# Chapter 8:VAPOUR CYCLE¶

## Example 8.1, Page No: 362¶

In :
#Variable declaration
TH=311.06; # Source temperature in degree celcius
pb=10; # Boiler pressure in MPa
TL=32.88; # Sink temperature in degree celcius
pc=5; # Condenser pressure in kPa
# From steam tables at pb
h2=1407.56; # specific enthalpy in kJ/kg
h3=2724.7; # specific enthalpy in kJ/kg
s2=3.3596; # specific entropy in kJ/kg K
s3=5.6141; # specific entropy in kJ/kg K
# From steam tables at pc
hf=137.82; hfg=2423.7; # specific enthalpy in kJ/kg
sf=0.4764; sfg=7.9187; # specific entropy in kJ/kg K

#Calculation
x1=(s2-sf)/sfg; # quality of steam at state 1
x4=(s3-sf)/sfg; # quality of steam at state 4
h1=hf+x1*hfg; # specific enthalpy at state 1
h4=hf+x4*hfg; # specific enthalpy at state 4
wT=h3-h4; # Turbine work
wC=h2-h1; # Compressor work
wnet=wT-wC; # Net work output
rw=wnet/wT; # Ratio of net work to trbine work
eff_th=wnet/qH; # Thermal efficiency
SSC=3600/wnet; # specific steam consumption

#Result
print "Ratio of net work to trbine work =",round(rw,3),"\nThermal efficiency =",round(eff_th*100,1),"%"
print "specific steam consumption =",round(SSC,3),"kg/kWh"

Ratio of net work to trbine work = 0.618
Thermal efficiency = 47.6 %
specific steam consumption = 5.741 kg/kWh


## Example 8.2, Page No:366¶

In :
#Variable declaration
TH=311.06; # Source temperature in degree celcius
p2=10; # Boiler pressure in MPa
TL=32.88; # Sink temperature in degree celcius
p1=5; # Condenser pressure in kPa
# From steam tables at p2
h3=2724.7; # specific enthalpy in kJ/kg
s3=5.6141; # specific entropy in kJ/kg K
# From steam tables at p1
hf=137.82; hfg=2423.7; # specific enthalpy in kJ/kg
sf=0.4764; sfg=7.9187; # specific entropy in kJ/kg K

#Calculations
x4=(s3-sf)/sfg; # quality of steam at state 4
h4=hf+x4*hfg; # specific enthalpy at state 4
h1=137.82; # specific enthalpy at state 1 in kJ/kg
s1=0.4764; # specific entropy at state in kJ/kgK
v1=0.001005; # specific volume in m^3/kg
wp=abs (v1*(p2*10**3-p1)); # Pump work (absolute value)
h2=h1+wp; # specific enthalpy at state 2
wT=h3-h4; # Turbine work
wnet=wT-wp; # Net work output
rw=wnet/wT; # Ratio of net work to trbine work
eff_th=wnet/qH; # Thermal efficiency
SSC=3600/wnet; # specific steam consumption

#Results
print "Ratio of net work to trbine work =",round(rw,2),"\nThermal efficiency =",round(eff_th*100,0),"%"
print "specific steam consumption =",round(SSC,3),"kg/kWh"

Ratio of net work to trbine work = 0.99
Thermal efficiency = 39.0 %
specific steam consumption = 3.585 kg/kWh


## Example 8.3, Page No:369¶

In :
#Variable declaration
TH=311.06; # Source temperature in degree celcius
p2=10; # Boiler pressure in MPa
TL=32.88; # Sink temperature in degree celcius
p1=5; # Condenser pressure in kPa
# (a).Actual carnot cycle
eff_Tur=0.8; # Efficiency of turbine
eff_com=0.6; # Efficiency of compressure
# From steam tables at p2
h3=2724.7; # specific enthalpy in kJ/kg
s3=5.6141; # specific entropy in kJ/kg K
s2=3.3596; # specific entropy in kJ/kg K
h2=1407.56; # specific enthalpy in kJ/kg
# From steam tables at p1
hf=137.82; hfg=2423.7; # specific enthalpy in kJ/kg
sf=0.4764; sfg=7.9187; # specific entropy in kJ/kg K

