A.Theoretical Vapour Compression Cycle with Dry Saturated Vapour after Compression
Example 1
In an ammonia vapour compression system, the pressure in the evaporator is 2 bar. Ammonia at exit is 0.85 dry and at entry its dryness fraction is 0.19. During compression, the work done per kg of ammonia is 150 kJ. Calculate the C.O.P. and the volume of vapour entering the compressor per minute, if the rate of ammonia circulation is 4.5 kg/min. The latent heat and specific volume at 2 bar are 1325 kJ/kg and 0.58 m3/kg respectively.
Solution
Given: $p_1=p_4=2 bar ;x_1=0.85 ;x_4=0.19 ; w=150 kJ/kg ; m_a=4.5 kg/min ;h_{fg}=1325 kJ/hg ;v_g=0.58 $m3/kg C.O.P.
Since the ammonia vapour at entry to the evaporator (i.e. at point 4) has dryness fraction (x4) equal to 0.19, therefore enthalpy at point 4,
$h_4=x_4\times h_{fg}=0.19\times 1325=251.75$kJ/kg
Similarly, enthalpy of ammonia vapour at exit i.e. at point 1,
$h_1=x_1\times h{fg}=1126.25$kJ/kg
Therefore heat extracted from the evaporator or refrigerating effect,
$R_E=h_1-h_4=1126.25-251.75=874.5$kJ/kg
We know that work done during compression,
W=150 kJ/kg
Therefore
C.O.P=$\frac{R_E}{w}=\frac{874.5}{150}=5.83$ (Ans)
Volume of vapour entering the compressor per minute
We know that volume of vapour entering the compressor per minute
= Mass of refrigerant/ min $\times$ Specific volume
=$m_a\times v_g=4.5\times0.58=2.61 m_3/min$ (Ans)

Example 2
The temperature limits of an ammonia refrigerating system are 25° C and -10° C. If the gas is dry at the end of compression, calculate the coefficient of performance of the cycle assuming no undercooling of the liquid ammonia. Use the following table for properties of ammonia:

Solution
Given: $T_2=T_3=25° C = 25+273 = 298 k ;T_1=T_4=-10° C =263 K ;h_{f3}=h_4=298.9 kJ/kg ;h_{fg2}=1166.94 kJ/kg ; s_{f2}=1.1242 kJ/kg K ;h_{f1}=135.37 kJ/kg ;h_{fg1}=1297.68 kJ/kg ;s_{f1}=0.5443 kJ/kg K$
The T-s and p-h diagrams are shown in Fig. 9.4 (a) and (b) respectively.
Let $x_1$= Dryness fraction at point 1.
We know that entropy at point 1,
$s_1=s_{f1}+\frac{x_1h_{fg1}}{T_1}=0.5443+\frac{x\times1297.68}{263}$
$=0.5443+4.934x_1.......(i)$
Similarly, entropy at point 2,
$s_2=s_{f2}+\frac{x_1h_{fg2}}{T_2}$
$=0.5443+\frac{1166.94}{298}=5.04…..(ii) $
Since the entropy at point 1 is equal to entropy at point 2, therefore equating equations (i) and (ii),

We know that enthalpy at point 1,
$h_1=h_{f1}+x_1h_{fg1}=135.37+0.91\times1297.68=1316.26kJ/kg$
and enthalpy at point 2,
$h_2=h_{f2}+h_{fg2}=298.9+1166.94=1465.84kJ/kg$
Therefore coefficient of performance of the cycle
$=\frac{h_1-h_{f3}}{h_2-h_1}=\frac{1316.26-298.9}{1465.84-1316.26}=6.8$ (Ans)

Example 3
A vapour compression refrigerator works between the pressure limits of 60 bar and 25 bar. The working fluid is just dry at the end of compression and there is no under¬cooling of the liquid before the expansion valve. Determine :
1. C.O.P. of the cycle ;and
2. Capacity of the refrigerator if the fluid flow is at the rate of 5 kg/min

Solution
Given : $p_2=p_3=60 bar ;p_1=p_4=25 bar ;T_2=T_3=295 K ;T_1=T_4=261 K ;h_{f3}=h_4=151.96 kJ/kg ;h_{f1}=56.32 kJ/kg ; h_{g2}=h_2=293.29 kJ/kg ;h_{g1}=322.58 kJ/kg ;s_{f2}=0.554 kJ/kg K ;s_{f1}=0.226 kJ/kg ;s_{g2}=1.0332 kJ/kg K ;s_{g1}=1.2464 kJ/kg K$
1. C.O.P. of the cycle
The T-s and p-h diagrams are:

Let $x_1$ Dryness fraction of the vapour refrigerant enterin the compressor at point 1.
We know that entropy at point 1, $s_1=s_{f1}+x_1s_{fg1}=s_{f1}+x_1(s_{g1}-s_{f1}) …….(s_{g1}=s_{f1}+s_{g1})$
$=0.226+x_1(1.2464-0.226)=0.226+1.0204x_1$ …..(i)
and entropy at point 2,
$s_2=1.0332 kJ/kgK$ …….(Given)…..(ii)
Since the entropy at point 1 is equal to entropy at point 2,
therefore equating equations (i) and (ii),
0.226+1.0204$x_1=1.0332$ (or)
$x_1$=0.791
We know that enthalpy at point 1,
$h_1=h_{f1}+x_1\times h_{fg1}=h_{f1}+x_1(h_{g1}-h_{f1})$ ……( $ h_{g1}=h_{f1}+h_{fg1}$)
=$56.32+0.761(322.58-56.32)=266.93kJ/kg$
Therefore C.O.P of the cycle
$=\frac{h_1-h_{f3}}{h_2-h_1}=\frac{266.93-151.96}{293.29-266.93}=4.36$ (Ans)
Capacity of the refrigerator
We know that the heat extracted or refrigerating effect produced per kg of refrigerant
$h_1-h_{f3}=266.93-11.93=114.97kJ/kg$
Since the fluid flow is at the rate of 5 kg/min, therefore total heat extracted
$=5\times114.97=574.85kJ/min$
Therefore capacity of the refrigerator
$\frac{574.85}{210}=2.74TR$ (Ans)

B.Theoretical Vapour Compression Cycle with Wet Vapour after Compression
Example 4
Find the theoretical C.O.P. for a CO2 machine working between the temperature range of 25°C and - 5°C. The dryness fraction of CO2 gas during the suction stroke is 0.6. Following properties of CO2 are given

Solution

Given: $T_2=T_3=25° C = 25+273 = 298 K ;T_1=T_4=-5° C = -5+273 = 268 K; x_1=0.6 ;h_{f3}=h_{f2}=164.77 kJ/kg ; h_{f1}=h_{f4}=72.57 kJ/kg ;s_{f2}=0.5978 kJ/kg K ;s_{f1}=0.2862 kJ/kg K ;h_2’=282.23 kJ/kg ;h_1’=321.33 kJ/kg ; s_2’=0.9918 kJ/kg K ;s_1’=1.2146 kJ/kg K ;h_{fg2}=117.46 kJ/kg ;h_{fg1}=248.76 kJ/kg$
The T-s and p-h diagrams are shown below.
First of all, let us find the dryness fraction at point 2, i.e. $x_2$.
We know that the entropy at point 1,
$s_1=s_{f1}+\frac{x_1h_{fg1}}{T_1}=0.2862+\frac{0.6\times248.76}{268}=0.8431$ ......(i)
Similarly, entropy at point 2,
$s_2=s_{f2}+\frac{x_2h_{fg2}}{T_2}=0.5978+\frac{x_2\times 117.64}{298}$
=$0.5978=0.3941x_2$ .....(ii)

Example 5
An ammonia refrigerating machine fitted with an expansion valve works between the temperature limits of-10°C and 30°C. The vapour is 95% dry at the end of isentropic compression and the fluid leaving the condenser is at 30°C. Assuming actual C.O.P. as 60% of the theoretical, calculate the kilograms of ice produced per kW hour at 0°C from water at 10°C. Latent heat of ice is 335 kJ/kg. Ammonia has the following properties

Solution
Given : $T_1=T_4=-10° C = +273 = 263 K ; T_2=T_3=30° C = 30+273 = 303 K; x_2=0.95 ; h_{f3}=h_{f2}=323.08 kJ/kg ; h_{f1}=h_{f4}=135.37 kJ/kg ; h_{fg2}=1145.8 kJ/kg ; h_{fg1}=1297.68 kJ/kg ; s_{f2}=1.2037 ; s_{f1}=0.5443 ; s_2’=4.9842 ; s_1’=5.4770$
The T-s and p-h diagrams are shown in Fig. 9.8 (a) and (b) respectively.
Let $x_1$= Dryness fraction at point 1.
We know that entropy at point 1,
$s_1=s_{f1}+\frac{x_1h_{fg1}}{T_1}=0.5443+\frac{x_1\times1297.68}{263}$
=$0.5443+4.934x_1$ ...(i)
Similarly the entropy at point 2
$s_2=s_{f2}+\frac{x_2h_{fg2}}{T_2}=1.2037+\frac{0.95\times1145.8}{303}=4.796$ .....(ii)
Since the entropy at point 1($s_1$) is equal to entropy at point 2 ($s_2$),
therefore equating equations (i) and (ii),
=0.5443+4.93$x_1$=4.796 (or) $x_1$=0.86

Therefore enthalpy at point 1,
$h_1=h_{f1}+x_1h_{fg1}=135.37+0.86\times1297.68=1251.4kJ/kg$
and enthalpy at point 2,
$h_2=h_{f2}+x_2h_{fg2}=323.08+0.95\times1145.8=1411.6kJ/kg$
We know that theoretical C.O.P.
$=\frac{h_1-h_{f3}}{h_2-h_1}=\frac{1251.4-323.08}{1411.6-1251.4}=5.8$
Therefore
actual C.O.P.=$0.6\times5.8=3.48$
Work to be spent corresponding to 1 kW hour,
W=3600kJ
Therefore actual heat extracted or refrigeration effect-produced per kW hour
$=W\times$ Actual C.O.P=3600$\times3.48=12528kJ$
We know that heat extracted from 1 kg of water at 10°C for the formation of 1 kg of ice at 0° C
$=1\times 4.187\times10+335=376.7kJ$
Therefore amount of ice produced
$=\frac{12528}{376.87}=33.2kg/kWhour$ (Ans)