# Chapter 14 : Refrigeration Cycles¶

### Example 14.1 page no : 717¶

In :
# Variables
T2 = 235; 			#K
P = 1.3; 			#kW

# Calculations and Results
COP = 14000./P/60./60.;
print ("(i) C.O.P. of Carnot refrigerator  = %.3f")% (COP)

T1 = T2/COP + T2;
t1 = T1-273;
print ("(ii) Higher temperature of the cycle  = %.3f")% (t1), ("0C")

print ("(iii) Heat delivered as heat pump")
Qabs = 14000./60; 			#Heat absorbed
W = P*60;
Q = Qabs+W;
print ("Q = %.3f")% (Q), ("kJ/min")

COP = Q/W;
print ("COP of heat pump  = %.3f")% (COP)

(i) C.O.P. of Carnot refrigerator  = 2.991
(ii) Higher temperature of the cycle  = 40.557 0C
(iii) Heat delivered as heat pump
Q = 311.333 kJ/min
COP of heat pump  = 3.991


### Example 14.2 page no : 718¶

In :
# Variables
T1 = 308.; 		    	    #K
T2 = 258.; 			        #K
capacity = 12.; 			#tonne

# Calculations and Results
COP = T2/(T1-T2);
print ("(i) Co-efficient of performance  = "), (COP)

print ("(ii) Heat rejected from the system per hour")
W = capacity*14000./5.16;
Q = capacity*14000.+W;
print ("Q = %.3f")% (Q), ("kJ/h")

P = W/60./60.;
print ("(iii) Power required  = %.3f")% (P),("kW")

(i) Co-efficient of performance  =  5.16
(ii) Heat rejected from the system per hour
Q = 200558.140 kJ/h
(iii) Power required  = 9.044 kW


### Example 14.3 page no : 718¶

In :
# Variables
T2 = 268.; 			#K
T1 = 308.; 			#K
Q = 29.; 			#Heat leakage from the surroundings into the cold storage in kW

# Calculations
COP_ideal = T2/(T1-T2);
COP_actual = 1./3*COP_ideal;
W = Q/COP_actual;

# Results
print ("Power required  = %.3f")%(W), ("kW")

Power required  = 12.985 kW


### Example 14.4 page no : 719¶

In :
# Variables
T1 = 293.; 			#K
T2 = 265.; 			#K
T0 = 273.; 			#K
L = 335.; 			#Latent heat of ice in kJ/kg
cpw = 4.18;

# Calculations
COP = T2/(T1-T2);
Rn = cpw*(T1-T0)+L;
m_ice = COP*3600./Rn;

# Results
print ("ice formed per kWh  = %.3f")% (m_ice), ("kg")

ice formed per kWh  = 81.394 kg


### Example 14.5 page no : 719¶

In :
# Variables
T1 = 291.; 			#K
T2 = 265.; 			#K
T0 = 273.; 			#K
cpw = 4.18; 		#kJ/kg
cpi = 2.09; 		#kJ/kg
L = 334.; 			#kJ/kg
m = 400.; 			#kg

# Calculations
COP = T2/(T1-T2);
Rn = cpw*(T1-T0) + L + cpi*(T0-T2);
W = Rn*m/COP/3600; 			#kJ/s

# Results
print ("Least power  = %.3f")% (W), ("kW")

Least power  = 4.644 kW


### Example 14.6 page no : 720¶

In :
# Variables
cpw = 4.18; 			#kJ/kg

print ("(i) Quantity of ice produced")
t = 20.; 			#0C
L = 335.; 			#kJ/kg
capacity = 280.; 	#tonnes

# Calculations and Results
Q1 = cpw*t + L; 			#Heat to be extracted per kg of water (to form ice at 0°C)
Rn = capacity*14000; 		#kJ/h
m_ice = Rn*24./Q1/1000;

print ("Quantity of ice produced in 24 hours  = %.3f")% (m_ice), ("tonnes")

print ("(ii) Minimum power required  = "),
T1 = 298.; 			#K
T2 = 263.; 			#K

COP = T2/(T1-T2);
W = Rn/COP/3600.; 			#kJ/s
print ("Power required  = %.3f")% (W), ("kW")

(i) Quantity of ice produced
Quantity of ice produced in 24 hours  = 224.749 tonnes
(ii) Minimum power required  =  Power required  = 144.909 kW


