# Chapter 11 Steam Boilers¶

## Example 1 Page No:228¶

In :
#Input data
ms=5000                 #Boiler produces wet steam in Kg/h
x=0.95                  #Dryness function
P=10                    #Operating pressure in bar
mf=5500                 #Bour in the furnace in Kg
Tw=40                   #Feed water temp in degree celsius

#calculation
#from steam table
hfw=167.45              #In KJ/Kg
hf=762.61               #In KJ/Kg
hfg=2031.6              #In KJ/Kg
hs=(hf+x*hfg)           #Enthalpy of wet stream in KJ/Kg
me=ms/mf                #Mass of evaporation
E=((me*(hs-hfw))/(2257))*10 #Equivalent evaporation in Kg/Kg of coal

#output

print("Enthalpy of wet stream=",round(hs,2),"KJ/Kg")
print("Mass of evaporation=",round(me,2),)
print("Equivalent evaporation=",round(E,2),"Kg/Kg of coal")

Enthalpy of wet stream= 2692.63 KJ/Kg
Mass of evaporation= 0.91
Equivalent evaporation= 10.17 Kg/Kg of coal


## Example 2 Page No:229¶

In :
#Input data
p=14                            #Boiler pressure in bar
me=9                            #Evaporates of water in Kg
Tw=35                           #Feed water entering in degree celsius
x=0.9                           #Steam stop value
CV=35000                        #Calorific value  of the coal

#Calculation
#From Steam Table
hfw=146.56                      #In KJ/Kg
hf=830.07                       #In KJ/Kg
hfg=1957.7                      #In KJ/Kg
hs=hf+x*hfg                     #Enthalpy of wet stream in KJ/Kg
E=((me*(hs-hfw))/2257)          #Equivalent evaporation in Kg/Kg of coal
etaboiler=((me*(hs-hfw))/CV)*100#Boiler efficiency in %

#Output
print("Enthalpy of wet stream=",hs,"KJ/Kg")
print("Equivalent evaporation=",round(E,2),"Kg/Kg of coal")
print("Boiler efficiency=",round(etaboiler,2),"%")

Enthalpy of wet stream= 2592.0 KJ/Kg
Equivalent evaporation= 9.75 Kg/Kg of coal
Boiler efficiency= 62.88 %


## Example 3 Page No:228¶

In :
#Input data
ms=2500                          #Saturated steam per bour in Kg
x=1
P=15                             #Boiler pressure in bar
Tw=25                            #Feed water entering in degree celsius
mf=350                           #Coal burnt in Kg/bour
CV=32000                         #Calorific value in Kj/Kg

#calculation
#steam table
hfw=104.77                       #In KJ/Kg
hf=844.66                        #In KJ/Kg
hfg=1945.2                       #In KJ/Kg
hg=2789.9                        #In KJ/Kg
hs=2789.9                        #Enthalpy of dry steam in KJ/Kg
me=ms/mf                         #mass of evaporation
E=((me*(hs-hfw))/2257)           #Equivalent evaporation in Kg/Kg ofcoal
etaboiler=((me*(hs-hfw))/CV)*100 #Boiler efficiency in %

#Output
print("mass of evaporation=",round(me,3),)
print("Equivalent evaporation=",round(E,2),"Kg/Kg ofcoal")
print("Boiler efficiency=",round(etaboiler,2),"%")

mass of evaporation= 7.143
Equivalent evaporation= 8.5 Kg/Kg ofcoal
Boiler efficiency= 59.94 %


## Example 4 Page No:231¶

In :
#Input data
mf=500                          #Boiler plant consumes of coal in Kg/h
CV=32000                        #Calorific value in Kj/Kg
ms=3200                         #plant generates in Kg/h
P=1.2                           #Absolute pressure MN/m**2
MN=12
Tsup=300                        #Absolute temperature in degree celsius
Tw=35                           #Feed water temperature
Cps=2.3

#calculation
hfw=146.56                      #In KJ/Kg
Ts=187.96                       #In Degree celsius
hf=798.43                       #In KJ/Kg
hfg=1984.3                      #In KJ/Kg
hg=2782.7                       #In KJ/Kg
hs=hg+Cps*(Tsup-Ts)             #Enthalpy of superheated steam in KJ/Kg
me=ms/mf                        #mass of evaporation
E=((me*(hs-hfw))/2257)          #Equivalent evaporation in Kg/Kg ofcoal
etaboiler=((me*(hs-hfw))/CV)*100#Boiler efficiency in %

