# Actual vapour-compression cycle

An ideal, simple, reversible vapour-compression cycle is not a practical proposition. The departure from reversibility arises from the irreversible nature of the throttling expansion process, pressure losses in the evaporator, condenser and pipelines, heat transfer through finite temperature differences and a measure of irreversibility in the compression process. For example, Figure 9.5 shows two of these irreversibilities: the throttling expansion (3^1) through the expansion valve and the desuperheating process (2-2′) when the vapour first enters the condenser. The condensation process (2′-3) and the evaporation process (4-1), on the other hand, are reversible.

Liquid refrigerant is incompressible and if it entered the suction valve of a reciprocating compressor it could damage the cylinder head or the valve plates at top dead centre. To avoid this risk it is arranged that evaporation is completed before the vapour leaves the evaporator and, consequently, the state then entering the suction valve has several degrees of superheat.

It is often also arranged that there is more than enough heat transfer surface in the condenser to change the refrigerant from a superheated vapour to a saturated liquid. In this case the liquid is sub-cooled to a temperature less than its saturated temperature for the prevailing pressure. A consequence of this is that any loss of position head (because the condenser might be at a lower level than the expansion valve), or any frictional pressure drop in the liquid line, is less likely to cause the liquid to flash to gas before it reaches the expansion valve. Figure 9.7 shows a pressure-enthalpy diagram of a simple cycle with superheat at the outlet from the evaporator and sub-cooling at the outlet from the condenser. Pressure drops through the piping, the evaporator and the condenser are ignored and isentropic compression is assumed. It can be seen that the dryness fraction is reduced and the refrigerating effect is improved, compared with the simple plant shown in Figure 9.1.

3

(0

Tfl

CD

°- 497.6 pe

Hq — h3 h3- hy hi h2

= h4

Enthalpy

Fig. 9.7 Pressure-enthalpy diagram of a simple, actual, vapour-compression cycle showing superheat at evaporator outlet and sub-cooling at condenser outlet.

EXAMPLE 9.6

A plant using R134a 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 outlet from the condenser. The duty is 352 kW of refrigeration. Assuming isentropic compression and ignoring pressure drops in the piping, evaporator and condenser, calculate the following: (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, (z) the COP, (/’) the corresponding Carnot COP, (k) the percentage Carnot cycle efficiency.

Answers

*(a) *Using equation (9.3), the notation of Figure 9.7 and referring to Tables 9.1 and 9.2 as necessary, it is established that h4> = 200.00 kJ kg-1 kg-1 and hy = 398.68 kJ kg-1. In Figure 9.7 it is seen that state 4 has the same enthalpy as state 3, which is 5 K cooler than state 3′. Table 9.1 quotes the enthalpy of saturated liquid at state 3′ as 248.94 kJ kg-1. Table 9.1 also gives the specific heat of saturated liquid at 35°C as the mean of the specific heats at 34°C and 36°C, namely, 1.4725 kJ kg-1 K-1 and as 1.447 kJ kg-1 K-1 at 30°C. The mean over the range 30°C to 35°C is then (1.4725 + 1.447)/2 = 1.460 kJ kg-1 K1. Hence h3 = h4 = 248.94 — 1.460 x 5 = 241.64 kJ kg-1. By equation (9.3)

/= (h4 — hA’)l(hy-hA-) = (241.64 — 200)/(398.68 — 200) = 0.21

*(b) *Equation (9.4) gives the refrigerating effect. In Table 9.1 the enthalpy of saturated vapour at 292.69 kPa and 0°C is given as 398.68 kPa kg-1 and the specific heat of saturated vapour as 0.883 kJ kg-1 K“1. Assuming the specific heat does not alter very much for the small temperature change of 5 K at constant pressure, the enthalpy of state 1, leaving the

evaporator, is calculated as hx = 398.68 + 0.883 x 5 = 403.10 kJ kg *. Since h4 = h3 the refrigerating effect is

Qt = (hi — h4) = (hi — h3) = (403.10 — 241.64) = 161.46 kJ kg“1

Compare this with 149.74 kJ kg-1 obtained as the answer to example 9.2(a). The increase is about 8 per cent. The benefit of 5 degrees of superheat at entry to the compressor is 0.883 x 5 = 4.415 kJ kg-1, which is less than 3 per cent of the refrigerating effect. Increasing the superheat to get more cooling capacity is seldom worthwhile.

*(c) *Equation (9.5) gives the mass flow rate of refrigerant:

M = 352/161.46 = 2.180 kgs’1

*(d) *In the absence of pressure drops and heat exchanges with the surroundings, the state into the suction side of the compressor is the same as at the outlet from the evaporator, namely, 5°C and 292.69 kPa. Taking the specific volume of saturated vapour at 0°C as 0.069 35 m3 kg-1 from Table 9.1 and making the assumption that, over a small temperature change near saturation, Charles’ law (see section 2.5) may be applied without much error, gives the required specific volume at state 1 (Figure 9.7):

Vi = 0.069 35 x (273 + 5)/(273 + 0) = 0.070 62 m3 kg’1

This corresponds to a density of 14.160 kg m-3. Reference to a pressure-enthalpy chart for R134a (Figure 9.10) verifies this. Hence the volumetric flow rate is 2.180 x 0.070 62 =

0. 153 95 m3 s-1.

*(e) *Equation (9.6) gives the work done in compression. From Table 9.1 the enthalpy and entropy of dry saturated vapour at 0°C and 292.69 kPa are 398.68 kJ kg-1 and 1.7274 kJ-1 kg-1 K“1, respectively. The enthalpy at 5 K of superheat and 292.69 kPa has already been established as 403.10 kJ kg’1. Hence the enthalpy change from 0°C saturated to 5 K of superheat at a constant pressure of 292.69 kPa is 403.10 — 398.68 = 4.42 kJ kg-1 and the mean temperature at which this enthalpy change occurs is 2.5°C. The principles discussed in section 9.2, leading to equation (9.2), allow a calculation of the corresponding change in entropy. Hence a reasonable calculation for the entropy at state 1, entering the compressor, is

Si = 1.7274 + 4.42/(273 + 2.5) = 1.7434 kJ kg"1 K"1

Reference to a pressure-enthalpy diagram for R134a verifies this. This is also the entropy at state 2, leaving the compressor. We must now refer to Table 9.2 and interpolate between 800 and 1000 kPa to determine the enthalpy and temperature of the gas discharged at state

2. First values of h and t must be established at the correct entropy of the adiabatic compression process for the two relevant tabulated pressures:

P = 1000 kPa |
P = 800 kPa |
||||

S |
H |
T |
S |
H |
T |

1.7310 |
415.45 |
45 |
1.7268 |
419.40 |
35 |

1.7434 |
426.60 |
48.60 |
1.7434 |
424.52 |
39.91 |

1.7482 |
430.91 |
50 |
1.7437 |
424.61 |
40 |

Finally, values of h and t must be determined from the above interpolations at the correct pressure and entropy:

P |
S |
H |
T |

1000 |
1.7434 |
426.60 |
48.60 |

887.11 |
1.7434 |
425.43 |
43.69 |

800 |
1.7434 |
424.52 |
39.91 |

By equation (9.6) the work done in compression is

Wr = (425.43 — 403.10) = 22.33 kJ kg”1

(/) By equation (9.7) the power absorbed by the compressor is

Wr = 2.180 x 22.33 = 48.68 kW

*(g) *From the final interpolation in part (e), above, t = 43.69°C, leaving the compressor.

(,h) The rate of heat rejection at the condenser is given by equation (9.9):

Gc = 2.180 x (425.43 — 241.64) = 400.7 kW

This equals the total refrigeration duty plus the power absorbed by the compressor: 352 + 48.68 = 400.7 kW.

(i) Equation (9.10) gives the COP:

161.46/22.33 = 7.23

Such high COPs are academic and would not be realised in practice because they assume isentropic compression and ignore inefficiencies, pressure drops and heat gains or losses.

(j) The Carnot COP, according to equation (9.11), is:

(273 + 35)/(35 — 0) = 8.8

(k) Percentage Carnot efficiency = (7.23 x 100)/8.8 = 82.1 per cent.

Posted in Air Conditioning Engineering

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