# Chapter 6 : Steam generators¶

## Ex: 6.1 Pg: 411¶

In :
#Input data
H=18##The length of furnace wall riser in m
O=76.2##The outer diameter of the furnace wall riser in mm
T=6.1##The thickness of the furnace wall riser in mm
P=80##Pressure at which saturated water is recieved in bar
V=1.5##The velocity of the saturated water in m/s
CR=12.5##Assuming circulation ratio
S=1.2##Assuming slip ratio
g=9.81##Gravitational force constant in m/s**2
pi=3.142##Mathematical constant

#Calculations
xt=1/CR##The quality of steam at the top of the riser
vf=0.001384##Specific volume of saturated liquid at 80 bar in m**3/kg
vfg=0.02214##Specific volume of Evaporation gas at 80 bar in m**3/kg
vg=0.02352##Specific volume of saturated gas at 80 bar in m**3/kg
pf=1/vf##Density of the saturated liquid at 80 bar in kg/m**3
vt=vf+(xt*vfg)##Specific volume of the steam at the top of the riser in m**3/kg
pt=1/vt##Density of steam at the top of the riser in kg/m**3
pm=(pt+pf)/2##Mean density in kg/m**3
Ph=(H*g*(pf-pm))/1000##The pressure head developed in kPa
C=(vf/vg)*S##The part of calculation for the void fraction
VF=1/(1+((1-xt)*C)/xt)##The void fraction at riser exit
hfg=1441.3##Enthalpy of the evaporation in kJ/kg
di=O-12.2##Inner diameter of the furnace wall riser in mm
A=(pi*di**2)/4##Inner area in m**2
w=pf*A*V##Mass flow rate of saturated water entering the riser in kg/s
ws=xt*w##The rate of steam formation in the riser tube in kg/s
h=((ws*hfg)/(O*H))/1000##Heat transfer rate per unit projected area in kW/m**2

#output
print " (a) The pressure head developed = %3.1f kPa \n (b)The void fraction at riser exit = %3.4f \n (c) The heat transfer rate per unit projected area = %3.1f kW/m**2 "%(Ph,VF,h)


 (a) The pressure head developed = 35.8 kPa
(b)The void fraction at riser exit = 0.5519
(c) The heat transfer rate per unit projected area = 293.1 kW/m**2


## Ex: 6.2 Pg: 413¶

In :
from __future__ import division
#Input data
t=60##The temperature of water while supplying it to desuperheater in degree centigrade
ws=200##The amount of steam carrying in a steam line in t/h
p=35##The pressure of steam in bar
ts=400##The temperature to be maintained by the steam in degree centigrade
to=450##The outlet temperature of the steam from boiler in degree centigrade
h1=3337.2##The enthalpy of steam at 450 degree C in kJ/kg
h2=252##The enthalpy of water at 60 degree C in kJ/kg
h3=3222.3##The enthalpy of steam at 400 degree C in kJ/kg

#Calculations
w=(ws*(h1-h3))/(h3-h2)##Mass flow rate of water in t/h
w1=w*(1000/3600)##Mass flow rate of water in kg/s

#Output
print "The amount of water that must be sprayed is %3.3f t/h  or  %3.3f kg/s "%(w,w1)

The amount of water that must be sprayed is 7.737 t/h  or  2.149 kg/s


## Ex: 6.3 Pg: 413¶

In :
from math import log
#Input data

H=15##The high of downcomer riser circuit in m
P=160##The pressure at which downcomer riser circuit operates in bar
xe=0.5##The exit quality of the steam
S=1.2##Slip factor
vf=0.001711##Specific volume of saturated liquid in m**3/kg
vg=0.009306##Specific volume of saturated gas in m**3/kg
g=9.806##Gravitational force constant in m/s**2

#Calculations
C=S*(vf/vg)##The part of calculation for the void fraction
VF=1/(1+((1-xe)*C)/xe)##The void fraction at riser exit
pf=1/vf##Density of the saturated liquid in kg/m**3
pg=1/vg##Density of the saturated gas in kg/m**3
pm=pf-(((pf-pg)/(1-C))*(1-((1/((VF)*(1-C)))-1)*log(1/(1-(VF*(1-C))))))##The average mixture density in the riser in kg/m**3
P1=g*(pf-pm)*H##Pressure head developed due to natural circulation in N/m**2
P2=P1/1000##ressure head developed due to natural circulation in kPa

#Output
print "The pressure head developed due to natural circulation is %3.0f N/m**2  or  %3.3f kPa"%(P1,P2)

The pressure head developed due to natural circulation is 38163 N/m**2  or  38.163 kPa


## Ex: 6.4 Pg: 414¶

In :
from math import ceil
#Input data

W=120##The amount of electricity produced in the power plant in MW
po=100##The pressure of the steam at the outlet of boiler in bar
to=500##The temperature of steam at the outlet of boiler in degree centigrade
p=0.1##The condenser pressure in bar
nb=0.9##The efficiency of the boiler
CV=25.7##The calorific value of the coal in MJ/kg
ti=160##The feed water temperature at boiler inlet in degree centigrade
H=40##The high of the risers in the furnace wall in m
xt=0.08##The quality of the steam at the top of the riser
v=2##The exit velocity of the riser and entering the drum in m/s
Do=60##The outer diameter of the risers in mm
T=3##The thickness of the wall in mm
pi=3.142##Mathematical constant
g=9.806##Gravitational force constant in m/s**2

#Calculations
h1=3374.8##Enthalpy at point 1 in kJ/kg
s1=6.6011##Entropy at point 1 in kJ/kgK
sf=0.6479##Entropy of the saturated liquid at point 1 in kJ/kgK
sg=7.5055##Entropy of the Saturated vapour at point 1 in kJ/kgK
x2=(s1-sf)/sg##The quality of the steam
h2=191.46+(x2*2393.29)##Enthalpy at point 2 in kJ/kg
h3=191.46##Enthalpy at point 3 in kJ/kg
h5=675.5##Enthalpy at point 5 in kJ/kg
ws=(W*1000)/(h1-h2)##Mass flow rate of steam in kg/s
wf=(ws*(h1-h5))/(nb*CV*1000)##Mass flow rate of fuel in kg/s
E=ws/wf##Evaporation factor
vf=0.0014523##The specific volume of saturated liquid in m**3/kg
vg=0.0165884##The specific volume of saturated vapour in m**3/kg
vt=vf+(xt*vg)##Specific volume at the top in m**3/kg
pt=1/vt##Density of the steam at the top in kg/m**3
pf=1/vf##The density of the steam in kg/m**3
pm=(pf+pt)/2##The average mixture density in kg/m**3
H1=(g*H*(pf-pm))/10**5##Pressure head available for natural circulation in bar
CR=1/xt##Circulation ratio
di=(Do-(2*T))/1000##The inner diameter of the riser in m
A=(pi*di**2)/4##Area for the inner diameter in m**2
w=(A*pt*v*xt)##The rate of steam formation in the riser in kg/s
Nr=ceil(ws)/w##The number of risers
hfg=1319.8##Enthalpy of the evaporation in kJ/kg
Ha=(w*hfg)/((Do/1000)*H)##Heat absorption rate per unit projected area of the riser in kW/m**2

#Output
print " (a)The steam generation rate = %3.3f kg/s \n (b) The fuel burning rate = %3.3f kg/s \n (c) The evaporation factor = %3.2f \n (d) The pressure head available for natural circulation = %3.4f bar \n (e) The circulation ratio = %3.1f \n (f)The number of risers required = %3.0f \n (g) The heat absorbtion rate per unit projected area of the riser = %3.2f kW/m**2"%(ws,wf,E,H1,CR,Nr,Ha)

 (a)The steam generation rate = 93.383 kg/s
(b) The fuel burning rate = 10.898 kg/s
(c) The evaporation factor = 8.57
(d) The pressure head available for natural circulation = 0.6448 bar
(e) The circulation ratio = 12.5
(f)The number of risers required = 713
(g) The heat absorbtion rate per unit projected area of the riser = 72.51 kW/m**2


## Ex: 6.5 Pg: 415¶

In :
#Input data
ws=64##The steam flow rate in kg/s
p=60##The pressure at which steam leaves the boiler in bar
m=0.02##Moisture contant in the steam
wf=62##The feedwater flow rate in kg/s
Pf=3##concentration of feedwater in ppm
wm=2##The flow rate of makeup water
Pm=50##concentration of makeup water in ppm
Ps=5##Leaving the drum water in ppm
Pw=1000##The concentration in the drum water in ppm
mf=7##The fuel burning rate in kg/m
CV=23##The heating value in MJ/kg
ta=30##The room temperature in degree centigrade
hf=1213.35##Enthalpy of saturated liquid at 60 bar in kJ/kg
ha=125.79##Enthalpy at ambient temperature in kJ/kg

#Calculations
BD=((wf*Pf)+(wm*Pm)-(m*ws*Ps))/1000##The rate of blowdown in kg/s
E=((BD*(hf-ha))/(mf*CV*1000))*100##The energy loss in blowdown in percentage
S=m*ws*Ps*10**-6*3600*24##Scale deposition in superheater tubes

#Output
print " (a)The blowdown required = %3.4f kg/s \n (b) Heat loss in blowdown as a percentage of total heat released in the furnace = %3.2f percentage \n (c) The deposition of scale in superheater tube = %3.3f kg/day "%(BD,E,S)

 (a)The blowdown required = 0.2796 kg/s
(b) Heat loss in blowdown as a percentage of total heat released in the furnace = 0.19 percentage
(c) The deposition of scale in superheater tube = 0.553 kg/day


## Ex: 6.6 Pg: 416¶

In :
from math import log
#Input data
ws=600##Mass flow rate of feedwater in kg/s
p=140##The inlet pressure of the feedwater in bar
t=170##The inlet temperature of the feedwater in degree centigrade
wg=1250##The mass flow rate of flue gases in kg/s
tg2=450##The temperature at which flue gases leave the economisers coils in degree centigrade
Vf=12##The velocity of the flue gas in m/s
Vw=1.2##The velocity of the water leaving the coil in m/s
Do=0.07##The outer diameter of the tube in m
Di=0.06##The inner diameter of the tube in m
U=70##The overall heat transfer coefficient in W/m**2K
Cp=1.12##The specific heat capacity of the flue gases in kJ/kgK
V=0.08##The vertical pitch of the coil in m
B=4.8##The width of the duct in m
C=0.005##The clearence on the both sides of the duct in m
pi=3.142##Mathematical constant

#Calculations
hf=1571.1##The enthalpy of the saturated liquid at 140 bar in kJ/Kg
ts=336.75##The saturated temperature at 140 bar in degree centigrade
vf=0.001611##The specific volume of the saturated liquid at 140 bar in m**3/kg
hf1=719.21##The enthalpy of the saturated liquid at 170 degree C in kJ/kg
vf1=0.001114##The specific volume of the saturated liquid at 170 degree C in m**3/kg
tg1=((ws*(hf-hf1))/(wg*Cp))+tg2##The temperature at which flue gases enters the economisers coils in degree centigrade
t1m=(478.25-280)/(log(478.25/280))##The mean temperature for inlet and exit temperature in degree centigrade
Q=ws*(hf-hf1)##The rate of heat transfer in the economiser in kW
Ao=(Q/(U*t1m))*10**3##The outer area in m**2
n=((ws*(vf/Vw)*(4/pi)*(1/Di**2)))##The number of coils needed in the economiser
l=Ao/(n*pi*Do)##The length of one coil in m
nt=l/(B-(2*C))##The number of turns in on ecoil
VH=nt*V##The vertical height of the duct occupied by the economiser coils

#Output
print " (a) The number of coils needed in the economiser = %3.0f \n (b) The length of one coil = %3.1f m \n (c) The verticle height of the duct occupied by the economiser coils = %3.2f m "%(n,l,VH)

 (a) The number of coils needed in the economiser = 285
(b) The length of one coil = 314.7 m
(c) The verticle height of the duct occupied by the economiser coils = 5.26 m


## Ex: 6.7 Pg: 418¶

In :
from math import log
#Input data

tg2=160##The temperature to which the flue gases are cooled in degree centigrade
ta1=35##The ambient temperature of the air in degree centigrade
wa=1167##The mass flow rate of air in kg/s
Vg=13##The inlet velocity of the flue gases in m/s
U=30##The overall heat transfer coefficient in W/m**2K
Cpg=1.10##The specific heat of the flue gas in kJ/kgK
Cpa=1.005##The specific heat of the air in kJ/kgK
R=0.287##Real gas constant in kJ/kgK
wg=1250##The mass flow rate of gas in kg/s
tg1=450##The temperature at the inlet of flue gas in degree centigrade
P=101.325##Atmospheric temperature in kPa
pi=3.1414##Mathematical constant
Di=0.06##The inner diameter of the tube in m
Do=0.065##The outer diameter of the tube in m

#Calculations
vg1=(R*(273+tg1))/P##Specific volume of the gas in m**3/kg
ta2=((wg*Cpg*(tg1-tg2))/(wa*Cpa))+ta1##The temperature of the heated air in degree centigrade
t1m=(75-125)/log(75/125)##The mean temperature of the inlet and exit temperature in degree centigrade
Q=wg*Cpg*(tg1-tg2)##The rate of heat transfer in the economiser in kW
Ao=(Q/(U*t1m))*10**3##The outer area in m**2
n=((wg*(vg1/Vg)*(4/pi)*(1/Di**2)))##The number of coils needed in the economiser
l=Ao/(n*pi*Do)##The length of one coil in m

#Output
print "(a)The length of the tubes = %3.2f m\n (b) The number of tubes = %3.0f  "%(l,n)

(a)The length of the tubes = 9.55 m
(b) The number of tubes = 69647


## Ex: 6.8 Pg: 419¶

In :
#Input data
di=0.05##The inner diameter of the superheater coil in m
T=0.005##The thickness of the coil in m
p=60##The pressure of the steam at the exit in bar
t=500##The temperature of the steam at the exit in degree centigrade
V2=10##The velocity of the steam at the exit in m/s
ws=80##The mass flow rate of steam in kg/s
H=140##The heat flux in the super heated coils in kW/m**2
pi=3.142##Mathematical constant
Do=0.06##The outer diameter of the tube in m

#Calculations
h1=2784.3##The enthalpy of the saturated gas at 60 bar in kJ/kg
h2=3422.2##The enthalpy of the saturated gas at 500 degreeC in kJ/kg
v2=0.05665##The specific volume of gas at 500 degreeC in m**3/kg
Q=ws*(h2-h1)##Heat absorption rate in superheater coil in kW
Ao=Q/H##Surface area required in m**2
n=((ws*(v2/V2)*(4/pi)*(1/di**2)))##The number of coils needed in the economiser
l=Ao/(n*pi*Do)##The length of one coil in m

#Output
print " (a)The length of the one coil = %3.2f m\n (b) The number of coils = %3.0f  "%(l,n)

 (a)The length of the one coil = 8.38 m
(b) The number of coils = 231