# Chapter 10 :- Refrigeration and Heat pump systems¶

## Example 10.1 Page no-459

In [1]:
# Given:-
Tc = 273.00                                                                        # temperature of cold region in kelvin
Th = 299.00                                                                        # temperature of hot region in kelvin
mdot = 0.08                                                                        # mass flow rate in kg/s

# Analysis
# At the inlet to the compressor, the refrigerant is a saturated vapor at 0C, so from Table A-10
h1 = 247.23                                                                        # in kj/kg
s1 = 0.9190                                                                        # in kj/kg.k

# The pressure at state 2s is the saturation pressure corresponding to 26C, or
p2 = 6.853                                                                         # in bars
# The refrigerant at state 2s is a superheated vapor with
h2s = 264.7                                                                        # in kj/kg
# State 3 is saturated liquid at 26C, so
h3 = 85.75                                                                         # in kj/kg
h4 = h3                                                                            # since The expansion through the valve is a throttling process

# Part(a)
Wcdot = mdot*(h2s-h1)                                                              # The compressor work input in KW
print '-> The compressor power, in kW, is:  '
print round(Wcdot,2)

# Part(b)
Qindot = mdot*(h1-h4)*60/211                                                       # refrigeration capacity in ton
print '-> The refrigeration capacity in tons is:  '
print round(Qindot,2)

# Part(c)
beta = (h1-h4)/(h2s-h1)
print '-> The coefficient of performance is:  '
print round(beta,2)

# Part(d)
betamax = Tc/(Th-Tc)
print '-> The coefficient of performance of a Carnot refrigeration cycle operating between warm and cold regions at 26 and 0C, respectively is:  '
print betamax

-> The compressor power, in kW, is:
1.4
-> The refrigeration capacity in tons is:
3.67
-> The coefficient of performance is:
9.24
-> The coefficient of performance of a Carnot refrigeration cycle operating between warm and cold regions at 26 and 0C, respectively is:
10.5

## Example 10.2 Page no-462

In [2]:
# Given:-
mdot = 0.08                                                                      # mass flow rate in kg/s
# Analysis
# At the inlet to the compressor, the refrigerant is a saturated vapor at 10C, so from Table A-10,
h1 = 241.35                                                                      # in kj/kg
s1 = .9253                                                                       # in kj/kg.k
# Interpolating in Table A-12 gives
h2s = 272.39                                                                     # in kj/kg.k
# State 3 is a saturated liquid at 9 bar, so
h3 = 99.56                                                                       # in kj/kg
h4 = h3                                                                          # since The expansion through the valve is a throttling process

# Part(a)
Wcdot = mdot*(h2s-h1)                                                            # The compressor power input in KW
# Result
print '-> The compressor power in kw is:  '
print round(Wcdot,2)

# Part(b)
Qindot = mdot*(h1-h4)*60/211                                                     # refrigeration capacity in tons
# Result
print '-> The refrigeration capacity in tons is:  '
print round(Qindot,2)

# Part(c)
beta = (h1-h4)/(h2s-h1)
# Result
print '-> The coefficient of performance is:  '
print round(beta,2)

-> The compressor power in kw is:
2.48
-> The refrigeration capacity in tons is:
3.23
-> The coefficient of performance is:
4.57

## Example 10.3 Page no-464

In [3]:
# Given:-
Tnot = 299                                                         #in kelvin
etac = .8                                                          #compressor efficiency of 80 percent
mdot = .08                                                         #mass flow rate in kg/s
#analysis
#State 1 is the same as in Example 10.2, so
h1 = 241.35                                                        #in kj/kg
s1 = .9253                                                         #in kj/kg.k
#from example 10.2
h2s = 272.39                                                       #in kj/kg
h2 =(h2s-h1)/etac + h1                                             #in kj/kg
#Interpolating in Table A-12,
s2 = .9497                                                         #in kj/kg.k

h3 = 91.49                                                         #in kj/kg
s3 = .3396
h4 = h3                                                            #since The expansion through the valve is a throttling process
#from data table
hf4 = 36.97                                                        #in kj/kg
hg4 = 241.36                                                       #in kj/kg
sf4 = .1486                                                        #in kj/kg.k
sg4 = .9253                                                        #in kj/kg.k
x4 = (h4-hf4)/(hg4-hf4)                                            #quality at state 4
s4 = sf4 + x4*(sg4-sf4)                                            #specific entropy at state 4 in kj/kg.k