### Example 14.7 page no : 720¶

In :
# Variables
cp1 = 1.25; 			#kJ/kg 0C
cp2 = 2.93; 			#kJ/kg 0C
L = 232.     			#kJ/kg
T1 = -3. 	    		#0C
T2 = -8. 		    	#0C
T3 = 25. 			    #0C

# Calculations and Results
Q1 = cp2*(T3-T1) + L + cp1*(T1-T2); 			#Heat removed in 8 hours from each kg of fish
Q = Q1*20*1000./8                    			#Heat removed by the plant /min

capacity = Q/14000.; 			#tonnes
print ("(i) Capacity of the refrigerating plant  = %.3f")% (capacity), ("tonnes")

print ("(ii) Carnot cycle C.O.P. between this temperature range.")
T1 = 298.; 			#K
T2 = 265.; 			#K

COP = T2/(T1-T2);
print ("COP of reversed carnot cycle  = %.3f")% (COP)

print ("(iii) Power required")
COP_actual = 1./3*COP;
W = Q/COP_actual/3600; 			#kJ/s

print ("Power  = %.3f")% (W), ("kW")

(i) Capacity of the refrigerating plant  = 57.195 tonnes
(ii) Carnot cycle C.O.P. between this temperature range.
COP of reversed carnot cycle  = 8.030
(iii) Power required
Power  = 83.094 kW


### Example 14.8 page no : 721¶

In :
# Variables
T1 = 1273.; 		#K
T2 = 298.; 			#K
T3 = 268.; 			#K
T4 = 298.; 			#K

# Calculations
#Let Q2/Q1 = r1, r2 = Q3/Q4;
r1 = 298./1273; 		#Q2/Q1
r2 = 268./298; 			#Q3/Q4

#Let Q4/Q1 = r
r = (1.-r1)/(1-r2);

# Results
print ("The ratio in which the heat pump and heat engine share the heating load  = %.3f")% (r)

The ratio in which the heat pump and heat engine share the heating load  = 7.608


### Example 14.9 page no : 724¶

In :
# Variables
y = 1.4;
n = 1.35;
cp = 1.003; 		#kJ/kg K
p2 = 1.; 			#bar
p1 = 8.; 			#bar
T3 = 282.; 			#K
T4 = 302.; 			#K
T1 = T4;

# Calculations
T4 = T3*(p1/p2)**((n-1)/n);
T2 = T1*(p2/p1)**((n-1)/n);
Q1 = cp*(T3-T2); 			#Heat extracted from cold chamber per kg of air
Q2 = cp*(T4-T1); 			#Heat rejected in the cooling chamber per kg of air
cv = cp/y;
R = cp-cv;
W = n/(n-1)*R*((T4-T3) - (T1-T2));
COP = Q1/W;

# Results
print ("COP = %.3f")% (COP)

COP = 1.270


### Example 14.10 page no : 726¶

In :
# Variables
p1 = 1000.; 			#kPa
p2 = 100.    			#kPa
p4 = p1;
p3 = p2;
E = 2000.    			# Refrigerating effect produced in kJ/min
T3 = 268.; 	    		#K
T1 = 303.; 		    	#K
y = 1.4;

# Calculations and Results
print ("(i) Mass of air circulated per minute")
T2 = T1*(p2/p1)**((y-1)/y);
e = cp*(T3-T2); 			#Refrigerating effect per kg; kJ/kg

m = E/e;
print ("m = %.3f")% (m), ("kg/min")

print ("(ii) Compressor work (Wcomp.), expander work (Wexp.) and cycle work (Wcycle)")
T4 = T3*(p4/p3)**((y-1)/y);

Wcomp = y/(y-1)*m*R*(T4-T3);
print ("Compressor work  = %.3f")% (Wcomp), ("kJ/min")

Wexp = y/(y-1)*m*R*(T1-T2);
print ("Expander work  = %.3f")% (Wexp), ("kJ/min")

W_cycle = Wcomp-Wexp;
print ("Wcycle = %.3f")% (W_cycle), ("kJ/min")

print ("(iii) C.O.P. and power required")
COP = E/W_cycle;
print ("COP  = %.3f")% (COP)

P = W_cycle/60;
print ("Power required  = %.3f")% (P), ("kW")

(i) Mass of air circulated per minute
m = 17.954 kg/min
(ii) Compressor work (Wcomp.), expander work (Wexp.) and cycle work (Wcycle)
Compressor work  = 4491.675 kJ/min
Expander work  = 2630.279 kJ/min
Wcycle = 1861.395 kJ/min
(iii) C.O.P. and power required
COP  = 1.074
Power required  = 31.023 kW


