Chapter 3 : Mass transfer coefficient and interphase mass transfer

Example 3.1

In [11]:
# Variable Declaration      
v=6.;       #velocity in m/s
l=6.;       #length in m
pa1=10.;     #pressure at 1 in atm
pa2=0;     #pressure at 2 in atm
t=373.;     # temperature in kelvin
p=1.;      #pressure of naphthalene in atm at 373kelvin
D=5.15*10**-6;    #diffusivity of naphthalene in C02 in m**2/s
d=0.946;        #density of air in kg/m**3
u=.021*10**-3;         #viscosity of air in Newton*s/m**2
ID=0.075;       #diameter in m

# Calculation
nre=(ID*v*d)/(u);     #calc. of reynolds no.
cf=2*0.023*(nre)**(-0.2);        #friction factor
nsc=(u)/(d*D);        #calc of schmidt no.
kc=(cf*v)/(2*(nsc)**(2./3));
na=(kc*10**5*(pa1/760-0))/(8314*t);   #difussion flux in kmol/m**2*s
sub=na*2*3.14*(ID/2)*l;          #rate of sublimation

# Result
print "\nrate of sublimation :%f *10**-6 kmol/s\n"%(sub/10**-6);
#End

rate of sublimation :4.298312 *10**-6 kmol/s

Example 3.2

In [12]:
 
# Variable Declaration   
v=0.30;           #velocity of parallelair in m/s
t=300;            #temperature of air in kelvin
p=10.0**5/760;                #pressure of air in pascal
Dab=5.9*10**-4;             #diffusivity of naphthalene in in air in m**2/s 
pa1=0.2*10**5/760;          #pressure of air at 1 in pascal
pa2=0;                     #pressure of air at 2 in pascal
d=1.15;                #density of air in kg/m**3
u=0.0185*10**-3;        #viscosity of air in Newton*s/m**2
D=1.;                   #length in m
a=1.;                   #area of plate in m**2

# Calculation   
Nsc=u/(d*Dab);           #schmidt no. calculation
Nre=(D*v*d)/u;        #reynolds no. calculation
                     #flow is turbulent 
f=0.072*(Nre)**-.25;        #friction factor using "chilton colburn" analogy
k_c=(f*v)/(2*(Nsc)**.667);    #mass transfer coefficient
NA=k_c*(pa1-pa2)/(8314*300);    #mass flux calc.
sub=NA*a;                        #rate  of sublimation in kmol/m**2*s

# Result
print "\nrate of sublimation :%f *10**-7 kmol/s\n"%(sub/10**-7)
#End

rate of sublimation :1.077674 *10**-7 kmol/s

Example 3.3

In [4]:
    

# a is CO2 and b is water
p=2.;              #total pressure at 1 in atm 
pa1=0.2*10**5;     #pressure of CO2 at pt 1 in atm
pa2=0;            #pressure of CO2 at pt 2 is 0 since air is pure
ya1=0.1;          #mole fraction of CO2 at 1 is 0.2/2
ya2=0;            #mole fraction of CO2 at 2 is 0 since air is pure
yb1=0.9;          #mole fraction of water at 1 is (1-0.1)
yb2=1.0;          #mole fraction of water at 2 is 1.0 since total pressure has to be constant.
k_y1=6.78*10**-5;        #mass transfer coefficient in kmol/m**2*s*molefraction
import math

# Calculation   
yb_ln=(yb2-yb1)/(math.log(yb2/yb1));    #log mean is represented by yb_ln

k_y=k_y1/yb_ln; 

# Result
print "\nvalue of mass transfer coefficient k_y is:%f *10**-5 kmol/m**2*s*(molefractin)"%(k_y/10.0**-5)
k_g=k_y/p;                        #mass ttransfer coefficient in lmol/m**2*s*atm
print "\nvalue  mass transfer coefficient k_g is:%f *10**-5 kmol/m**2*s*(atm)"%(k_g/10**-5)

NA=k_y*(ya1-ya2);                #mass flux in kmol/m**2*s
print "\nvalue of rate of mass transfer :%f *10**-6 kmol/m**2*s"%(NA/10.0**-6)
#end

value of mass transfer coefficient k_y is:7.143443 *10**-5 kmol/m**2*s*(molefractin)

value  mass transfer coefficient k_g is:3.571721 *10**-5 kmol/m**2*s*(atm)

value of rate of mass transfer :7.143443 *10**-6 kmol/m**2*s

Example 3.4

In [5]:
 
NA=7.5*10**-7;       #mass flux in gmol/cm**2*s
Dab=1.7*10**-5;     #diffusivity if SO2 in water in cm**2/s
c=1./18.02;          #concentration is density/molecular weight in gmol/cm**2*s
#SO2 is absorbed from air into water

xa1=0.0025;        #liquid phase mole fraction at 1
xa2=0.0003;        #liquid phase mole fraction at 2
    #NA=kc(Ca1-Ca2)=Dab*(Ca1-Ca2)/d

# Calculation   
k_c=NA/(c*(xa1-xa2));    #k_c=Dab/d=NA/c(xa1-xa2)

# Result    
print "\nmass transfer coefficient k_c is:%f cm/s"%k_c

d=Dab/k_c;
print "\nfilm thickness d is :%f cm"%d
#end

mass transfer coefficient k_c is:0.006143 cm/s

film thickness d is :0.002767 cm

Example 3.5

In [13]:
Kg=2.72*10**-4;            #overall gas phase mass transfer coefficient in kmol/m**2*S*atm
r_gas=0.85*(1/Kg);        #given that gas phase resisitance is 0.85 times overall resistance
kg=1.0/r_gas;            
m=9.35*10**-3;            #henry's law constant in atm*m**3/kmol
kl=m/(1./Kg-1/kg);        #liquid phase mass transfer coefficient in m/s
print "\nthe value of liquid film coefficient kl : %f*10**-5 m/s"%(kl/10**-5)
print "\nthe value of gas film coefficient kg : %f*10**-5 m/s"%(kg/10**-5);
p=1.;                     #overall pressure in atm

        #NA=Kg(pag-pa*)=kg(pag-pai)=kl(Cai-Cal)
# Calculation   
Yag=0.1;                #molefraction of ammonia
Cal=6.42*10**-2;         #liquid phase concentration 
Pag=Yag*p;              #pressure of ammonia
       
#Pai=m*C_ai;
#NA=kg(pag-pai)=kl(Cai-Cal)
        
#Cai=poly([0],'Cai');        #calc. of conc. in gas phase
x =  1.6588481  #x=roots((Pag-m*Cai)*(kg/kl)-(Cai-Cal));

# Result    
print "\nthe value of interphase conc.cai :%f kmol/m**3"%x
Pai=m*x;
print "\nthe value of interphase pressure pai is:%f atm"%Pai
#end

the value of liquid film coefficient kl : 1.695467*10**-5 m/s

the value of gas film coefficient kg : 32.000000*10**-5 m/s

the value of interphase conc.cai :1.658848 kmol/m**3

the value of interphase pressure pai is:0.015510 atm

Example 3.6

In [14]:
         
m=0.3672;
t=.9;
        #Pai and Cai indicates interfacial pressure and conc.
        #Pal and Cal indicates bulk pressure and conc.
Yag=0.15;                #molefraction of ammonia
Cal=0.147;         #liquid phase concentration in kmol/m**3
p=1;                    #overall pressure

# Calculation   
Pag=Yag*p;              #pressure of ammonia

        #Pai=m*C_ai;
        #kg/kl=(Cai-Cal)/(Pag-Pai);
        
#Cai=poly([0],'Cai');        #calc. of conc. in gas phase
x =  0.211954 #x=roots((Pag-m*Cai)*(t)-(Cai-Cal));

