Actual Vapour Compression Cycle

The actual vapour compression cycle differs from the theoretical vapor compression cycle in many ways, some of which are unavoidable and cause losses. The main deviations between the theoretical cycle and actual cycle are as follows:

1. The vapour refrigerant leaving the evaporator is in superheated state.
2. The compression of refrigeration is neither isentropic nor polytropic.
3. The liquid refrigerant before entering the expansion valve is sub-cooled in the condenser.
4. The pressure drops in the evaporator and condenser.

The actual vapour compression cycle on T-s diagram is shown . The various processes are discussed below

(a) Process 1-2-3: This process shows the flow of refrigerant in the evaporator. The point 1 represents the entry of refrigerant into the evaporator and the point 3 represents the exit of refrigerant from evaporator in a superheated state. The point 3 also represents the entry of refrigerant into the compressor in a superheated condition. The superheating of vapour refrigerant from point 2 to point 3 may be due to

(i) automatic control of expansion valve so that the refrigerant leaves the evaporator as the superheated vapour.

(ii) picking up of larger amount of heat from the evaporator through pipes located within the cooled space.

(iii) picking up of heat from the suction pipe, i.e., the pipe connecting the evaporator delivery and the compressor suction valve.

In the first and second case of superheating the vapour refrigerant, the refrigerating effect as well as the compressor work is increased. The coefficient of performance, as compared to saturation cycle at the same suction pressure may be greater, less or unchanged. The superheating also causes increase in the required displacement of compressor and load on the compressor and condenser. This is indicated by 2-3 on T-s diagram as shown.

(b) Process 3-4-5-6-7-8: This process represents the flow of refrigerant through the compressor. When the refrigerant enters the compressor through the suction valve at point 3, the pressure falls to point 4 due to frictional resistance to flow. Thus the actual suction pressure (Ps) is lower than the evaporator pressure (Pe). During suction and prior to compression, the temperature of the cold refrigerant vapour rises to point 5 when it comes in contact with the compressor cylinder walls. The actual compression of the refrigerant is shown by 5-6 , which is neither isentropic nor polytropic. This is due to the heat transfer between the cylinder walls and the vapour refrigerant.

The temperature of the cylinder walls is some-what in between the temperatures of cold suction vapour refrigerant and hot discharge vapour refrigerant. It may be assumed that the heat absorbed by the vapour refrigerant from the cylinder walls during the first part of the compression stroke is equal to heat rejected by the vapour refrigerant to the cylinder walls. Like the heating effect at suction given by 4-5, there is a cooling effect at discharge as given by 6-7. These heating and cooling effects take place at constant pressure. Due to the frictional resistance of flow, there is a pressure drop i.e., the actual discharge pressure (Pd) is more than the condenser pressure (Pc).

(c) Process 8-9-10-11: This process represents the flow of refrigerant though the condenser. The process 8-9 represents the cooling of superheated vapour refrigerant to the dry saturated state. The process 9-10 shows the removal of latent heat which changes the dry saturated refrigerant into liquid refrigerant. The process 10-11 represents the sub-cooling of liquid refrigerant in the condenser before passing through the expansion valve.

This is desirable as it increases the refrigerating effect per kg of the refrigerant flow. It also reduces the volume of the refrigerant partially evaporated from the liquid refrigerant while passing through the expansion valve. The increase in refrigerating effect can be obtained by large quantities of circulating cooling water which should be at a temperature much lower than the condensing temperatures.

(d) Process 11-1: This process represents the expansion of subcooled liquid refrigerant by throttling from the condenser pressure to the evaporator pressure.

Effect of Suction Pressure

We have discussed in the previous article that in actual practice, the suction pressure (or evaporator pressure) decreases due to the frictional resistance of flow of the refrigerant. Let us consider a theoretical vapour compression cycle 1'-2'-3-4' when the suction pressure decreases from ps to ps' as shown on p-h diagram. It may be noted that the decrease in suction pressure

1. decreases the refrigerating effect from (h1 - h4) to (h1'- h4') and
2. increases the work required for compression from (h2 – h1) to (h2' – h1')

Since the C.O.P. of the system is the ratio of refrigerating effect to the work done, therefore with the decrease in suction pressure, the net effect is to decrease C.O.P. of the refrigerating system for the same amount of refrigerant flow. Hence with the decrease in suction pressure, the refrigerating capacity of the system decreases and the refrigeration cost increases.

Effect of Discharge Pressure

We have already discussed that in actual practice, the discharge pressure (or condenser pressure) increases due to frictional resistance of flow of the refrigerant. Let us consider a theoretical vapour compression cycle 1-2'-3'-4' when the discharge pressure increases from Pd to Pd' as shown on p-h diagram.

