Chapter 11 : Particle to Gas Mass and Heat Transfer¶
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%matplotlib inline
Example 1, Page 265¶
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db=0.37; #Equilibrium bubble size in cm
dp=0.028; #Particle size in cm
rhos=1.06; #Density of solids in g/cc
ephsilonmf=0.5; #Void fraction at minimum fluidization condition
phis=0.4; #Sphericity of solids
gammab=0.005; #Ratio of volume of dispersed solids to that of bubble phase
rhog=1.18E-3; #Density of air in g/cc
myu=1.8E-4; #Viscosity of gas in g/cm s
D=0.065; #Diffusion coefficient of gas in cm**2/s
Sc=2.35; #Schmidt number
etad=1; #Adsorption efficiency factor
y=1;
umf=1.21; #Velocity at minimum fluidization condition in cm/s
ut=69; #Terminal velocity in cm/s
g=980; #Acceleration due to gravity in square cm/s**2
uo=[10,20,30,40,50];#Superficial gas velocity in cm/s
#CALCULATION
n=len(uo);
i=0;
Rept=(dp*ut*rhog)/myu;
Shstar=2+(0.6*(Rept**0.5)*(Sc**(1/3)));#Sherwood no. from Eqn.(1)
Kbc=4.5*(umf/db)+5.85*((D**0.5*g**0.25)/db**(5/4));#Gas interchange coefficient between bubble and cloud from Eqn.(10.27)
ubr=0.711*(g*db)**0.5;#Rise velocity of the bubble
x = [0,0,0,0,0]
Shbed = [0,0,0,0,0]
Rep = [0,0,0,0,0]
while i<n:
x[i]=(uo[i]-umf)/(ubr*(1-ephsilonmf));#The term delta/(1-epshilonf) after simplification
Shbed[i]=x[i]*((gammab*Shstar*etad)+((phis*dp**2*y)/(6*D))*Kbc);#Sherwood no. from Eqn.(11)
Rep[i]=(dp*uo[i]*rhog)/myu;#Reynolds of the particle
i=i+1;
#OUTPUT
print 'The desired result is the relationship between Shbed and Rep The points gives a straight line of the form y=mx+c'
print 'Rep',
print '\t\tShbed'
i=0;
while i<n:
print '%f'%Rep[i],
print '\t%f'%Shbed[i]
i=i+1;
import matplotlib.pyplot as plt
plt.plot(Rep,Shbed);
plt.xlabel("Rep");
plt.ylabel("Shbed");
plt.show()
Example 2, Page 267¶
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from scipy.optimize import fsolve
import math
#INPUT
umf=0.12 #Velocity at minimum fluidization condition in cm/s
uo=40.; #Superficial gas velocity in cm/s
ub=120; #Velocity of the bubble in cm/s
D=0.7; #Diffusion coefficient of gas in cm**2/s
abkbe1=1.; #Bubble-emuslion interchange coefficient for non absorbing particles(m=0)
abkbe2=18.; #Bubble-emuslion interchange coefficient for highly absorbing particles(m=infinity)
g=980.; #Acceleration due to gravity in square cm/s**2
#CALCULATION
#For non absorbing particles m=0,etad=0
Kbc=(ub/uo)*(abkbe1);
dbguess=2;#Guess value of db
def solver_func(db): #Function defined for solving the system
return abkbe1-(uo/ub)*(4.5*(umf/db)+5.85*(D**0.5*g**0.25)/(db**(5/4.)));#Eqn.(10.27)
d=fsolve(solver_func,dbguess)
#For highly absorbing particles m=infinity, etad=1
M=abkbe2-(uo/ub)*Kbc;
#For intermediate condition
alpha=100.;
m=10.;
etad=1./(1+(alpha/m));#Fitted adsorption efficiency factor from Eqn.(23)
abkbe3=M*etad+(uo/ub)*Kbc;
#OUTPUT
print 'For non absorbing particles:\tDiameter of bubble=%fcm\tBubble-cloud interchange coefficient=%fs**-1'%(d,Kbc);
print 'For highly absorbing partilces:\tM=%f'%(M);
print 'For intermediate condition:\tFitted adsorption efficiency factor:%f\tBubble-emuslion interchange coefficient:%fs**-1'%(etad,abkbe3);
#====================================END OF PROGRAM ======================================================
Example 3, Page 273¶
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rhos=1.3; #Density of solids in g/cc
phis=0.806; #Sphericity of solids
gammab=0.001; #Ratio of volume of dispersed solids to that of bubble phase
rhog=1.18E-3; #Density of air in g/cc
Pr=0.69; #Prandtl number
myu=1.8E-4; #Viscosity of gas in g/cm s
Cpg=1.00; #Specific heat capacity of gas in J/g K
ephsilonmf=0.45; #Void fraction at minimum fluidization condition
kg=2.61E-4; #Thermal concuctivity of gas in W/cm k
dp=0.036; #Particle size in cm
umf=6.5; #Velocity at minimum fluidization condition in cm/s
ut=150.; #Terminal velocity in cm/s
db=0.4; #Equilibrium bubble size in cm
etah=1; #Efficiency of heat transfer
uo=[10.,20.,30.,40.,50.];#Superficial gas velocity in cm/s
g=980.; #Acceleration due to gravity in square cm/s**2
#CALCULATION
Nustar=2+(((dp*ut*rhog)/myu)**0.5*Pr**(1./3));#Nusselt no. from Eqn.(25)
Hbc=4.5*(umf*rhog*Cpg/db)+5.85*((kg*rhog*Cpg)**0.5*g**0.25/db**(5./4));#Total heat interchange across the bubble-cloud boundary from Eqn.(32)
ubr=0.711*(g*db)**0.5;#Rise velocity of the bubble from Eqn.(6.7)
n=len(uo);
i=0;
x = [0,0,0,0,0]
Nubed = [0,0,0,0,0]
Rep = [0,0,0,0,0]
while i<n:
x[i]=(uo[i]-umf)/(ubr*(1-ephsilonmf));#The term delta/(1-epshilonf) after simplification
Nubed[i]=x[i]*(gammab*Nustar*etah+(phis*dp**2/(6*kg))*Hbc);#Nusselt no. from Eqn.(36)
Rep[i]=(dp*uo[i]*rhog)/myu;#Reynolds of the particle
i=i+1;
#OUTPUT
print 'The desired result is the relationship between Nubed and Rep which is in the form of a straight line y=mx+c'
print 'Rep',
print '\t\tNubed'
i=0;
while i<n:
print '%f'%Rep[i],
print '\t%f'%Nubed[i]
i=i+1;
import matplotlib.pyplot as plt
plt.plot(Rep,Nubed);
plt.xlabel("Rep");
plt.ylabel("Nubed");
plt.show()
Example 4, Page 274¶
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import math
#Variable declaration
rhog=1.2; #Density of air in kg/m**3
myu=1.8E-5; #Viscosity of gas in kg/m s
kg=2.6E-2; #Thermal concuctivity of gas in W/m k
dp=1E-4; #Particle size in m
rhos=8920; #Density of solids in kg/m**3
Cps=390; #Specific heat capacity of the solid in J/kg K
ephsilonf=0.5; #Void fraction of the fluidized bed
umf=0.1; #Velocity at minimum fluidization condition in m/s
uo=0.1; #Superficial gas velocity in m/s
pi=3.14
#CALCULATION
to=0; #Initial temperature of the bed
T=100; #Temperature of the bed
t=0.99*T; #Particle temperature i.e. when it approaches 1% of the bed temperature
mp=(pi/6)*dp**3*rhos; #Mass of the particle
A=pi*dp**2; #Surface area of the particle
Rep=(dp*uo*rhog)/myu; #Reynold's no. of the particle
Nubed=0.0178; #Nusselt no. from Fig.(6)
hbed1=(Nubed*kg)/dp; #Heat transfer coefficient of the bed
t1=(mp*Cps/(hbed1*A))*math.log((T-to)/(T-t));#Time needed for the particle approach 1 percentage of the bed temperature in case(a)
hbed2=140*hbed1;#Since from Fig.(6) Nup is 140 times Nubed
t2=(mp*Cps/(hbed2*A))*math.log((T-to)/(T-t));#Time needed for the particle approach 1 percentage of the bed temperature in case(b)
#OUTPUT
print 'Case(a):Using the whole bed coefficient from Fig.(6)'
print '\tTime needed for the particle approach 1 percentage of the bed temperature is %.0fs'%t1
print 'Case(b):Uisng the single-particle coefficient of Eqn.(25),also shown in Fig.(6)'
print '\tTime needed for the particle approach 1 percentage of the bed temperature is %.2fs'%t2
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