Chapter 9 : Solid Movement Mixing Segregation and Staging

Example 1, Page 218

In [2]:
umf=0.015;                 #Velocity at minimum fluidization condition in m/s
ephsilonmf=0.5;            #Void fraction at minimum fluidization condition
uo=0.1;                    #Superficial gas velocity in m/s
delta=0.2;                 #Bed fraction in bubbles
db=0.06;                   #Equilibrium bubble size in m
dt=[0.1,0.3,0.6,1.5];      #Various vessel sizes in m
ub=[0.4,0.75,0.85,1.1];    #Bubble velocity in m/s
Dsv=[0.03,0.11,0.14,0.23]; #Reported values of vertical dispersion coefficient

#CALCULATION
n=len(ub);
i=0;
fw1=2;#Wake fraction from Hamilton et al.
fw2=0.32;#Wake fraction from Fig.(5.8)
fw=(fw1+fw2)*0.5;#Average value of wake fraction
Dsv1 = []
Dsv2 = []
while i<n:
    Dsv1.append(12*((uo*100)**0.5)*((dt[i]*100)**0.9));#Vertical distribution coefficient from Eqn.(3)
    Dsv2.append((fw**2*ephsilonmf*delta*db*ub[i]**2)/(3*umf));#Vertical distribution coefficient from Eqn.(12)
    i=i+1;

print Dsv1

#OUTPUT
print '\t\tVertical dispersion coefficient(m**2/s)'
print 'Vessel Size(m)',
print '\tFrom Experiment',
print '\tFrom Eqn.(3)',
print '\tFrom Eqn.(12)'
i=0;
while i<n:
    print '%.2f'%dt[i],
    print '\t%.3f'%Dsv[i],
    print '\t%.2f'%(Dsv1[i]/10**4),
    print '\t%.2f'%Dsv2[i]
    i=i+1;   

[301.42637178114967, 810.1965234492834, 1511.8801720132121, 3448.7632274996104]
		Vertical dispersion coefficient(m**2/s)
Vessel Size(m) 	From Experiment 	From Eqn.(3) 	From Eqn.(12)
0.10 	0.030 	0.03 	0.03
0.30 	0.110 	0.08 	0.10
0.60 	0.140 	0.15 	0.13
1.50 	0.230 	0.34 	0.22

Example 2, Page 222

In [3]:
Lmf=0.83;                          #Length of bed at minimum fluidization condition in m
dp=450.0;                          #Average particle size in micrometer
ephsilonmf=0.42;                   #Void fraction at minimum fluidization condition
umf=0.17;                          #Velocity at minimum fluidization condition in m/s
uo=[0.37,0.47,0.57,0.67];          #Superficial gas velocity in m/s
Dsh=[0.0012,0.0018,0.0021,0.0025]; #Horizontal Drift Coefficient from Experiment in m**2/s
db=[0.10,0.14];                    #Equilibrium bubble size in m
g=9.81;                            #Acceleration due to gravity in m/s**2


#CALCULATION
n=len(uo);
m=len(db);
k=0;
alpha=0.77;#Since we are not dealing with Geldart A or AB solids
uf=umf/ephsilonmf;
ubr = []
ub = []
delta = []
Dshc = []
for j in range(m):
    for i in range(n):
        ubr.append(0.711*(db[j]*g)**0.5);#Rise velocity of a single bubble in m/s
        ub.append(uo[i]-umf+ubr[k]);#Rise velocity of bubbles in a bubbling bed
        delta.append((uo[i]-umf)/(ub[k]+umf));#Bed fraction in bubbles
        if ubr[i]>uf:
            Dshc.append((3/16.0)*(delta[k]/(1-delta[k]))*((alpha**2*db[j]*ubr[k]*((((ubr[k]+2*uf)/(ubr[k]-uf))**(1.0/3))-1))));
                #Horizontal Distribution coeff. from Eqn.(14)
        else:
            Dsh.append((3.0/16)*(delta/(1-delta))*(alpha**2*umf*db/ephsilonmf))
                #Horizontal Distribution coeff. from Eqn.(15)
        Dshc.append((3/16.0)*(delta[k]/(1-delta[k]))*((alpha**2*db[j]*ubr[k]*((((ubr[k]+2*uf)/(ubr[k]-uf))**(1/3.0))-1))));#Horizontal Distribution coeff. from Eqn.(14)
        k=k+1;
i=0;
j=0;
k=0;
while k<m*n:
    print 'Snce we do not have ub=%fm/s>>uf=%fm/s we use Eqn.(14).'%(ub[k],uf)
    print 'Gas Velocity(m/s)'
    print '\tHorizontal Drift Coefficient Calculated(m**2/s)'
    print '\tHorizontal Drift Coefficient from Experiment(m**2/s)'
    while j<m:
        print 'db=%fm'%db[j]
        while i<n:
            print '%f'%uo[i],
            print '\t\t%f'%Dshc[k],
            print '\t\t\t\t\t%f'%Dsh[i]
            i=i+1;   
            k=k+1;
        i=0;
        j=j+1;

Snce we do not have ub=0.904213m/s>>uf=0.404762m/s we use Eqn.(14).
Gas Velocity(m/s)
	Horizontal Drift Coefficient Calculated(m**2/s)
	Horizontal Drift Coefficient from Experiment(m**2/s)
db=0.100000m
0.370000 		0.001283 					0.001200
0.470000 		0.001283 					0.001800
0.570000 		0.001924 					0.002100
0.670000 		0.001924 					0.002500
db=0.140000m
0.370000 		0.002566 					0.001200
0.470000 		0.002566 					0.001800
0.570000 		0.003207 					0.002100
0.670000 		0.003207 					0.002500

Example 3, Page 232

In [4]:
import math

#Variable declaration
Gsup=1.5;               #Solid interchange rate in kg/m**2plate s
dor=19.1;               #Orifice diameter in mm
dp=210;                 #Particle size in micrometer
uo=0.4;                 #Superficial gas velocity in m/s
fopen=[0.12,0.17,0.26]; #Open area fraction 
pi=3.14;

#CALCULATION
n=len(fopen);
uor = []
ls1 = []
i=0
while i<n:
    uor.append(uo/fopen[i]);  #Gas velocity through the orifice
    ls1.append(Gsup/fopen[i]);#Flux of solids through the holes
    i=i+1;

ls2=[12,20,25];                #Flux of solids through holes from Fig.13(c) for different uor values
fopen1=0.12;                   #Open area fraction which gives reasonable fit
lor=math.sqrt(((math.pi/4)*dor**2)/fopen1);#Orifice spacing

#OUTPUT
print 'fopen',
print '\t\tuor(m/s)',
print '\tls from Eqn.(18)',
print '\tls from Fig.13(c)'
i=0;
while i<n:
    print '%f'%fopen[i],
    print '\t%f'%uor[i],
    print '\t%f'%ls1[i],
    print '\t\t%f'%ls2[i]
    i=i+1;   

print '\nFor square pitch, the orifice spacing should be %fmm'%lor

fopen 		uor(m/s) 	ls from Eqn.(18) 	ls from Fig.13(c)
0.120000 	3.333333 	12.500000 		12.000000
0.170000 	2.352941 	8.823529 		20.000000
0.260000 	1.538462 	5.769231 		25.000000

For square pitch, the orifice spacing should be 48.863850mm
In [ ]: