# Chapter 3 : Combined Cycle Power Generation¶

## Ex: 3.1 Pg: 143¶

In :
from __future__ import division
#Input data
p=40#Pressure in bar
T1=400+273#Temperature in K
T2=40+273#Temperature in K
x=[0,10,515.5,72.23,363.0,0.1478,0.5167,80.9*10**-6,0.0333]#Property values from table p(bar),t(degree C), hf,hg(kJ/kg),sf,sg(kJ/kg.K),vf,vg(m**3/kg)
y=[0,0.2,277.3,38.35,336.55,0.0967,0.6385,77.4*10**-6,1.163]#Property values from table p(bar),t(degree C), hf,hg(kJ/kg),sf,sg(kJ/kg.K),vf,vg(m**3/kg)

#Calculations
h1=3216#Enthalpy in kJ/kg
s1=6.7690#Entropy in kJ/kg.K
s2=s1#Entropy in kJ/kg.K
x2=(s2-0.5725)/(8.2570-0.5725)#Dryness fraction
h2=167.57+x2*2406.7#Enthalpy in kJ/kg
h3=167.57#Enthalpy in kJ/kg
h4=(167.57+p*100*1.008*10**-3)#Enthalpy in kJ/kg
h5=1087.31#Enthalpy in kJ/kg
h6=2801.4#Enthalpy in kJ/kg
ha=x[(4)]#Enthalpy in kJ/kg
sa=x[(6)]#Entropy in kJ/kg.K
sb=sa#Entropy in kJ/kg.K
xb=(sb-y[(5)])/(y[(6)]-y[(5)])#Dryness fraction
hb=(y[(3)]+xb*(y[(4)]-y[(3)]))#Enthalpy in kJ/kg
hc=y[(3)]#Enthalpy in kJ/kg
hd=hc#Enthalpy in kJ/kg
m=(h6-h5)/(hb-hc)#Mass of mercury circulated per kg of steam
Q1=m*(ha-hd)+(h1-h6)+(h5-h4)#Heat supplied in kJ/kg
Q2=(h2-h3)#Heat rejected in kJ/kg
nc=(1-(Q2/Q1))*100#Efficiency in percent

#Output
print " (a) The amount of mercury circulated per kg of water is %3.4f kg \n (b) The efficiency of the combined cycle is %3.1f percent"%(m,nc)

 (a) The amount of mercury circulated per kg of water is 7.4151 kg
(b) The efficiency of the combined cycle is 48.1 percent


## Ex: 3.2 Pg: 145¶

In :
from __future__ import division
#Input data
m=5#Mass flow rate in kg/s
p1=40#Pressure in bar
T1=440+273#Temperature in K
p2=1.5#Pressure in bar
p3=1#Pressure in bar
T3=60+273#Temperature in K
p4=16#Pressure in bar
T4=100+273#Temperature in K
p5=9#Pressure in bar

#Calculations
h1=3307.1#Enthalpy in kJ/kg
s1=6.9041#Entropy in kJ/kg.K
s2=s1#Entropy in kJ/kg.K
h2=2570.8#Enthalpy in kJ/kg
h3=417.46#Enthalpy in kJ/kg
h6=(251.13+(1.0172*10**-3)*(p3-0.1994)*100)#Enthalpy in kJ/kg
m3=(m/2)#Mass flow rate in kg/s
m6=m3#Mass flow rate in kg/s
h4=(m3*h3+m6*h6)/m#Enthalpy in kJ/kg
h5=(h4+(1.0291*10**-3)*(p1-p3)*100)#Enthalpy in kJ/kg
ha=241.58#Enthalpy in kJ/kg
sa=0.7656#Entropy in kJ/kg.K
sb=sa#Entropy in kJ/kg.K
hb=229.43#Enthalpy in kJ/kg
hc=71.93#Enthalpy in kJ/kg
hd=hc+(0.7914*10**-3*(p4-p5)*100)#Enthalpy in kJ/kg
Q1=(m*(h1-h5))/1000#Heat supplied in kW
Wnets=(m*((h1-h2)-(h5-h4)))#Net workdone by steam in kW
mR12=(m3*(h2-h3))/(ha-hd)#Mass of R12 in kg/s
WnetR=(mR12*((ha-hb)-(hd-hc)))#Net workdone by R12 in kW
T=Wnets+WnetR#Total output in kW
Qh=(m6*(h2-h6))#Heat rejected in kW

#Output
print "(a) Rate of heat transfer in the steam generator is %3.3f kW \n (b) The net power output of the binary cycle is %d kW \n (c) The rate of heat transfer to the industrial process is %3.0f kW"%(Q1,T,Qh)

(a) Rate of heat transfer in the steam generator is 14.844 kW
(b) The net power output of the binary cycle is 4030 kW
(c) The rate of heat transfer to the industrial process is 5799 kW


## Ex: 3.3 Pg: 146¶

In :
from numpy import mat
#Input data
rp=7.5#Pressure ratio
T1=15+273#Inlet air temperature in K
T3=750+273#Maximum temperature in K
T6=100+273#Temperature in K
p1=50#Pressure in bar
T7=600+273#Temperature in K
p2=0.1#Pressure in bar
P=200#Total power in MW
CV=43.3#calorific value in MJ/kg
cpg=1.11#Specific heat for gas in kJ/kg.K
g=1.33#Ratio of specific heats for gas
cpa=1.005#Specific heat for air in kJ/kg.K
g1=1.4#Ratio of specific heats for air

#Calculations
T2=(T1*rp**((g1-1)/g1))#Temperature in K
T4=(T3/rp**((g-1)/g))#Temperature in K
ha=3670#Enthalpy in kJ/kg
hb=2305#Enthalpy in kJ/kg
hc=192#Enthalpy in kJ/kg
hd=hc#Enthalpy in kJ/kg
#ma*cpg*(T3-T6)=ms*(ha-hd)
#ma*cpg*(T3-T4)-ma*cpa*(T2-T1)+ms*(ha-hb)=P*1000
#Solving these two equations
A=mat([[cpg*(T3-T6), (hd-ha)],[cpg*(T3-T4)-cpa*(T2-T1), (ha-hb)]])#Coefficient matrix
B=mat([,[(P*10**3)]])#Constant matrix
X=(A**-1)*B#Variable matrix

Wgt=(cpg*(T3-T4)-cpa*(T2-T1))*X*10**-3#Net workdone by Gas turbine in MW
Wst=(P-Wgt)#Net workdone by steam turbine in MW
Q1=(X*cpg*(T3-T2+T3-T4))#Heat supplied in MW
nth=(P/(Q1*10**-3))*100#Thermal efficiency in percent
af=(CV*10**3)/(cpg*(T3-T2+T3-T4))#Air fuel ratio

#Output
print " (a) The flow rate of air is %3.2f kg/s and steam is %3.2f kg/s \n (b) The power outputs of the gas turbine is %3.2f MW and steam turbine is %3.2f MW \n (c) The thermal efficiency of the combined plant is %3.0f percent \n (d) The air fuel ratio is %3.1f"%(X,X,Wgt,Wst,nth,af)

 (a) The flow rate of air is 396.33 kg/s and steam is 82.22 kg/s
(b) The power outputs of the gas turbine is 87.77 MW and steam turbine is 112.23 MW
(c) The thermal efficiency of the combined plant is  50 percent
(d) The air fuel ratio is 42.7


## Ex: 3.4 Pg: 148¶

In :
from __future__ import division
from math import log
#Input data
p1=1#Pressure in bar
T1=25+273#Temperature in K
rp=8#Pressure ratio of compressor
Tm=900+273#Maximum temperature in K
pd=3#pressure drop in combustion chamber in percent
nc=0.88#Efficiency of compressor
nt=0.88#Efficiency of turbine
CV=44.43#Calorific value of fuel in MJ/kg
cpa=1.006#Specific heat of air in kJ/kg.K
cpg=1.148#Specific heat of gas in kJ/kg.K
g1=1.333#Specific heat ratio of gas
g=1.4#Specific heat ratio of air
T3=425+273#Temperature in K
p2=40#Pressure in bar
p3=0.04#Condensor pressure in bar
Th=170.4+273#Temperature of feed water to the HRSG in K
nst=0.82#Efficiency of steam turbine
pdh=5#Pressure drop in HRSG in kPa
m=29.235#Steam flow rate in kg/s
A=1.0401#si=1.0401+0.1728*(h/c)
B=0.1728#si=1.0401+0.1728*(h/c)

