# Chapter 2 : Analysis of Steam Cycles¶

## Ex: 2.1 Pg: 82¶

In [2]:
#Input data
p1=40#Initial pressure of steam in bar
T1=500#Initial temperature of steam in degree C
m1=5500#Rate of steam in kg/h
p2=2#Pressure of steam after expansion in bar
n1=0.83#Isentropic efficiency
q=0.87#Quality
m2=2700#Mass flow rate in kg/h
p3=0.1#Pressure of steam after expansion in l.p turbine in bar
n2=0.78#Isentropic efficiency

#Calculations
h1=3445.3#Enthalpy in kJ/kg
s1=7.0901#Entropy in kJ/kg.K which is 1.5301+x2s*5.5970
x2s=(5.5600/5.5970)#dryness fraction
h2s=(504.7+(x2s*2201.9))#Enthalpy in kJ/kg
h2=h1-(n1*(h1-h2s))#Enthalpy in kJ/kg
h3=(504.7+(q*2201.9))#Enthalpy in kJ/kg
h4=((m2*h3+m1*h2)/(m1+m2))#Enthalpy in kJ/kg
x4=(2183.78/2201.9)#dryness fraction
s4=(1.5301+x4*5.5970)#Entropy in kJ/kg.K
x5s=0.8574#dryness fraction
h5s=(191.84+x5s*2392.5)#Enthalpy in kJ/kg
dh4h5=(n2*(h4-h5s))#Difference in enthalpy (h4-h5) in kJ/kg
h6=191.83#Enthalpy in kJ/kg
W1=((m1*(h1-h2))+((m1+m2)*dh4h5))/3600#Power output of the plant in kW
Q1=(m1*(h1-h6))/3600#Heat input in kW
n1=(W1/Q1)*100#Efficiency in percent
WT=(m1*(h1-h2))/3600#Power output without the geothermal heat supply in kW
Q2=(m1*(h1-h6))/3600#Heat input without the geothermal heat supply in kW
n2=(WT/Q2)*100#Efficiency of the cycle without the geothermal heat supply in percent

#Output
print "(a)Power output of the cycle is %3.1f kW \n Efficiency of the cycle is %3.1f percent \n\n (b)Without geothermal heat supply \n Power output of the cycle is %3.2f kW \n Efficiency of the cycle is %3.2f percent"%(W1,n1,WT,n2)

(a)Power output of the cycle is 1746.3 kW
Efficiency of the cycle is 35.1 percent

(b)Without geothermal heat supply
Power output of the cycle is 955.17 kW
Efficiency of the cycle is 19.22 percent


## Ex: 2.2 Pg: 83¶

In [3]:
#Input data
p1=90#Initial pressure of steam in bar
T1=500#Initial temperature of steam in degree C
O=(500*1000)#Output in kW
T2=40#Condensation temperature in degree C
nhp=0.92#Efficiency of h.p turbine
nlp=0.9#Efficiency of l.p turbine
np=0.75#Isentropic efficiency of the pump
TTD=-1.6#Temperature in degree C

#Calculations
p2=(0.2*p1)#Optimum reheat pressure in bar
h1=3386.1#Enthalpy in kJ/kg
s1=6.6576#Entropy in kJ/kg.K
s2s=s1#Entropy in kJ/kg.K
h2s=2915#Enthalpy in kJ/kg
h3=3469.8#Enthalpy in kJ/kg
s3=7.4825#Entropy in kJ/kg.K
x4s=(s3-0.5725)/7.6845#Dryness fraction
h4s=(167.57+x4s*2406.7)#Enthalpy in kJ/kg
h5=167.57#Enthalpy in kJ/kg
h7=883.42#Enthalpy in kJ/kg
Wps=(0.001008*p1*10)#Workdone by the pump in kJ/kg
h6s=176.64#Enthalpy in kJ/kg
dh1h2=(nhp*(h1-h2s))#Difference in enthalpy (h1-h2) in kJ/kg
h2=h1-dh1h2#Enthalpy in kJ/kg
dh3h4=(nlp*(h3-h4s))#Difference in enthalpy (h3-h4) in kJ/kg
h4=h3-dh3h4#Enthalpy in kJ/kg
Wp=(Wps/np)#Workdone by the pump in kJ/kg
h6=(Wp+h5)#Enthalpy in kJ/kg
tsat=207.15#Saturation temperature at 18 bar in degree C
t9=(tsat-TTD)#Temperature in degree C
h9=875#Enthalpy in kJ/kg
m=((h9-h6)/(h2-h7))#Mass of steam in kg
WT=(dh1h2+(1-m)*dh3h4)#Workdone by the turbine in kJ/kg
Wnet=(WT-Wp)#Net workdone in kJ/kg
ws=(O/Wnet)#Mass flow rate of steam at turbine inlet in kg/s
Q1=((h1-h9)+(1-m)*(h3-h2))#Heat input in kJ/kg
n=(Wnet/Q1)*100#Efficiency of the cycle in percent
Wr=(Wnet/WT)#Work ratio

#Output
print "(a)Mass flow rate of steam at turbine inlet is %3.0f kg/s \n (b)The cycle efficiency is %3.2f percent \n (c)Work ratio is %3.3f"%(ws,n,Wr)

(a)Mass flow rate of steam at turbine inlet is 452 kg/s
(b)The cycle efficiency is 38.82 percent
(c)Work ratio is 0.999


## Ex: 2.3 Pg: 86¶

In [4]:
#Input data
p1=70#Pressure at which an ideal seam power plant operates in bar
T1=550#Temperature at which an ideal seam power plant operates in degrees C
p2=0.075#Pressure at which an ideal seam power plant operates in bar

#Calculations
TB=285.9#Saturation temperature at 70 bar in degree C
TC=40.3#Saturation temperature at 0.075 bar in degree C
Tr=(TB-TC)/(7+1)#Temperature rise per heater for maximum cycle efficiency in degree C
t1=(TB-Tr)#Temperature at heater 1 in degree C
P1=4.33#Pressure at heater 1 in MPa
t2=(t1-Tr)#Temperature at heater 2 in degree C
P2=2.5318#Pressure at heater 2 in MPa
t3=(t2-Tr)#Temperature at heater 3 in degree C
P3=1.367#Pressure at heater 3 in MPa
t4=(t3-Tr)#Temperature at heater 4 in degree C
P4=0.6714#Pressure at heater 4 in MPa
t5=(t4-Tr)#Temperature at heater 5 in degree C
P5=0.2906#Pressure at heater 5 in MPa
t6=(t5-Tr)#Temperature at heater 6 in degree C
P6=0.108#Pressure at heater 6 in MPa
t7=(t6-Tr)#Temperature at heater 7 in degree C
P7=32.65#Pressure at heater 7 in kPa

#Output
print "The optimum pressure and temperature at different heaters are: \n Heater 1: t1 = %3.1f degree C and p1 = %3.2f MPa\n Heater 2: t2 = %3.1f degree C and p2 = %3.4f MPa\n Heater 3: t3 = %3.1f degree C and p3 = %3.3f MPa\n Heater 4: t4 = %3.1f degree C and p4 = %3.4f MPa\n Heater 5: t5 = %3.1f degree C and p5 = %3.4f MPa\n Heater 6: t6 = %3.1f degree C and p6 = %3.3f MPa\n Heater 7: t7 = %3.1f  degree C and p7 = %3.2f kPa"%(t1,P1,t2,P2,t3,P3,t4,P4,t5,P5,t6,P6,t7,P7)

The optimum pressure and temperature at different heaters are:
Heater 1: t1 = 255.2 degree C and p1 = 4.33 MPa
Heater 2: t2 = 224.5 degree C and p2 = 2.5318 MPa
Heater 3: t3 = 193.8 degree C and p3 = 1.367 MPa
Heater 4: t4 = 163.1 degree C and p4 = 0.6714 MPa
Heater 5: t5 = 132.4 degree C and p5 = 0.2906 MPa
Heater 6: t6 = 101.7 degree C and p6 = 0.108 MPa
Heater 7: t7 = 71.0  degree C and p7 = 32.65 kPa


## Ex: 2.4 Pg: 87¶

In [5]:
#Input data
ng=0.97#Efficiency of electric generator
nt=0.95#Efficiency of turbine
nb=0.92#Efficiency of boiler
nc=0.42#Efficiency of cycle
no=0.33#Efficiency of overall plant

#Calculations
na=(no/(ng*nt*nb*nc))#Efficiency of auxiliaries
n=(1-na)*100#Percentage of total electricity generated which is consumed in running the auxiliaries

#Output
print "Percentage of total electricity generated which is consumed in running the auxiliaries is %3.2f percent"%(n)

Percentage of total electricity generated which is consumed in running the auxiliaries is 7.32 percent


## Ex: 2.5 Pg: 87¶

In [6]:
#Input data
T1=140#Temperature with which feed water enters into economiser in degree C
T2=[25,250]#Temperature from air is preheated to in degree C
P1=60#Pressure with which steam leaves the drum in bar
x1=0.98#Dryness fraction
T3=450#Temperature with which steam leaves the superheater in degree C
cc=25.2#Calorific value of coal in MJ/kg
r=8.5#Rate of evaporation of steam per kg coal
wf=1#Mass of coal in kg
R=15#Air fuel ratio by mass
Cpa=1.005#Specific heat of air at constant pressure in kJ/kg.K
Cpw=4.2#Specific heat of water at constant pressure in kJ/kg.K

#Calculations
h1=(T1*Cpw)#Enthalpy in kJ/kg
hf=1213.35#Enthalpy in kJ/kg
h2=hf#Enthalpy in kJ/kg
hfg=1571#Enthalpy in kJ/kg
h4=3301.8#Enthalpy in kJ/kg
h3=(hf+x1*hfg)#Enthalpy in kJ/kg
n=((r*(h4-h1))/(wf*cc*1000))*100#Efficiency
he=(r*(h2-h1))/wf*10**-3#Heat transfer in the economiser in MJ/kg
hb=(r*(h3-h2))/wf*10**-3#Heat transfer in the boiler in MJ/kg
hs=(r*(h4-h3))/wf*10**-3#Heat transfer in the superheater in MJ/kg
ha=(R*Cpa*(T2[1]-T2[0]))/wf*10**-3#Heat transfer in the air preheater in MJ/kg
pe=((h2-h1)/(h4-h1))*100#Percentage of total heat absorbed in the economiser in percent
pb=((h3-h2)/(h4-h1))*100#Percentage of total heat absorbed in the boiler in percent
ps=((h4-h3)/(h4-h1))*100#Percentage of total heat absorbed in the superheater in percent

#Output
print "Efficiency of steam generator is %3.2f percent \n\n Heat transfer per kg fuel in \n (i)economiser is %3.4f MJ/kg \n (ii)boiler is %3.3f MJ/kg \n (iii)superheater is %3.3f MJ/kg \n (iv)air pre-heater is %3.3f MJ/kg \n\n Percentage of total heat absorption taking place in \n (i)economiser is %3.2f percent \n (ii)boiler is %3.2f percent \n (iii)superheater is %3.2f percent"%(n,he,hb,hs,ha,pe,pb,ps)

Efficiency of steam generator is 91.54 percent

Heat transfer per kg fuel in
(i)economiser is 5.3155 MJ/kg
(ii)boiler is 13.086 MJ/kg
(iii)superheater is 4.665 MJ/kg
(iv)air pre-heater is 3.392 MJ/kg

Percentage of total heat absorption taking place in
(i)economiser is 23.04 percent
(ii)boiler is 56.73 percent
(iii)superheater is 20.23 percent


## Ex: 2.6 Pg: 88¶

In [7]:
#Input data
p1=150#Pressure of inlet steam in bar
T1=550#Temperature of steam in degree C
p2=20#Pressure after expansion in bar
T2=500#Reheat temperature in degree C
pc=0.075#Condenser pressure in bar
php=50#Pressure of steam in h.p turbine in bar
pip=[10,5,3]#Pressure of steam in i.p turbines in bar
plp=1.5#Pressure of steam in l.p turbine in bar
m=300*1000#Steam flow rate in kg/h