#Calculations for (a)
x1=(s2-sf)/sfg; # quality of steam at state 1
x4=(s3-sf)/sfg; # quality of steam at state 4
h1=hf+x1*hfg; # specific enthalpy at state 1
h4=hf+x4*hfg; # specific enthalpy at state 4
wTs=h3-h4; # Turbine work
wT=eff_Tur*wTs; # Actual turbine work
wCs=h2-h1; # Compressor work
wC=wCs/eff_com; # Actual compressor work
wnet=wT-wC; # Net work output
h2a=h1+wC; # specific enthalpy
rw=wnet/wT; # Ratio of net work to trbine work
eff_th=wnet/qH; # Thermal efficiency
SSC=3600/wnet; # specific steam consumption

#Results for (a)
print "(a).Actual carnot cycle","\nRatio of net work to trbine work =",round(rw,4),"\nThermal efficiency =",round(eff_th*100,2),"%"
print "specific steam consumption =",round(SSC,1),"kg/kWh"

#Variable declaration for (b)
#   (b).Actual Rankine cycle
eff_Tur=0.8; # Efficiency of turbine
eff_pump=0.9; # Efficiency of Pump
# From steam tables at p1
h1=137.82; # specific enthalpy at state 1 in kJ/kg
s1=0.4764; # specific entropy at state in kJ/kgK
v1=0.001005; # specific volume in m^3/kg

#Calculations for (b)
wps=abs (v1*(p2*10**3-p1)); # Pump work (absolute value)
wp=wps/eff_pump; # Actual pmp work
h2a=h1+wp; # # specific enthalpy at state 2
wnet=wT-wp; # Net work output
rw=wnet/wT; # Ratio of net work to trbine work
eff_th=wnet/qH; # Thermal efficiency
SSC=3600/wnet; # specific steam consumption

#Results for (b)
print "\n\n(b).Actual Rankine cycle","\nRatio of net work to trbine work =",round(rw,3),"\nThermal efficiency =",round(eff_th*100,0),"%"
print "specific steam consumption =",round(SSC,1),"kg/kWh"

(a).Actual carnot cycle
Ratio of net work to trbine work = 0.2046
Thermal efficiency = 15.68 %
specific steam consumption = 21.7 kg/kWh

(b).Actual Rankine cycle
Ratio of net work to trbine work = 0.986
Thermal efficiency = 31.0 %
specific steam consumption = 4.5 kg/kWh


## Example 8.4, Page No:379¶

In :
#Variable declaration
T3=450; # temperature of steam at state 3 in degree celcius
p3=10; # Boiler pressure in MPa
p6=5; # Condenser pressure in kPa
# From steam tables at state 1
hf=137.82; hfg=2423.7; # specific enthalpy in kJ/kg
sf=0.4764; sfg=7.9187; # specific entropy in kJ/kg K
h1=137.82; # specific enthalpy at state 1 in kJ/kg
s1=0.4764; # specific entropy at state in kJ/kgK
v1=0.001005; # specific volume in m^3/kg

#Calculations for (a)
wp=abs (v1*(p3*10**3-p6)); # Pump work (absolute value)
h2=h1+wp; # specific enthalpy at state 2
#   (a).Rankine cycle with superheat
# From steam tables at state 3
h3=3240.9;# specific enthalpy in kJ/kg
s3=6.419; # specific entropy in kJ/kg K
# State 4_1
x4_1=(s3-sf)/sfg; # Quality of steam at state 4_1
h4_1=hf+x4_1*hfg; # specific enthalpy at state 4_1
wT=h3-h4_1; # Turbine work
wnet=wT-wp; # Net work output
eff_th=wnet/qH; # Thermal efficiency
SSC=3600/wnet; # specific steam consumption

#Results for (a)
print "(a).Rankine cycle with superheat","\nQuality of steam at exhaust = ",round(x4_1,2),"\nThermal efficiency =",round(eff_th*100,1),"%"
print "specific steam consumption =",round(SSC,3),"kg/kWh"

#Variable declaration for (b)
#   (b).Reheat cycle
s4=s3; # isentropic expansion
x4=0.975; # Quality of steam at state 4
# from steam table intermediate pressure is selected for s4 & x4 by interpolation and assumed by round value
p4=1.2; # Intermediate pressure in MPa
# From steam tables at state 4
hf4=798.6; hfg4=1986.2; # specific enthalpy in kJ/kg
h4=hf4+x4*hfg4; # specific enthalpy in kJ/kg
# From steam tables at state 5
h5=3368.5;# specific enthalpy in kJ/kg
s5=7.5266; # specific entropy in kJ/kg K

#Calculations for (b)
x6=(s5-sf)/sfg; # Quality of steam at state 6
h6=hf+x6*hfg; # specific enthalpy in kJ/kg
wT=(h3-h4)+(h5-h6); # Turbine work
wnet=wT-wp; # Net work output
eff_th=wnet/qH; # Thermal efficiency
SSC=3600/wnet; # specific steam consumption

