Chapter 10 : Refrigeration

Example 10.1 Page No : 503

In [2]:
T1 = 70+460; 			#70F = 70+460 R 			#Energy flows into the system at reservoir at constant temperature T1(unit:R)
T2 = 32+460; 			#32F = 32+460 R 			#Heat is rejected to the constant temperature T2(Unit:R)
print "Solution for a"
COP = T2/(T1-T2); 			#Coefficient of performance
print "Coefficient of performanceCOP) of the cycle is %.2f"%(COP);
print "Solution for b"
Qremoved = 1000; 			#Unit:Btu/min 			#heat removal
WbyJ = Qremoved/COP; 			#The power required 			#Unit:Btu/min
print "The power required is %.2f Btu/min"%(WbyJ);
print "Solution for c"
Qrej = Qremoved+WbyJ; 			#The rate of heat rejected to the room  			#Unit:Btu/min
print "The rate of heat rejected to the room is %.2f Btu/min"%(Qrej);

Solution for a
Coefficient of performanceCOP) of the cycle is 12.00
Solution for b
The power required is 83.00 Btu/min
Solution for c
The rate of heat rejected to the room is 1083.00 Btu/min

Example 10.2 Page No : 504

In [4]:
T1 = 20+273; 			#20C = 20+273 R 			#Energy flows into the system at reservoir at constant temperature T1(unit:R)
T2 = -5+273; 			#-5C = -5+273 R 			#Heat is rejected to the constant temperature T2(Unit:R)
print "Solution for a"
COP = T2/(T1-T2); 			#Coefficient of performance
print "Coefficient of performanceCOP) of the cycle is %.2f"%(COP);
print "Solution for b"
Qremoved = 30; 			#Unit:kW 			#heat removal 
W = Qremoved/COP; 			#power required 			#unit:kW
print "The power required is %.2f kW "%(W);
print "Solution for c"
Qrej = Qremoved+W; 			#The rate of heat rejected to the room  			#Unit:kW
print "The rate of heat rejected to the room is %.2f kW"%(Qrej);

Solution for a
Coefficient of performanceCOP) of the cycle is 10.00
Solution for b
The power required is 3.00 kW 
Solution for c
The rate of heat rejected to the room is 33.00 kW

Example 10.3 Page No : 505

In [6]:
T1 = 70+460; 			#70F = 70+460 R 			#Energy flows into the system at reservoir at constant temperature T1(unit:R)
T2 = 20+460; 			#20F = 20+460 R 			#Heat is rejected to the constant temperature T2(Unit:R)
print "Solution for a"
COP = T2/(T1-T2); 			#Coefficient of performance
print "Coefficient of performanceCOP) of the cycle is %.2f"%(COP);
print "Solution for b"
HPperTOR = 4.717/COP; 			#Horsepower per ton of refrigeration 			#Unit:hp/ton
COPactual = 2; 			#Actual Coefficient of performance(COP) is stated to be 2
HPperTORactual = 4.717/COPactual; 			#Horsepower per ton of refrigeration(actual) 			#Unit:hp/ton
print "The horsepower required by the actual cycle over the minimum is %.2f hp/ton"%(HPperTORactual-HPperTOR);

Solution for a
Coefficient of performanceCOP) of the cycle is 9.00
Solution for b
The horsepower required by the actual cycle over the minimum is 1.83 hp/ton

Example 10.4 Page No : 506

In [1]:
from numpy.linalg import inv

COP = 4.5; 			#Coefficient of performance 			#From problem 10.1
HPperTOR = 4.717/COP; 			#Horsepower per ton of refrigeration 			#Unit:hp/ton
Qremoved = 1000; 			#Unit:Btu/min 			#From problem 10.1
#1000 Btu/min /200 Btu/min ton  =  5 tons of refrigeration
HPrequired = HPperTOR*5; 			#The horsepower required 			#unit:hp
print "The horsepower required is %.2f hp"%(HPrequired);
#In problem 10.1, 77.2 Btu/min was required
print "The power required is %.2f hp"%(77.2*778*inv([[33000]])); 			#1 Btu = 778 ft*lbf 			#1 min*hp  =  33000 ft*lbf
#The ratio of the power required in each problem is the same as the inverse ratio of the COP value
#Therefore,
print "The power required is %.2f hp"%((COP/12.95)*HPrequired); 			#COPin problem 10.1) = 12.95
print "This checks our results"

The horsepower required is 5.24 hp
The power required is 1.82 hp
The power required is 1.82 hp
This checks our results

Example 10.5 Page No : 506

In [8]:
# given data
COP = 10.72; 			#In the problem 10.2 			#Coefficient of performance
P = 2.8; 			#In the problem 10.2 			#The power was 2.8 kW
COPactual = 3.8; 			#Actual Coefficient of performance(COP)

# calculation
power = P*COP/COPactual; 			#The power required  			#unit:kW

# results
print "The power required is %.2f kW"%(power)

The power required is 7.90 kW

Example 10.6 Page No : 509

In [10]:
#From Appendix 3,at 120psia,the corresponding saturation temperature is 66 F, enthalpies are
h1 = 116.0; 			#Unit:Btu/lbm 			#enthalpy
h2 = 116.0; 			#Unit:Btu/lbm 			#Throttling gives h1 = h2 			#enthalpy
h3 = 602.4; 			#Unit:Btu/lbm 			#enthalpy
#From the consideration that s3 = s4,h4 is found at 15 psia,
s3 = 1.3938; 			#s = entropy 			#Unit:Btu/(lbm*F)
#Therefore by interpolation in the superheat tables at 120 psia,
t4 = 237.4; 			#Unit:fahrenheit 			#temperature
h4 = 733.4; 			#Unit:Btu/lbm 			#enthalpy
print "Solution for a"
COP = (h3-h1)/(h4-h3); 			#Coefficient of performance
print "Coefficient of performance is %.2f"%(COP);
print "Solution for b"
print "The work of compression is %.2f Btu/lbm"%(h4-h3);
print "Solution for c"
print "The refrigatering effect is %.2f Btu/lbm"%(h3-h1);
print "Solution for d"
tons = 30; 			#capacity of 30 tons is desired
print "The pounds per minute of ammonia required for ciculation is %.2f lbm/min"%((200*tons)/(h3-h1));
print "Solution for e"
print "The ideal horsepower per ton of refrigeration is %.2f hp/ton"%((4.717*h4-h3)/(h3-h1));

Solution for a
Coefficient of performance is 3.71
Solution for b
The work of compression is 131.00 Btu/lbm
Solution for c
The refrigatering effect is 486.40 Btu/lbm
Solution for d
The pounds per minute of ammonia required for ciculation is 12.34 lbm/min
Solution for e
The ideal horsepower per ton of refrigeration is 5.87 hp/ton

Example 10.7 Page No : 510

In [12]:
#From Appendix 3,110 psig corresponds to 96 F, enthalpies are
h1 = 30.14; 			#Unit:Btu/lbm 			#enthalpy
h2 = 30.14; 			#Unit:Btu/lbm 			#Throttling gives h1 = h2 			#enthalpy
h3 = 75.110; 			#Unit:Btu/lbm 			#enthalpy
#From the consideration that s3 = s4,at -20F,
s3 = 0.17102; 			#Unit:Btu/(lbm*F) 			#s = entropy
#Therefore by interpolation in the Freon-12 superheat table at these values,
h4 = 89.293; 			#Unit:Btu/lbm 			#enthalpy
print "Solution for a"
COP = (h3-h1)/(h4-h3); 			#Coefficient of performance
print "Coefficient of performance is %.2f"%(COP);
print "Solution for b"
print "The work of compression is %.2f Btu/lbm"%(h4-h3);
print "Solution for c"
print "The refrigatering effect is %.2f Btu/lbm"%(h3-h1);
print "Solution for d";
tons = 30; 			#capacity of 30 tons is desired
print "The pounds per minute of ammonia required for ciculation is %.2f lbm/min"%((200*tons)/h3-h1);
print "Solution for e"
print "The ideal horsepower per ton of refrigeration is %.2f hp/ton"%((4.717*h4-h3)/(h3-h1))

Solution for a
Coefficient of performance is 3.17
Solution for b
The work of compression is 14.18 Btu/lbm
Solution for c
The refrigatering effect is 44.97 Btu/lbm
Solution for d
The pounds per minute of ammonia required for ciculation is 49.74 lbm/min
Solution for e
The ideal horsepower per ton of refrigeration is 7.70 hp/ton

