Chapter 12 : Refrigeration

Example 12.1 Page No : 216

In [1]:
#Given
m = 500.0;#mass of ice produced per hour  in Kg
T1 = 15.0;#Initial temperature of water
T2 = -5.0;#Final temperature of ice
Ci = 0.5;#Specific  heat of ice in Kcal/Kg deg celsius
Cw = 1.0;#Specific heat of water in Kcal/Kg deg celsius
L_f = 79.71;#Latent heat of fusion in Kcal/Kg
Tf = 0.0;#Frezzing point of ice in deg celsius

#To Calculate the theoretical horse power required
Q2 = m*(Cw*(T1-Tf)+L_f+Ci*(Tf-T2));#Heat to be extracted per hour in Kcal
#From equation 12.1 (page no 220)
COP = (T2+273)/((T1+273)-(T2+273));
W = Q2/COP;#Work in Kcal/hr
W1 = W*(427/(60*4500.0));
print "The therotical horse power required is %f hp"%(W1);
#end
The therotical horse power required is 5.736411 hp

Example 12.2 Page No : 217

In [3]:
import math
from scipy.integrate import quad

#Given
Ta = 298.0;#Initial temperature in K
Tb = 203.0;#Final temperature in k
T1 = 298.0;#Water temperature in K
n = 1.0;#Kgmole of CO2
#Cp = 5.89+0.0112T ; Specific heat of CO2 in Kcal/Kgmole K

#To Calculate the compressor load
#From equation 12.2a and b (page no 221)
def f(T):
    y = ((T1-T)/T)*n*(5.89+0.0112*T);
    return y
W = quad(f,Ta,Tb)[0];
print "The compressor load is %f Kcal/Kgmole"%(W);
#end
The compressor load is -164.797031 Kcal/Kgmole

Example 12.3 Page No : 221

In [4]:
#Given
#Consider the figure 12.4 (page no 226) 
m = 5.0;#tonnes of refrigeration
T1 = 273-10.0;#temperature of the saturated vapour in K
T2 = 273+35.0;#temperature of the super heated vapour in K
T3 = 273+25.0;#temperature of the saturated liquid in K
T4 = 273+25.0;#temperature of the wet vapour in K
H1 = 341.8;#enthalpy of the saturated vapour in Kcal/Kg
H2 = 409.0;#enthalpy of the super heated vapour in Kcal/Kg
H3 = 350.0;#enthalpy of the saturated liquid in Kcal/Kg
H4 = 71.3;#enthalpy of the wet vapour in Kcal/Kg

#To Calculate the C.O.P, mass of refrigerant required, compressor horse power required and the C.O.P & compressor horse power for a reversed Carnot cycle
#(i)Calculation of the C.O.P of the compression cycle
#From equation 12.6 (page no 226)
COP = (H1-H4)/(H2-H1);
print "i)C.O.P of the compression cycle is %f"%(COP);

#(ii)Calculation of mass of refrigerant required
#From equation 12.7 (page no 226)
M = (m*50.4)/(H1-H4);
print " ii)The mass of refrigerant required is %f Kg/mt"%(M);

#(iii)Calculation of the compressor horse power
#From equation 12.5 (page no 226)
C_hp = (H2-H1)*M*(427/4500.0);
print " iii)The compressor horse power is %f hp"%(C_hp);

#(iv)Calculation for reversed Carnot cycle
#From equation 12.1 (page no 220)
COP = T1/(T3-T1);
print " iv)C.O.P for the reversed Carnot cycle is %f"%(COP);
Q2 = m*50.4*(427/4500.0);#in hp
C_hp = Q2/COP
print "    Compressor horse power for the reversed Carnot cycle is %f hp"%(C_hp);
#end
i)C.O.P of the compression cycle is 4.025298
 ii)The mass of refrigerant required is 0.931608 Kg/mt
 iii)The compressor horse power is 5.940430 hp
 iv)C.O.P for the reversed Carnot cycle is 7.514286
    Compressor horse power for the reversed Carnot cycle is 3.182205 hp

Example 12.4 Page No : 225

In [5]:
#Given
#Water at 20 deg cel is chilled to 10 deg cel by flash evaporation
Pv = 0.012;#Vapour pressure of water at 10 deg celsius in Kgf/sq.cm
H1 = 20.03;#Enthalpy of liquid water at 20 deg cel in Kcal/Kg
H2 = 10.4;#Enthalpy of liquid water at 10 deg cel in Kcal/Kg
Hv = 601.6;#Enthalpy of saturated vapour at 10 deg cel in Kcal/kg

#To calculate the pressure in the math.tank and the amount of make up water required
P = Pv;#pressure in the math.tank = vapour pressure of water
print "The pressure in the math.tank is %f Kgf/sq.cm"%(P);
#From equation 12.8 (page no 234)
x = (H1-H2)/(Hv-H2);
print " The amount of make up water required is %f Kg"%(x);
#end
The pressure in the math.tank is 0.012000 Kgf/sq.cm
 The amount of make up water required is 0.016289 Kg