# Result
print "\nthe value of  conc. of ammonia cai is  :%f kmol/m**3"%x
Pai=m*x;
print "\nthe value of interphase pressure Pai is  :%f atm"%Pai

the value of  conc. of ammonia cai is  :0.211954 kmol/m**3

the value of interphase pressure Pai is  :0.077830 atm

Example 3.7

In [8]:
 
# Variable Declaration   
D=.1;
l=3;             # l is length of bubble in cm
a=3.14*D*l;      # area in cm**2
Ca_o=0.0001;    #pure conc. of gas in g*mol/cc*atm
Ca=0;
NA=.482*10**-5;   # molar rate of absorption in g*moles/s
        #Pa_o and Ca_o indicates pure pressure and conc.
# Calculation   
kl=NA/(a*(Ca_o-Ca));    #mass transfer coefficient acc. to higbie's penetration theory
Q=4;                    #volumetric flow rate in cc/s
A=3.14*.1*.1/4;         #area of flow
v=Q/A;                 #velocity of flow in cm/s

#timt t=bubble length/linear velocity;
t=l/v;
DAB=(kl**2)*3.14*t/4;    #diffusivity in cm**2/s
D_new=0.09;              #revised diameter reduced to.09
a_new=3.14*l*D_new;        #revised area
A_new=3.14*0.09*0.09/4;    #revised flow area
v_new=Q/A_new;            #revised velocity

# Result    
print "\nthe value of diffusivity of gas DAB is :%f cm/s"%(DAB/10**-5)

t_new=l/v_new;            #revised time
kl_new=2*(DAB/(3.14*0.0047))**0.5;    #revised mass transfer coefficient 
NA_new=kl_new*a_new*(Ca_o-Ca);       #revised molar rate absorption in g*moles/s
print "\nthe value of NA_new is  :%f*10**-6 kmol/m**3"%(NA_new/10**-6)
#end

the value of diffusivity of gas DAB is :1.210021 cm/s

the value of NA_new is  :4.855188*10**-6 kmol/m**3

Example 3.8

In [9]:
 
# Variable Declaration  
Kg=7.36*10**-10;
p=1.013*10**5;
Ky=Kg*p;
#resistance in gas phase is 0.45 of total resistance & .55 in liquid phase
#(resistance in gas phase)r_gas=1/ky and (resistance in liq phase)r_liq=m'/kx
r_gas=0.45*(1./Ky);
ky=1./r_gas;
r_liq=0.55*(1./Ky);
print "film based liq phase mass transfer coeff.ky is  :%f "%ky

# Calculation
m1=86.45;
kx=m1/r_liq;
yag=.1;
xal=(.4/64)/((99.6/18)+(.4/64));
print "\n film based gas phase mass transfer coeff.ky is  :%f "%kx
#slope of the line gives -kx/ky=-70.61
m2=m1;                           # since equilibrium line a straigth line m'=m''
Kx=1/(1/kx+(1/(m2*ky)));        #overall liquid phase mass transfer coefficient
print "\n overall  liq phase mass transfer coefficient Kx is  :%f "%Kx
# equillibrium relation is given under
p = [0.2,0.3,0.5,0.7];
a = [29,46,83,119];
i=0;
x = [0,0,0,0]
y = [0,0,0,0]
    #looping for calcullating mole fraction
while (i<4):
    x[i]= (p[i]/64.)/(p[i]/64.+100/18);
    y[i]= a[i]/760.;                    #mole fraction plotted on y-axis
    i=i+1;                     
           #mole fraction plotted on x-axis
%matplotlib inline
from matplotlib.pyplot import *

# Result
plot(x,y);
title("Fig.3.17,Example 8");
xlabel("X-- Concentration of SO2 in liquid phase, X(10**4)(molefraction)");
ylabel("Y-- Concentration of SO2 in gas phase, Y(molefraction)");
show()
    #from the graph we get these values
yao=.083;        #corresponding to the value of xao=0.001128
xao=.00132;      #corresponding to the value of yag=.1
yai=.0925;      #corresponding to the perpendicular dropped from the pt(.001128,0.1)  
xai=.00123;
       