#Calculations
h1=3448.6#Enthalpy in kJ/kg
h4=3467.6#Enthalpy in kJ/kg
s1=6.5119#Entropy in kJ/kg.K
s2=s1#Entropy in kJ/kg.K
s3=s1#Entropy in kJ/kg.K
s4=7.4317#Entropy in kJ/kg.K
s5=s4#Entropy in kJ/kg.K
s6=s5#Entropy in kJ/kg.K
s7=s6#Entropy in kJ/kg.K
s8=s7#Entropy in kJ/kg.K
s9=s8#Entropy in kJ/kg.K
t2=370#Temperature in degree C
t3=245#Temperature in degree C
t5=400#Temperature in degree C
t6=300#Temperature in degree C
t7=225#Temperature in degree C
t8=160#Temperature in degree C
h2=3112#Enthalpy in kJ/kg
h3=2890#Enthalpy in kJ/kg
h5=3250#Enthalpy in kJ/kg
h6=3050#Enthalpy in kJ/kg
h7=2930#Enthalpy in kJ/kg
h8=2790#Enthalpy in kJ/kg
x9=(s9-0.5764)/7.6751#Dryness fraction
h9=168.79+x9*2406##Enthalpy in kJ/kg
h10=168.79#Enthalpy in kJ/kg
h11=h10+0.001*pip[1]*100#Enthalpy in kJ/kg
h12=467.11#Enthalpy in kJ/kg
t14=111.37#Temperature in degree C
h14=467#Enthalpy in kJ/kg
h13=h12#Enthalpy in kJ/kg
h14=h13#Enthalpy in kJ/kg
h15=h14#Enthalpy in kJ/kg
h16=561.47#Enthalpy in kJ/kg
h17=h16#Enthalpy in kJ/kg
h18=640.23#Enthalpy in kJ/kg
h19=h18+0.001*(p1-pip[1])*100#Enthalpy in kJ/kg
h20=762.8#Enthalpy in kJ/kg
h21=h20#Enthalpy in kJ/kg
h22=1154.23#Enthalpy in kJ/kg
h23=h22#Enthalpy in kJ/kg
m1=((h23-h21)/(h2-h22))#Mass in kg
m2=((h21-h19)-(m1*(h22-h20)))/(h5-h20)#Mass in kg
m3=(((1-m1-m2)*(h18-h17))-((m1+m2)*(h20-h18)))/(h6-h18+h18-h17)#Mass in kg
m4=((1-m1-m2-m3)*(h17-h15))/(h7-h16)#Mass in kg
m5=(((1-m1-m2-m3-m4)*(h14-h11))-(m4*(h16-h12)))/(h8-h12+h14-h11)#Mass in kg
WT=(h1-h2)+(1-m1)*(h2-h3)+(1-m1)*(h4-h5)+(1-m1-m2)*(h5-h6)+(1-m1-m2-m3)*(h6-h7)+(1-m1-m2-m3-m4)*(h7-h8)+(1-m1-m2-m3-m4-m5)*(h8-h9)#Workdone by turbine in kJ/kg
Wp=(0.5+14.5+0.15)#Workdone in kJ/kg
Wnet=(WT-Wp)#Net workdone in kJ/kg
Q1=(h1-h23)+(1-m1)*(h4-h3)#Heat supplied in kJ/kg
ncy=(Wnet/Q1)*100#Cycle efficiency in percent
t23=264#Temperature in degree C
sr=(3600/Wnet)#Steam rate in kJ/kWh
hr=((Q1/Wnet)*3600)#Heat rate in kJ/kWh
P=((Wnet*m)/3600)/10**3#Power output in MW

#Output
print "(a) The cycle efficiency is %3.2f percent \n (b) The feedwater temperature is %d degree C \n (c) The steam rate is %3.2f kJ/kWh \n (d) The heat rate is %3.0f kJ/kWh \n (e) The quality of steam at turbine exhaust is %3.4f \n (f) The power output is %3.2f MW"%(ncy,t23,sr,hr,x9,P)

(a) The cycle efficiency is 48.58 percent
(b) The feedwater temperature is 264 degree C
(c) The steam rate is 2.69 kJ/kWh
(d) The heat rate is 7411 kJ/kWh
(e) The quality of steam at turbine exhaust is 0.8932
(f) The power output is 111.58 MW


## Ex: 2.7 Pg: 92¶

In [8]:
#Input data
m=10000#Mass flow rate of steam in kg/h
p=3#Pressure of steam in bar
P=1000#Power in kW
n=0.7#Internal efficiency of turbine

#Calculations
dh=(P*3600)/m#Change in enthalpy in kJ/kg
h2=2725.3#Enthalpy in kJ/kg from Fig. E2.7
h1=dh+h2#Enthalpy in kJ/kg
dh1h2s=dh/n#Change in enthalpy in kJ/kg
h2s=h1-dh1h2s#Enthalpy in kJ/kg
x2s=(h2s-561.47)/2163.8#Dryness fraction
s2s=1.6718+x2s*(6.999-1.6718)#Entropy in kJ/kg.K
s1=s2s#Entropy in kJ/kg.K
p1=37.3#Pressure in bar from Mollier diagram
t1=344#Temperature in degree C

#Output
print "The steam condition required at inlet of the turbine: \n Enthalpy is %3.1f kJ/kg \n Entropy is %3.4f kJ/kg.K \n Pressure is %3.1f bar \n Temperature is %d degree C"%(h1,s1,p1,t1)

The steam condition required at inlet of the turbine:
Enthalpy is 3085.3 kJ/kg
Entropy is 6.6192 kJ/kg.K
Pressure is 37.3 bar
Temperature is 344 degree C


## Ex: 2.8 Pg: 93¶

In [9]:
from __future__ import division
#Input data
p1=40#Pressure in bar
T1=500+273#Temperature in K
p2=0.06#Pressure in bar
p3=2#Pressure in bar
CV=25#Calorific value in MJ/kg
n=88#Boiler efficiency in percent
T=6#Temperature rise in degree C

#Calculations
h1=3445.3#Enthalpy in kJ/kg
s1=7.0901#Entropy in kJ/kg.K
s2=s1#Entropy in kJ/kg.K
s3=s1#Entropy in kJ/kg.K
x2=(s2-1.5301)/5.5970#Dryness fraction
h2=2706.7#Enthalpy in kJ/kg
h26=2201.9#Difference in enthalpy in kJ/kg
w=(Hl*10**3)/h26#Rate of steam extraction in kg/h
x3=(s1-0.52)/7.815#Dryness fraction
h3=(149.79+x3*2416)#Enthalpy in kJ/kg
h4=149.79#Enthalpy in kJ/kg
ws=((Pl*10**3+(w*(h2-h3)))/((h1-h2)+(h2-h3)))#Steam generation capacity in kg/s
ws1=(ws*3600)/1000#Steam generation capacity in t/h
h7=(504.7+(1.061*10**-3*(p1-p3)*100))#Enthalpy in kJ/kg
h5=(149.79+(1.006*100*p1*10**-3))#Enthalpy in kJ/kg
Q1=(((ws-w)*(h1-h5))+(w*(h1-h7)))#Heat input in kW
wf=((Q1/1000)/((n/100)*CV))*(3600/1000)#Fuel burning rate in t/h
Q2=((ws-w)*(h3-h4))#Heat rejected to the condensor in kW
wc=(Q2/(4.187*T))/1000#Rate of flow of cooling water in m**3/s

#Output
print "(a) the steam generation capacity of the bolier is %3.2f t/h \n (b) the heat input to the boiler is %3.1f kW \n (c) the fuel burning rate of the bolier is %3.3f t/h \n (d) the heat rejected to the condensor is %3.0f kW \n (e) the rate of flow of cooling water in the condensor is %3.3f m**3/s"%(ws1,Q1,wf,Q2,wc)

(a) the steam generation capacity of the bolier is 16.74 t/h
(b) the heat input to the boiler is 15113.7 kW
(c) the fuel burning rate of the bolier is 2.473 t/h
(d) the heat rejected to the condensor is 8369 kW
(e) the rate of flow of cooling water in the condensor is 0.333 m**3/s


## Ex: 2.9 Pg: 94¶

In [10]:
#Input data
m=21000#Steam rate in kg/h
p1=17#Pressure in bar
T1=230+273#Temperature in K
P=132.56#Power in kW
x2=0.957#Dryness fraction
p2=3.5#Pressure in bar
Pl=1337.5#Power in l.p turbine in kW
p3=0.3#Pressure in bar
x3=0.912#Dryness fraction

#Calculations
h1=2869.7#Enthalpy in kJ/kg
s1=6.5408#Entropy in kJ/kg.K
h2=(870.44+x2*1924.7)#Enthalpy in kJ/kg
h3=h2#Enthalpy in kJ/kg
h56=(Pl*3600)/m#Difference in Enthalpy in kJ/kg
h6=(289.23+x3*2336.1)#Enthalpy in kJ/kg
h5=2649.04#Enthalpy in kJ/kg
s4s=s1#Entropy in kJ/kg.K
x4s=(s4s-1.7275)/5.2130#Dryness fraction
h4s=584.33+x4s*2148.1#Enthalpy in kJ/kg
w=(P/(h1-h2))#Flow rate in kg/s
ws=(m/3600)#Steam flow rate in kg/s
h4=((ws*h5)-(w*h3))/(ws-w)#Enthalpy in kJ/kg
x4=(h4-584.33)/2148.1#Dryness fraction
W=(ws-w)*(h1-h4)#Power developed by h.p turbine in kW
n=((h1-h4)/(h1-h4s))*100#Isentropic efficiency in percent

#Output
print "(a) the steam quality at the exhaust of the h.p turbine is %3.3f \n (b) the power developed by the h.p turbine is %3.2f kW \n (c) the isentropic efficiency of the h.p turbine is %3.2f percent"%(x4,W,n)

(a) the steam quality at the exhaust of the h.p turbine is 0.956
(b) the power developed by the h.p turbine is 1154.62 kW
(c) the isentropic efficiency of the h.p turbine is 76.61 percent