Question

Click Submit to complete this assessment. Question 25 In a vapor compression cycle if the refrigeration effect is 482 KJ/kg, while energy supplied to the compressor is 119 Kj/kg, the COP of the system is A Click Submit to complete this assessment. MacBook Pro esc & @ % 5 0 6 7 V 8 3 ۳ 2 4 E 1

Answer 1

Question 25

In vapor compression refrigeration cycle if the refrigeration effect is 482 kJ/kg, while energy supplied to the compressor is 119 kJ/kg , the COP of the system is ?

Answer :

The COP of the vapor compression refrigeration system is given by the following formula :

COP RE Winput

Where

RE=
Refrigeration effect
$W_{input}=$ Energy supplied to the compressor

Substitute the values in the above equation :

$COP=\frac{482}{119}$

COP = 4.050

Therefore the COP of the system is :  $\mathbf{COP=4.050}$

Question 12

Given information :

Evaporator temperature : $-20^0C$

Condensor temperature : $10^0C$

Answer :

The properties of refrigerant at the respective temperatures are :

Evaporator temperature : $-20^0C$

Sg 0.94575 kJ/kg-K

Condensor temperature : $10^0C$

Since the given cycle is ideal vapor compression refrigeration cycle

• The state of refrigerant at compressor exit is saturated vapor
• The state of refrigerant at condensor exit is saturated liquid
• Compressor is isentropic.

Hence :

Enthalpy of refrigerant at the compressor exit is :  $\mathbf{h_2=256.22\ kJ/kg}$

Since the compressor is isentropic

$s_1=s_2$

$s_1=s_f+x*(s_g-s_f)$

$s_1=0.10456+x*(0.94575-0.10456)$

$0.92661=0.10456+x*(0.84119)$

$x=0.977$

Therefore the dryness fraction is equal to :
x= 0.977

Calculating the enthalpy of refrigerant at the entry to compressor :

$h_1=h_f+x*(h_g-h_f)$

$h_1=25.47+0.977*(238.43-25.47)$

Since the state of refrigerant at the condensor exit is saturated liquid

Since the throttling process is isenthalpic process

Tabulating the Specific enthalpies

Calculating the compressor work :

$W_{in}=h_2-h_1$

Substitute the values

Therefore the Compressor work is : $\mathbf{W_{in}=22.699\ kJ/kg}$

Calculating the refrigeration effect :

$RE=h_1-h_4$

Substitute the values

Therefore the Refrigeration effect is : $\mathbf{RE=168.101\ kJ/kg}$

Calculating the COP

The COP of the vapor compression refrigeration system is given by the following formula :

Substitute the values

COP RE Winput

$COP=7.405$

Therefore the COP of the system is : $\mathbf{COP=7.405}$

Enthalpy at compressor exit	256.22 kJ/kg
Refrigeration effect	168.101 kJ/kg
Compressor work	22.699 kJ/kg
COP of the cycle	7.405
Dryness fraction	0.977