#part(a)
Wcdot = mdot*(h2-h1)                                                #compressor power in kw
print 'The compressor power in kw is: ',round(Wcdot,2),'kW'

#part(b)
Qindot = mdot*(h1-h4)*60/211                                        #refrigeration capacity in ton
print 'The refrigeration capacity in ton is: ',round(Qindot,2),'ton'

#part(c)
beta = (h1-h4)/(h2-h1)                                              #coefficient of performance
print 'The coefficient of performance is: ',round(beta,2)

#part(d)
Eddotc = mdot*Tnot*(s2-s1)                                           #in kw
Eddotv = mdot*Tnot*(s4-s3)                                           #in kw
print 'The rate of exergy destruction within the compressor is:',round(Eddotc,2),'kw.'
print 'The rate of exergy destruction within the valve is:',round(Eddotv,2),'kw'

The compressor power in kw is:  3.1 kW
The refrigeration capacity in ton is:  3.41 ton
The coefficient of performance is:  3.86
The rate of exergy destruction within the compressor is: 0.58 kw.
The rate of exergy destruction within the valve is: 0.39 kw

## Example 10.4 Page no-475

In [4]:
# Given:-
p1 = 1.00                                                                          # in bar
T1 = 270.00                                                                        # in kelvin
AV = 1.4                                                                           # in m^3/s
r = 3.00                                                                           # compressor pressure ratio
T3 = 300.00                                                                        # turbine inlet temperature in kelvin

# Analysis
# From Table A-22,
h1 = 270.11                                                                        # in kj/kg
pr1 = 0.9590
# Interpolating in Table A-22,
h2s = 370.1                                                                        # in kj/kg
# From Table A-22,
h3 = 300.19                                                                        # in kj/kg
pr3 = 1.3860
# Interpolating in Table A-22, we obtain
h4s = 219.00                                                                       # in kj/kg
# Calculations
pr2 = r*pr1
pr4 = pr3/r

# Part(a)
R = 8.314                                                                          # universal gas constant, in SI units
M = 28.97                                                                          # molar mass of air in grams

# Results
mdot = (AV*p1)/((R/M)*T1)*10**2                                                    # mass flow rate in kg/s
Wcycledot = mdot*((h2s-h1)-(h3-h4s))
print '-> The net power input in kw is:  '
print round(Wcycledot,2)

# Part(b)
Qindot = mdot*(h1-h4s)                                                             # refrigeration capacity in kw
print '-> The refregeration capacity in kw is:  '
print round(Qindot,2)

# Part(c)
beta = Qindot/Wcycledot                                                            # coefficient of performance
print 'The coefficient of performance is:  '
print round(beta,2)

-> The net power input in kw is:
33.97
-> The refregeration capacity in kw is:
92.34
The coefficient of performance is:
2.72

## Example 10.5 Page no-477

In [5]:
# Given:-
# Part(a)
wcdots = 99.99                                                              # work per unit mass for the isentropic compression determined with data from the solution in Example 10.4 in kj/kg
mdot = 1.807                                                                # mass flow rate in kg/s from 10.4
etac = 0.8                                                                  # isentropic efficiency of compressor
Wcdot = (mdot*wcdots)/etac                                                  # The power input to the compressor in kw

# Using data form the solution to Example 10.4 gives
wtdots =81.19                                                               # in kj/kg
etat = 0.8                                                                  # isentropic efficiency of turbine
# Calculations
Wtdot = mdot*etat*wtdots                                                    # actual turbine work in kw
Wdotcycle = Wcdot-Wtdot                                                     # The net power input to the cycle in kw
# Result
print '-> The net power input in kw is:  '
print round(Wdotcycle,2)

# Part(b)
h3 = 300.19                                                                 # in kj/kg
# From table A-22
h1 = 270.11                                                                 # in kj/kg
# Calculations
h4 = h3 -Wtdot/mdot
Qindot = mdot*(h1-h4)                                                       # refrigeration capacity in kw
# Result
print '-> The refrigeration capacity in kw is:  '
print round(Qindot,2)

# Part(c)
beta = Qindot/Wdotcycle                                                     # coefficient of performance
# Result
print '-> The coefficient of performance is:  '
print round(beta,2)

-> The net power input in kw is:
108.48
-> The refrigeration capacity in kw is:
63.01
-> The coefficient of performance is:
0.58