    # flux based on overall coefficient
NAo_gas=Ky*(yag-yao);
NAo_liq=Kx*(xao-xal);
print "overall gas phase mass transfer flux -NAo_gas is  :%f*10**-6 kmol/m**2*s "%(NAo_gas/10**-6)
print "overall liq phase mass transfer flux -NAo_liq is  :%f*10**-6 kmol/m**2*s "%(NAo_liq/10**-6);

    # flux based on film coefficient 
NAf_gas=ky*(yag-yai);
NAf_liq=kx*(xai-xal);
print "film based gas phase mass transfer flux-NAf_gas is  :%f *10**-6 kmol/m**2*s"%(NAf_gas/10**-6);
print "film based liq phase mass transfer flux-NAf_liq is  :%f *10**-6 kmol/m**2*s"%(NAf_liq/10**-6);
#end

film based liq phase mass transfer coeff.ky is  :0.000166 

 film based gas phase mass transfer coeff.ky is  :0.011719 

 overall  liq phase mass transfer coefficient Kx is  :0.006445 
Populating the interactive namespace from numpy and matplotlib

WARNING: pylab import has clobbered these variables: ['f']
`%pylab --no-import-all` prevents importing * from pylab and numpy

overall gas phase mass transfer flux -NAo_gas is  :1.267466*10**-6 kmol/m**2*s 
overall liq phase mass transfer flux -NAo_liq is  :1.235953*10**-6 kmol/m**2*s 
film based gas phase mass transfer flux-NAf_gas is  :1.242613 *10**-6 kmol/m**2*s
film based liq phase mass transfer flux-NAf_liq is  :1.192479 *10**-6 kmol/m**2*s

Example 3.9

In [10]:
 
# Variable Declaration   
Cas=1.521*10**-7;
v=1525;       #velocity in m/s
D=0.0516;    #diffusivity  in cm**2/s
d=1.25*10**-3;        #density of air in g/cm**3
u=1.786*10**-4;         #viscosity of air in n*s/m**2
Dia=2.54;       #diameter in cm

# Calculation   
nre=(Dia*v*d)/(u);     #calc. of reynolds no.
cf=2*0.036*(nre)**(-0.25);        #friction factor
nsc=(u)/(d*D);        #calc of schmidt no.
kc=(cf*v)/(2*(nsc)**(2./3));        #cf/2=kc/uo*(sc)**2/3

#rate of mass transfer=(cross sectional area)*(air velocity)*dc =(3.14*d**2/4)*v*dc  -----1 eqn
                                        
#flux for mass transfer from the surface=kc*(Cas-C)
#                  rate of mass transfer=(flux)*mass transfer
#                                       =kc*(Cas-C)*3.14*dx*D------2 eqn
#                           solving ----1 & -----2 we get
#                           
#                           (3.14*d**2/4)*v*dc=kc*(Cas-C)*3.14*dx*d;
#                           dc/(Cas-C)=(kc*3.14*d*v)/(3.14*d**2/4)*dx
#                           solving this we get
#                           ln[(Cas-C)/(Cas-C_in)]=(kc*4*x)/(d*v)
import math
x=183;            #upper limit of x
C_in=0;            #C=C_in=0;
t=(kc*4*x)/(Dia*v);    #variable to take out the exponential
z=math.e**t;
C_final=Cas-(z*(Cas-C_in));        #value of c_final in g*mol/cc;

# Result
print "\t conc. of acid at outlet   :%f *10**-8 g*mol/cc\n"%(abs(C_final/10**-8))
rate=(3.14*Dia**2/4)*v*(C_final-C_in);
print "\trate of mass transfer  :%f *10**-4 g*mol/s\n"%(abs(rate/10**-4));
#End

	 conc. of acid at outlet   :7.709387 *10**-8 g*mol/cc

	rate of mass transfer  :5.954246 *10**-4 g*mol/s