#Results for (b)
print "\n\n(b).Reheat cycle","\nQuality of steam at exhaust = ",round(x6,2),"\nThermal efficiency =",round(eff_th*100,1),"%"
print "specific steam consumption =",round(SSC,3),"kg/kWh"

(a).Rankine cycle with superheat
Quality of steam at exhaust =  0.75
Thermal efficiency = 41.2 %
specific steam consumption = 2.825 kg/kWh

(b).Reheat cycle
Quality of steam at exhaust =  0.89
Thermal efficiency = 42.1 %
specific steam consumption = 2.295 kg/kWh


## Example 8.5, Page No:386¶

In :
#Variable declaration
T5=311.06; # temperature of steam at state 5  in degree celcius
p5=10; # Boiler pressure in MPa
p4=p5;
T7=32.88; # temperature of steam at state 7  in degree celcius
p7=5; # Condenser pressure in kPa
p1=p7;
# From steam tables at p7
h1=137.82; # specific enthalpy at state 1 in kJ/kg
s1=0.4764; # specific entropy at state 1 in kJ/kgK
v1=0.001005; # specific volume in m^3/kg

#Calculations
wp=abs (v1*(p5*10**3-p7)); # Pump work (absolute value)
h2=h1+wp; # specific enthalpy at state 2
T6=(T5+T7)/2; # Temperature of bleed system
h5=2724.7; # specific enthalpy at state 5 in kJ/kg
s5=5.6141; # specific entropy at state 5 in kJ/kgK
# From steam tables at state 6
p6=791.5; # bleed steam pressure in kPa
p2=p6; p3=p6;
vf6=0.00114; # specific volume in m^3/kg
v3=vf6;
hf6=719.21; hfg6=2049.5; # specific enthalpy in kJ/kg
sf6=2.0419; sfg6=4.6244; # specific entropy in kJ/kg K
x6=(s5-sf6)/sfg6; # quality of steam at state 6
h6=hf6+x6*hfg6; # specific enthalpy at state 6
h3=hf6; # specific enthalpy at state 3
m1=(h3-h2)/(h6-h2); # Fraction of bleed steam
wLP=abs (v1*(p2-p1)); # LP work
wHP=abs (v3*(p4*10**3-p3)); # HP work
wp=(1-m1)*wLP+wHP; # Total pump work
h2=h1+wp; h4=h3+wp; # Specific Enthalpies of water
# From steam tables at pc
hf7=137.82; hfg7=2423.7; # specific enthalpy in kJ/kg
sf7=0.4764; sfg7=7.9187; # specific entropy in kJ/kg K
x7=(s5-sf7)/sfg7; # quality of steam at state 7
h7=hf7+x7*hfg7; # specific enthalpy at state 4
wT=(h5-h6)+(1-m1)*(h6-h7); # Turbine work
wnet=wT-wp; # Net work output
eff_th=wnet/qH; # Thermal efficiency
SSC=3600/wnet; # specific steam consumption

#Result
print "Thermal efficiency =",round(eff_th*100,1),"%","\nspecific steam consumption =",round(SSC,3),"kg/kWh"

Thermal efficiency = 42.4 %
specific steam consumption = 4.254 kg/kWh


## Example 8.6, Page No:389¶

In :
#Variable declaration
ps=6.89+1; # Pressure of steam produced in bar (Absolute)
x=0.96; # Quality of steam produced
f=75; # Steady flow of water in litres
t=9.5; # Time consumption of water in minutes
tf=685; #Time consumption of 10 litre fuel in seconds
Vf=10; # consumption of fuel in litres
Sf=0.85; # specific gravity of water
CV=43125; # Calorific value of fuel in kJ/kg

#Calculation
ms=f/(t*60);# Steam generation
mf=Vf*Sf/tf; # consumption of fuel
# From steam tables at ps
hf=718.5; hfg=2050; # specific enthalpy in kJ/kg
hs=hf+x*hfg; # specific enthalpy of steam produced
hFW=146.7; # Enthalpy of feed water at 35 degree celcius
eff_boiler=(ms*(hs-hFW))/(mf*CV); # Boiler Efficiency

#Result
print "Boiler Efficiency = ",round(eff_boiler*100,2),"%"

