Volumetric efficiency

The capacity of a refrigeration compressor is a function of the mass flow rate of gas handled. However, the full volume of the cylinder swept by the piston is not available for pumping gas: some clearance volume must be left at the top of the cylinder, otherwise there would be a danger of the piston damaging the cylinder head or the valves as the stroke increased with wear on the crankshaft and bearings. This is illustrated in Figure 9.8. Since the actual amount of gas pumped is less than the theoretical amount, there is a concept of volumetric efficiency, defined by:

Actual volume of fresh gas, „ .

Riv =—————— :—j————- — x 100

Swept volume

V — V* = ./ x 100 (9.23) *a m: Volume Fig. 9.8 Pressure-volume relationship for an ideal compression cycle.

The small amount of gas trapped in the clearance volume at the top of the stroke re — expands as the piston moves downward and reduces the volume available for the entry of fresh gas. Fresh gas cannot enter the cylinder until re-expansion has reduced the pressure within the cylinder to less than the suction pressure.

The ratio of the absolute condensing pressure to the absolute suction pressure is termed the compression ratio and this clearly has an effect on the volumetric efficiency. Table 9.5 gives some typical values of volumetric efficiency.

Table 9.5 Some typical, approximate volumetric efficiencies, according to Trane Air Conditioning Manual (1961)

Compression ratio 2.0 3.0 4.0 5.0 6.0

Volumetric efficiency 78% 74% 70% 66% 62%

Answer

Compression ratio = 887.11/292.69 = 3.031 From Table 9.5 riv = 73.9 per cent.

Denoting the cylinder diameter by d, also equal to the bore, the required swept volume

Is

8 x 7Tx J3 x 1425 _ Q’j/73rn3 „-i

0. 739 x4x60 m s

This can be equated to the volumetric flow rate at the suction state, to yield the value of d: 201.93c?3 = 0.154 whence

D = 0.995 m = 995 mm and is the diameter of the cylinders and their stroke.

EXAMPLE 9.9

A plant using R717 (ammonia) has a duty of 352 kW of refrigeration with a four cylinder compressor running at 1425 rpm. It evaporates at 0°C and condenses at 35°C, with 5 K of superheat at the evaporator outlet and 5 K of sub-cooling at the condenser outlet. The mean specific heat capacity of saturated liquid ammonia at 32.5°C is given as 4.867 kJ kg-1 K”1 (see ASHRAE Handbook (1997)). Assuming isentropic compression and ignoring pressure drops in the piping, the evaporator and condenser, determine the following, making use of Tables 9.3 and 9.4: (a) the dryness fraction at entry to the evaporator, (b) the refrigerating effect, (c) the mass flow rate of refrigerant, (d) the volumetric flow rate at the suction state,

(e) the work done in compression, (/) the compressor power, (g) the temperature of the superheated vapour at discharge from the compressor, (h) the rate of heat rejection at the condenser, (i) the COP, (j) the corresponding Carnot COP, (k) the percentage Carnot efficiency, (/) the stroke and bore of the compressor cylinders, assuming these to be equal.

Answer

(a) Making use of the value given for the mean specific heat of saturated liquid refrigerant at 32.5°C the enthalpy of sub-cooled liquid at 30°C is 347.1 — 5 x 4.867 = 322.8 kJ kg-1 K1. Using the notation of Figure 9.7 and equation (9.3):

/= (322.8 — 181.1)/(1443.1 — 181.1) = 0.11

(b) By equation (9.4):

Qr = (1456.5 — 322.8) = 1133.7 kJ kg~‘

(c) By equation (9.5):

M = 352/1133.7 kg s"1 = 0.3105 kg s“1

(d) At compressor suction (5°C and 429.4 kPa) v = 0.2966 m3 kg-1 and hence v = 0.3105 x 0.2966 = 0.0921 m3 s~‘.

(e) At compressor discharge the entropy, s2 = S] = 5.3842 kJ kg-1 K_1. Interpolating for this entropy in Table 9.4 at a pressure of 1350 kPa yields:

S	T-h	H
5.3632	О О	1615.1
5.3842	52.9°	1622.7	Whence h2 = 1622.7 kJ kg
5.3995	55°	1628.2

By equation (9.6):

Wr = (1622.7 — 1456.5) = 166.2 kJ kg’1 (/) By equation (9.7) the compressor power is Wr = 0.3105 x 166.2 = 51.60 kW

(g) From the interpolation in part (e) the temperature of the vapour at discharge from the compressor is 35°C + 52.9 К = 87.9°C.

(h) Using the mass flow rate and equation (9.8):

Rate of heat rejection at the condenser = 0.3105 x (1622.7 — 322.8) = 403.6 kW

(i) By equation (9.10):

COP = (1456.5 — 322.8)/(l622.7 — 1456.5) = 6.82

(j) By equation (9.11):

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

(k) Percentage Carnot efficiency = 6.82 x 100/7.8 = 87.4 per cent.

(/) The compression ratio is 1350/429.4 = 3.14 and from Table 9.5 the volumetric efficiency is 73.4 per cent. The swept volume is (4 x n x d3 x 1425)/(0.734 x 4 x 60) = 101,65d3. This is equated to the volumetric flow rate at the suction state (part (d), above):

101.65J3 = 0.0921 whence d 3 = 0.000 906 05 m3 and d = 0.0968 m.

The stoke and bore are each equal to 96.8 mm.

Posted in Air Conditioning Engineering