It may be noted that the increase in discharge pressure:

1. decreases the refrigerating effect from (h1 - h4) to (h1 - h4')
2. increases the work required for compression from (h2-h1) to (h2' - h1)

From above, we see that the effect of increase in discharge pressure is similar to the effect of decrease in suction pressure. But the effect of increase in discharge pressure is not as severe on the refrigerating capacity of the system as that of decrease in suction pressure.

Example 1 Simple ammonia-compression system operates with a capacity of 150 tonnes. The condensation temperature in the condenser is 35°C. The evaporation temperature in brine cooler is - 25°C. The ammonia leaves the evaporator and enters the compressor at - 8°C. Ammonia enters the expansion valve at 30°C. Wire drawing through the compressor valves Suction = 0.118 bar ; Discharge = 0.23 bar ; Compression index = 1.22 ; Volumetric efficiency = 0.75.

Calculate :

1. Power ;
2. Heat transferred to cylinder
3. Piston displacement ;
4. Heat transfer in condenser; and

5. Coefficient of performance.
   Solution: Given :
   150 TR ; 35º C = 35+273 = 308 K ; 25º C = -25+273 = 248 K ; 8 + 273 = 265 K ; 1.22 ; 0.75

   From p-h diagram, we find that the pressure corresponding to evaporation temperature of - 25° C,
   $p_1$=$p^"_1$=$p_4$=1.518 bar
   Since there is a suction pressure drop of 0.118 bar due to wire drawing, therefore pressure point 1',
   $p_1'$=1.518-0.118=1.14 bar=1.4 $\times$$10^5$ $\frac{N}{m^2}$ Pressure corresponding to condensation temperature of 35°C =13.5 bar ![][4] Since there is a discharge pressure drop of 0.23 bar due to wire drawing, therefore pressure at point 2', $p_2'$=13.5+0.23=13.73 bar=13.73 $\times$$10^5$$\frac{N}{m^2}$ ***Note***: Point 1 represents the inlet of the suction valve and point 1' is the outlet of the suction valve. The point 2' represents the inlet of discharge valve and point 2 is the outlet of the discharge valve. From p-h diagram, we also find that enthalpy of superheated ammonia vapours at point I or 1', $h_1$=$h_1'$=1440$\frac{kJ}{Kg}$
   Specific volume at point 1',
   $v_1'$=0.8$\frac{m^3}{Kg}$
   Temperature at point 1',
   $T_1'$= -9 º C==9+273=264 K
   Let $v_2'$= Specific volume at point 2'.
   Since the compression is according to $pv^{1.2}$=C therefore $p_1'$$(v_1')^n$=$p_2'$$(v_2')^n$ or $v_2'$=$v_1'(\frac{p_1'}{p_2'})^{\frac{1}{n}}$=0.8$(\frac{1.4}{13.73})^{\frac{1}{1.22}}$=0.123$\frac{m^3}{Kg}$
   Now plot a point 2' on the p-h diagram corresponding to $p_2'$=13.73 bar and $v_2'$=0.123$\frac{m^3}{Kg}$ .
   From the p-h diagram, we find that
   Enthalpy of superheated ammonia vapours at point 2 or 2',
   $h_2=h_2'$=1620$\frac{kJ}{Kg}$

   Temperature at point 2', $T_2'=90$ º Cand enthalpy of liquid ammonia at point 3,
   $h_{f3}$=$h_4$=320$\frac{kJ}{Kg}$

1. Power

We know that refrigerating effect per kg,
$R_E=h_1-h_{f3}$=1440$-$320=1120$\frac{kJ}{Kg}$ and refrigerating capacity =150TR=150$\times$210=31500$\frac{kJ}{min}$ $\cdots$(given) Mass flow of the refrigerant, $m_R=\frac{3150}{1120} $ kg/min We know that work done by the compressor per minute =$m_R\times\frac{n}{n-1}(p_2'v_2'-p_1'v1')$ $28.12\times\frac{1.22}{1.22-1}(13.73\times10^5\times0.123'-1.5\times10^4\times0.8)$J/min =89$\times10^5$J/min Therefore power=$frac{8900}{60}=148.93kW(Ans)

2. Heat transferred to cylinder water jacket
We know that actual work done by the compressor
=$m_R(h_2'-h_1')=28.12(1620-1440)=5062\frac{Kj}{min}$
Therefore heat transferred to cylinder water jacket
=8900-5062=3838 kJ/min (Ans)

3. Piston displacement
We know that piston displacement =$\frac{m_R\text v_1'}{\eta _v}$=$\frac{28.12\text 0.8}{0.75}$ (Ans)
4. Heat transfer in condenser
We know that heat transfer in condenser
=$m_R(h_2-h_{f3})=28.12(1620-320)=36556 \frac{kJ}{min}$ (Ans)

5. Coefficient of performance
We know that coefficient of performance,
C.O.P=$\frac{refrigating capacity}{work done}=\frac{31500}{8900}$=3.54 (Ans)