### Example 14.11 page no : 727¶

In :
import math

# Variables
y = 1.4;
cp = 1.003; 		#kJ/kg K
T3 = 289.; 			#K
T1 = 314.; 			#K
p1 = 5.2; 			#bar
p2 = 1.; 			#bar
capacity = 6.; 		#tonnes
R = 287.; 			#J/kg K
l = 0.2; 			#m

# Calculations and Results
T4 = T3*(p1/p2)**((y-1)/y);
T2 = T1*(p2/p1)**((y-1)/y);

COP = T2/(T1-T2);
print ("(i) C.O.P.  = %.3f")%(COP)

e = cp*(T3-T2); 			#Refrigerating effect per kg of air
E = capacity*14000; 		#Refrigerating effect produced by the refrigerating machine in kJ/h

m = E/e/60;
print ("(ii)mass of air in circulation  = %.3f")%(m),("kg/min")

V3 = m*R*T3/p2/10**5;
print ("(iii)Piston displacement of compressor"),(" = %.2f")%(V3),("m^3/min")

V_swept = V3/2/240;

V2 = m*R*T2/p2/10**5;
V_swept = V2/2/240;
print ("Piston displacement of expander"),(" = %.3f")%(V2),("m^3/min")

d_c = math.sqrt(V_swept/l/math.pi*4);
print ("(iv)Diameter or bore of the expander cylinder  = %.0f")%(d_c*1000), ("mm")

d_c = math.sqrt(V_swept/l/math.pi*4);
print ("Diameter or bore of the compressor cylinder  = %.0f")%(d_c*1000),("mm")

W = capacity*14000/COP/3600;
print ("(v) Power required to drive the unit"),(" = %.3f")%(W),("kW")

# Answers are slightly different because of rounding error.

(i) C.O.P.  = 1.662
(ii)mass of air in circulation  = 15.016 kg/min
(iii)Piston displacement of compressor  = 12.45 m^3/min
Piston displacement of expander  = 8.449 m^3/min
(iv)Diameter or bore of the expander cylinder  = 335 mm
Diameter or bore of the compressor cylinder  = 335 mm
(v) Power required to drive the unit  = 14.039 kW


### Example 14.12 page no : 744¶

In :
# Variables
m = 6.; 			#kg/min
n_relative = 0.50;
cpw = 4.187; 			#kJ/kg K
L = 335.; 			#kJ/kg

h_f2 = 31.4; 			#kJ/kg
h_fg2 = 154.; 			#kJ/kg
h_f3 = 59.7; 			#kJ/kg
h_fg3 = 138.; 			#kJ/kg
h_f4 = 59.7; 			#kJ/kg
x2 = 0.6;
s_f3 = 0.2232; 			#kJ/kg K
s_f2 = 0.1251; 			#kJ/kg K
T2 = 268.; 			#K
T3 = 298.; 			#K

# Calculations
h2 = h_f2+x2*h_fg2;
x3 = ((s_f2-s_f3)+x2*(h_fg2/T2))*T3/h_fg3;
h3 = h_f3+x3*h_fg3;
h1 = h_f4;
COP_th = (h2-h1)/(h3-h2); 			#Theoritical COP
COP = n_relative*COP_th;
Q = cpw*(20-0) + L; 			#Heat extracted from 1 kg of water at 20°C for the formation of 1 kg of ice at 0°C
m_ice = COP*m*(h3-h2)/Q*60*24/1000; 			#in 24 hours

# Results
print ("m_ice = %.3f")%(m_ice), ("tonnes")

m_ice = 0.661 tonnes


### Example 14.13 page no : 745¶

In :
# Variables
L = 335.; 			    #kJ/kg
h3 = 1319.22; 			#kJ/kg
h1 = 100.04; 			#kJ/kg
h4 = h1;
s_f2 = -2.1338; 		#kJ/kg K
s_g2 = 5.0585; 			#kJ/kg K
s_g3 = 4.4852; 			#kJ/kg K
h_f2 = -54.56; 			#kJ/kg
h_g2 = 1304.99; 		#kJ/kg

