The work done in compression

In the ideal case, where there is no heat lost from the cylinders and no friction, the process of compression is adiabatic, reversible and isentropic. This is shown in Figure 9.2 by the line joining the points 1 and 2. It is assumed that there is no pressure drop in the pipeline between the outlet from the evaporator and the suction port on the compressor and that no heat gains or losses occur between the suction line and its environment. The compressor sucks dry saturated vapour from the evaporator and power must be provided to effect this process. The work done in compression (vvr) for this ideal case is given by the enthalpy increase across the compressor:

Wr = (h2 — hi) (9.6)

Multiplying this by the mass flow rate of refrigerant handled, m, yields the power needed for compression, wr, in W or kW:

Wr=m(h2-h]) (9.7)

State 1 (Figure 9.2) is known because, with the simplifying assumptions made, it is that of the dry saturated vapour leaving the evaporator. State 2 is that of superheated vapour leaving the compressor, which can be identified by two known facts for the ideal case considered:

(i) its pressure is the same as the condensing pressure and

(ii) it has the same entropy as that of the dry saturated vapour entering the suction port. Hence h and h2 can be established.

EXAMPLE 9.3

Calculate the work done in compression and the corresponding compressor power, for the plant used in example 9.2.

Answer

Figure 9.3 illustrates the answer. From Table 9.1, the enthalpy of dry saturated vapour at 0°C leaving the evaporator is 398.68 kJ kg-1 and the entropy is 1.7274 kJ kg-1 K-1. At state

2, leaving the compressor, the pressure of the superheated vapour is 887.11 kPa and the entropy is also 1.7274 kJ kg-1 K_1. Unfortunately, Table 9.2 is not particularly user — friendly and reference to it shows that this entropy, for the relevant pressure, occurs somewhere between 800 kPa and 1000 kPa. A double linear interpolation is necessary to identify the enthalpy and temperature of the superheated vapour discharged from the compressor at state 2. First, values of enthalpy and temperature having entropies above and below 1.7274 kJ kg-1 K1 must be determined for the pressures each side of the condensing pressure of 887.11 kPa:

P = 1000 kPa p = 800 kPa

S

H

T

S

H

T

1.7310

425.45

45

1.7437

424.61

40

1.7274

424.30

43.95

1.7274

419.58

35.18

1.7139

419.99

40

1.7268

419.40

35

It is seen that linear interpolations yield a pair of values of h and t at an entropy of 1.7274 kJ kg’1 K-1. The values of h and t obtained for an entropy of 1.7274 kJ kg-1 K_1 are then interpolated between 1000 and 800 kPa to yield the required values with a pressure of

887.11 kPa, at compressor discharge.

P

S

H2

H

1000

1.121A

424.30

43.95

887.11

1.7274

421.64

39.00

800

1.7274

419.58

35.18

At compressor discharge h2 = 421.64 kJ kg-1 K_1 and, from example 9.1 and Figure 9.3, hx at compressor suction is 398.68 kJ kg-1. Hence, by equation (9.6), the work done in compression, wr, is (421.64 — 398.68) = 22.96 kJ kg-1 and, by equation (9.7), the corresponding compressor power, Wr, is 2.351 x 22.96 = 53.98 kW.

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