#Output
print("Enthalpy of superheated steam=",round(hs,2),"KJ/Kg")
print("mass of evaporation=",me,)
print("Equivalent evaporation=",round(E,1),"Kg/Kg ofcoal")
print("Boiler efficiency",round(etaboiler,2),"%")


Enthalpy of superheated steam= 3040.39 KJ/Kg
mass of evaporation= 6.4
Equivalent evaporation= 8.2 Kg/Kg ofcoal
Boiler efficiency 57.88 %


## Example 5 Page No:232¶

In :
#Input data
ms=5000                           #Steam generted in Kg/h
mf=700                            #Coal burnt in Kg/h
CV=31402                          #Cv of coal in KJ/Kg
x=0.92                            #quality of steam
P=1.2                             #Boiler pressure in MPa
Tw=45                             #Feed water temperature in degree celsius

#calculation
hfw=188.35                        #In KJ/Kg
hf=798.43                         #In KJ/Kg
hfg=1984.3                        #In KJ/Kg
hs=hf+x*hfg                       #Enthalpy of wet stream in KJ/Kg
me=ms/mf                          #mass of evaporation
E=((me*(hs-hfw))/2257)            #Equivalent evaporation in Kg/Kg of coal
etaboiler=((me*(hs-hfw))/CV)*100  #Boiler efficiency in %

#Output
print("Enthalpy of wet stream=",round(hs,2),"KJ/Kg")
print("mass of evaporation=",round(me,2),"")
print("Equivalent evaporation=",round(E,1),"Kg/Kg of coal")
print("Boiler efficiency=",round(etaboiler,2),"%")


Enthalpy of wet stream= 2623.99 KJ/Kg
mass of evaporation= 7.14
Equivalent evaporation= 7.7 Kg/Kg of coal
Boiler efficiency= 55.4 %


## Example 6 Page No:233¶

In :
#Input data
ms=6000                   #Boiler produce of steam Kg/h
P=25                      #Boiler pressure in bar
Tsup=350                  #Boiler temperature in degree celsius
Tw=40                     #Feed water temperature indegree celsius
CV=42000                  #Calorific value in Kj/Kg
etaboiler=75/100          #Expected thermal efficiency in %

#Calculation
hfw=167.45                #In KJ/Kg
Ts=223.94                 #In degree celsius
hf=961.96                 #In KJ/Kg
hfg=1839.0                #In KJ/Kg
hg=2800.9                 #In KJ/Kg
Cps=2.3
hs=((hg)+(Cps)*(Tsup-Ts))         #Enthalpy of superheated steam KJ/Kg
mf=((ms*(hs-hfw))/(CV*etaboiler)) #Boiler efficiency in %
me=ms/mf                          #Equivalent mass of evaporation
E=((me*(hs-hfw))/2257)            #Equivalent evaporation in Kg/Kg of oil

#Output
print("Enthalpy of superheated steam=",hs,"KJ/Kg")
print("Boiler efficiency=",round(mf,1),"%")
print("Equivalent mass of evaporation=",round(me,3),)
print("Equivalent evaporation=",round(E,2),"Kg/Kg of oil")

Enthalpy of superheated steam= 3090.838 KJ/Kg
Boiler efficiency= 556.8 %
Equivalent mass of evaporation= 10.775
Equivalent evaporation= 13.96 Kg/Kg of oil


## Example 7 Page No:234¶

In :
#Input data
E=12                        #Boiler found steam in Kg/Kg
CV=35000                    #Calorific value in KJ/Kg
ms=15000                    #Boiler produces in Kg/h
P=20                        #Boiler pressure in bar
Tw=40                       #Feed water in degree celsius
mf=1800                     #Fuel consumption

#calculation
#R=me(hs-hfw)
hfw=167.45                  #In KJ/Kg
hg=2797.2                   #In KJ/Kg
Ts=211.37                   #In degree celsius
Cps=2.3
R=E*2257                    #Equivalent evaporation in KJ/Kg of coal
etaboiler=(R/CV)*100        #Boiler efficiency in %
me=ms/mf                    #Equivalent mass evaporation in KJ/Kg of coal
hs=(R/me)+hfw               # In KJ/Kg
Tsup=((hs-hg)/Cps)+Ts       #Enthalpy of superheated steam in degree celsius

