Chapter 15 : Circulation Systems

Example 1, Page 369

In [1]:
import math

#Variable declaration
thalf=1;   #Half life of catalyst in s
F=960;     #Feed rate of oil in tons/day
W=50;      #Weight of the bed in tons
a=0.5;     #Activity after time equal to half life
abar=0.01; #Average activity of the catalyst

#CALCULATION
Ka=-math.log(a)/thalf;#Rate constant is s**-1, assuming I order kinetics from Eqn.(12)
Fs=Ka*W*abar/(1-abar);#Circulation rate of solids from Eqn.(16)
x=(Fs*60*60*24.0)/F;  #Circulation rate per feed of oil

#OUTPUT
print '\nSolid recirculation per feed of oil =%ftons of solid circulated/ton feed oil'%x
Solid recirculation per feed of oil =31.506690tons of solid circulated/ton feed oil

Example 2, Page 370

In [2]:
#Variable declaration
deltaHr1=1260.;                               #Enthalpy change during endothermic reaction in kJ/kg
deltaHr2=-33900.;                             #Enthal[y change during exothermic reaction in kJ/kg
H1=703.;                                      #Enthalpy of feed oil in kJ/kg
T1=260.;                                      #Temperature of feed oil in degree celcius
H3=1419.;                                     #Enthalpy of cracked product in kJ/kg
T3=500.;                                      #Temperature of cracked product in degree celcius
Ta=20.;                                       #Temperature of entering air in degree celcius
Cpa=1.09;                                     #Specific heat of entering air in kJ/kg K
Cpf=1.05;                                     #Specific heat of flue gases in kJ/kg K
Cps=1.01;                                     #Specific heat of solids in kJ/kg K
Cpv=3.01;                                     #Specific heat of vaporized feed in kJ/kg K
T4=[520.,540.,560.,580.,600.,620.,640.,660.]; #Temperature of flue gas in degree celcius
V=22.4;                                       #Volume of 1 mole of Carbon dioxide gas in N-m**3
M=12.;                                        #Molecular weight of carbon in kg
rho=1.293;                                    #Density of carbon dioxide gas in kg/N-m**3
xa=0.21;                                      #Mass fraction of oxygen in air
betac=0.07;                                   #Mass fraction of carbon

#CALCULATION
n=len(T4);
i=0;
x1 = [0,0,0,0,0,0,0,0]
x2 = [0,0,0,0,0,0,0,0]
excess_air = [0,0,0,0,0,0,0,0]

x2min=betac*(V*rho/(M*xa));#Minimum amount of air required for complete combustion
while i<n:
    x1[i]=(deltaHr1+0.93*H3-H1)/(Cps*(T4[i]-T3));#Fs/F1 by simplifying the overall energy balance
    x2[i]=((0.07*(-deltaHr2)-(deltaHr1+0.93*H3-H1))/(Cpf*(T4[i]-Ta)))-0.07;#F2/F1 by simplifying the energy balance for regenerator
    if x2[i]>x2min:
        excess_air[i]=(x2[i]-x2min)/x2min; #Excess air used
    else:
        excess_air[i]=0;
    i=i+1;

#OUTPUT    
print 'T4(degree celcius)',
print '\tFs/F1',
print '\t\tF2/F1',
print '\t\tExcess air(percentage)'
i=0;
while i<n:
    print '%f'%T4[i],
    print '\t\t%f'%x1[i],
    print '\t%f'%x2[i],
    print '\t%f'%(excess_air[i]*100);
    i=i+1;

#Disclaimer: The values of F2/F1 obtained by manual calculation has close correspondance to the ones obtained as the output, whereas it deviates largely from the values given in textbook.
T4(degree celcius) 	Fs/F1 		F2/F1 		Excess air(percentage)
520.000000 		92.904455 	0.875390 	8.807235
540.000000 		46.452228 	0.839029 	4.287699
560.000000 		30.968152 	0.805362 	0.102944
580.000000 		23.226114 	0.774099 	0.000000
600.000000 		18.580891 	0.744992 	0.000000
620.000000 		15.484076 	0.717825 	0.000000
640.000000 		13.272065 	0.692412 	0.000000
660.000000 		11.613057 	0.668586 	0.000000