Example 10.8 Page No : 517

In [13]:
#From Appendix 3,Using the Freon-12 tables, enthalpies are
h1 = 28.713; 			#Unit:Btu/lbm 			#enthalpy
h2 = 28.713; 			#Unit:Btu/lbm 			#Throttling gives h1 = h2 			#enthalpy
h3 = 78.335; 			#Unit:Btu/lbm 			#enthalpy
#From the consideration that s3 = s4,
s3 = 0.16798; 			#Unit:Btu/(lbm*F) 			#s = entropy
#Therefore by interpolation in the superheat tables at 90 F,
s = 0.16798; 			#entropy at 90F 			#Btu/lbm*F
h4 = 87.192; 			#Unit:Btu/lbm 			#enthalpy
print "The heat extracted is %.2f Btu/lbm"%(h3-h1);
print "The work required is %.2f Btu/lbm"%(h4-h3);
COP = (h3-h1)/(h4-h3); 			#Coefficient of performance
print "The Coefficient of performanceCOP) of this ideal cycle is %.2f"%(COP);

The heat extracted is 49.62 Btu/lbm
The work required is 8.86 Btu/lbm
The Coefficient of performanceCOP) of this ideal cycle is 5.60

Example 10.9 Page No : 518

In [14]:
#From Appendix 3,Using the HFC-134a tables, enthalpies are
h1 = 41.6; 			#Unit:Btu/lbm 			#enthalpy
h2 = 41.6; 			#Unit:Btu/lbm 			#Throttling gives h1 = h2 			#enthalpy
h3 = 104.6; 			#Unit:Btu/lbm 			#enthalpy
#From the consideration that s3 = s4,
s3 = 0.2244; 			#Unit:Btu/(lbm*F) 			#s = entropy
h4 = 116.0; 			#Unit:Btu/lbm 			#enthalpy
print "The heat extracted is %.2f Btu/lbm"%(h3-h1);
print "The work required is %.2f Btu/lbm"%(h4-h3);
COP = (h3-h1)/(h4-h3); 			#Coefficient of performance
print "The Coefficient of performanceCOP) of this ideal cycle is %.2f"%(COP);

The heat extracted is 63.00 Btu/lbm
The work required is 11.40 Btu/lbm
The Coefficient of performanceCOP) of this ideal cycle is 5.53

Example 10.10 Page No : 518

In [15]:
print "Solution for a";
#By defination,the efficiency of the compressor is the ratio of the ideal compression work to actual compression work.
#Based on the points on fig.10.12, 			#n = (h4-h3)/(h4'-h3);
#There is close correspondence between 5.3 psia and -60F for saturated conditions.Therefore,state 3 is a superheated vapour at 5.3 psia and approximately -20F,because the problem states
#that state 3 has a 40F superheat.Interpolation in the Freon tables in Appendix 3 yields
T = -20; 			#Unit:F  			#temperature
# p      h        s
#7.5  75.719  0.18371
#5.3  76.885  0.18985             h3 = 75.886 Btu/lbm
#5.0  75.990  0.19069

#At 100 psia and s = 0.18985,
#  t           s      h
# 170F     0.18996  100.571
# 169.6F   0.18985  100.5          h4 = 100.5 Btu/lbm
# 160F     0.18726  98.884

#The weight of refrigerant is given by
# 200(tons)/(h3-h1)  =  (200*5)/(75.886-h1)
#In the saturated tables,h1 is
#    p        h 
# 101.86   26.832
# 100psia  26.542
# 98.87    26.365

#m = mass flow/min
h1 = 26.542; 			#enthalpy 			#Unit:Btu/lbm
n = 0.8; 			#Efficiency
h4 = 100.5; 			#enthalpy 			#Unit:Btu/lbm
h3 = 75.886; 			#enthalpy 			#Unit:Btu/lbm
m = (200*5)/(75.886-h1); 			#mass
h4dashminush3 = (h4-h3)/n; 
#Total work of compression = m*(h4minush3)
J = 778; 			#J = Conversion factor
work = (h4dashminush3*m*J)/33000; 			#1 horsepower  =  33,000 ft*LBf/min 			#Unit:hp 			#work
print "%.2f horsepower is required to drive the compressor if it has a mechanical efficiency 100percentage"%(work);

print "Solution for b";
#Assuming a specific heat of the water as unity,we obtain
#From part (a),
#h4'-h3 = h4minush3
h4dash = h4dashminush3+h3; 			#Unit:Btu/lbm
mdot = (m*(h4dash-h1))/(70-60); 			#water enters at 60F and leaves at 70F 			#the required capacity in lbm/min
print "%.2f lbm/min of cooling water i.e. %.2f gal/min is the required capacity of cooling water to pump"%(mdot,mdot/8.3);

