# Chapter 11: Refrigeration Cycles¶

## Example 11-1 ,Page No.613¶

In [3]:
#given values
P1=0.14;#intial pressure in MPa
P2=0.8;#final pressure in MPa
m=0.05;#mass flow rate in kg/s

#from refrigerant-134a tables
h1=239.16;
s1=0.94456;
h2=275.39;
h3=95.47;

#calculation
s2=s1;#isentropic process
h4=h3;#throttling
QL=(h1-h4)*m;
Wm=m*(h2-h1);
Qh=m*(h2-h3);
COPR=QL/Wm;
print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);
print'the power input to the compressor %f kW'%round(Wm,2);
print'the rate of heat rejection to the environment %f kW'%round(Qh);
print'the COP of the refrigerator is %f'%round(COPR,2);

the rate of heat removal from the refrigerated space 7.180000 kW
the power input to the compressor 1.810000 kW
the rate of heat rejection to the environment 9.000000 kW
the COP of the refrigerator is 3.970000


## Example 11-2 ,Page No.615¶

In [5]:
#given data
m=0.05;#mass flow rate in kg/s
P1=0.14;#inlet pressure in MPa
T1=-10;#inlet temperature in C
P2=0.8;#outlet pressure in MPa
T2=50;#outlet temperature in C
P3=0.72;#condensor pressure in MPa
T3=26;#condensor temperature in C

#from refrigerant tables
h1=246.36;
h2=286.69;
h3=87.83;
h2S=284.21;#at isentropic conditions

#calculations
h4=h3;#throttling
QL=m*(h1-h4);
Wm=m*(h2-h1);
nC=(h2S-h1)/(h2-h1);
COPR=QL/Wm;
print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);
print'the power input to the compressor %f kW'%round(Wm,2);
print'the isentropic efficiency of the compressor is %f'%round(nC,3);
print'the COP of the refrigerator is %f'%round(COPR,2);

the rate of heat removal from the refrigerated space 7.930000 kW
the power input to the compressor 2.020000 kW
the isentropic efficiency of the compressor is 0.939000
the COP of the refrigerator is 3.930000


## Example 11-3 ,Page No.621¶

In [8]:
#given data
mA=0.05;#mass flow rate in kg/s
P1=0.14;#inlet pressure in MPa
P5=0.32;#pressure at heat exchanger in MPa
P7=0.8;#oultet pressure in MPa

#from tables
h1=239.16;
h2=255.93;
h3=55.16;
h5=251.88;
h6=270.92;
h7=95.47;

#calculations
h4=h3;#throttling
h8=h7;#throttling
# E out = E in
# mA*h5 + mB*h3 = mA*h8 + mB*h2
mB=mA*(h5-h8)/(h2-h3);
QL=mB*(h1-h4);
# W in = Wcomp I,in + Wcomp II,in
Win=mA*(h6-h5)+mB*(h2-h1);
COPR=QL/Win;
print'the mass flow rate of the refrigerant through the lower cycle %f kg/s'%round(mB,4);
print'the rate of heat removal from the refrigerated space %f kW'%round(QL,2);
print'the power input to the compressor %f kW'%round(Win,2);
print'the coefficient of performance of this cascade refrigerator is %f'%round(COPR,2);

the mass flow rate of the refrigerant through the lower cycle 0.039000 kg/s
the rate of heat removal from the refrigerated space 7.170000 kW
the power input to the compressor 1.610000 kW
the coefficient of performance of this cascade refrigerator is 4.460000


## Example 11-4 ,Page No.624¶

In [10]:
#given data
P1=0.14;#inlet pressure in MPa
P5=0.32;#throttled pressure in MPa
P7=0.8;#oultet pressure in MPa

#from tables
h1=239.16;
h2=255.93;
h3=251.88;
h5=95.47;
h7=55.16;

#from saturated liquid-vapour table
#at P=0.32 MPa
hf=55.16;
hfg=196.71;

#calculations
h8=h7;#throttling
h6=h5;#throttling
#the quality at state 6
x6=(h6-hf)/hfg;
qL=(1-x6)*(h1-h8);
# W in = Wcomp I,in + Wcomp II,in
#enthalaoy at state 9
# E out = E in
h9=x6*h3+(1-x6)*h2;
# s9 = s4 i.e isentropic process
#at 0.8MPa and s4=0.9416 kJ/kg
h4=274.48;
Win=(1-x6)*(h2-h1)+(1)*(h4-h9);
COPR=qL/Win;
print'the fraction of the refrigerant that evaporates as it is throttled to the flash chamber is %f'%round(x6,4);
print'the amount of heat removed from the refrigerated space %f kJ/kg'%round(qL,1);
print'the compressor work per unit mass of refrigerant flowing through the condensor %f kJ/kg'%round(Win,2);
print'the coefficient of performance is %f'%round(COPR,2)

the fraction of the refrigerant that evaporates as it is throttled to the flash chamber is 0.204900
the amount of heat removed from the refrigerated space 146.300000 kJ/kg
the compressor work per unit mass of refrigerant flowing through the condensor 32.710000 kJ/kg
the coefficient of performance is 4.470000


## Example 11-5 ,Page No.630¶

In [13]:
#given data
m=0.1;#mass flow rate in lbm/s
T1=0+460;#refigerated space at temperature in R
T3=80+460;#temperature of surrounding medium in R

#from Table A–17E
# at T1
h1=109.90;
Pr1=.7913;
#pressure ratio at compressor is 4
Pr2=4*Pr1;
#at Pr2
h2=163.5;
T2=683;
#at T3
h3=129.06;
Pr3=1.3860;
#pressure ratio at compressor is 4
Pr4=Pr3/4;
#at Pr4
h4=86.7;
T4=363;

#calculations
qL=h1-h4;
Wout=h3-h4;
Win=h2-h1;
COPR=qL/(Win-Wout);
Qrefrig=m*qL;
print'the minimum temperatures in the cycle %f F'%round(T4-460);
print'the maximum temperatures in the cycle %f F'%round(T2-460);
print'the coefficient of performance is %f'%round(COPR,2)
print'the rate of refrigeration for a mass flow rate of 0.1 lbm/s. %f Btu/s'%round(Qrefrig,2)

the minimum temperatures in the cycle -97.000000 F
the maximum temperatures in the cycle 223.000000 F
the coefficient of performance is 2.060000
the rate of refrigeration for a mass flow rate of 0.1 lbm/s. 2.320000 Btu/s


## Example 11-6 ,Page No.636¶

In [14]:
#given data
COPR=0.1;#refrigerator COP
T1=20;#intial temp in C
T2=4;#final temp in C
t=30*60;#total time required in sec
V=0.350;#volumne of can in L

#constants used
p=1;#on kg/L
c=4.18;#in kJ/kg-C from Table A-3

#calculations
m=p*V;
Qcooling=m*c*(T1-T2)/t*1000;#converted in W by multiplying by 1000
Win=Qcooling/COPR;
print'the average electric power consumed by the thermoelectric refrigerator %i W'%round(Win)

the average electric power consumed by the thermoelectric refrigerator 130 W