# Calculations
x2 = (s_g3-s_f2)/(s_g2-s_f2);
h2 = h_f2+x2*(h_g2-h_f2);
COP_theoritical = (h2-h1)/(h3-h2);
COP_actual = 0.62*COP_theoritical;
RE = COP_actual*(h3-h2); 			#Actual refrigerating effect per kg
Q = 28.*1000.*L/24./3600; 			#Heat to be extracted per second
m = Q/RE; 			#Mass of refrigerant circulated per second
W = m*(h3-h2);

# Results
print ("Power required  = %.3f")%(W), ("kW")

Power required  = 19.577 kW


### Example 14.14 page no : 747¶

In :
# Variables
h_f2 = 158.2; 			#kJ/kg
x2 = 0.62;
h_fg2 = 1280.8;
h1 = 298.9; 			#kJ/kg
h_f4 = h1;
s_f2 = 0.630; 			#kJ/kg K
T2 = 268.; 			    #K
T3 = 298.; 			    #K
s_f3 = 1.124; 			#kJ/kg K
h_fg3 = 1167.1; 		#kJ/kg
m = 6.4; 			    #kg/min
cp = 4.187;
L = 335.; 			    #kJ/kg
h_f3 = 298.9; 			#kJ/kg

# Calculations
h2 = h_f2+x2*h_fg2;
x3 = ((s_f2-s_f3)+x2*h_fg2/T2)/h_fg3*T3;
h3 = h_f3+x3*h_fg3;
COP_theoritical = (h2-h1)/(h3-h2);
COP_actual = 0.55*COP_theoritical;
W1 = 81.87 #h3-h2; 			#Work done per kg of refrigerant

W = m*W1/60; 			#Work done per second kJ/s
Q = round(15*cp+L,1);
m_ice = W*3600.*24/Q;

# Results
print ("Amount of ice formed in 24 hours  = %.3f")% (m_ice), ("kg")

# Note : Answer is slightly different because of rounding error.

Amount of ice formed in 24 hours  = 1896.717 kg


### Example 14.15 page no : 748¶

In :
# Variables
RE = 5*14000./3600; 	#Total refrigeration produced in kg/s
h2 = 183.19; 			#kJ/kg
h3 = 209.41; 			#kJ/kg
h4 = 74.59; 			#kJ/kg
h1 = h4;

# Calculations and Results
RE_net = h2-h1; 			#Net refrigerating effect produced per kg
m = RE/RE_net;
print ("(i) The refrigerant flow rate"),(" = %.3f")% (m), ("kg/s")

COP = (h2-h1)/(h3-h2);
print ("(ii) The C.O.P.  = %.3f")% (COP)

P = m*(h3-h2);
print ("(iii) The power required to drive the compressor  = %.3f")%(P), ("kW")

rate = m*(h3-h4);
print ("(iv) The rate of heat rejection to the condenser  = %.3f")%(rate),("kW")

(i) The refrigerant flow rate  = 0.179 kg/s
(ii) The C.O.P.  = 4.142
(iii) The power required to drive the compressor  = 4.695 kW
(iv) The rate of heat rejection to the condenser  = 24.139 kW


### Example 14.16 page no : 749¶

In :
print ("(iii)")

# Variables
h2 = 344.927; 			#kJ/kg
h4 = 228.538; 			#kJ/kg
h1 = h4;
cpv = 0.611; 			#/kJ/kg0C
# s2 = s3
t3 = 39.995; 			#0C

# Calculations
h3 = 363.575+cpv*(t3-30);
Rn = h2-h1;
W = h3-h2;
COP = Rn/W;

# Results
print ("COP  = %.3f")% (COP)

cp = 2.0935; 			#kJ/kg 0C
Q = 2400./24./3600*(4.187*(15-0)+335+cp*(0-(-5)))

W = Q/COP;
print ("Work required  = %.3f")% (W), ("kW")

(iii)
COP  = 4.702
Work required  = 2.412 kW


### Example 14.17 page no : 751¶

In :
# Variables
h2 = 352.; 			#kJ/kg
h3 = 374.; 			#kJ/kg
h4 = 221.; 			#kJ/kg
h1 = h4;
v2 = 0.08; 			#m**3/kg
rpm = 500.;
D = 0.2;
L = 0.15;
n_vol = 0.85;

# Calculations
RE = h2-h1;
V = math.pi/4*D**2*L*rpm*2*n_vol;
m = V/v2;

# Results
print ("(ii)Mass of refrigerant circulated per minute  =  %.3f")% (m), ("kg/min")

cc = 50.*(h2-h1)*60./14000.;
print ("(iii) Cooling capacity in tonnes of refrigeration  = %.3f")%(cc), ("TR")

