# Chapter 5 : Combustion Mechanism, Combustion Equipment And Firing Methods¶

## Ex 5.1 Page 308¶

In :
#Input data
Vs=2500##The mass of a bed of solid particles in kg
p=2650##The density of the solid in kg/m**3
d=800*10**-6##The mean particle size in m
s=0.84##The sphericity of the particle

#Calculations
As=(6*Vs)/(p*d*s)##The total surface area of the particles in the bed

#Output
print " The total surface area of the particles in the bed As = %3.0f m**2 "%(As)

 The total surface area of the particles in the bed As = 8423 m**2


## Ex 5.2 Page 309¶

In :
#Input data
d=427*10**-6##The mean particle size in m
pg=1.21##The density of air in kg/m**3
v=1.82*10**-5##The viscosity of air in kg/ms
pl=1620##The density of the loosely packed bed in kg/m**3
ps=2780##The density of the solids in kg/m**3
c1=27.2##(Grace,1982)constant value.
c2=0.0408##(Grace,1982)constant value
g=9.812##Gravitational forc constant in m/s**2

#Calculations
E=1-(pl/ps)##The voidage of the bed
Ar=((pg)*(ps-pg)*g*(d**3))/v**2##Archimedes number
Re=(c1**2+(c2*Ar))**(0.5)-c1##Reynolds number
Umf=Re*v/(pg*d)##Minimum superficial velocity in m/s

#Output
print " (a) The voidage of the bed = %3.3f \n (b) The minimum fluidization velocity Umf = %3.3f m/s "%(E,Umf)

 (a) The voidage of the bed = 0.417
(b) The minimum fluidization velocity Umf = 0.187 m/s


## Ex 5.3 Page 309¶

In :
from scipy.optimize import fsolve
#Input data
d=427*10**-6##The mean particle size in m
pg=1.21##The density of air in kg/m**3
v=1.82*10**-5##The viscosity of air in kg/ms
Umf=0.14##Minimum superficial velocity in m/s
Ar=7753##Archimedes number from previous example problem

#Calculations

Re=(Umf*pg*d)/v##Reynolds number
def F(x):##function definition
f = 7753*x**2- 381.1*x -4793#
return f
x = 100##Initial guss
y = fsolve(F,x)#

#Output
print "The sphericity of particles is = %3.3f "%(y)

The sphericity of particles is = 0.811


## Ex 5.4 Page 310¶

In :
from __future__ import division
#Input data
O=35##The output of the fluidized bed combustion system in MW
n=0.80##Efficiency of the fluidized bed combustion system
H=26##The heating value of coal in MJ/kg
S=3.6##Sulphur content in the coal in %
C=3##The calcium sulphur ratio
Ca=85##The amount of calcium carbonate in the limestone in %
CaCO3=100##The molecular weight of CaCO3

#Calculations
Cb=O/(n*H)##Coal burning rate in kg/s
Cb1=Cb*3600##Coal burning rate in kg/h
Sf=(Cb1*(S/100))/32##Flow rate of sulphur in Kmol/h
Cf=Sf*C##The flow rate of calcium in Kmol/h
Caf=Cf*CaCO3##Mass flow rate of CaCO3 in kg/h
L=Caf/(Ca/100)##Mass flow rate of limestone in kg/h

#Output
print "The required flow rate of limestone is %3.1f kg/h "%(L)

The required flow rate of limestone is 2405.3 kg/h


## Ex 5.5 Page 310¶

In :
#Input data
CV=24##The calorific value of the fuel in MJ/kg
C=0.65##The amount of calorific value released in the bed in %
to=850##Temperature at which products leave in degree centigrade
ti=30##The inlet temperature in degree centigrade
tb=850##The bed temperature in degree centigrade
A=14.5##The air fuel ratio by mass
Cp=1.035##The specific heat of the products leaving the bed surface in kJ/kgK
B=7000##The burning rate of coal in kg/h

#Calculations
H=(C*CV*1000)-(A*Cp*(to-ti))##Heat removal from the bed per kg fuel in kJ/kg fuel
Hr=(H*B)/3600##Rate of heat removal from the bed in kW
Hb=(B/3600)*(1-C)*CV*1000##The rate of heat removal from the above bed zone in kW

#Output
print " (a) The rate of heat removal from the bed = %3.0f kW \n (b) The rate of heat removal from the above bed zone = %3.0f kW "%(Hr,Hb)

 (a) The rate of heat removal from the bed = 6405 kW
(b) The rate of heat removal from the above bed zone = 16333 kW


## Ex 5.6 Page 311¶

In :
#Input data
tb=850##The bed temperature in degree centigrade
CV=25##The calorific value of the fuel in MJ/kg
A=9.5##The stoichiometric air fuel ratio by mass
E=20##The amount of excess air used in %
F=4.8##The total fueling rate in MW
p=0.3145##The density of air at bed temperature in kg/m**3
f=2##The firing rate in MW/m**2
v=2.7##The fluidizing velocity in m/s

#Calculations
P=F/f##Planform area in m**2
m=(F*1000)/(CV*1000)##Fuel burning rate in kg/s
ma=A*(1+(E/100))*m##Mass flow rate of air in kg/s
Pa=ma/(p*v)##Planform area in m**2

#Output
print " (a) The planform area = %3.1f m**2 \n (b) Fuel burning rate = %3.3f kg/s \n    Air flow rate = %3.4f kg/s \n    Planform area = %3.2f m**2 "%(P,m,ma,Pa)

 (a) The planform area = 2.4 m**2
(b) Fuel burning rate = 0.192 kg/s
Air flow rate = 2.1888 kg/s
Planform area = 2.58 m**2