In [1]:

```
# 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)
```

In [3]:

```
# 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
```

In [3]:

```
# 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)
```

In [4]:

```
# 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)
```

In [5]:

```
# 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)
```

In [6]:

```
# 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)
```

In [4]:

```
# 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
```

In [8]:

```
# 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)
```

In [9]:

```
# 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)
```

In [10]:

```
# 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)
```

In [8]:

```
# 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
```

In [12]:

```
# 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)
```

In [11]:

```
# 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
```

In [3]:

```
# 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)
```

In [15]:

```
# 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)
```

In [12]:

```
# 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
```

In [17]:

```
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)
```