# Chapter 13 : Diffusion Mass Transfer¶

## Example 13.1 Page No : 544¶

In [1]:
# Variables
ro2 = 0.21;			#Ratio of O2 in the mixture
rn2 = 0.79;			#Ratio of N2 in the mixture
T = (25+273);			#Temperature of container in degree C
p = 1;			#Total pressure in atm

# Calculations
Co2 = (ro2*10**5)/(8314*T);			#Molar concentration of O2 in K.mol/m**3
Cn2 = (rn2*10**5)/(8314*T);			#Molar concentration of N2 in K.mol/m**3
po2 = (32*Co2);			#Mass density in kg/m**3
pn2 = (28*Cn2);			#Mass density in kg/m**3
p = (po2+pn2);			#Overall mass density in kg/m**3
mo2 = (po2/p);			#Mass fraction of O2
mn2 = (pn2/p);			#Mass fraction of N2
M = (ro2*32)+(rn2*28);			#Average molecular weight

# Results
print 'Molar concentration of O2 is %3.4f K.mol/m**3  \n \
Molar concentration of N2 is %3.3f K.mol/m**3  \n \
Mass density of O2 is %3.3f kg/m**3  \n \
Mass density of N2 is %3.3f kg/m**3  \n \
Mole fraction of O2 is %3.2f  \n \
Mole fraction of N2 is %3.2f  \n \
Mass fraction of O2 is %3.3f  \n \
Mass fraction of N2 is %3.3f  \n \
Average molecular weight is %3.2f'%(Co2,Cn2,po2,pn2,ro2,rn2,mo2,mn2,M)

Molar concentration of O2 is 0.0085 K.mol/m**3
Molar concentration of N2 is 0.032 K.mol/m**3
Mass density of O2 is 0.271 kg/m**3
Mass density of N2 is 0.893 kg/m**3
Mole fraction of O2 is 0.21
Mole fraction of N2 is 0.79
Mass fraction of O2 is 0.233
Mass fraction of N2 is 0.767
Average molecular weight is 28.84


## Example 13.2 Page No : 545¶

In [2]:
# Variables
yh2 = 0.4;			#Mole fraction og H2
yo2 = 0.6;			#Mole fraction of O2
vh2 = 1;			#velocity of H2 in m/s
vo2 = 0;			#velocity of O2 in m/s

# Calculations
V = (yh2*vh2)+(yo2*vo2);		#Molar average velocity in m/s
M = (yh2*2)+(yo2*32);			#Molecular weight of the mixture
mh2 = (yh2*2)/M;	    		#Mass fraction of H2
mo2 = (yo2*32)/M;		    	#Mass fraction of O2
v = (mh2*vh2)+(mo2*vo2);		#Mass average velocity in m/s
x1 = (mh2*vh2);			        #Mass flux
x2 = (mo2*vo2);			        #Mass flux
y1 = (v*vh2);		    	    #Molar flux
y2 = (yo2*vo2);	    		    #Molar flux
jh2 = (mh2*(vh2-v));			#Mass diffusion flux
jo2 = (mo2*(vo2-v));			#Mass diffusion flux
Jh2 = (yh2*(vh2-V));			#Molar diffusion flux
Jo2 = (yo2*(vo2-V));			#Molar diffusion flux

# Results
print 'Molar average velocity is %3.1f m/s \n \
Mass average velocity is %3.2f m/s  \n \
Mass flux of H2 when it is stationary is %3.2fp kg/m2.s3 \n \
Mass flux of O2 when it is stationary is %3.0f kg/m**2.s \n \
Molar flux of H2 when it is stationary is %3.2fC k.mol/m**2.s \n \
Molar flux of O2 when it is stationary is %3.0f k.mol/m**2.s \n \
Mass diffusion flux of H2 across a surface moving with mass average velocity is %3.4fp kg/m**2.s \n \
Mass diffusion flux of O2 across a surface moving with mass average velocity is %3.4fp kg/m**2.s  \n \
Molar diffusion flux across a surface moving with molar average velociy for H2 is %3.2fC k.mol/m**2.s \n \
Molar diffusion flux across a surface moving with molar average velociy for O2\
is %3.2fC k.mol/m**2.s'%(V,v,x1,x2,y1,y2,jh2,jo2,Jh2,Jo2)

Molar average velocity is 0.4 m/s
Mass average velocity is 0.04 m/s
Mass flux of H2 when it is stationary is 0.04p kg/m2.s3
Mass flux of O2 when it is stationary is   0 kg/m**2.s
Molar flux of H2 when it is stationary is 0.04C k.mol/m**2.s
Molar flux of O2 when it is stationary is   0 k.mol/m**2.s
Mass diffusion flux of H2 across a surface moving with mass average velocity is 0.0384p kg/m**2.s
Mass diffusion flux of O2 across a surface moving with mass average velocity is -0.0384p kg/m**2.s
Molar diffusion flux across a surface moving with molar average velociy for H2 is 0.24C k.mol/m**2.s
Molar diffusion flux across a surface moving with molar average velociy for O2 is -0.24C k.mol/m**2.s


## Example 13.3 Page No : 557¶

In [3]:
# Variables
t = 0.001;			#Thickness of the membrane in m
CA1 = 0.02;			#Concentration of helium in the membrane at inner surface in k.mol/m**3
CA2 = 0.005;			#Concentration of helium in the membrane at outer surface in k.mol/m**3
DAB = 10**-9;			#Binary diffusion coefficient in m**2/s

# Calculations
NAx = ((DAB*(CA1-CA2))/t)/10**-9;			#Diffusion flux of helium through the plastic in k.mol/sm**2 *10**-9

# Results
print 'Diffusion flux of helium through the plastic is %3.0f*10**-9 k.mol/sm**2'%(NAx)

Diffusion flux of helium through the plastic is  15*10**-9 k.mol/sm**2


## Example 13.4 Page No : 557¶

In [4]:
# Variables
T = 273+25;			#Temperature of Helium gas in K
p = 4;			#Pressure of helium gas in bar
Di = 0.1;			#Inner diamter of wall in m
Do = 0.003;			#Outer diamter of wall in m
DAB = (0.4*10**-13);			#Binary diffusion coefficient in m**2/s
S = (0.45*10**-3);			#S value for differentiation

# Calculations
A = (3.14*Di**2);			#Area in m**2
V = (3.14*Di**3)/6;			#Volume in m**3
R = 0.08316			#Gas consmath.tant in m**3 bar/kmol.K
d = ((-6*R*T*DAB*S*p)/(Do*Di))/10**-11;			#Decrease of pressure with time in bar/s*10**-11

# Results
print 'Initial rate of leakage for the system is provided by the decrease of pressure \
with time which is %3.2f*10**-11 bar/s'%(d)

Initial rate of leakage for the system is provided by the decrease of pressure  with time which is -3.57*10**-11 bar/s


## Example 13.5 Page No : 558¶

In [6]:
import math

# Variables
po2 = 2;			#Pressure of O2 in bar
Di = 0.025;			#inside diamter of the pipe in m
L = 0.0025;			#Wall thickness in m
a = (0.21*10**-2);			#Diffusivity of O2 in m**2/s
S = (3.12*10**-3);			#Solubility of O2 in k.mol/m**3.bar
DAB = (0.21*10**-9);			#Binary diffusion coefficient in m**2/s

# Calculations
CAi = (S*po2);			#Concentration of O2 on inside surface in kmol/m**3
RmA = ((math.log((Di+(2*L))/Di))/(2*3.14*DAB));			#Diffusion resismath.tance in sm**2
Loss = (CAi/RmA)/10**-11;			#Loss of O2 by diffusion per meter length of pipe *10**-11

# Results
print 'Loss of O2 by diffusion per meter length of pipe is %3.2f*10**-11 kmol/s'%(Loss)

Loss of O2 by diffusion per meter length of pipe is 4.51*10**-11 kmol/s


## Example 13.6 Page No : 560¶

In [7]:
# Variables
p = 1;			#Pressure of system in atm
T = 25+273;			#Temperature of system in K
pco2 = (190./760);			#Partial pressure of CO2 at one end in atm
pco2o = (95./760);			#Partial pressure of CO2 at other end in atm
DAB = (0.16*10**-4);			#Binary diffusion coefficient in m**2/s from Table 13.3
R = 0.08205			#Gas constant in m**3 atm/kmol.K

# Calculations
NAx = (DAB*(pco2-pco2o))/(R*T*p);			#Equimolar counter diffusion in kmol/m**2s
M = (NAx*3.14*(0.05**2/4)*3600);			#Mass transfer rate in kmol/h
MCO2 = (M*44)/10**-5;			#Mass flow rate of CO2 in kg/h *10**-5
Mair = (29*-M)/10**-5;			#Mass flow rate of air in kg/h *10**-5

# Results
print 'Mass transfer rate of CO2 is %3.2f*10**-5 kg/h Mass transfer rate of air is %3.2f*10**-5 kg/h'%(MCO2,Mair)

Mass transfer rate of CO2 is 2.54*10**-5 kg/h Mass transfer rate of air is -1.68*10**-5 kg/h


## Example 13.7 Page No : 563¶

In [8]:
import math
# Variables
T = 27+273;			#Temperature of water in K
D = 0.02;			#Diameter of the tube in m
L = 0.4;			#Length of the tube in m
DAB = (0.26*10**-4);			#Diffusion coefficient in m**2/s

# Calculations
p = 1.0132;			#Atmospheric pressure in bar
pA1 = 0.03531;			#Vapour pressure in bar
m = ((p*10**5*3.14*(D/2)**2*18*DAB)/(8316*T*L))*(1000*3600)*math.log(p/(p-pA1));			#Diffusion rate of water in gram per hour

# Results
print 'Diffusion rate of water is %3.4f gram per hour'%(m)

Diffusion rate of water is 0.0019 gram per hour


## Example 13.8 Page No : 564¶

In [9]:
import math
# Variables
T = 25+273;			#Temperature of water in K
D = 0.02;			#Diameter of the tube in m
L = 0.08;			#Length of the tube in m
m = (8.54*10**-4);			#Diffusion coefficient in kg/h

# Calculations
p = 1.0132;			#Atmospheric pressure in bar
pA1 = 0.03165;			#Vapour pressure in bar
DAB = (((m/3600)*8316*T*L)/(p*10**5*3.14*(D/2)**2*18*math.log(p/(p-pA1))*10**2))/10**-4;			#Diffusion coefficient of water in m**2/s *10**-4

# Calculations
print 'Diffusion coefficient of water is %3.3f*10**-4 m**2/s'%(DAB)

Diffusion coefficient of water is 0.259*10**-4 m**2/s


## Example 13.9 Page No : 569¶

In [1]:
# Variables
CAs = 0.02;			#Carbon mole fraction
CAo = 0.004;		#Content of steel
CA = 0.012;			#Percet of depth
d = 0.001;			#Depth in m
H = (6*10**-10);	#Diffusivity of carbon in m**2/s

# Calculations
X = (CA-CAs)/(CAo-CAs);			#Calculation for erf function
n = 0.48;           			#erf(n) = 0.5; n = 0.48
t = ((d/(n*2.))**2/(3600.*H))*3600;			#Time required to elevate the carbon content of steel in s

# Results
print 'Time required to elevate the carbon content of steel is %3.2f s'%(t)

# note : rounding error.

Time required to elevate the carbon content of steel is 1808.45 s