Chapter 5 : Bubbles in Dense Beds

Example 1, Page 126

In [1]:
dt=60.0;        #ID of tube in cm 
dp=300;         #Size of particles of bed in micrometers
umf=3;          #Velocity at minimum fluidization condition in cm/s
ephsilonmf=0.5; #Void fraction of bed at minimum fluidization condition
db=5;           #Diameter of bubble in cm
g=980;          #Acceleration due to gravity in cm/s**2

#Computation of rise velocity of bubble
if (db/dt)<0.125:
    ubr=(0.711*((g*db)**0.5));#Rise velocity by Eqn.(3)
elif (db/dt)<0.6:
    ubr=(0.711*((g*db)**0.5))*1.2*exp(-1.49*(db/dt));#Rise velocity by Eqn.(4)   

#Computation of cloud thickness
Rb=db/2.0;          #Radius of bubble
uf=umf/ephsilonmf;  #Velocity of emulsion gas
Rc=Rb*((ubr+(2*uf))/(ubr-uf))**(1/3.0);#Radius of cloud by Eqn.(6)

print 'The rise velocity of the bubble=%.1f cm/s'%ubr
print 'The cloud thickness=%.2f cm'%(Rc-Rb)
print 'From Fig.8(page 124)comparing fw vs dp, for dp = %.0f micrometer, wake fraction = 0.24'%dp
The rise velocity of the bubble=49.8 cm/s
The cloud thickness=0.30 cm
From Fig.8(page 124)comparing fw vs dp, for dp = 300 micrometer, wake fraction = 0.24

Example 2, Page 132

In [2]:
import math

#Variable declaration
uo=15;     #Superificial gas velocity in cm/s
umf=1;     #Velocity at minimum fluidization condition in cm/s
lor=2.0;   #Pitch of perforated plate in cm
g=980;     #Acceleration due to gravity in cm/s**2
#Case(a) For porous plate
dbo1=(2.78/g)*(uo-umf)**2;#Initial bubble size using Eqn.(19)

#Case(b) For Perforated plate
Nor=(2/math.sqrt(3))*(1/lor)**2;#Number of orifices in cm**-2
dbo2=(1.30/(g**0.2))*((uo-umf)/Nor)**0.4;#Initial bubble size using Eqn.(15) assuming inital bubble size is smaller than hole spacing

print 'Case(a) For porous plate'
print '\tInitial bubble size=%.2fcm'%dbo1
print 'Case(b) For Perforated plate'
print '\tInitial bubble size=%.2fcm'%dbo2
print '\tSince %f<%f, the equation used is correct.'%(dbo2,lor)
Case(a) For porous plate
	Initial bubble size=0.56cm
Case(b) For Perforated plate
	Initial bubble size=1.55cm
	Since 1.548765<2.000000, the equation used is correct.
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