the potential refrigeration effect is lost, because an isentropic expander would have dropped the fluid to a lower quality at state 4 than the throttle does, leaving more liquid available to absorb heat in the evaporator; and the small amount of work an expander could have recovered is thrown away rather than fed back into the compressor. Large industrial systems occasionally install an expander for exactly this reason. Almost nothing else does, because the fraction recovered is small, the extra rotating machinery is expensive and one more thing to maintain, and the throttle valve it replaces costs almost nothing and never breaks.

Both mechanisms are instances of a single idea that runs through every real cycle discussed in this course: irreversibility costs you exactly where it occurs, and it can be priced in the same currency: either as extra work paid in, as at the compressor, or as potential effect never collected, as at the throttle. Whether that price is worth paying to avoid (a better compressor, an expander, tighter superheat control) is an economic question resting on a thermodynamic answer, and the thermodynamic answer is what this section has been computing.