Chapter 17 : Design of Catalytic Reactors

Example 1, Page 434

In [1]:
dt=[0.081,0.205,3.6]; #Reactor diameter for the three reactors in m
dte=[0.04,0.12,0.70]; #Equivalent diameters for the three reactors in m
db=[0.05,0.057,0.07]; #Estimated bubble size in the three reactors in m
Kr1=1.3889;           #Kinet1ic constant for Reaction 1 in s**-1
Kr2=0.6111;           #Kinetic constant for Reaction 2 in s**-1
Kr3=0.022;            #Kinetic constant for Reaction 3 in s**-1
dp=60.;               #Particle size in micrometer
ephsilonm=0.50;       #Void fraction of fixed bed
ephsilonmf=0.55;      #Void fraction at minimum fluidized condition
umf=0.006;            #Velocity at minimum fluidization condition in m/s
D=2E-5;               #Diffusion coefficient of gas in m**2/s
gammab=0.005;         #Ratio of volume of dispersed solids to that of bubble phase
uo=0.2;               #Superficial gas velocity in m/s
XA=0.9;               #Conversion
g=9.81;               #Acceleration due to gravity in square m/s**2

#CALCULATION
Kr12=Kr1+Kr2;
n=len(dt);
i=0;
ubr = [0,0,0]
ub = [0,0,0]
delta = [0,0,0]
ephsilonf = [0,0,0]
gammac = [0,0,0]
gammae = [0,0,0]
Kbc = [0,0,0]
Kce = [0,0,0]
Kf12 = [0,0,0]
Kf3 = [0,0,0]
KfA = [0,0,0]
KfAR = [0,0,0]
KfAR1 = [0,0,0]
tou = [0,0,0]
y = [0,0,0]
SR = [0,0,0]
XA1 = [0,0,0]
y1 = [0,0,0]
y2 = [0,0,0]
tou2 = [0,0,0]
Lf = [0,0,0]
Lm = [0,0,0]
XA2 = [0,0,0]

