The vapour-compression refrigeration cycle

Engineering · Second- and third-year engineering · about 12 min

Fridges, freezers, air conditioners and heat pumps all use the same idea: make a refrigerant boil at a low temperature so it soaks up heat, then compress it so it can dump that heat somewhere warmer. The coefficient of performance (COP) says how much cooling you get for each unit of work you pay for.

The key idea

For the ideal cycle, with enthalpies $h$ at the four states (in kJ/kg):

q_L = h_1 - h_4 \qquad w_{in} = h_2 - h_1 \qquad \text{COP}_R = \frac{q_L}{w_{in}}

The heat rejected in the condenser is $q_H = h_2 - h_3 = q_L + w_{in}$ .

The four components

Compressor (state 1 to 2): saturated vapour is compressed to high pressure and temperature. This is where the work goes in.
Condenser (2 to 3): the hot vapour gives heat to the surroundings and condenses to liquid.
Expansion valve (3 to 4): the liquid's pressure drops suddenly. No work or heat is exchanged, so $h_4 = h_3$ ; it leaves as a cold, mostly liquid mixture.
Evaporator (4 to 1): the cold refrigerant boils, absorbing heat from the space being cooled.

A closed rectangular loop of pipe with arrows showing flow. The condenser is a long flat block on the top side and the evaporator is a long flat block on the bottom side. The compressor is a circle on the right side. The expansion valve is a bow-tie symbol on the left side. Flow runs from the evaporator to the right, up through the compressor, left through the condenser, and down through the valve back to the evaporator. Numbers mark the pipe: 1 between evaporator and compressor, 2 between compressor and condenser, 3 between condenser and valve, 4 between valve and evaporator. A dark red arrow leaves the condenser upwards, a dark blue arrow enters the evaporator from below, and a dark red arrow enters the compressor from the right. — The refrigerant runs round a loop through four components.

Why the COP is usually bigger than 1

A refrigerator doesn't create cold; it moves heat. Moving heat takes much less energy than producing it, so a typical fridge removes $3$ to $5$ kJ of heat for every kJ of electrical work. The ideal limit, from the Carnot cycle in reverse, is

\text{COP}_{R,\,max} = \frac{T_L}{T_H - T_L}

with temperatures in kelvin.

Worked example: COP and refrigerant flow rate

In an ideal vapour-compression cycle using R-134a, the property tables give $h_1 = 244.5$ kJ/kg leaving the evaporator, $h_2 = 275.0$ kJ/kg leaving the compressor, and $h_3 = 101.6$ kJ/kg leaving the condenser. Find the COP, and the refrigerant mass flow rate needed for $5$ kW of cooling.

A graph with pressure on the vertical axis and enthalpy on the horizontal axis. A thin dome-shaped curve shows the saturation line. On it, a closed loop with arrows: 4 to 1 is a horizontal line running right, low down inside the dome, with 1 on the right side of the dome. 1 to 2 rises steeply up and to the right, ending outside the dome. 2 to 3 is a horizontal line running left, near the top of the dome, with 3 on the left side of the dome. 3 to 4 is a straight vertical line running down. No values on the axes. — The same cycle on a pressure-enthalpy diagram.

Show the solution

The expansion valve doesn't change enthalpy, so $h_4 = h_3 = 101.6$ kJ/kg.

Refrigerating effect: $q_L = h_1 - h_4 = 244.5 - 101.6 = 142.9$ kJ/kg

Compressor work: $w_{in} = h_2 - h_1 = 275.0 - 244.5 = 30.5$ kJ/kg

\text{COP}_R = \frac{142.9}{30.5} = 4.69

Mass flow for $5$ kW of cooling:

\dot m = \frac{\dot Q_L}{q_L} = \frac{5}{142.9} = 0.035 \text{ kg/s}

Common mistakes

Forgetting that the expansion valve keeps enthalpy constant

Throttling is isenthalpic: $h_4 = h_3$ . That's often the missing value.

Using the heat rejected instead of the heat absorbed

A refrigerator's COP uses the cooling, $q_L$ . A heat pump's COP uses the heating, $q_H$ .

Thinking a COP above 1 breaks energy conservation

The work isn't turned into cooling; it moves heat that was already there. Energy still balances: $q_H = q_L + w_{in}$ .

Check your understanding

1. Which component is the work input in a vapour-compression cycle?

2. A fridge removes $400$ W of heat using $100$ W of work. What is its COP?

3. Across the expansion valve, which property stays the same?

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