#Calculations
#Gas turbine plant
T2=(rp**((g-1)/(g*nt)))*T1#Temperature in K
#Combustor
pc=((pd/100)*rp)#Pressure loss in bar
pcx=(rp-pc)#Pressure in bar
f=((cpg*(Tm-T1))-(cpa*(T2-T1)))/((CV*10**3)-(cpa*(T2-T1)))#Fuel flow rate in kg/s
af=(1-f)/f#Air fuel ratio
#C8H18+12.5O2->8CO2+9H2O
afc=(12.5*32)/(0.232*114)#Air fuel ratio for stoichiometric combustion
ea=((af-afc)/afc)*100#Excess air in percent
#Gas turbine
p4=p1+0.05#Pressure in bar
T4=(Tm/(pcx/p4)**(((g1-1)*nt)/g1))#Temperature in K
#HRSG
T5=250+30#Temeprature in K
ha=3272#Enthalpy in kJ/kg
hf=1087.31#Enthalpy in kJ/kg
ws=(cpg*((T4-273)-T5))/(ha-hf)#Flow rate in kg/s
he=721.1#Enthalpy in kJ/kg
T6=(T4-273)-((ws*(ha-he))/cpg)#Temperature in degree C
#Power output
sa=6.853#Entropy in kJ/kg.K
sbs=sa#Entropy in kJ/kg.K
xbs=(sbs-0.4266)/8.052#Dryness fraction
hbs=(121.46+xbs*2432.9)#ENthalpy in kJ/kg
Wst=(m*(ha-hbs)*nst)#Workdone in kW
wg=(m/ws)#gas flow rate in kg/s
wa=(1-f)*wg#Air flow rate entering the compressor in kg/s
Wgt=(wg*cpg*(Tm-T4))-(wa*cpa*(T2-T1))#Power output of gas turbine in kW
TO=Wst+Wgt#Total power output in kW
wf1=(f*wa)#Fuel mass flow rate in kg/s
wf=4.466#Rounding off of wf1 for exact answers
no=(TO/(wf*(CV*10**3)))*100#Overall efficiency of the combined plant in percent
ns=((ha-hbs)/(ha-he))*nst#Efficiency of steam plant
ngtp=(Wgt/(wf*(CV*10**3)))#Efficiency of the GT plant
xL=((wg*cpg*(T6-(T1-273)))/(wf*(CV*10**3)))#Lost heat coefficient
nov=(ns+ngtp-ns*ngtp-ngtp*xL)#The overall efficiency
#Energy fluxes and irreversibilities
si=(A+B*((18*1)/(8*12)))#si for octane C8H18
dHo=(wf*CV*10**3)#Power in kW
dGo=(si*dHo)#Power in kW
TS=(dGo-dHo)#Power in kW
#Compressor
dS=(cpa*log(T2/T1))-(((cpa*(g-1))/g)*log(rp))#change in entropy in kJ/kg.K
Ic=(wa*T1*dS)#power in kW
Icx=((wg*T1*((cpg*log(Tm/T1))-(((cpg*(g1-1))/g1)*log(pcx))))-(wa*T1*((cpa*log(T2/T1))-(((cpa*(g-1))/g)*log(rp))))+TS)#Compressor in kW
Icg=(-cpg*log(Tm/T4))-(((cpg*(g1-1))/g1)*log(p4/pcx))#Difference in entropy in kJ/kg.K
IGT=(Icg*T1*wg)#Gas turbine in kW
se=2.046#Enntropy in kJ/kg.K
sae=(sa-se)#Difference in entropy in kJ/kg.K
s64=(cpg*log((T6+273)/T4))-(((cpg*(g1-1))/g1)*log(p4/p1))#Difference in entropy in kJ/kg.K
Ih=(T1*m*sae)+(wg*T1*s64)#For HRSG in kW
hb=(ha-(nst*(ha-hbs)))#Enthalpy in kJ/kg
xb=(hb-121.46)/2432.9#Dryness Fraction
sb=(0.4226+xb*8.052)#Entropy in kJ/kg.K
Ist=(m*(sb-sa)*T1)#For steam turbine in kW
Iexh=(wg*cpg*((T6-(T1-273))-(T1*log((T6+273)/T1))))#For exhaust in kW
Tl=Icx+Icg+IGT+Ih+Ist+Iexh#Exergy losses in kW
T=Tl+Wgt+Wst#Total exergy output and exergy destruction in kW
ee=((Wst+Wgt)/T)*100#Exergy efficiency in percent

#Output
print "(a) Total power output is %3.2f kW and overall efficiency is %3.2f percent lost heat coefficient is %3.3f\n Exergy efficiency is %3.0f percent \n\n Input is %3.0f kW \n Total Output is %3.0f kW \n Total losses is %3.0f kW \n Exergy outut + exergy destruction = %3.0f kW which is 1.3 percent gretter than the exergy input"%(TO,no,xL,ee,dGo,(Wgt+Wst),Tl,T)

(a) Total power output is 82422.08 kW and overall efficiency is 41.54 percent lost heat coefficient is 0.351
Exergy efficiency is  40 percent

Input is 212810 kW
Total Output is 82422 kW
Total losses is 123309 kW
Exergy outut + exergy destruction = 205731 kW which is 1.3 percent gretter than the exergy input


## Ex: 3.5 Pg: 154¶

In :
#Input data
n1=0.5#Efficiency of mercury
n2=0.4#Efficiency of steam
n3=0.25#Efficiency of composite cycle

#Calculations
n=(1-(1-n1)*(1-n2)*(1-n3))*100#Overall efficiency of the combined cycle in percent

#Output
print "The overall efficiency of the combined cycle is %3.1f percent"%(n)

The overall efficiency of the combined cycle is 77.5 percent


## Ex: 3.6 Pg: 156¶

In :
#Input data
z=30.0#Percentage of total energy of fuel
n=40.0#Cycle efficiency in percent

#Calculations
on=((z/100)+(1-(z/100))*(n/100))*100#Overall efficiency in percent

#Output
print "The overall efficiency of the combined plant is %3.0f percent"%(on)

The overall efficiency of the combined plant is  58 percent


## Ex: 3.7 Pg: 158¶

In :
from __future__ import division
from math import exp
#Input data
Tc=1250+273#Cathode temperature in K
Ta=500+273#Anode temperature in K
e=1.602*10**-19#Charge in coloumb
K=1.38*10**-23#Boltzmann constant in J/molecule.K
b=18#Constant

#Calculations
Va=((b*K*Ta)/e)#Voltage of anode in V
Vc=((b*K*Tc)/e)#Voltage of cathode in V
Vo=Vc-Va#Output voltage in V
Ja=(120*Ta**2*exp(-b))#Current density in Cathode in A/cm**2
Jc=(120*Tc**2*exp(-b))#Current density in Anode in A/cm**2
P=Vo*(Jc-Ja)#Power output per unit area in /cm**2
nth=(((Tc-Ta)/Tc)*(b/(b+2)))*100#Thermal efficiency in percent

#Output
print "(a) The output voltage is %3.4f V \n (b) The current density in the cathode is %3.3f A/cm**2 and anode is %3.3f A/cm**2 \n (c) Power output per unit area is %3.2f W/cm**2 \n (d) Thermal efficiency is %3.1f percent"%(Vo,Jc,Ja,P,nth)

(a) The output voltage is 1.1629 V
(b) The current density in the cathode is 4.239 A/cm**2 and anode is 1.092 A/cm**2
(c) Power output per unit area is 3.66 W/cm**2
(d) Thermal efficiency is 44.3 percent


## Ex: 3.8 Pg: 159¶

In :
from __future__ import division
from math import sqrt
#Input data
P=100#Power in kW
V=115#Voltage in V
To=1500#Outer temperature in K
Te=1000#Exit temperature in K
Ta=350#Ambient temperature in K
nth=30#Thermal efficiency in percent
nge=92#Generator efficiency in percent
#Properties of thermoelectrons
a=0.0012#At 1250K in V/K
kp=0.02#In W/cm.K
kn=0.03#In W/cm.K
dp=0.01#In ohm.cm
dn=0.012#In ohm.cm
J=20#Current density in A/cm**2

#Calculations
zmax=(a**2/(sqrt(dp*kp)+sqrt(dn*kn))**2)#Maximum value of figure of merit in K**-1
mo=sqrt(1+(zmax*((To+Te)/2)))#Optimum value of the resistance ratio
nmax=(((To-Te)/To)*((mo-1)/(mo+(Te/To))))*100#Maximum thermal efficiency in percent
Vl=(a*(To-Te)*(mo/(mo+1)))#Voltage per couple in V
nc=(V/Vl)#Number of couples in series
L=((a*(To-Te))/((1+mo)*(dp+dn)))/J#Length in cm
A=((P*Te)/V)/J#Area in cm**2
I=(J*A)#Current in A
Vo=(a*(To-Te))#Voltage in V
Q1=((a*I*To)-((1/2)*(L/A)*I**2*(dp+dn))+((A/L)*(kp+kn)*(To-Te)))/1000#Heat input to the thermoelectric generator in kW
Q2=((a*I*Te)+((A/L)*(kp+kn)*(To-Te))+P)/1000#Heat rejected at full load in kW
Q1n=(((A/L)*(kp+kn)*(To-Te)))/1000#At no load heat input in kW
Q2n=Q1n#At no load heat rejected in kW
no=((nmax/100)+(1-(nmax/100))*(nth/100)*(nge/100))*100#Overall efficiency in percent

#Output
print "(a) The thermal efficiency of thermocouple generator is %3.1f percent \n (b) The number of thermo couples in series is %d \n (c) The lenght of the thermal elements is %3.3f cm and area is %3.2f cm**2 \n (d) The output open-circuit voltage is %3.1f V \n (e) At full load: \n The heat input is %3.3f kW \n The heat rejected is %3.3f kW \n At no load: \n The heat input is %3.3f kW \n The heat rejected is %3.3f kW \n (f) The overall efficiency of the combined thermo-electric steam power plant is %3.2f percent"%(nmax,nc,L,A,Vo,Q1,Q2,Q1n,Q2n,no)

(a) The thermal efficiency of thermocouple generator is 9.1 percent
(b) The number of thermo couples in series is 309
(c) The lenght of the thermal elements is 0.519 cm and area is 43.48 cm**2
(d) The output open-circuit voltage is 0.6 V