Vapour Compression Refrigeration Cycle and COP with Solved Examples
A refrigerator moves heat from a cold space to a warmer surrounding, which is the opposite of the natural direction of heat flow. The second law says this needs a work input, and the vapour compression cycle is how almost every domestic fridge, split air conditioner and cold storage plant does it. Its performance is measured by the coefficient of performance (COP). This article explains the four components, gives the Carnot limit, and works through a cycle using refrigerant property values.
The Four Components
| State change | Component | What happens |
|---|---|---|
| 1 to 2 | Compressor | Low-pressure vapour is compressed to high pressure and temperature |
| 2 to 3 | Condenser | Hot vapour rejects heat to outside air and condenses to liquid |
| 3 to 4 | Expansion device (capillary tube or valve) | Liquid pressure drops; some liquid flashes to vapour; enthalpy stays constant |
| 4 to 1 | Evaporator | Cold refrigerant absorbs heat from the space and boils to vapour |
The refrigerant boils at a low temperature in the evaporator because its pressure is low. The compressor then raises the pressure so that the same refrigerant condenses at a temperature above the outdoor air, allowing heat to be dumped outside. In a split AC, the evaporator is the indoor unit and the condenser plus compressor make up the outdoor unit.
Key Formulas
Refrigerating effect q_L = h1 - h4 (kJ/kg) Compressor work w = h2 - h1 (kJ/kg) Heat rejected q_H = h2 - h3 = q_L + w Throttling h4 = h3 COP (refrigerator) COP_R = q_L / w COP (heat pump) COP_HP = q_H / w = COP_R + 1 Carnot COP COP_R,max = T_L / (T_H - T_L) (kelvin) 1 tonne of refrigeration (TR) = 3.517 kW = 211 kJ/min
The tonne of refrigeration comes from the heat needed to freeze one short ton of water at 0 degrees C in 24 hours. It is still the standard unit for AC capacity in India, so a 1.5 ton split AC is rated for about 5.3 kW of cooling.
Example 1: The Carnot Limit
A refrigerator evaporator runs at minus 10 degrees C and the condenser at 40 degrees C. The best possible COP is:
T_L = 263.15 K, T_H = 313.15 K COP_max = 263.15 / (313.15 - 263.15) = 5.26
Lowering the condensing temperature to 30 degrees C would raise this to 6.58. This is why a fridge works harder in a hot kitchen and why condenser coils should be kept clean and well ventilated.
Example 2: Ideal Cycle with R-134a
An ideal vapour compression cycle using R-134a operates between 0.14 MPa and 0.8 MPa with a refrigerant flow of 0.05 kg/s. Saturated vapour leaves the evaporator and saturated liquid leaves the condenser. From R-134a property tables the enthalpies are approximately as below (always confirm against the tables your course or code uses, as reference states differ between sources).
| State | Condition | h (kJ/kg) |
|---|---|---|
| 1 | Saturated vapour at 0.14 MPa | 239.16 |
| 2 | Superheated vapour at 0.8 MPa, same entropy as 1 | 275.39 |
| 3 | Saturated liquid at 0.8 MPa | 95.47 |
| 4 | After throttling, h4 = h3 | 95.47 |
q_L = 239.16 - 95.47 = 143.69 kJ/kg w = 275.39 - 239.16 = 36.23 kJ/kg Cooling capacity Q_L = 0.05 * 143.69 = 7.18 kW = 2.04 TR Compressor power W = 0.05 * 36.23 = 1.81 kW Heat rejected Q_H = 7.18 + 1.81 = 8.99 kW COP_R = 7.18 / 1.81 = 3.97 COP as heat pump = 8.99 / 1.81 = 4.97
The condenser must reject more heat than the evaporator absorbs, because the compressor work also ends up as heat. Outdoor units are sized for Q_H, not Q_L.
The ideal cycle COP of 3.97 is well below the Carnot value for the same saturation temperatures. Two losses explain the gap: the throttling process wastes the energy that an expansion turbine could have recovered, and the compressor discharges superheated vapour, so part of the heat is rejected at temperatures far above the condensing temperature. Real systems lose more again through compressor inefficiency, pressure drops in the lines and heat exchanger temperature differences.
Example 3: Power Drawn by a Split AC
A 1.5 TR air conditioner delivers 1.5 x 3.517 = 5.28 kW of cooling. If its actual COP at the operating condition is 3.5, the compressor and fans draw about 5.28 / 3.5 = 1.51 kW. Running for 8 hours a day for 30 days uses about 362 kWh. An older unit with COP 2.8 would draw 1.88 kW and use about 452 kWh in the same period, which is why BEE star ratings matter on the electricity bill.
Common Mistakes
- Using Celsius in the Carnot COP formula.
- Assuming COP must be less than 1 like an efficiency; refrigerator COP is normally well above 1.
- Forgetting h4 = h3 for the expansion device.
- Sizing the condenser for cooling load only and ignoring compressor heat.
- Mixing enthalpy values from different tables that use different reference states.
Summary
The vapour compression cycle uses a compressor, condenser, expansion device and evaporator to pump heat uphill. COP is refrigerating effect divided by compressor work, the Carnot COP sets the upper limit, and one TR equals 3.517 kW. With four enthalpy values you can find capacity, power, heat rejection and COP for any simple refrigeration system.