# Chapter 6 : Refrigeration Cycles¶

## Example 6.1 Page no : 308¶

In :
# Variables
COP = 8.5;			#Co-efficient of performance
T1 = 300.;			#Room temperature in K
T2 = 267.;			#Refrigeration temperature in K

# Calculations
COPmax = T2/(T1-T2);			#Maximum COP possible

# Results
print 'Maximum COP possible is %3.2f  \
\nSince the COP claimed by the inventor is more than the maximum possible COP\
his claim is not correct'%(COPmax)

Maximum COP possible is 8.09
Since the COP claimed by the inventor is more than the maximum possible COP his claim is not correct


## Example 6.2 Page no : 309¶

In :
# Variables
TL = 268.;			#Low temperature in K
TH = 293.;			#High temperature in K
t = 24.;			#time in hrs
C = 2100.;			#Capacity of refrigerator in kJ/s
Tw = 10.;			#Water temperature in oC
L = 335.;			#Latent heat of ice in kJ/kg

# Calculations
COP = TL/(TH-TL);			#Co-efficient of performance
Pmin = C/COP;			#Minimum power required in kW
Qr = (4.187*(Tw-0))+L;			#Heat removed from water in kJ/kg
m = C/Qr;			#mass of ice formed in kg/s
W = (m*t*3600)/1000;			#Weight of ice formed in tons

# Results
print 'Minimum power required is %3.2f kW  \
\nWeight of ice formed in 24 hours is %3.2f tons'%(Pmin,W)

# rounding off error

Minimum power required is 195.90 kW
Weight of ice formed in 24 hours is 481.44 tons


## Example 6.3 Page no : 309¶

In :
# Variables
TL = -10.;			#Temperature of brine in oC
TH = 20.;			#Temperature of water in oC
L = 335.;			#Latent heat of ice in kJ/kg

# Calculations
Qr = (4.187*(TH-0))+L;			#Heat removed from water in kJ/kg
COP = (TL+273)/(TH-TL);			#Co-efficient of performance
mi = (COP*3600)/Qr;			#mass of ice formed per kWh in kg

# Results
print 'Mass of ice formed per kWh is %3.1f kg'%(mi)

Mass of ice formed per kWh is 75.4 kg


## Example 6.4 Page no : 310¶

In :
# Variables
P1 = 1.2;			#Pressure at point 1 in bar
P2 = 7.;			#Pressure at point 2 in bar
m = 0.05;			#mass flow rate of refrigerant in kg/s
h1 = 340.1;			#Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
s1 = 1.57135;			#Entropy at point 1 from refrigerant-12 tables in kJ/kg-K
s2 = 1.57135;			#Entropy at point 2 from refrigerant-12 tables in kJ/kg-K
h2 = 372.;			#Enthalpy at point 2 from refrigerant-12 tables in kJ/kg
h3 = 226.575;			#Enthalpy at point 3 from refrigerant-12 tables in kJ/kg
h4 = 226.575;			#Enthalpy at point 4 from refrigerant-12 tables in kJ/kg

# Calculations
Q2 = m*(h1-h4);			#Rate of heat removed from the refrigerated space in kW
W = m*(h2-h1);			#Power input to the compressor in kW
Q1 = m*(h2-h3);			#Rate of heat rejection to the environment in kW
COP = Q2/W;			#Co-efficient of performance

# Results
print 'Rate of heat removed from the refrigerated space is %3.2f kW  \
\nPower input to the compressor is %3.3f kW  \
\nRate of heat rejection to the environment is %3.2f kW  \
\nCo-efficient of performance is %3.2f'%(Q2,W,Q1,COP)

Rate of heat removed from the refrigerated space is 5.68 kW
Power input to the compressor is 1.595 kW
Rate of heat rejection to the environment is 7.27 kW
Co-efficient of performance is 3.56


## Example 6.5 Page no : 311¶

In :
# Variables
T2 = 40.;			#Temperature at point 2 in oC
T1 = -10.;			#Temperature at point 1 in oC
h2 = 367.155;			#Enthalpy at point 2 from refrigerant-12 tables in kJ/kg
s2 = 1.54057;			#Entropy at point 2 from refrigerant-12 tables in kJ/kg-K
s1 = 1.54057;			#Entropy at point 1 from refrigerant-12 tables in kJ/kg-K
sg = 1.56004;			#Entropy from refrigerant-12 tables in kJ/kg-K
sf = 0.96601;			#Entropy from refrigerant-12 tables in kJ/kg-K
hf = 190.822;			#Enthalpy from refrigerant-12 tables in kJ/kg-K
hfg = 156.319;			#Enthalpy from refrigerant-12 tables in kJ/kg-K
h3 = 238.533;			#Enthalpy at point 3 from refrigerant-12 tables in kJ/kg-K
h4 = h3;			#Enthalpy at point 4 from refrigerant-12 tables in kJ/kg-K

# Calculations
x1 = (s1-sf)/(sg-sf);			#Quality factor
h1 = hf+(x1*hfg);			#Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
COP = (h1-h4)/(h2-h1);			#Co-efficient of performance

# Results
print 'COP of the system is %3.2f'%(COP)

COP of the system is 4.12


## Example 6.6 Page no : 311¶

In :
# Variables
Tc = 35.;			#Temperature of condenser in oC
Te = -15.;			#Temperature of evaporator in oC
m = 10.;			#Mass of ice per day in tons
Tw = 30.;			#Temperature of water in oC
Ti = -5.;			#Temperature of ice in oC
nv = 0.65;			#Volumetric efficiency
N = 1200.;			#Speed in rpm
x = 1.2;			#Stroke to bore ratio
na = 0.85;			#Adiabatic efficiency
nm = 0.95;			#Mechanical efficiency
S = 4.187;			#Specific heat of water in kJ/kg
L = 335.;			#Latent heat of ice in kJ/kg
h1 = 1667.24;			#Enthalpy at Te from Ammonia chart in kJ/kg
h2 = 1925.;			#Enthalpy at Te from Ammonia chart in kJ/kg
h4 = 586.41;			#Enthalpy at Tc from Ammonia chart in kJ/kg
v1 = 0.508;			#Specific humidity at Te from Ammonia chart in (m**3)/kg

# Calculations
Qr = (((m*1000)/24)*((S*(Tw-0))+L+(1.94*(0-Ti))))/3600;			#Refrigerating capacity in kW
mr = Qr/(h1-h4);			#Refrigerant mass flow rate in kg/s
T2 = 112;			#Discharge temperature in oC
D = ((mr*v1*4*60)/(nv*3.14*x*N))**(1./3);			#Cylinder diameter in m
L = x*D;			#Stroke length in m
W = (mr*(h2-h1))/(na*nm);			#Compressor motor power in kW
COPth = (h1-h4)/(h2-h1);			#Theoretical COP
COPact = Qr/W;			#Actual COP

# Results
print 'Refrigerating capacity of plant is %3.2f kW  \
\nRefrigerant mass flow rate is %3.4f kg/s  \
\nDischarge temperature is %3.0f oC  \
\nCylinder diameter is %3.3f m  \
\nStroke length is %3.3f m  \
\nCompressor motor power is %3.2f kW  \
\nTheoretical COP is %3.2f  \
\nActual COP is %3.2f'%(Qr,mr,T2,D,L,W,COPth,COPact)

Refrigerating capacity of plant is 54.43 kW
Refrigerant mass flow rate is 0.0504 kg/s
Discharge temperature is 112 oC
Cylinder diameter is 0.128 m
Stroke length is 0.153 m
Compressor motor power is 16.08 kW
Theoretical COP is 4.19
Actual COP is 3.39


