Chapter 14 : Refrigeration Cycles

Example 14.1 Page No : 587

In [1]:
# Variables
T2 = 268.; 
T1 = 308.;

# Calculation
COP = T2/(T1-T2);
ACOP = COP/3; 			# Actual COP
Q2 = 29.; 			# in kW
W = Q2/ACOP;

# Results
print "Power required to derive the plane is %.0f kW"%W

Power required to derive the plane is 13 kW

Example 14.2 Page No : 589

In [5]:
# Variables
h1 = 236.04; s1 = 0.9322; s2 = s1;
P2 = 0.8; 			# in MPa
h2 = 272.05; h3 = 93.42; h4 = h3;
m = 0.06; 			# mass flow rate

# Calculation
Q2 = m*(h1-h4);
Wc = m*(h2-h1);
Q1 = m*(h2-h4);
COP = Q2/Wc;

# Results
print "The rate of heat removal is %.2f kW"%Q2
print "Power input to the compressor is %.2f kW"%Wc
print "The heat rejection rate in the condenser is %.2f kW"%Q1
print "COP is %.3f"%COP

The rate of heat removal is 8.56 kW
Power input to the compressor is 2.16 kW
The heat rejection rate in the condenser is 10.72 kW
COP is 3.961

Example 14.3 Page No : 589

In [1]:
# Variables
h1 = 183.19; 
h2 = 209.41; 
h3 = 74.59; 
h4 = h3;
T1 = 313.; 
T2 = 263.;
W = 70000./3600; 			# Plant capacity in kW

# Calculation
w = W/(h1-h4); 			# Refrigerant flow rate
v1 = 0.077;
VFR = w*v1;
T = 48.; 			# in degree
P2 = 9.6066; P1 = 2.1912;
rp = P2/P1; 			# Pressure ratio
Q1 = w*(h2-h3);
hf = 26.87; hfg = 156.31;
x4 = (h4-hf)/hfg;
COP = (h1-h4)/(h2-h1);
PI = w*(h2-h1);
COP = T2/(T1-T2);
COP_v = 4.14;
r = COP_v/COP;

# Results
print "Refrigerant flow rate is %.2f Kg/s"%w
print "Volume flow rate is %.4f m3/s"%VFR
print "Compressor discharge temperature is",T,"degree"
print "Pressure ratio is %.2f"%rp
print "Heat rejected to the condenser is %.2f kW"%Q1
print "Flash gas percentage is %.1f %%"%(x4*100)
print "COP is",COP
print "Power required to drive the compressor is %.2f kW"%PI
print "Ratio of COP of carnot refrigerator is %.3f"%r

# note : rounding off error is there.

Refrigerant flow rate is 0.18 Kg/s
Volume flow rate is 0.0138 m3/s
Compressor discharge temperature is 48.0 degree
Pressure ratio is 4.38
Heat rejected to the condenser is 24.14 kW
Flash gas percentage is 30.5 %
COP is 5.26
Power required to drive the compressor is 4.69 kW
Ratio of COP of carnot refrigerator is 0.787

Example 14.4 Page No : 590

In [16]:
import math 

# Variables
h3 = 882.
h2 = 1034.;
h6 = 998.
h1 = 1008.;
v1 = 0.084;

# Calculation and Results
h4 = h3-h1+h6; h5 = h4;
t4 = 25.+273;
print "Refrigeration effect is",h6-h5,"kJ/kg"

m = 10.;
w = (m*14000)/((h6-h5)*3600); 			# in kg/s
print "Refrigerant flow rate is %.2f kg/s"%w

v1 = 0.084;
VFR = w*3600*v1; 			# in kg/h
ve = 0.8; 			# volumetric efficiency
CD = VFR/(ve*60); 			# in m3/min
N = 900;
n = 2;
D = ((CD*4)/(math.pi*1.1*N*n))**(1./3); 			# L = 1.1D L = length D = diameter
L = 1.1*D;
print "Diameter of cylinder is %.1f cm"%(D*100)
print "Length of cylinder is %.2f cm"%(L*100)
COP = (h6-h5)/(h2-h1);
PI = w*(h2-h1);
print "Power required to drive the compresor is %.2f kW"%PI
print "COP is %.2f"%COP

Refrigeration effect is 126.0 kJ/kg
Refrigerant flow rate is 0.31 kg/s
Diameter of cylinder is 10.8 cm
Length of cylinder is 11.85 cm
Power required to drive the compresor is 8.02 kW
COP is 4.85

Example 14.5 Page No : 592

In [3]:
import math 

# Variables
P2 = 1554.3;
P1 = 119.5;
Pi = math.sqrt(P1*P2);
h1 = 1404.6; 
h2 = 1574.3; 
h3 = 1443.5; 
h4 = 1628.1;
h5 = 371.7; h6 = h5; h7 = 181.5;
w = 30.; 			# capacity of plant

# Calculation
m2_dot = (3.89*30)/(h1-h7);
m1_dot = m2_dot*((h2-h7)/(h3-h6));
Wc_dot = m2_dot*(h2-h1)+m1_dot*(h4-h3);
COP = w*3.89/Wc_dot;
			# math.single stage
h1_ = 1404.6; h2_ = 1805.1;
h3_ = 371.1; h4_ = h3_;
m_dot = (3.89*30)/(h1_-h4_);
Wc = m_dot*(h2_-h1_);
COP_ = w*3.89/Wc;
IW = (Wc-Wc_dot)/Wc_dot;
ICOP = (COP-COP_)/COP_

# Results
print "Increase in work of compression is %.2f"%(IW*100)
print "Increase in COP for 2 stage compression is %.2f %%"%(ICOP*100)

# note : rounding error is there.

Increase in work of compression is 15.72
Increase in COP for 2 stage compression is 15.72 %

Example 14.6 Page No : 593

In [20]:
# Variables
tsat = 120.2+273
hfg = 2201.9;
T1 = 120.2+273;
T2 = 30.+273;
Tr = -10.+273;

# Calculation
COP_max = ((T1-T2)*Tr)/((T2-Tr)*T1);
ACOP = 0.4*COP_max;
Qe = (20.*14000)/3600; 			# in KW
Qg = Qe/ACOP;
x = 0.9; 
H = x*hfg;
SFR = Qg/H;

# Results
print "Steam flow rate required is %.4f kg/s"%SFR

Steam flow rate required is 0.0651 kg/s

Example 14.7 Page No : 594

In [4]:
# Variables
T1 = 277.; 
T3 = 273.+55;
rp = 3.; 			# Pressure ratio
g = 1.4; 
cp = 1.005;

# Calculation and Results
T2s = T1*(rp**((g-1)/g));
T2 = T1+(T2s-T1)/0.72
T4s = T3/(rp**((g-1)/g));
T34 = 0.78*(T3-T4s); 			# T3-T4
T4 = T3-T34;
COP = (T1-T4)/((T2-T1)-(T3-T4));
print "COP of the refrigerator is %.2f"%COP
P = (3.*14000)/(COP*3600)
print "Driving power required is %.1f kW"%P
m = (3.*14000)/(cp*(T1-T4));
print "Mass flow rate is %.2f kg/s"%(m/3600)

# note: rounding off error is there.

COP of the refrigerator is 0.25
Driving power required is 47.5 kW
Mass flow rate is 0.65 kg/s

Example 14.8 Page No : 596

In [26]:
# Variables
P1 = 2.4; T1 = 0+273;
h1 = 188.9; s1 = 0.7177; v1 = 0.0703;
P2 = 9; T2 = 60+273;
h2 = 219.37;
h2s = 213.27;
h3 = 71.93; h4 = h3;
v1 = 0.0703;
A1V1 = 0.6/60;

# Calculation
m_dot = A1V1/0.0703;
Wc_dot = m_dot*(h2-h1);
Q1_dot = m_dot*(h2-h3);
COP = Q1_dot/Wc_dot;
nis = (h2s-h1)/(h2-h1);

# Results
print "Power input is %.2f kW"%Wc_dot
print "Heating capacity is %.2f kW"%Q1_dot
print "COP is %.3f"%COP
print "The isentropic compressor efficiency is %.2f %%"%(nis*100)

Power input is 4.33 kW
Heating capacity is 20.97 kW
COP is 4.839
The isentropic compressor efficiency is 79.98 %

Example 14.9 Page No : 597

In [14]:
# Variables
T1 = 275.; 
T3 = 310.;
P1 = 1. ; P2 = 4.;

# Calculation
T2s = T1*(P2/P1);
nc = 0.8;
T2 = T1 + (T2s-T1)*nc;
pr = 0.1;
P3 = P2-0.1;
P4 = P1+0.08;
PR = P3/P4;

# Results
print "Pressure ratio for the turbine is %.3f"%PR
T4s = T3*(1./PR)**(0.286);
nt = 0.85;
T4 = T3-(T3-T4s)*nt;
COP = (T1-T4)/((442.26-T3)-(T1-T4));
print "COP is %.3f"%COP

# note : rounding off error is there.

Pressure ratio for the turbine is 3.611
COP is 0.533