Boiler Efficiency =  62.45 %


## Example 8.7, Page No:394¶

In :
#Variable declaration
TL=-15; # Source temperature in degree celcius
TH=40; # Sink temperature in degree celcius
# From the table of properties of Freon - 12
h2=203.05; h3=74.53; hf4=22.31; hg4=180.85; # specific enthalpy in kJ/kg
s1=0.682; s3=0.2716; sf4=0.0906; sg4=0.7046; sfg4=sg4-sf4;# specific entropy in kJ/kg K

#Calculations
sf1=sf4; sfg1=sfg4; s2=s1; s4=s3; hf1=hf4; hg1=hg4; hfg4=hg4-hf4; hfg1=hfg4;# refer figure 8.20
x1=(s1-sf1)/sfg1; # Dryness fraction at state 1
x4=(s4-sf4)/sfg4; # dryness fraction at state 4
h1=hf1+x1*hfg1; h4=hf4+x4*hfg4; # specific enthalpy in kJ/kg
wc=h2-h1; # work of compression
wE=h3-h4; # Work of expansion
qL=h1-h4; # Refrigerating effect
qH=h2-h3; # Heat rejected
wnet=wc-wE; # Net work
COPc=qL/wnet; # COP of the cycle

#Results
print "Work of compression = ",round(wc,2),"kJ/kg","\nWork of expansion =",round(wE,2),"kJ/kg"
print "Refrigerating effect = ",round(qL,2),"kJ/kg"
print "Heat rejected =",qH,"kJ/kg","\nCOP of the cycle =",round(COPc,3)

Work of compression =  28.04 kJ/kg
Work of expansion = 5.48 kJ/kg
Refrigerating effect =  105.97 kJ/kg
Heat rejected = 128.52 kJ/kg
COP of the cycle = 4.699


## Example 8.8, Page No:400¶

In :
#Variable declaration
TL=-15; # Source temperature in degree celcius
TH=40; # Sink temperature in degree celcius
# From the table of properties of Freon - 12
h3=74.53; h1=180.85; h4=h3; # specific enthalpy in kJ/kg
s1=0.7046; s2=0.682; # specific entropy in kJ/kg K

# (a)

#Calculation for (i)
# (i).Condensor and Evaporator pressure
pc=0.9607; # Saturation pressure at TH in MPa
pE=0.1826; # Saturation pressure at TL in MPa

#Results for (i)
print "(a)","\n(i).Condensor and Evaporator pressure","\nCondensor pressure = ",pc,"MPa"
print "Evaporator pressure = ",pE,"MPa"

#Calculation for (ii)
# (ii).Compressor discharge temperature  & Enthalpy
p2=pc; # Condensor pressure
s2=s1; # refer figure 8.25
# From the table of properties of Freon - 12 at pc
t2=46.8; # Compressor discharge temperature in degree celcius
h2=208.3; # specific enthalpy in kJ/kg

#Results for (ii)
print "\n(ii).Compressor discharge temperature  & Enthalpy","\nCompressor discharge temperature = ",t2,"oC"
print "Enthalpy = ",h2,"kJ/kg"

#Calculation for (iii)
# (iii).Ratio of COP of the cycle to Carnot COP
w=h2-h1; # Compressor work
qL=h1-h4; # Refrigeration effect
COP=qL/w; # COP of the cycle
COPc=4.68; # COP of carnot cycle from example 8.7
r=COP/COPc; # Ratio of COP of the cycle to Carnot COP

#Results for (iii)
print "\n(iii).Ratio of COP of the cycle to Carnot COP = ",round(r,2)

#Calculation for (b)
# (b)
QL=0.440; # Capacity of refrigerator in kW (1/8 ton of refrigeration)
m=QL/qL; # Mass flow rate of refrigerant
W=m*w; # Power consumption of compressor
QH=QL+W; # Heat rejected

#Results for (b)
print "\n\n(b)","\nMass flow rate of refrigerant = ",round(m,5),"kg/s","\nPower consumption of compressor = ",round(W,4),"kW"
print "Heat rejected to surroundings = ",round(QH,4),"kW   (Error in textbook answer)"

(a)
(i).Condensor and Evaporator pressure
Condensor pressure =  0.9607 MPa
Evaporator pressure =  0.1826 MPa

(ii).Compressor discharge temperature  & Enthalpy
Compressor discharge temperature =  46.8 oC
Enthalpy =  208.3 kJ/kg

(iii).Ratio of COP of the cycle to Carnot COP =  0.83

(b)
Mass flow rate of refrigerant =  0.00414 kg/s
Power consumption of compressor =  0.1136 kW
Heat rejected to surroundings =  0.5536 kW   (Error in textbook answer)