## Example 6.7 Page no : 313¶

In :
# Variables
T1 = -5.;			#Temperature at point 1 in oC
T2 = 30.;			#Temperature at point 2 in oC
m = 13500.;			#mass of ice per day in kg
Tw = 20.;			#Temperature of water in oC
COP = 0.6;			#Co-efficient of performance
h2 = 1709.33;			#Enthalpy at point 2 in kJ/kg
s2 = 6.16259;			#Entropy at point 2 in kJ/kg-K
s1 = 6.16259;			#Entropy at point 1 in kJ/kg-K
sf = 1.8182;			#Entropy in kJ/kg-K
sg = 6.58542;			#Entropy in kJ/kg-K
hf = 400.98;			#Enthalpy in kJ/kg
hfg = 1278.35;			#Enthalpy in kJ/kg
h4 = 562.75;			#Enthalpy at point 4 in kJ/kg
S = 4.187;			#Specific heat of water in kJ/kg
L = 336.;			#Latent heat of ice in kJ/kg

# Calculations
x1 = (s1-sf)/(sg-sf);			#Quality factor
h1 = hf+(x1*hfg);			#Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
COPi = (h1-h4)/(h2-h1);			#Ideal COP
COPact = COP*COPi;			#Actual COP
Qr = ((m*S*(Tw-0))+(m*L))/(24*3600);			#Total amount of heat removed in kJ/s
mr = Qr/(h1-h4);			#Circulation rate of ammonia in kg/s
W = mr*(h2-h1);			#Power required in kW

# Results
print 'Circulation rate of ammonia is %3.3f kg/s  \
\nPower required is %3.3f kW  \
\nCOP is %3.3f'%(mr,W,COPact)

# rounding off error

Circulation rate of ammonia is 0.065 kg/s
Power required is 9.374 kW
COP is 4.198


## Example 6.8 Page no : 314¶

In :
# Variables
Tc = 20.;			#Temperature of condenser in oC
Te = -25.;			#Temperature of evaporator in oC
m = 15.;			#Mass of ice per day in tons
Ts = 5.;			#Subcooled temperature in oC
Tsh = 10.;			#Superheated temperature in oC
n = 6.;			#No. of cylinders
N = 950.;			#Speed of compressor in rpm
x = 1.;			#Stroke to bore ratio
h1 = 402.;			#Enthalpy at point 1 from R-22 tables in kJ/kg
h2 = 442.;			#Enthalpy at point 2 from R-22 tables in kJ/kg
h3 = 216.;			#Enthalpy at point 3 from R-22 tables in kJ/kg
h4 = 216.;			#Enthalpy at point 4 from R-22 tables in kJ/kg
v1 = 2.258;			#Specific volume at point 1 in (m**3)/min

# Calculations
Re = h1-h4;         			#Refrigerating effect in kJ/kg
mr = (m*14000)/(Re*60);			#Mass flow of refrigerant in kg/min
Pth = (mr*(h2-h1))/60;			#Theoretical power in kW
COP = (h1-h4)/(h2-h1);			#Co-efficient of performance
Dth = v1/n;			            #Theoretical print lacement per cylinder
D = (((Dth*4)/(3.147*N))**(1./3))*1000;			#Theoretical bore of compressor in mm
L = D;              			#Theoretical stroke of compressor in mm

# Results
print 'Refrigerating effect is %3.0f kJ/kg  \
\nMass flow of refrigerant per minute is %3.2f kg/min  \
\nTheoretical input power is %3.2f kW  COP is %3.2f  \
\nTheoretical bore of compressor is %3.2f mm  \
\nTheoretical stroke of compressor is %3.2f mm'%(Re,mr,Pth,COP,D,L)

Refrigerating effect is 186 kJ/kg
Mass flow of refrigerant per minute is 18.82 kg/min
Theoretical input power is 12.54 kW  COP is 4.65
Theoretical bore of compressor is 79.56 mm
Theoretical stroke of compressor is 79.56 mm


## Example 6.9 Page no : 316¶

In :
# Variables
T2 = 40.;			#Temperature at point 2 in oC
T1 = -5.;			#Temperature at point 1 in oC
h2 = 367.155;			#Enthalpy at point 2 from F-12 tables in kJ/kg
sg = 1.55717;			#Entropy from F-12 tables in kJ/kg-K
s1 = 1.54057;			#Entropy at point 1 from F-12 tables in kJ/kg-K
sf = 0.98311;			#Entropy from F-12 tables in kJ/kg-K
hf = 195.394;			#Enthalpy from F-12 tables in kJ/kg
hfg = 153.934;			#Enthalpy from F-12 tables in kJ/kg
h4 = 238.533;			#Enthalpy at point 4 from F-12 tables in kJ/kg
h4s = 218;			#Enthalpy at point 4 with subcooling from F-12 tables in kJ/kg

# Calculations
x1 = (s1-sf)/(sg-sf);			#Quality factor
h1 = hf+(x1*hfg);			#Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
COPns = (h1-h4)/(h2-h1);			#Co-efficient of performance with no subcooling
COPs = (h1-h4s)/(h2-h1);			#Co-efficient of performance with subcooling

# Results
print 'COP with no subcooling is %3.3f  \
\nCOP with subcooling is %3.3f'%(COPns,COPs)

COP with no subcooling is 4.773
COP with subcooling is 5.695


## Example 6.10 Page no : 309¶

In :
# Variables
Tg = 470.;			#Heating temperature in K
T0 = 290.;			#Cooling temperature in K
TL = 270.;			#Refrigeration temperature in K

# Calculations
COP = ((Tg-T0)/Tg)*(TL/(T0-TL));			#Ideal COP of absorption refrigeration system

# Results
print 'Ideal COP of absorption refrigeration system is %3.2f'%(COP)

Ideal COP of absorption refrigeration system is 5.17


## Example 6.11 Page no : 317¶

In :
# Variables
T1 = -18.;			#Temperature at point 1 in oC
T3 = 27.;			#Temperature at point 3 in oC
rp = 4.;			#Pressure ratio
m = 0.045;			#mass flow rate in kg/s
y = 1.4;			#Ratio of specific heats
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K

# Calculations
x = (y-1)/y;			#Ratio
T2 = (rp**x)*(273+T1);			#Temperature at point 2 in K
Tmax = T2-273;			#Maximum temperature in oC
T4 = ((1/rp)**x)*(273+T3);			#Temperature at point 4 in K
Tmin = T4-273;			#Minimum temperature in oC
qL = Cp*(T1-Tmin);			#Heat rejected
Wcin = Cp*(Tmax-T1);			#Compressor work
Wtout = Cp*(T3-Tmin);			#Turbine work
Wnet = Wcin-Wtout;			#Net work done
COP = qL/Wnet;			#Co-efficient of performance
Qref = m*qL;			#Rate of refrigeration in kW

# Results
print 'Maximum temperature in the cycle is %3.0f oC  \
\nMinimum temperature in the cycle is %3.0f oC  \
\nCOP is %3.2f  \
\nRate of refrigeration is %3.2f kW'%(Tmax,Tmin,COP,Qref)

# rounding off error

Maximum temperature in the cycle is 106 oC
Minimum temperature in the cycle is -71 oC
COP is 2.06
Rate of refrigeration is 2.40 kW


## Example 6.12 Page no : 318¶

In :
# Variables
P1 = 1.;			#Pressure at point 1 in bar
T1 = 268.;			#Temperature at point 1 in K
P2 = 5.;			#Pressure at point 2 in bar
T3 = 288.;			#Temperature at point 3 in K
n = 1.3;			#Adiabatic gas constant
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K

# Calculations
x = (n-1)/n;			#Ratio
T2 = ((P2/P1)**x)*T1;			#Temperature at point 2 in K
T4 = ((P1/P2)**x)*T3;			#Temperature at point 4 in K
W = Cp*(T3-T4);			#Work developed per kg of air in kJ/kg
Re = Cp*(T1-T4);			#Refrigerating effect per kg of air in kJ/kg
Wnet = Cp*((T2-T1)-(T3-T4));			#Net work output in kJ/kg
COP = Re/Wnet;			#Co-efficient of performance

# Results
print 'Work developed per kg of air is %3.3f kJ/kg  \
\nRefrigerating effect per kg of air is %3.3f kJ/kg  \
\nCOP of the cycle is %3.2f'%(W,Re,COP)