import math
while i<n:
    #Preliminary Calcualtions
    ubr[i]=0.711*(g*db[i])**0.5;#Rise velocity of bubble from Eqn.(6.7)
    ub[i]=1.55*((uo-umf)+14.1*(db[i]+0.005))*dte[i]**0.32+ubr[i];#Bubble velocity for Geldart A particles from Equation from Eqn.(6.11)
    delta[i]=uo/ub[i];#Fraction of bed in bubbles from Eqn.(6.29)
    ephsilonf[i]=1-(1-delta[i])*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)
    fw=0.6;#Wake volume to bubble volume from Fig.(5.8)
    gammac[i]=(1-ephsilonmf)*((3/(ubr[i]*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)
    gammae[i]=((1-ephsilonmf)*((1-delta[i])/delta[i]))-gammab-gammac[i];#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)
    Kbc[i]=4.5*(umf/db[i])+5.85*((D**0.5*g**0.25)/db[i]**(5/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)
    Kce[i]=6.77*((D*ephsilonmf*0.711*(g*db[i])**0.5)/db[i]**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)
    #Effective rate constant from Eqn.(12.32)
    Kf12[i]=(gammab*Kr12+1/((1/Kbc[i])+(1/(gammac[i]*Kr12+1/((1/Kce[i])+(1/(gammae[i]*Kr12)))))))*(delta[i]/(1-ephsilonf[i]));
    #Rate of reaction 2 for fluidized bed from Eqn.(12.14)
    Kf3[i]=(gammab*Kr3+1/((1/Kbc[i])+(1/(gammac[i]*Kr3+1/((1/Kce[i])+(1/(gammae[i]*Kr3)))))))*(delta[i]/(1-ephsilonf[i]));
    #Rate of raection with respect to A from Eqn.(12.35)
    KfA[i]=((Kbc[i]*Kce[i]/gammac[i]**2+(Kr12+Kce[i]/gammac[i]+Kce[i]/ \
    gammae[i])*(Kr3+Kce[i]/gammac[i]+Kce[i]/gammae[i]))*delta[i]*Kbc[i] \
    *Kr12*Kr3/(1-ephsilonf[i]))/(((Kr12+Kbc[i]/gammac[i])* \
    (Kr12+Kce[i]/gammae[i])+Kr12*Kce[i]/gammac[i])*((Kr3+Kbc[i]/gammac[i])* \
    (Kr3+Kce[i]/gammae[i])+Kr3*Kce[i]/gammac[i]));
    KfAR[i]=((Kr1/Kr12)*Kf12[i])-KfA[i];#Rate of reaction from Eqn.(12.34)
    KfAR1[i]=((Kr1/Kr12)*Kf12[i]);#Since KfA is small
    #(b)Relate Selectivity with conversion in three reactors
    x=-math.log(1-XA);#The term Kf12*tou in Eqn.(12.26)
    tou[i]=x/Kf12[i];#Residence time from Eqn.(12.26)
    y[i]=(KfAR1[i]/(Kf3[i]-Kf12[i]))*(math.exp(-x)-math.exp(-tou[i]*Kf3[i]));#CR/CAi from Eqn.(12.27)
    SR[i]=y[i]/XA;#Selectivity of R
    #(c)Relate exit composition to space time
    tou1=5;#Space time in s
    XA1[i]=1-math.exp(-Kf12[i]*tou1);#Conversion from Eqn.(12.26)
    y1[i]=((KfAR1[i]/(Kf12[i]-Kf3[i]))*(math.exp(-Kf3[i]*tou1)-math.exp(-Kf12[i]*tou1)));#CR/CAi R from Eqn.(12.27)
    #(d)Calculate height of bed needed to maximize production
    y2[i]=(KfAR1[i]/Kf12[i])*(Kf12[i]/Kf3[i])**(Kf3[i]/(Kf3[i]-Kf12[i]));#CRmax/CAi R from Eqn.(12.37)
    tou2[i]=math.log(Kf3[i]/Kf12[i])/(Kf3[i]-Kf12[i]);#Space time from Eqn.(38)
    Lf[i]=(uo/(1-ephsilonf[i]))*tou2[i];#Length of bed at fully fluidized condition from Eqn.(12.5)
    Lm[i]=Lf[i]*(1-ephsilonf[i])/(1-ephsilonm);#Length of bed when settled
    XA2[i]=1-math.exp(-Kf12[i]*tou2[i]);#Conversion from Eqn.(12.26)
    i=i+1;

#OUTPUT
print 'Let Laboratory, Pilot plant, Semicommercial unit be Reactor 1,2 & 3 respectively'
print '(a)Relation between effective rate constant(Kf12) to the gas flow rate(uo)',
print '\tReactor No.\tKf12(s**-1)\tuo(m/s)'
i=0;
while i<n:
     print '\t%1.0f'%i
     print '\t\t%f'%Kf12[i],
     print '\t%f'%uo
     i=i+1;

print '\n(b)Relation between selectivity with conversion'
print '\n\tReactor No.\tKf12(s**-1)\tSR(mol R formed/mol A reacted)'
i=0
while i<n:
     print '\t%1.0f'%i,
     print '\t\t%f'%Kf12[i],
     print '\t%f'%SR[i]
     i=i+1;

print '(c)Relation between exit compostion and space time',
print '\tReactor No.\tXA\t\tCR/CAi'
i=0;
while i<n:
     print '\t%1.0f'%i,
     print '\t\t%f'%XA1[i],
     print '\t%f'%y1[i]
     i=i+1;

print '(d)Height of bed needed to maximize the production of acrylonitrile',
print '\tReactor No.\tLm(m)\t\tXA'
i=0;
while i<n:
     print '\t%1.0f'%i,
     print '\t\t%f'%Lm[i],
     print '\t%f'%XA2[i]
     i=i+1;
Let Laboratory, Pilot plant, Semicommercial unit be Reactor 1,2 & 3 respectively
(a)Relation between effective rate constant(Kf12) to the gas flow rate(uo) 	Reactor No.	Kf12(s**-1)	uo(m/s)
	0
		0.410042 	0.200000
	1
		0.270620 	0.200000
	2
		0.128980 	0.200000