#Output
print("Equivalent evaporation=",R,"KJ/Kg of coal")
print("Boiler efficiency=",round(etaboiler,2),"%")
print("Equivalent mass evaporation=",round(me,2),"KJ/Kg of coal")
print("hs=",round(hs,2),"KJ/Kg")
print("Enthalpy of superheated steam=",round(Tsup,2),"degree celsius")

Equivalent evaporation= 27084 KJ/Kg of coal
Boiler efficiency= 77.38 %
Equivalent mass evaporation= 8.33 KJ/Kg of coal
hs= 3417.53 KJ/Kg
Enthalpy of superheated steam= 481.08 degree celsius


## Example 8 Page No:236¶

In :
#Input data
ms=6000                           #Steam generated in Kg/h
mf=700                            #Coal burnt in Kg/h
CV=31500                          #Cv of coal in KJ/Kg
x=0.92                            #Dryness in fraction of steam
P=12                              #Boiler pressure in bar
Tsup=259                          #Temperature of steam in degree celsius
Tw=45                             #Hot well temperature in degree celsius

#calculation
hfw=188.35                        #In KJ/Kg
Ts=187.96                         #In degree celsius
hf=798.43                         #In KJ/Kg
hfg=1984.3                        #In KJ/Kg
hg=2782.7                         #In KJ/Kg
Cps=2.3
me=ms/mf                          #Equivalent mass evaporation
hs=hf+x*hfg                       #Enthalpy of wet steam in KJ/Kg
E=((me*(hs-hfw))/2257)            #Equivalent evaporation in Kg/Kg of coal
hs1=(hg+Cps*(Tsup-Ts))            #Enthalpy of superheated steam in KJ/Kg
E1=((me*(hs1-hfw))/2257)          #Equivalent evaporation(with superheater) in Kg/Kg of coal
etaboiler=((me*(hs-hfw))/CV)*100  #Boiler efficiency without superheater in %
etaboiler1=((me*(hs1-hfw))/CV)*100#Boiler efficiency with superheater in %

#Output
print("Equivalent mass evaporation=",round(me,2),)
print("Enthalpy of wet steam=",hs,"KJ/Kg")
print("Equivalent evaporation=",round(E,2),"Kg/Kg of coal")
print("Enthalpy of superheated steam=",round(hs1,2),"KJ/Kg")
print("Equivalent evaporation(with superheater)=",round(E1,2),"Kg/Kg of coal")
print("Boiler efficiency without superheater=",round(etaboiler,2),"%")
print("Boiler efficiency without superheater=",round(etaboiler1,2),"%")

Equivalent mass evaporation= 8.57
Enthalpy of wet steam= 2623.986 KJ/Kg
Equivalent evaporation= 9.25 Kg/Kg of coal
Enthalpy of superheated steam= 2946.09 KJ/Kg
Equivalent evaporation(with superheater)= 10.47 Kg/Kg of coal
Boiler efficiency without superheater= 66.28 %
Boiler efficiency without superheater= 75.04 %


## Example 9 Page No:237¶

In :
#Input data
P=15                             #Boiler produces steam in bar
Tsup=250                         #Boiler temperature in degree celsius
Tw=35                            #Feed water in degree celsius
MWh=1.5                          #steam supplied to the turbine
CV=32000                         #Coal of calorific value in KJ/Kg
etaboiler=80/100                 #Thermal efficiency in %
fr=210                           #Firing rate in Kg/m**2/h
#From steam table(temp basis at 35 degree celsius)
hfw=146.56                       #In KJ/Kg
Ts=198.29                        #In degree celsius
hfg=1945.2                       #In KJ/Kg
hg=2789.9                        #In KJ/Kg
Cps=2.3

#calculator
hs=hg+Cps*(Tsup-Ts)               #Enthalpy of superheated steam(with superheater) in KJ/Kg
ms=9000/MWh                       #Steam rate in Kg/MWh
mf=((ms*(hs-hfw))/(etaboiler*CV)) #Mass of steam consumption in Kg/h
GA=mf/fr                          #Grate rate in m**2

#Output
print("Enthalpy of superheated steam(with superheater)=",hs,"KJ/Kg")
print("Steam rate=",ms,"Kg/h")
print("ass of steam consumption=",round(mf,1),"Kg/h")
print("Grate rate=",round(GA,3),"m**2")

Enthalpy of superheated steam(with superheater)= 2908.833 KJ/Kg
Steam rate= 6000.0 Kg/h
ass of steam consumption= 647.4 Kg/h
Grate rate= 3.083 m**2


## Example 10 Page No:242¶

In :
#Input data
ma=18            #Boileruses of per Kg of fuel in Kg/Kg
hw=25*10**-3     #Chimney height to produce draught in mm
Tg=315+273       #Temperature of chimney gases in degree celsius
Ta=27+273        #Out side air temp in degree celsius

#Calculation
#Draught produce in terms of water column in m
H=(hw/(353*(1/Ta-1/Tg*((ma+1)/ma))))*1000

#Output
print("Draught produce in terms of water column=",round(H,2),"m")

Draught produce in terms of water column= 46.04 m


## Example 11 Page No:242¶

In :
#Input data
H=40                             #High discharge in m
ma=19                            #Fuel gases per Kg of fuel burnt
Tg=220+273                       #Average temp of fuel gases in degree celsius
Ta=25+273                        #Ambient temperature in degreee celsius

#calculation
hw=353*H*(1/Ta-1/Tg*((ma+1)/ma)) #Draught produce in terms of water column in mm
H1=H*((Tg/Ta)*(ma/(ma+1))-1)     #Draught produce in terms of hot gas column in m

#output
print("Draught produce in terms of water column=",round(hw,2),"mm")
print("Draught produce in terms of hot gas column=",round(H1,2),"m")

Draught produce in terms of water column= 17.23 mm
Draught produce in terms of hot gas column= 22.87 m


## Example 12 Page No:243¶

In :
#Input data
H=27                                #Chimney height in m
hw=15                        #Draught produces of water column in mm
ma=21                               #Gases formed per Kg of fuel burnt in Kg/Kg
Ta=25+273                           #Temperature of the ambient air in degree celsius

#calculation
Tg=-(((ma+1)/ma)/((hw/(353*H))-(1/Ta))) #Mean temperature of fuel gases in K

#Output
print("Mean temperature of fuel gases",Tg,"k")

Mean temperature of fuel gases 587.9248031162673 k


## Example 13 Page No:244¶

In :
#Input data
hw=20                   #Static draught of water in mm
H=50                    #Chimney height in m
Tg=212+273              #Temperature of the fuel degree celsius
Ta=27+273               #Atmospheric air in degree celsius

#calculation
ma=(-((hw/(353*H))-Ta*Tg))*10**-4   #Air-fuel ratio in Kg/Kg of fuel burnt-3

#Output
print("Air-fuel ratio",round(ma,1),"Kg/Kg of fuel burnt")

Air-fuel ratio 14.5 Kg/Kg of fuel burnt


## Example 14 Page No:245¶

In :
#Input data
import math
H=24            #Chimney height in m
Ta=25+273       #Ambient temperature in degree celsius
Tg=300+273      #Temperature of fuel gases in degree celsius
ma=20           #Combustion space of fuel burnt in Kg/Kgof fuel
g=9.81

#calculation
hw=((353*H)*((1/Ta)-((1/Tg)*((ma+1)/ma))))#Theoretical draught in millimeters of water in mm
H1=H*((Tg/Ta)*(ma/(ma+1))-1)              #Theoretical draught produced in hot gas column in m
H2=H1-9.975                               #Draught lost in friction at the grate and passage in m
V=math.sqrt(2*g*H2)                       #Actual draught produced in hot gas column in m

#Output
print("Theoretical draught in millimeters of water=",round(hw,2),"mm")
print("Theoretical draught produced in hot gas column=",round(H1,2),"m")
print("Draught lost in friction at the grate and passage=",round(H2,3),"m")
print("Actual draught produced in hot gas column=",round(V,),"m")

Theoretical draught in millimeters of water= 12.9 mm
Theoretical draught produced in hot gas column= 19.95 m
Draught lost in friction at the grate and passage= 9.975 m
Actual draught produced in hot gas column= 14 m


## Example 15 Page No:246¶

In :
#Input data
import math
H=38                          #Stack height in m
d=1.8                         #Stack diameter discharge in m
ma=17                         #Fuel gases per Kg of fuel burnt Kg/Kg
Tg=277+273                    #Average temperature of fuel gases in degree celsius
Ta=27+273                     #Temperature of outside air in degree celsius
h1=0.4                        #Theoretical draught is lost in friction in
g=9.81
pi=3.142

#calculation
H1=H*(((Tg/Ta)*(ma/(ma+1))-1))#Theoretical draught produce in hot gas column in m
gp=0.45*27.8                  #Draught lost in friction at the grate and pasage in m
C=H1-gp                       #Actual draught produce in hot gas column in m
V=math.sqrt(2*9.81*C)         #Velocity of the flue gases in the chimney in m/s
rhog=((353*(ma+1))/(ma*Tg))   #Density of flue gases in Kg/m**3
mg=(rhog*((pi/4)*(d**(2))*V)) #Mass of gas flowing through the chimney in Kg/s

#Output
print("Theoretical draught produce in hot gas column=",round(H1,1),"m")
print("Draught lost in friction at the grate and pasage=",gp,"m")
print("Actual draught produce in hot gas column=",round(C,2),"m")
print("Velocity of the flue gases in the chimney =",round(V,2),"m/s")
print("Density of flue gases=",round(rhog,3),"Kg/m**3")
print("Mass of gas flowing through the chimney=",round(mg,),"Kg/s")

Theoretical draught produce in hot gas column= 27.8 m
Draught lost in friction at the grate and pasage= 12.51 m
Actual draught produce in hot gas column= 15.29 m
Velocity of the flue gases in the chimney = 17.32 m/s
Density of flue gases= 0.68 Kg/m**3
Mass of gas flowing through the chimney= 30 Kg/s


## Example 16 Page No:247¶

In :
#Input data
import math
hw=1.9           #Drauhgt water in cm
Tg=290+273       #Temp of flue gases in degree celsius
Ta=20+273        #Ambient temp in degree celsius
ma=22            #Flue gases formed in kg/Kg of coal
d=1.8            #Fuel burnt in m
pi=3.142
g=9.81

#calculation
H=(hw/(353*(1/Ta-1/Tg*((ma+1)/ma))))*10 #Theoretical draught produced in water column in m
H1=H*(((Tg/Ta)*(ma/(ma+1))-1))          #Theoretical draught produced in hot gas column n m
V=math.sqrt(2*g*H1)                     #Velocity of tthe flue gases in the chimney in m/s
rhog=((353*(ma+1))/(ma*Tg))             #Density  of flue gases in Kg/m**3
mg=rhog*((pi/4)*d**2)*V                 #Mass of gas flowing through the chimney in Kg/s

#Output
print("Theoretical draught produced in water column=",round(H,1),"m")
print("Theoretical draught produced in hot gas column=",round(H1,),"m")
print("Velocity of tthe flue gases in the chimney=",round(V,2),"m")
print("Density  of flue gases=",round(rhog,4),"Kg/m**3")
print("Mass of gas flowing through the chimney=",round(mg,1),"Kg/s")

Theoretical draught produced in water column= 34.6 m
Theoretical draught produced in hot gas column= 29 m
Velocity of tthe flue gases in the chimney= 23.85 m
Density  of flue gases= 0.6555 Kg/m**3
Mass of gas flowing through the chimney= 39.8 Kg/s


## Example 17 Page No:248¶

In :
#Input data
import math
mf=8000           #Average coal consumption in Kg/h
ma=19            #Flue gases formed in Kg/Kg
Tg=270+273       #Average temperature of the chimney in degree celsius
Ta=27+273        #Ambient temperature in degree celsius
hw=18            #Theoretical draught produced by the chimney in mm
h1=0.6           #Draught is lost in friction H1
g=9.81
pi=3.142

#calculation
H=(hw/(353*(1/Ta-1/Tg*((ma+1)/ma)))) #Theoretical draught produced in water column in m
H1=H*(((Tg/Ta)*(ma/(ma+1)))-1)       #Theoretical draught produced in hot gas column in m
gp=h1*H1                             #Draught is lost in friction at the grate and passing in m
hgc=H1-gp                            #Actual draught produced in hot gas column in m
V=math.sqrt(2*g*(hgc))             #Velocity of the flue gases in the chimney in m/s
rhog=((353*(ma+1))/(ma*Tg))          #Density of flue gases in Kg/m**3
mg=((mf/3600)*ma)                    #Mass of gas fowing throgh the chimney in Kg/s
d=math.sqrt(mg/(rhog*(pi/4)*V))        #Diameter of the chimney in m