Work developed per kg of air is 89.795 kJ/kg
Refrigerating effect per kg of air is 69.695 kJ/kg
COP of the cycle is 2.22


## Example 6.13 Page no : 319¶

In :
# Variables
T1 = 277.;			#Temperature at point 1 in K
T3 = 328.;			#Temperature at point 3 in K
P1 = 0.1;			#Pressure at point 1 in MPa
P2 = 0.3;			#Pressure at point 2 in MPa
nc = 0.72;			#Isentropic efficiency of compressor
nt = 0.78;			#Isentropic efficiency of turbine
y = 1.4;			#Adiabatic gas constant
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K
m = 3.;			#Cooling load in tonnes

# Calculations
x = (y-1)/y;			#Ratio
T2s = T1*((P2/P1)**x);			#Temperature at point 2s in K
T2 = ((T2s-T1)/nc)+T1;			#Temerature at point 2 in K
T4s = T3*((P1/P2)**x);			#Temperature at point 4s in K
T4 = T3-((T3-T4s)*nt);			#Temperature at point 4 in K
Re = Cp*(T1-T4);			#Refrigerating effect in kJ/kg
Wnet = Cp*((T2-T1)-(T3-T4));			#Net work output in kJ/kg
COP = Re/Wnet;			#Co-efficient of performance
P = (m*3.52)/COP;			#Driving power required in kW
ma = (m*3.52)/Re;			#Mass flow rate of air in kg/s

# Results
print 'COP of refrigerator is %3.2f  \
\nDriving power required is %3.0f kW  \
\nMass flow rate of air is %3.2f kg/s'%(COP,P,ma)

# rounding off error

COP of refrigerator is 0.25
Driving power required is  43 kW
Mass flow rate of air is 0.59 kg/s


## Example 6.14 Page no : 321¶

In :
# Variables
P1 = 2.5;			#Pressure at point 1 in bar
P3 = 9.;			#Pressure at point 3 in bar
COPr = 0.65;			#Ratio of actual COP to the theoretical COP
m = 5.;			#Refrigerant flow in kg/min
T1 = 309;			#Temperature at point 1 in K
T2s = 300;			#Temperature at point 2s in K
h1 = 570.3;			#Enthalpy at P1 from the given tables in kJ/kg
h4 = 456.4;			#Enthalpy at P3 from the given tables in kJ/kg
h2g = 585.3;			#Enthalpy at P3 from the given tables in kJ/kg
s2 = 4.76;			#Entropy at P1 from the given tables in kJ/kg-K
s2g = 4.74;			#Entropy at P3 from the given tables in kJ/kg-K
Cp = 0.67;			#Specific heat at P3 in kJ/kg-K

# Calculations
T2 = (2.718**((s2-s2g)/Cp))*T2s;			#Temperature at point 2 in K
h2 = h2g+(Cp*(T2-T2s));			#Enthalpy at point 2 in kJ/kg
COPR = (h1-h4)/(h2-h1);			#Refrigerant COP
COPact = COPr*COPR;			#Actual COP
qL = COPact*(h2-h1);			#Heat rejected in kJ/kg
QL = ((m*qL*60)/3600)/3.516;			#Cooling produced per kg of refrigerant in tonnes of refrigeration

# Results
print 'Theoretical COP is %3.2f  \
\nNet cooling produced per hour is %3.2f TR'%(COPR,QL)

Theoretical COP is 5.40
Net cooling produced per hour is 1.75 TR


## Example 6.15 Page no : 322¶

In :
# Variables
T2 = 298.;			#Temperature at point 2 in K
T1 = 268.;			#Temperature at point 1 in K
hf1 = -7.54;			#Liquid Enthalpy at T1 in kJ/kg
x1 = 0.6;			#Quality factor 1
hfg1 = 245.3;			#Latent heat at T1 in kJ/kg
sf1 = 0.251;			#Liquid Entropy at T1 in kJ/kg-K
s1 = 0.507;			#Entropy at point 1 in kJ/kg-K
hfg2 = 121.4;			#Latent heat at T2 in kJ/kg
hf2 = 81.3;			#Liquid Enthalpy at T2 in kJ/kg
h4 = hf2;			#Enthalpy at point 4 in kJ/kg