Example 3, Page 379

In [3]:
Fs=100.;        #Solid flowrate in kg/s
ephsilon1=0.55;
ephsilon2=0.5;
p1=120.;         #Pressure at upper level in kPa
rhos=1000.;      #Density of solid in kg/m**3
rhog=1.;         #Density of gas in kg/m**3
gc=1.;           #Conversion factor
g=9.81;          #Acceleration due to gravity in m/s**2
di=0.34;         #Diameter of downcomer in m
pi=3.14;

#CALCULATION
x=(ephsilon1/ephsilon2)*((1-ephsilon2)/(1-ephsilon1));#To find pressure at lower level using Eqn.(30)
p2=x*p1;#Pressure at lower level using Eqn.(30)
deltap=p2-p1;
ephsilonbar=0.5*(ephsilon1+ephsilon2);
deltah=(deltap*10**3*gc)/(rhos*(1-ephsilonbar)*g);#Static head height from Eqn.(28)
At=0.25*pi*di**2;#Area of downcomer
Gs=Fs/At;#Flux of solids in downcomer
Gg=Gs*(ephsilon1/(1-ephsilon1))*(rhog/rhos)*(x-1);#Required gas aeration rate from Eqn.(31)
Fg=Gg*At;#Flow rate of gas required

#OUTPUT
print '\nThe required flow rate of gas required for location of %fm below downcomer is %.4fkg/s'%(deltah,Fg)
The required flow rate of gas required for location of 5.722768m below downcomer is 0.0272kg/s

Example 4, Page 380

In [4]:
import math

#Variable declaration
Fs=600;#Solid circulation rate in kg/s
dpbar=60;#Mean size of solids in micrometer
pA=120;#Pressure in vessel A in kPa
pB=180;#Pressure in vessel B in kPa
LfA=8;#Bed height in vessel A in m
LfB=8;#Bed height in vessel B i m
#Bulk densities in kg/m**3
rho12=100;
rho34=400;
rho45=550;
rho67=200;
rho78=200;
rho910=400;
rho1011=400;
rho1112=550;
rho13=100;
deltapdA=7;#Pressure drop across the distributor in regenerator in kPa
deltapdB=8;#Pressure drop across the distributor in reactor in kPa
deltap12=(9+4);#Friction loss and pressure difference required to accelerate the solids in transfer lines in kPa
deltap78=(15+3);#Friction loss and pressure difference required to accelerate the solids in transfer lines in kPa
deltap45=20;#Friction loss across the reactor's stripper downcomer in kPa
deltap1112=4;#Friction loss across the regenerator's downcomer in kPa
deltapvA=5;#Pressure drop assigned for the control valve in regenerator in kPa
deltapvB=15;#Pressure drop assigned for the control valve in reactor in kPa
deltah12=15;#Height of the riser in m
deltah86=30;#Height of the riser in m
deltah1011=7;#Height difference h10-h11 in m
g=9.81;#Acceleration due to gravity in m/s**2
gc=1;#Conversion factor
pi=3.14;

#CALCULATION
Gs=900;#From Fig.(8), to find dt
dt=math.sqrt((4/math.pi)*Fs/Gs);#Diameter of the downcomer
#Height of downcomer A from Eqn.(7)
deltahA=(1/(rho1112*g))*((pB-pA)*gc*(10**3)+(deltap12+deltapdB+deltap1112+deltapvA)*gc*10**3-rho12*g*(-deltah12)-rho34*g*(-LfB)-rho1011*g*deltah1011);
#Height of downcomer B from Eqn.(8)
deltahB=(1/(rho45*g))*(-(pB-pA)*gc*10**3+(deltap45+deltapvB+deltap78+deltapdA)*gc*10**3+rho78*g*deltah86+rho910*g*LfA)

#OUTPUT
print 'Height of downcomer for:'
print '\tRegenerator:%d m'%deltahA
print '\tReactor:%.1f m'%deltahB
Height of downcomer for:
	Regenerator:20 m
	Reactor:16.7 m
In [ ]: