# Coefficient of performance

The coefficient of performance of a refrigeration machine is the ratio of the energy removed at the evaporator (refrigerating effect) to the energy supplied to the compressor. Thus, using the notation of Figure 9.2, we have the formula:

C0P = V^

The best possible performance giving the highest COP would be obtained from a system operating on a Carnot cycle. Under such conditions the refrigeration cycle would be thermodynamically reversible, and both the expansion and compression processes would be isentropic. A Carnot cycle is shown on a temperature-entropy diagram in Figure 9.4.

Point 3 indicates the condition of the liquid as it leaves the condenser. Line 34 represents the passage of the refrigerant through an expansion engine, the work of which helps to

 Entropy Fig. 9.4 Carnot refrigeration cycle on a temperature-entropy diagram.

Drive the refrigeration machine. During the expansion process the pressure of the refrigerant drops from pc to pe, its temperature is reduced from Tc to Tc and part of the refrigerant vaporises. The expansion process, being reversible and adiabatic, takes place at constant entropy, with the result that the line 34 is vertical.

During the next part of the cycle, refrigerant is evaporated at constant pressure pt and temperature Tc, this being represented on the diagram by the line 41. The process comes to an end at the point 1, where the refrigerant still consists of a liquid-vapour mixture but the proportion of liquid is small.

The mixture then enters another engine, the compressor, which is driven by some external source of power supplemented by the output of the expansion engine already described. In the compressor, the refrigerant is increased in pressure from pe to pc, the temperature rises from Te to Tc and the remaining liquid evaporates, so that the refrigerant leaves the machine as a saturated vapour—point 2′ on the diagram. The compression process is adiabatic, reversible and, therefore, isentropic.

The final part of the cycle is shown by the line 2’3, which represents the liquefaction of the refrigerant in the condenser at constant pressure pc and temperature Tc.

Since an area under a process line on a temperature-entropy diagram represents a quantity of energy (see equation (9.2)) the quantities involved in the Carnot refrigeration cycle are as follows:

Heat rejected at condenser rc(s12′ — S34)

Heat received at evaporator Te(s? — 534)

Work supplied to machines (Tc — Te)(s]2′ — ^34)

Since the coefficient of performance is the energy received at the evaporator divided by the energy supplied to the machine, we have for the Carnot cycle:

Using the data of examples 9.1 to 9.4 calculate the coefficient of performance. What COP would be obtained if a Carnot cycle were used?

From equation (9.10)

COP = (398.68 — 248.94)/(421.64 — 398.68)

= 149.74/22.96 = 6.52

And from equation (9.11)

Carnot COP = (0 + 273)/[(35 + 273) — (0 + 273)] = 7.8

It will be seen from this example that the simple saturation cycle under consideration had a coefficient of performance which was 83.7 per cent of that attainable with a Carnot cycle. Comparisons of this kind are sometimes used as an index of performance (see Table 9.6).

The simple, basic saturation cycle shown in Figure 9.2 is illustrated in Figure 9.5 as an absolute temperature-entropy diagram. The principal reasons why the coefficient of performance of this cycle is less than that of the corresponding Carnot cycle (as in Figure 9.4) are: the excursion of the compression process line into the superheated region and the irreversible nature of the method of throttling expansion used for the simple cycle.

 Fig. 9.5 Simple saturation refrigeration cycle on a temperature-entropy diagram.

 H# h3 h-1 hy hp Enthalpy Fig. 9.6 Carnot refrigeration cycle on a pressure-enthalpy diagram.

Posted in Engineering Fifth Edition