(b)Relation between selectivity with conversion

	Reactor No.	Kf12(s**-1)	SR(mol R formed/mol A reacted)
	0 		0.410042 	0.641507
	1 		0.270620 	0.618358
	2 		0.128980 	0.558283
(c)Relation between exit compostion and space time 	Reactor No.	XA		CR/CAi
	0 		0.871292 	0.564802
	1 		0.741562 	0.484243
	2 		0.475286 	0.313823
(d)Height of bed needed to maximize the production of acrylonitrile 	Reactor No.	Lm(m)		XA
	0 		3.056064 	0.956404
	1 		4.137401 	0.939139
	2 		7.049378 	0.897005

Example 2, Page 438

In [2]:
deltaHr=5.15E8; #Heat of reaction in J/k mol
W=5E4;          #Weight of acrylonitirle produced per 334-day year in tonnes
db=0.07;        #Estimated bubble size in m
dte=0.7;        #Equivalent diameter in m
Kf12=0.35;      #Effective rate constant in s**-1 from Example 1
dp=60;          #Particle size in micrometer
ephsilonm=0.50; #Void fraction of fixed bed
ephsilonmf=0.55;#Void fraction at minimum fluidized condition
T=460;          #Temperature in reactor in degree C
Pr=2.5;         #Pressure inside reactor in bar
#Feed gas composition
x1=1;           #Propylene
x2=1.1;         #Ammonia
x3=11;          #Air
do1=0.08;       #OD of heat exchanger tubes in m\
L=7;            #Length of tubes in m
ho=300;         #Outside heat transfer coefficient in W/m**2 K
hi=1800;        #Inside heat transfer coefficient in W/m**2 K
Tc=253.4;       #Temperature of coolant in degree C
pi=3.14;

#CALCULATION
#Preliminary calculation
uo=0.46;#Superficial gas velocity from Fig.E1(a) for the value of Kf12 & db
tou=8;#Space time from Fig.E2(b) for highest concentraion of product R
Lm=uo*tou/(1-ephsilonm);
y=0.58;#CR/CAi from Fig.E1(c) for the value of tou & Kf12
XA=0.95#From Fig.E1(c) for the value of tou & Kf12
SR=y/XA;#Selectivity of R

#Cross-sectional area of the reactor
P=W*10**3/(334*24*3600);#Production rate of acrylonitrile
F=(P/0.053)/(SR*XA/0.042);#Feed rate of propylene
V=((F*22.4*(T+273)*(x1+x2+x3))/(42*273*Pr));
At=V/uo;#Cross-sectional area of reactor needed for the fluidized bed

#Heat exchanger calculation
q=F*XA*deltaHr/42;#Rate of heat liberation in the reactor
U=(ho**-1+hi**-1)**-1;#Overall heat transfer coefficient
deltaT=T-Tc;#Driving force for heat transfer
Aw=q/(U*deltaT);#Heat exchanger area required to remove q
Nt=Aw/(pi*do1*L);
li1=(At/Nt)**0.5;#Pitch for square pitch arrangement
dte1=4*(li1**2-(pi/4)*do1**2)/(pi*do1);
if dte1>dte:
    li=(pi/4*dte*do1+pi/4*do1**2)**0.5;#Pitch if we add dummy tubes
import math
f=li**2-pi/4*do1**2;#Fraction of bed cross section taken up by tubes
dt1=math.sqrt(4/pi*At/(1-f));#Reactor diameter including all its tubes

#OUTPUT
print 'Superficial gas velocity=%fm/s'%uo
print 'No. of %1.0fm tubes required=%1.0f'%(L,Nt);
print 'Reactor diameter=%fm'%dt1
Superficial gas velocity=0.460000m/s
No. of 7m tubes required=295
Reactor diameter=7.173176m

Example 3, Page 444

In [3]:
db=0.08;       #Estimated bubble size in m
dte=2;         #Equivalent diameter in m
F1=55.6;       #Feed rate of oil in kg/s
XA=0.63;       #Conversion
uo=0.6;        #Superficial gas velocity in m/s
T1=500.0;      #Temperature of reactor in degree C
T2=580.0;      #Temperature of regenerator in degree C
Fs=F1*23.3;    #Solid circulation rate from Ex.(15.2)
rhos=1200.0;   #Density of catalyst in kg/m**3
dpbar=60.0;    #Average particle size in micrometer
ephsilonm=0.50;#Void fraction of fixed bed
ephsilonmf=0.55;#Void fraction at minimum fluidized condition
umf=0.006;      #Velocity at minimum fluidization condition in m/s
dt=8.0;         #Diameter of reactor in m
D=2E-5;         #Diffusion coefficient of gas in m**2/s
Kr=8.6;         #Rate constant for reaction at 500 degree C in s**-1
Ka1=0.06;       #Rate constant for deactivatiion at 500 degree C in s**-1
Ka2=0.012;      #Rate constant for regeneration at 580 degree C in s**-1
gammab=0.005;   #Ratio of volume of dispersed solids to that of bubble phase
g=9.81;         #Acceleration due to gravity in square m/s**2
pi=3.14;

#CALCULATION
#Parameters for the fluidized reactor
ubr=0.711*(g*db)**0.5;#Rise velocity of bubble from Eqn.(6.7)
ub=1.55*((uo-umf)+14.1*(db+0.005))*dte**0.32+ubr;#Bubble velocity for Geldart A particles from Equation from Eqn.(6.11)
delta=uo/ub;#Fraction of bed in bubbles from Eqn.(6.29)
ephsilonf=1-(1-delta)*(1-ephsilonmf);#Void fraction of fixed bed from Eqn.(6.20)
fw=0.6;#Wake volume to bubble volume from Fig.(5.8)
gammac=(1-ephsilonmf)*((3/(ubr*ephsilonmf/umf-1))+fw);#Volume of solids in cloud to that of the bubble from Eqn.(6.36)
gammae=((1-ephsilonmf)*((1-delta)/delta))-gammab-gammac;#Volume of solids in emulsion to that of the bubble from Eqn.(6.35)
Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5.0/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)
Kce=6.77*((D*ephsilonmf*0.711*(g*db)**0.5)/db**3)**0.5;#Gas interchange coefficient between emulsion and cloud from Eqn.(10.34)
import math
#Bed height versus catalyst activity in reactor
a1bar=0.07;#Guess value for average activity in reactor
x=Kr*a1bar;#Value of Kra1 to be used in the following equation
Kf=(gammab*x+1/((1/Kbc)+(1/(gammac*x+1/((1/Kce)+(1/(gammae*x)))))))*(delta/(1-ephsilonf));#Effective rate constant from Eqn.(12.14)
tou=-math.log(1-XA)/Kf;#Space time from Eqn.(12.16)
Lm=tou*uo/(1-ephsilonm);#Length of fixed bed for guess value of a1bar
a1bar1=[0.0233,0.0465,0.0698,0.0930,0.116,0.140];#Various activity values to find Lm
x1 = [0,0,0,0,0,0]
Kf1 = [0,0,0,0,0,0]
tou1 = [0,0,0,0,0,0]
Lm1 = [0,0,0,0,0,0]

n=len(a1bar1);
i=0;
while i<n:
    x1[i]=Kr*a1bar1[i];
    Kf1[i]=(gammab*x1[i]+1/((1/Kbc)+(1/(gammac*x1[i]+1/((1/Kce)+ \
    (1/(gammae*x1[i])))))))*(delta/(1-ephsilonf));
    #Effective rate constant from Eqn.(12.14)
    
    tou1[i]=-math.log(1-XA)/Kf1[i];#Space time from Eqn.(12.16)
    Lm1[i]=tou1[i]*uo/(1-ephsilonm);
    #Length of fixed bed for guess value of a1bar...Condition [i]
    i=i+1;