COP = (h2-h1)/(h3-h2);
print ("(iv)COP  = %.3f")% (COP)

(ii)Mass of refrigerant circulated per minute  =  50.069 kg/min
(iii) Cooling capacity in tonnes of refrigeration  = 28.071 TR
(iv)COP  = 5.955


### Example 14.18 page no : 751¶

In :
# Variables
te = -10.; 			#0C
tc = 40.; 			#0C
h3 = 220.; 			#kJ/kg
h2 = 183.1; 			#kJ/kg
h1 = 74.53; 			#kJ/kg
h_f4 = 26.85; 			#kJ/kg
m = 1.;      			#kg

# Calculations and Results
COP = (h2-h1)/(h3-h2);
print ("(i) The C.O.P. the cycle  = %.3f")%(COP)

RC = m*(h2-h1);
print ("(ii) Refrigerating capacity  = %.3f")%(RC),("kJ/min")

CP = m*(h3-h2)/60;
print ("Compressor power  = %.3f")% (CP), ("kJ/s")

(i) The C.O.P. the cycle  = 2.942
(ii) Refrigerating capacity  = 108.570 kJ/min
Compressor power  = 0.615 kJ/s


### Example 14.19 page no : 752¶

In :
import math

# Variables
h2 = 178.61; 			#kJ/kg
h3a = 203.05; 			#kJ/kg
h_f4 = 74.53; 			#kJ/kg
h1 = h_f4;
s3a = 0.682; 			#kJ/kg K
s2 = 0.7082; 			#kJ/kg K
cp = 0.747; 			#kJ/kg K
T3a = 313.; 			#K
CE = 20.; 			    #Cooling effect
C = 0.03;
v_g = 0.1088;
p_d = 9.607;
p_s = 1.509;
n = 1.13;

# Calculations
m = CE/(h2-h1);
T3 = T3a*math.e**((s2-s3a)/cp)
h3 = h3a+cp*(T3-T3a);
P = m*(h3-h2);

# Results
print ("Power required by the machine  = %.3f")%(P), ("kW")

n_vol = 1+C-C*(p_d/p_s)**(1./n); 			#Volumetric efficiency
V1 = m*v_g; 			                    #volume of refrigerant at the intake conditions
V_swept = V1/n_vol;

V = V_swept*60./300;
print ("Piston print lacement  = %.5f")% (V), ("m**3")

Power required by the machine  = 6.300 kW
Piston print lacement  = 0.00478 m**3


### Example 14.20 page no : 754¶

In :
import math

# Variables
h2 = 1450.22; 			#kJ/kg
h3a = 1488.57; 			#kJ/kg
h_f4 = 366.072; 		#kJ/kg
cpl2 = 4.556; 			#kJ/kg K
cpv1 = 2.492; 			#kJ/kg K
cpv2 = 2.903; 			#kJ/kg K
T1 = 303.; 			    #K
T2 = 308.; 			    #K
s3a = 5.2086; 			#kJ/kg K
s2 = 5.755; 			#kJ/kg K
T3a = 308.; 			#K
N = 1000.;

# Calculations
h_f4a = h_f4-cpl2*(T2-T1);
h1 = h_f4a;
T3 = T3a*math.e**((s2-s3a)/cpv2);
h3 = h3a+cpv2*(T3-T3a);
m = 50./(h2-h1);

# Results
P = m*(h3-h2);
print ("(i) Power required  = %.3f")%(P), ("kW")

print ("(ii) Cylinder dimensions ")
D = (m*4*60/math.pi/1.2/N/0.417477)**(1./3);
print ("Diameter of cylinder  = %.3f")% (D), ("m")

L = 1.2*D;
print ("Length of the cylinder = %.3f")% (L), ("m")

(i) Power required  = 10.096 kW
(ii) Cylinder dimensions
Diameter of cylinder  = 0.190 m
Length of the cylinder = 0.228 m


### Example 14.21 page no : 756¶

In :
# Variables
cooling_load = 150.; #W
n_vol = 0.8;
N = 720.; 			#rpm
h2 = 183.; 			#kJ/kg
h1 = 74.5; 			#kJ/kg
v2 = 0.08; 			#m**3/kg

# Calculations
m = cooling_load/(108.5*1000);
d = m*v2/n_vol;

# Results
print ("Mass flow rate of the refrigerant  = %.6f")% (m),("kJ/s")

print ("Displacement volume of the compressor  = %6f")% (d), ("m**3/s")

Mass flow rate of the refrigerant  = 0.001382 kJ/s
Displacement volume of the compressor  = 0.000138 m**3/s


### Example 14.22 page no : 757¶

In :
# Variables
h2 = 183.2; 			#kJ/kg
h3 = 222.6; 			#kJ/kg
h4 = 84.9    			#kJ/kg
v2 = 0.0767; 			#m**3/kg
v3 = 0.0164; 			#m**3/kg
v4 = 0.00083; 			#m**3/kg

# Calculations
V = 1.5*1000*10**(-6); 			#Piston print lacement volume m**3/revolution
n_vol = 0.80;

print ("(i) Power rating of the compressor (kW)")
discharge = V*1600*n_vol; 			#Compressor discharge
m = discharge/v2;

P = m/60*(h3-h2); 			#kW
print ("Power  = %.3f")% (P), ("kW")

RE = m/60*(h2-h4);
print ("(ii) Refrigerating effect  = %.3f")% (RE), ("kW")

(i) Power rating of the compressor (kW)
Power  = 16.438 kW
(ii) Refrigerating effect  = 41.012 kW


### Example 14.23 page no : 757¶

In :
# Variables
COP = 6.5;
W = 50.; 			#kW
h3a = 201.45; 			#kJ/kg
h_f4 = 69.55; 			#kJ/kg
h1 = h_f4;
h2 = 187.53; 			#kJ/kg
cp = 0.6155; 			#kJ/kg
t3a = 35.; 			#0C

# Calculations
RC = W*COP; 			#Refrigerating capacity
Q1 = h2-h_f4; 			#Heat extracted per kg of refrigerant
rate = RC/Q1; 			#Refrigerant flow rate
Q2 = W/rate; 			#Heat input per kg
h = h2+Q2; 			#Enthalpy of vapour after compression
Q = h-h3a; 			#Superheat

t3 = Q/cp+t3a;

# Results
print ("The refrigerant temperature = %.3f")% (t3), ("°C")

The refrigerant temperature = 41.874 °C


### Example 14.24 page no : 758¶

In :
import math

# Variables
Q1 = 500.; 			#total heating requirement of 500 kJ/min
n_compressor = 0.8;
s1 = 0.7035; 			#kJ/kg K
s2 = 0.6799; 			#kJ/kg K
T2 = 322.31; 			#K
cp = 0.7; 			    #kJ/kg K
h_v2 = 206.24; 			#kJ/kg
h_l2 = 84.21; 			#kJ/kg
h_v1 = 182.07 			#kJ/kg

# Calculations
Q2 = Q1/n_compressor; 			#Heat rejected by the cycle

#Entropy of dry saturated vapour at 2 bar =  Entropy of superheated vapour at 12 bar
T = T2*math.e**((s1-s2)/cp);

H = h_v2+cp*(T-T2); 			#Enthalpy of superheated vapour at 12 bar
Q3 = H-h_l2; 			#Heat rejected per cycle
m = Q2/Q3; 			#kg/min
W = m*(H-h_v1)/60; 			#kW
W_actual = W/n_compressor;

# Results
print ("Power  = %.3f")% (W_actual), ("kW")

Power  = 3.201 kW


### Example 14.25 page no : 759¶

In :
# Variables
h2a = 183.2; 			#kJ/kg K
cpv = 0.733; 			#Vapour specific heat in kJ/kg K
cpl = 1.235; 			#Liquid specific heat in kJ/kg K
s2a = 0.7020; 			#Entropy of vapour in kJ/kg K
s3a = 0.6854; 			#Entropy of vapour in kJ/kg K
T2 = 270.; 			    #K
T2a = 263.; 			#K
T3a = 303.; 			#K
h3a = 199.6; 			#kJ/kg
h_f4 = 64.6; 			#kJ/kg
dT4 = 6.; 			    #dT4 = T4-T4a
v2a = 0.0767;
n = 2.; 			    #number of cylinder

# Calculations and Results
h2 = h2a+cpv*(T2-T2a);
s2 = s2a+cpv*math.log(T2/T2a);
T3 = T3a*math.e**((s2-s3a)/cpv);
h3 = h3a+cpv*(T3-T3a);
h_f4a = h_f4-cpl*dT4;
h1 = h_f4a;
v2 = v2a/T2a*T2;

RE = h2-h1;
print ("(i) Refrigerating effect per kg  = "), (RE), ("kJ/kg")

m = 2400/RE;
print ("(ii) Mass of refrigerant to be circulated per minute  = %.3f")% (m), ("kg/min")

v = m*v2;
print ("(iii) Theoretical piston print lacement per minute  = %.3f")%(v), ("m**3/min")

P = m/60*(h3-h2);
print ("(iv) Theoretical power required to run the compressor  =  %.3f")% (P), ("kW")

Q = m*(h3-h_f4a);
print ("(v) Heat removed through the condenser per min  = %.3f")% (Q), ("kJ/min")

print ("(vi) Theoretical bore (d) and stroke (l)")
d = (v/n/math.pi*4/1.25/1000)**(1./3)*1000;
print ("Theroritical bore  = %.3f")% (d), ("mm")

l = 1.25*d;
print ("stroke  = %.3f")% (l), ("mm")

(i) Refrigerating effect per kg  =  131.141 kJ/kg
(ii) Mass of refrigerant to be circulated per minute  = 18.301 kg/min
(iii) Theoretical piston print lacement per minute  = 1.441 m**3/min
(iv) Theoretical power required to run the compressor  =  6.833 kW
(v) Heat removed through the condenser per min  = 2809.995 kJ/min
(vi) Theoretical bore (d) and stroke (l)
Theroritical bore  = 90.202 mm
stroke  = 112.752 mm


### Example 14.26 page no : 761¶

In :
# Variables
h2 = 1597.; 			#kJ/kg
h3 = 1790.; 			#kJ/kg
h4 = 513.; 			#kJ/kg
h1 = h4;
t3 = 58.; 			#0C
x1 = 0.13;
tc = 27.; 			#0C
capacity = 10.5; 			#tonnes

# Calculations and Results
t = t3-tc;
print ("(i) Condition of the vapour at the outlet of the compressor  = "), (t), ("C")

print ("(ii) Condition of vapour at entrance to evaporator  = "), (x1)

COP = (h2-h1)/(h3-h2);
print ("(iii)COP  = %.3f") %(COP)

P = capacity*14000./COP/3600;
print ("(iv) Power required  = %.3f")% (P), ("kW")

(i) Condition of the vapour at the outlet of the compressor  =  31.0 C
(ii) Condition of vapour at entrance to evaporator  =  0.13
(iii)COP  = 5.617
(iv) Power required  = 7.270 kW


### Example 14.27 page no : 762¶

In :
import math

# Variables
h2 = 615.; 			#kJ/kg
h3 = 664.; 			#kJ/kg
h4 = 446.; 			#kJ/kg
h1 = h4;
v2 = 0.14; 			#m**3/kg
capacity = 20.; 	#tonnes
n = 6.; 			#number of cylinder

# Calculations and Results
RE = h2-h1;
print ("(i) Refrigerating effect per kg  = "), (RE), ("kJ/kg")

m = capacity*14000./RE/60.;
print ("(ii) Mass of refrigerant to be circulated per minute  = %.3f")% (m), ("kg/min")

v = v2*m;
print ("(iii) Theoretical piston print lacement  = %.3f")% (v), ("m**3/min")

P = m/60*(h3-h2);
print ("(iv) Theoretical power  = %.3f")% (P), ("kW")

COP = (h2-h1)/(h3-h2);
print ("(v)COP  = %.3f")% (COP)

Q = m*(h3-h4);
print ("(vi) Heat removed through the condenser  = %.3f")% (Q), ("kJ/min")

print ("(vii) Theoretical print lacement per minute per cylinder")

d = (v/n*4/math.pi/950.)**(1./3)*1000.;
print ("Diameter of cylinder  = %.3f")% (d), ("mm")

l = d;
print ("Stroke length  = %.3f")% (l), ("mm")

(i) Refrigerating effect per kg  =  169.0 kJ/kg
(ii) Mass of refrigerant to be circulated per minute  = 27.613 kg/min
(iii) Theoretical piston print lacement  = 3.866 m**3/min
(iv) Theoretical power  = 22.551 kW
(v)COP  = 3.449
(vi) Heat removed through the condenser  = 6019.724 kJ/min
(vii) Theoretical print lacement per minute per cylinder
Diameter of cylinder  = 95.227 mm
Stroke length  = 95.227 mm