Solution for a
14.70 horsepower is required to drive the compressor if it has a mechanical efficiency 100percentage
Solution for b
162.35 lbm/min of cooling water i.e. 19.56 gal/min is the required capacity of cooling water to pump

Example 10.11 Page No : 521

In [17]:
print "Solution for a";
#From appendix3,reading the p-h diagram directly,we have
h3 = 76.2; 			#Unit:Btu/lbm 			#Enthalpy
h4 = 100.5; 			#Unit:Btu/lbm 			#Enthalpy
n = 0.8; 			#Efficiency 			#From 10.10
work = (h4-h3)/n; 			#Work of compression 			#Unit:Btu/lbm
#The enthalpy of saturated liquid at 100 psia is given at 26.1 Btu/lbm.Proceeding as before yields
m = (200*5)/(h3-26.1); 			#Unit:lbm/min 			#m = massflow/min
J = 778; 			#J = Conversion factor
totalwork = (m*work*J)/33000; 			#1 horsepower  =  33,000 ft*LBf/min 			#total ideal work 			#unit:hp
print "Total ideal work of compression is %.2f hp"%(totalwork);

print "Solution for b";
h4dash = h3+work; 			#Btu/lbm
mdot = (m*(h4dash-26.5))/(70-60); 			#water enters at 60F and leaves at 70F 			#the required capacity in lbm/min
print "%.2f lbm/min of cooling water i.e. %.2f gal/min is the required capacity of cooling water to pump"%(mdot,mdot/8.3);

Solution for a
Total ideal work of compression is 14.29 hp
Solution for b
159.83 lbm/min of cooling water i.e. 19.26 gal/min is the required capacity of cooling water to pump

Example 10.12 Page No : 526

In [21]:
# given data
COP = 2.5; 			#Coefficient of performance
cp = 0.24; 			#Unit:Btu/(lbm*R) 			#Specific heat constant for constant pressure process
T1 = -100+460; 			#temperatures converted to absolute temperatures; 			#Unit:R 			#lowest temperature of the cycle
T3 = 150+460; 			#temperatures converted to absolute temperatures; 			#Unit:R 			#Upper temperature of the cycle
#T1/T4-T1  =  COP
T4 = (3.5*T1)/COP; 			#Unit:R 			#temperature at point 4
#T2/T3-T2  = COP
T2 = (COP*T3)/3.5; 			#Unit:R 			#temperature at point 2
print "The work of the expander is %.2f Btu/lbm of air"%(cp*T4-T1);
print "The work of the compressor is %.2f Btu/lbm of air"%(cp*T3-T2);
print "The net work required by the cycle is %.2f Btu/lbm"%(((cp*T3-T2))-cp*(T4-T1));
print "Per ton of refrigeration, the required airflow is %.2f lbm/min per ton"%((200/cp*T2-T1))

The work of the expander is -239.04 Btu/lbm of air
The work of the compressor is -289.31 Btu/lbm of air
The net work required by the cycle is -323.87 Btu/lbm
Per ton of refrigeration, the required airflow is 362735.24 lbm/min per ton

Example 10.13 Page No : 536

In [22]:
#A VACUUM REFRIGERATION SYSTEM 
#A vacuum refrigeration system is used to cool water from 90F to 45F
h1 = 58.07; 			#Unit:Btu/lbm 			#enthalpy
h2 = 13.04; 			#Unit:Btu/lbm  			#enthalpy
h3 = 1081.1; 			#Unit:Btu/lbm 			#enthalpy
m1 = 1; 			#mass 			#unit:lbm
#m2 = 1-m3 			#unit:lbm
#Now, m1*h1  =  m2*h2 + m3*h3
#Putting the values and arranging the equation,
m3 = (m1*h1-h2)/(h3+h2); 			#The mass of vapour that must be removed per pound 			#unit:lbm

