Example 1
A vapour compression refrigeration plant works between pressure limits of 5.3 bar and 2.1 bar. The vapour is superheated at the end of compression, its temperature being 37° C The vapour is superheated by 5°C before entering the compressor. If the specific heat of superheated vapour is 0.63 kJ/kg.K, find the coefficient of performance of the plant. Use the data given below:

Solution
Given :
$p_2=5.3 bar ;p_1=2.1 bar ;T_2=37°C = 37+273 = 310 K ;T_1-T_1’= 5°C ; c_p=0.63 kJ/kg K ; T_2’=15.5°C = 15.5+273 = 288.5 K ;T_1’=-14°C = -14+273 = 259 K ;h_{f3}=h_{f2}’=56.15 kJ/kg ;h_{f1}’=25.12 kJ/kg ;h_{fg2}’=144.9 kJ/kg ; h_{fg1}’=158.7 kJ/kg$
We know that enthalpy of vapour at point 1,

$h_1=h_1’+c_p(T_1-T_1’)=(h_{f1}’+h_{fg1}’)+c_p(T_1-T_1’)$

=$(25.12+158.7)+0.63\text5=186.97$kJ/kg

Similarly, enthalpy of vapour at point 2,

$h_2=h_2’+c_p(T_2-T_2’)=(h_{f2}’+h_{fg2}’)+c_p(T_2-T_2’)$

=$(56.15+144.9)+0.63(310-288.5) =214.6$kJ/kg

Therefore coefficient of performance of the plant,

C.O.P=
$\frac{h_1-h_{f3}}{h_2-h_1}=\frac{186.97-56.15}{214.6-186.97}=\frac{130.82}{27.63}=4.735$ (Ans)