#Output
print("Theoretical draught produced in water column=",round(H,1),"m")
print("Theoretical draught produced in hot gas column=",round(H1,3),"m")
print("Draught is lost in friction at the grate and passing=",round(gp,2),"m")
print("Actual draught produced in hot gas column=",round(hgc,3),"m")
print("Velocity of the flue gases in the chimney=",round(V,2),"")
print("Density of flue gases=",round(rhog,3),"Kg/m**3")
print("Mass of gas fowing throgh the chimney=",round(mg,3),"Kg/s")
print("Diameter of the chimney=",round(d,3),"m")

Theoretical draught produced in water column= 36.6 m
Theoretical draught produced in hot gas column= 26.304 m
Draught is lost in friction at the grate and passing= 15.78 m
Actual draught produced in hot gas column= 10.522 m
Velocity of the flue gases in the chimney= 14.37
Density of flue gases= 0.684 Kg/m**3
Mass of gas fowing throgh the chimney= 42.222 Kg/s
Diameter of the chimney= 2.338 m


## Example 18 Page No:251¶

In :
#Input data
import math
H=24                  #Chimney height in m
Ta=25+273             #Ambient temperature in degree celsius
Tg=300+273            #Temp of flue gases passing through the chimney in degree celsius
ma=20                 #Combustion space of fuel burnt in Kg/kg of fuel
g=9.81

#calculation
hw=((353*H)*((1/Ta)-((1/Tg)*((ma+1)/ma)))) #Theoretical draught produced in water column in m
##Calculation mistake in book of hw it is correct according to data &calculation

H1=H*(((Tg/Ta)*(ma/(ma+1))-1))             #Theoretical draught produced in hot gas column in m
H2=0.5*H1                                  #Draught is lost in friction at the grate and passing in m
hgc=H1-H2                                  #Actual draught produced in hot gas column in m
V=math.sqrt(2*g*H2)                        #Velocity of the flue gases in the chimney in m/s

#Output
print("Theoretical draught produced in water column=",round(hw,1),"m")
print("Theoretical draught produced in hot gas column=",round(H1,2),"m")
print("Draught is lost in friction at the grate and passing=",round(H2,3),"m")
print("Actual draught produced in hot gas column=",round(hgc,3),"m")
print("Velocity of the flue gases in the chimney=",round(V,),"m/s")

Theoretical draught produced in water column= 12.9 m
Theoretical draught produced in hot gas column= 19.95 m
Draught is lost in friction at the grate and passing= 9.975 m
Actual draught produced in hot gas column= 9.975 m
Velocity of the flue gases in the chimney= 14 m/s


## Example 19 Page No:252¶

In :
#Input data
H=38              #Stack height in m
d=1.8             #Stack diameter in m
ma=18             #Flue gases per kg of the fuel burnt
Tg=277+273        #Average temp of the flue gases in degree celsius
Ta=27+273         #Temperature of outside air in degree celsius
h1=0.4            #Theorical draught is lost in friction in %
g=9.81

#calculation
H1=H*(((Tg/Ta)*(ma/(ma+1))-1)) #Theoretical draught produced in hot gas column in m
gp=0.40*H1                     #Draught is lost in friction at the grate and passing in m
hgc=H1-gp                      #Actual draught produced in hot gas column in m
V=math.sqrt(2*g*hgc)           #Velocity of the flue gases in the chimney in m/s
rhog=((353*(ma+1))/(ma*Tg))    #Density of flue gases in Kg/m**3
mg=rhog*((pi/4)*d**2)*V        #Mass of gas fowing throgh the chimney in Kg/s

#Output
print("Theoretical draught produced in hot gas column=",round(H1,3),"m")
print("Draught is lost in friction at the grate and passing=",round(gp,2),"m")
print("Actual draught produced in hot gas column=",round(hgc,2),"m")
print("Velocity of the flue gases in the chimney=",round(V,2),"m/s")
print("Density of flue gases=",round(rhog,2),"Kg/m**3")
print("Mass of gas fowing throgh the chimney=",round(mg,1),"Kg/s")

Theoretical draught produced in hot gas column= 28.0 m
Draught is lost in friction at the grate and passing= 11.2 m
Actual draught produced in hot gas column= 16.8 m
Velocity of the flue gases in the chimney= 18.16 m/s
Density of flue gases= 0.68 Kg/m**3
Mass of gas fowing throgh the chimney= 31.3 Kg/s


## Example 20 Page No:253¶

In :
#Input data
import math
hw=19           #Draught produced water in cm
Tg=290+273      #Temperature of flue gases in degree celsius
Ta=20+273       #Ambient temperature  in degree celsius
ma=22           #Flue gases formed  per kg of fuel burnt in kg/kg of coal
d=1.8           #Diameter of chimney
g=9.81

#calculation
H=(hw/((353)*((1/Ta)-((1/Tg)*((ma+1)/ma))))) #Theoretical draught produced in hot gas column in m
H1=H*(((Tg/Ta)*(ma/(ma+1))-1))               #Draught is lost in friction at the grate and passing in m
V=math.sqrt(2*g*H1)                          #Velocity of the flue gases in the chimney in m/s
rhog=((353*(ma+1))/(ma*Tg))                  #Density of flue gases in Kg/m**3
mg=rhog*((pi/4)*d**2)*V                      #Mass of gas fowing throgh the chimney in Kg/s

#Output
print("Theoretical draught produced in hot gas column=",round(H,),"m")
print("Draught is lost in friction at the grate and passing=",round(H1,1),"m")
print("Velocity of the flue gases in the chimney=",round(V,2),"m/s")
print("Density of flue gases=",round(rhog,4)," Kg/m**")
print("Mass of gas fowing throgh the chimney=",round(mg,1),"Kg/s")

Theoretical draught produced in hot gas column= 35 m
Draught is lost in friction at the grate and passing= 29.0 m
Velocity of the flue gases in the chimney= 23.85 m/s
Density of flue gases= 0.6555  Kg/m**
Mass of gas fowing throgh the chimney= 39.8 Kg/s


## Example 21 Page No:254¶

In :
#Input data
import math
mf=8000                #Average coal consumption in m
ma=18                  #Fuel gases formed ccoal fired in m
Tg=270+273             #Average temp of the chimney of water in degree celsius
Ta=27+273              #Ambient temp in degree celsius
hw=18                  #Theoretical draught produced by the chimney in mm
h1=0.6                 #Draught is lost in friction in H1
g=9.81
pi=3.142

#calculation
H=(hw/((353)*((1/Ta)-((1/Tg)*((ma+1)/ma))))) #Theoretical draught produced in water column in m
H1=H*(((Tg/Ta)*(ma/(ma+1))-1))               #Theoretical draught produced in hot gas column in m
gp=0.6*H1                                    #Draught is lost in friction at the grate and passing in m
hgc=H1-gp                                    #Actual draught produced in hot gas column in m
V=math.sqrt(2*g*hgc)                         #Velocity of the flue gases in the chimney in m/s
rhog=((353*(ma+1))/(ma*Tg))                  #Density of flue gases in Kg/m**3
mg=mf/3600*(ma+1)                            #Mass of gas fowing throgh the chimney in Kg/s
d=math.sqrt(mg/(rhog*(pi/4)*V))              #Diameter of flue gases in Kg/m**3

#Output
print("Theoretical draught produced in water column=",round(H,1),"m")
print("Theoretical draught produced in hot gas column=",round(H1,2),"m")
print("Draught is lost in friction at the grate and passing=",round(gp,2),"m")
print("Actual draught produced in hot gas column=",round(hgc,2),"")
print("Velocity of the flue gases in the chimney=",round(V,2),"m/s")
print("Density of flue gases=",round(rhog,3),"Kg/m**3")
print("Mass of gas fowing throgh the chimney=",round(mg,2),"Kg/s")
print("Diameter of flue gases=",round(d,3),"Kg/m**3")

Theoretical draught produced in water column= 36.7 m
Theoretical draught produced in hot gas column= 26.23 m
Draught is lost in friction at the grate and passing= 15.74 m
Actual draught produced in hot gas column= 10.49
Velocity of the flue gases in the chimney= 14.35 m/s
Density of flue gases= 0.686 Kg/m**3
Mass of gas fowing throgh the chimney= 42.22 Kg/s
Diameter of flue gases= 2.337 Kg/m**3


## Example 22 Page No:256¶

In :
#Input data
H=45          #Chimney height in m
Tg=370+273    #Temperature of flue gases in degree celsius
T1=150+273    #Temperature of flue gases in degree celsius
ma=25         #Mass of the flue gas formed in Kg/kg of a cosl fired
Ta=35+273     #The boiler temperature in degree celsius
Cp=1.004      #fuel gas

#calculation
#Efficeincy of chimney draught in %
A=(H*(((Tg/Ta)*(ma/(ma+1)))-1))/(Cp*(Tg-T1))*100

#Output
print("Efficeincy of chimney draught=",round(A,2),"%")

Efficeincy of chimney draught= 20.52 %