# results
print "The mass of vapour that must be removed per pound of entering water is %.2f lbm"%(m3);

The mass of vapour that must be removed per pound of entering water is 0.04 lbm

Example 10.14 Page No : 536

In [23]:
#In problem 10.13,
#A VACUUM REFRIGERATION SYSTEM 
#A vacuum refrigeration system is used to cool water from 90F to 45F
h1 = 58.07; 			#Unit:Btu/lbm 			#enthalpy
h2 = 13.04; 			#Unit:Btu/lbm  			#enthalpy
h3 = 1081.1; 			#Unit:Btu/lbm 			#enthalpy
m1 = 1; 			#mass 			#lbm
#m2 = 1-m3 			#unit:lbm
#Now, m1*h1  =  m2*h2 + m3*h3
#Putting the values and arranging the equation,
m3 = (m1*h1-h2)/(h3+h2); 			#The mass of vapour that must be removed per pound 			#unit:lbm
m2 = 1-m3; 			#mass 			#unit:lbm
print "The mass of vapour that must be removed per pound of entering water is %.2f lbm"%(m3);
#Now,in problem 10.14,
#The refrigeration effect can be determined as m3*(h3-h1) or m2*(h1-h2)
print "The refrigeration effect Using eqn m3*h3-h1) is %.2f Btu/lbm"%(m3*h3-h1)
print "The refrigeration effect Using eqn m2*h1-h2) is %.2f Btu/lbm"%(m2*h1-h2)

The mass of vapour that must be removed per pound of entering water is 0.04 lbm
The refrigeration effect Using eqn m3*h3-h1) is -13.58 Btu/lbm
The refrigeration effect Using eqn m2*h1-h2) is 42.64 Btu/lbm

Example 10.15 Page No : 539

In [24]:
#THE HEAT PUMP
T1 = 70.+460; 			#70F = 70+460 R 			#Energy flows into the system at reservoir at constant temperature T1(unit:R) 			#from problem 10.1
T2 = 32+460; 			#32F = 32+460 R 			#Heat is rejected to the constant temperature T2(Unit:R) 			#from problem 10.1
COP = T1/(T1-T2); 			#Coefficient of performance for carnot heat pump
print "Coefficient of performanceCOP) of the carnot cycle is %.2f"%(COP);
print "The COP can also be obtained from the energy items solved for in problem 10.1"
#In problem 10.1, The power was found to be 77.2 Btu/min and the total tare of heat rejection was 1077.2 Btu/min
#Therefore,
print "Coefficient of performanceCOP) of the cycle is %.2f"%(1077.2/77.2); 

Coefficient of performanceCOP) of the carnot cycle is 13.95
The COP can also be obtained from the energy items solved for in problem 10.1
Coefficient of performanceCOP) of the cycle is 13.95

Example 10.16 Page No : 539

In [27]:
#Let us first consider the cycle as a refrigeration cycle
#In problem 10.1
T1 = 70+460; 			#70F = 70+460 R 			#Energy flows into the system at reservoir at constant temperature T1(unit:R)
T2 = 0+460; 			#0F = 32+460 R 			#Heat is rejected to the constant temperature T2(Unit:R)
COP = T2/(T1-T2); 			#Coefficient of performance
print "Coefficient of performanceCOP) of the cycle is %.2f"%(COP);
Qremoved = 1000; 			#Unit:Btu/min 			#heat removal
WbyJ = Qremoved/COP; 			#the power input 			#unit:Btu/min
print "The power input is %.2f Btu/min"%(WbyJ);
Qrej = Qremoved+WbyJ; 			#The rate of heat rejected to the room 			#Unit:Btu/min
print "The rate of heat rejected to the room is %.2f Btu/min"%(Qrej);
print "The COP as a heat pump is %.2f"%(Qrej/WbyJ);
print "As a check, COP of heat pump is %.2f  =  1 + COP of carnot cycle %.2f"%(Qrej/WbyJ,COP);

Coefficient of performanceCOP) of the cycle is 6.00
The power input is 166.00 Btu/min
The rate of heat rejected to the room is 1166.00 Btu/min
The COP as a heat pump is 7.00
As a check, COP of heat pump is 7.00  =  1 + COP of carnot cycle 6.00