#Find the optimum size ratio for various a1bar
Lm=[5,6,7,8,10,12];
W1 = [0,0,0,0,0,0]
t1bar = [0,0,0,0,0,0]
t2bar = [0,0,0,0,0,0]
a1bar2 = [0,0,0,0,0,0]
m=len(Lm);
i=0;
while i<m:
    W1[i]=(pi/4)*dt**2*rhos*(1-ephsilonm)*Lm[i];#Bed weight
    t1bar[i]=W1[i]/Fs;#Mean residence time of solids in reactor
    t2bar[i]=t1bar[i]*(Ka1/Ka2)**0.5;#Mean residence time of soilds at optimum from Eqn.(16)
    a1bar2[i]=(Ka2*t2bar[i])/(Ka1*t1bar[i]+Ka1*t1bar[i]*Ka2*t2bar[i]+Ka2*t2bar[i]);#From Eqn.(15)...Condition (ii)
    i=i+1;

#Final design values
Lm4=7.3;#For satisfying condition [i] & (ii)
a1bar3=0.0744;#By interpolation
x2=a1bar3*Kr;
W11=(pi/4)*dt**2*rhos*(1-ephsilonm)*Lm4;#Bed weight for reactor
t1bar1=W11/Fs;#Mean residence time of solids in reactor
a2bar=(1+Ka1*t1bar1)*a1bar3;#Average activity in regenrator from Eqn.(10)
t2bar1=t1bar1*(Ka1/Ka2)**0.5;#Mean residence time of solids in regenerator from Eqn.(16)
W2=W11*(t2bar1/t1bar1);#Bed weight for regenerator
dt2=dt*(W2/W11)**0.5;#Diameter of regenerator assuming same static bed height for reactor and regerator

#OUTPUT
print 'Bed height versus catalyst activity in reactor'
print '\tAverage activity',
print '\tLength of fixed bed(m)'
i=0;
while i<n:
    print '\t%f'%a1bar1[i],
    print '\t\t%f'%Lm1[i];
    i=i+1;

print 'Optimum size ratio for various activity in reactor'
print '\tLength of fixed bed(m)',
print '\tAverage activity'
i=0
while i<m:
    print '\t%f'%Lm[i],
    print '\t\t%f'%a1bar2[i]
    i=i+1;

print 'Final design values'
print '\tDiameter of reactor(m):%.0f'%dt
print '\tBed weight for reactor(tons):%.0f'%(W11/10**3)
print '\tBed weight for regenerator(tons):%.0f'%(W2/10**3)
print '\tDiameter of regenerator(m):%.0f'%(dt2);
print '\tSolid circulation rate(tons/hr):%f'%(Fs*3.6);
Bed height versus catalyst activity in reactor
	Average activity 	Length of fixed bed(m)
	0.023300 		11.059747
	0.046500 		7.911053
	0.069800 		6.756202
	0.093000 		6.118750
	0.116000 		5.696470
	0.140000 		5.372072
Optimum size ratio for various activity in reactor
	Length of fixed bed(m) 	Average activity
	5.000000 		0.097879
	6.000000 		0.086112
	7.000000 		0.076871
	8.000000 		0.069420
	10.000000 		0.058149
	12.000000 		0.050026
Final design values
	Diameter of reactor(m):8
	Bed weight for reactor(tons):220
	Bed weight for regenerator(tons):492
	Diameter of regenerator(m):12
	Solid circulation rate(tons/hr):4663.728000
In [ ]: