problems on vapour compression cycle
1 Answer

Example 1

A vapour compression refrigerator uses methyl chloride (R-40) and operates between temperature limits of -10°C and 45°C. At entry to the compressor, the refrigerant is dry saturated and after compression it acquires a temperature of 60°C. Find the C.O.P. of the refrigerator. The relevant properties of methyl chloride are as follows

Solution Given : $T_1=T_4$=-10° C = $ -10+273 = 263$ K ;
$T_2'$ =$T_3=45$° C = 45+273 = 318 K ;$T_2$= 60° C = 60+273 = 333 K ; $h_{f1}$=45.4 kJ/kg ; $h_{f3}$=133 kJ/kg ; $h_1$=460.7 ; $h_2’=$483.6 kJ/kg ; $s_{f1}$=0.183 kJ/kg K ; $s_{f3}$=0.485 kJ/kg K ;$s_1=s_2=$1.637 kJ/kg K ; $s_2’=$1.587 kJ/kg K

Let $c_p=$ Specific heat at constant pressure for superheated vapour.
We know that entropy at point 2,
$1.637 =1.587 +2.3c_p\log\frac{338 }{318}$
1.637 =1.587+2.3$c_p\times$0.02=1.587+0.046$c_p$
Therefore $c_p$=1.09
and enthalpy at point 2,
$h_2=h_2’+c_p\times Degree of superheat$ =$h_2’+c_p(T_2-T_2’)$
$=483.6+1.09(333-318)=500$ kJ/kg
C.O.P. of refrigerator
=$\frac{h_1-h_{f3}}{h_2-h_1} = \frac{640.7-133}{500-460.7}=3.77$ (Ans)

Example 2
A refrigeration machine using R-12 as refrigerant operates between the pressures 2.5 bar and 9 bar. The compression is isentropic and there is no undercooling in the condenser. The vapour is in dry saturated condition at the beginning of the compression. Estimate the theoretical coefficient of performance. If the actual coefficient of performance is 0.65 of theoretical value, calculate the net cooling produced per hour. The refrigerant flow is 5 kg per minute. Properties of refrigerant are

Take $c_p$ for superheated vapour at 9 bar as 0.64 kJ/kg K.
Given : $T_2’=T_3=36° C = 36+273 = 309 K ;T_1=T_4=-7° C =-7+273 = 266 K;$ $(C.O.P)_{actual}=0.65(C.O.P)_{th}$; $m=5 kg /min ;h_{f3}=h_4=70.55 kJ/kg ;h_{f1}=h_{f4}=29.62 kJ/kg ;h_2’=201.8 kJ/kg ;h_1=184.5 kJ/kg ; s_2’=0.6836 kJ/kg K ;s_1=s_2=0.7001 kJ/kg K ;c_p= 0.64 kJ/kg K$

Theoretical coefficient of performance

First of all, let us find the temperature at point 2 ($T_2$ )
We know that entropy at point 2,
$s_2=s_2’+2.3 c_p\log\frac{T_2}{T_2’}$
0.7001=0.6836+2.3$\times 0.64\log\frac{T_2}{309}$
$\log\frac{T_2}{309}=\frac{0.7001-06836}{2.3\times 0.64}=0.0112$ $\frac{T_2}{309}=1.026$ ....(Taking antilog of 0.0112)
$T_2=1.026\text 309=317$K
We know that enthalpy of super heated vapour at point 2,
$h_2=h_2’+c_p(T_2-T_2’)$ $=201.8+0.64(317-309)$=206.92kJ/kg

Therefore theoretical coefficient of performance,
$(C.O.P)_{th}=\frac{h_1-h_{f3}}{h_2-h_1}=\frac{184.5-70.55}{206.92-184.5}= 5.1$ (Ans)
Net cooling produced per hour
We also know that actual C.O.P. of the machine,
$(C.O.P)_{actual}=0.65\times (C.O.P)_{th}=0.65\times 5.1=3.315$
and actual work done, $w_{actual}=h_2-h_1=206.92-184.5=22.42$ kJ/kg
We know that net cooling (or refrigerating effect) produced per kg of refrigerant
=$w_{actual}\times (C.O.P)_{actual}=22.42\times 3.315=74.3$kJ/kg
Therefore net cooling produced per hour
=$\frac{371.5}{210}=1.77$TR (Ans) .....(1TR=210kJ/min)

Example 3 A water cooler using R-12 works on the condensing and evaporating temperatures of 26°C and 2°C respectively. The vapour leaves the evaporator saturated and dry. The average output of cold water is 100 kg /hr cooled from 26°C to 6°C. Allowing 20% of useful heat into water cooler and the volumetric efficiency of the compressor as 80% and mechanical efficiency of the compressor and the electric motor as 85% and 95% respectively, find: 1. volumetric displacement of the compressor; and 2. power of the motor. Data for R-12 is given below


Given : $T_2’=T_3=26° C = 26+273 = 299 K ;T_1=T_4= 2° C = 2+273 = 275 K ;$
$m_W=100 kg/h T_{w1}=26° C = 26+273 = 299 K ;T_{w2}=6° C = 6+273 = 279 K ;$ $\eta_v=80% = 0.80 ;\eta_{m1}=85% = 0.85 ;\eta{m2}=95% = 0.95 ;h_{f3}=60.64 kJ/kg ;h_{f1}=37.92 kJ/kg ;h_2’=198.10 kJ/kg ;h_1=188.39 kJ/kg K ;s_{f3}=0.2270 kJ/kg K;s _{f1}=0.1487 kJ/kg K ;s_2’=0.6865 kJ/kg K ; s_1=s_2=0.6956 kJ/kg K ;$ $c_{p3}=0.996 kJ/kg K ; c_{p4}=1.067 kJ/kg K ;c_{p2}’= 0.674 kJ/kg K ;c_{p1}=0.620 kJ/kg K ;$ $v_2’=0.026 m^3/kg ;v_1=0.052 m^3/kg$

1. Volumetric displacement of the compressor
Since 20% of the useful heat is lost into water cooler, therefore actual heat extracted from the water cooler,
$h_E=1.2\text m_W\text c_W(T_{w1}-T{w2})$
=1.2$\text 100\text 4.187(299-279)$ $=10050$kJ/h=167.5kJ/min .....(Sp.heat of water,$c_W$=4.187kJ/kg K)
We know that heat extracted or the net refrigerating effect per kg of the refrigerant =$h_1-h_{f3}=188.39-60.64=127.75$kJ/kg
Therefore mass flow of the refrigerant,
and volumetric displacement of the compressor
=$\frac{m_R\text v_1}{\eta_v}=\frac{1.3\text 0.052}{0.80}=0.085\text m^3/min$

2. power of the motor
First of all, let us find the temperature at point 2 ($T_2$).
We know that entropy at point 2
0.6956=0.6865+2.3$\text 0.674 \log\frac{T_2}{299} $
$\log\frac{T_2}{299}$=0.00587 (or) $\frac{T_2}{299}=1.0134$.....(Taking antilog of 0.00587)
Therefore $T_2=299\text 1.0134=303$K
We know that enthalpy at point 2,
$ =198.10+0.674(303-299)=200.8kJ/kg$
We also know that work done by the compressor per kg of the refrigerant
and work done per minute
=$m_R\text12.41=1.3\text12.41=16.133kJ/min=0.27kJ/s=0.27kW $
Therefore power required for the compressor
and power of the motor
$=\frac{0.317}{\eta_{m2}}=\frac{0.317}{0.95}=0.334kW$ (Ans)

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