# Calculations
h1 = hf1+(x1*hfg1);			#Enthalpy at point 1 in kJ/kg
x2 = ((s1-sf1)*T2)/hfg2;			#Quality factor 2
h2 = hf2+(x2*hfg2);			#Enthalpy at point 2 in kJ/kg
COP = (h1-h4)/(h2-h1);			#COP of the machine

# Results
print 'COP of the machine is %3.2f'%(COP)

COP of the machine is 3.25


## Example 6.16 Page no : 323¶

In :
# Variables
P1 = 25.;			#Pressure at point 1 in bar
P2 = 60.;			#Pressure at point 2 in bar
h2 = 208.1;			#Vapour enthalpy at P2 in kJ/kg
h3 = 61.9;			#Liquid enthalpy at P2 in kJ/kg
h4 = h3;			#Liquid enthalpy at P2 in kJ/kg
s2 = 0.703;			#Vapour entropy at P2 in kJ/kg-K
sf1 = -0.075;			#Liquid entropy at P1 in kJ/kg-K
sfg1 = 0.971;			#Entropy in kJ/kg-K
hf1 = -18.4;			#Liquid Enthalpy at P1 in kJ/kg
hfg1 = 252.9;			#Latent heat at P1 in kJ/kg
m = 5.;			#Refrigerant flow in kg/min

# Calculations
x1 = (s2-sf1)/sfg1;			#Quality factor 1
h1 = hf1+(x1*hfg1);			#Enthalpy at point 1 in kJ/kg
COP = (h1-h4)/(h2-h1);			#Co-efficient of performance
QL = (m*(h1-h4))/60;			#Capacity of the refrigerator in kW

# Results
print 'COP of refrigerator is %3.2f  \
\nCapacity of refrigerator is %3.2f kW'%(COP,QL)

# rounding off error

COP of refrigerator is 5.13
Capacity of refrigerator is 10.19 kW


## Example 6.17 Page no : 324¶

In :
import math

# Variables
T1 = 271.;			#Temperature at point 1 in K
T = 265.;			#Temperature at point 1' in K
Ta = 303.;			#Temperature at point 2' in K
Cpv = 0.733;			#Specific heat of vapour in kJ/kg
Cpl = 1.235;			#Specific heat of liquid in kJ/kg
h = 184.07;			#Liquid enthalpy at T in kJ/kg
s = 0.7;			#Entropy at point 1' in kJ/kg-K
sa = 0.685;			#Vapour entropy at Ta in kJ/kg-K
ha = 199.62;			#Enthalpy at point 2' in kJ/kg
hfb = 64.59;			#Liquid enthalpy at Ta in kJ/kg
DT3 = 5.;			#Temperature difference in oC
Q = 2532.;			#Refrigeration capacity in kJ/min

# Calculations
s2 = s+(Cpv*((math.log(T1/T))/(math.log(2.718))));			#Entropy at point 1 in kJ/kg-K
h1 = h+(Cpv*(T1-T));			#Enthalpy at point 1 in kJ/kg-K
T2 = (2.718**((s2-sa)/Cpv))*Ta;			#Temperature at point 2 in K
h2 = ha+(Cpv*(T2-Ta));			#Enthalpy at point 2 in kJ/kg
h4 = hfb-(Cpl*DT3);			#Enthalpy at point 4 in kJ/kg
COP = (h1-h4)/(h2-h1);			#Co-efficient of performance
m = Q/(h1-h4);			#Mass flow rate of refrigerant in kJ/min
P = (m*(h2-h1))/(60*12);			#Power required in kW/TR

# Results
print 'COP is %3.2f  \
\nTheoretical power required per tonne of refrigeration is %3.3f kW/TR'%(COP,P)

COP is 6.23
Theoretical power required per tonne of refrigeration is 0.564 kW/TR