# Chapter 4: Diffusion In Solids¶

## Ex4.1: Page 89¶

In :
# Illustration 4.1
# Page: 89

import math
print'Illustration 4.1 - Page: 89\n\n'

# solution

#***Data****#
P = 2;# [atm]
a1 = 0.025;# [m]
a2 = 0.050;# [m]
solub = 0.053*P;# [cubic m H2 (STP)/(cubic m rubber)]
Ca1 = solub/22.41;# inner surface of the pipe
Ca2 = 0;# resistance to difusion of H2 away from the surface is negligible.
Da = 1.8*10**(-10);# [square m/s]
l = 1;# [m]
#********#

z = (a2-a1)/2;# [m]
# Using Eqn. 4.4
Sav = (2*(math.pi)*l*(a2-a1))/(2*math.log(a2/a1));# [square m]
# Using Eqn. 4.3
w = (Da*Sav*(Ca1-Ca2))/z;# [kmol H2/s for 1m length]
w = w*2.02*10**3*3600;# [g H2/m.h]
print'The rate of loss of H2 by diffusion per m of pipe length:',round(w,6),' g H2/m.h'

Illustration 4.1 - Page: 89

The rate of loss of H2 by diffusion per m of pipe length: 5.6e-05  g H2/m.h


## Ex4.2: Page 92¶

In :
# Illustration 4.2
# Page: 92

print'Illustration 4.2 - Page: 92\n\n'
print'Illustration 4.2 (a)\n\n'

# solution (a)

# Given
a = 3.0/2;# [cm]
thetha = 68*3600;# [s]
# Ca can e calculated in terms of g/100 cubic cm
Cao = 5.0;# [g/100 cubic cm]
Ca_thetha = 3.0;# [g/100 cubic cm]
Ca_Inf = 0.0;# [g/100 cubic cm]
#**********#

E = (Ca_thetha-Ca_Inf)/(Cao-Ca_Inf);
# E = 0.6;
# From Fig. 4.2 (Pg 91): For diffusion from only one exposed surface D*thetha/(4*a^2) = 0.128
D = 0.128*4*(a**2)/thetha;# [square cm/s]
D = D*10**(-4);# [square m/s]
print'Diffusivity of urea in gel from only one exposed durface:',round(D,12),'square m/s\n\n'

print'Illustration 4.2 (b)\n\n'

# Solution (b)

#****Data****#
# Ca can e calculated in terms of g/100 cubic cm
Cao = 5.0;# [g/100 cubic cm]
Ca_thetha = 1.0;# [g/100 cubic cm]
Ca_Inf = 0.0;# [g/100 cubic cm]
#*********#

E = (Ca_thetha-Ca_Inf)/(Cao-Ca_Inf);
# E = 0.2;
# From Fig. 4.2 (Pg 91): For diffusion from only one exposed surface D*thetha/(4*a**2) = 0.568
D = 4.70*10**(-6);# From Illusration 4.2(a) [square cm/s]
thetha = 0.568*4*a**2/D;# [s]
thetha = thetha/3600.0;# [h]
print'The time taken for the avg. conc. to fall to 1g/100 cubic cm is:',round(thetha),' hours'

print'Illustration 4.2 (c)\n\n'

# solution (c)

#****Data*****#
Cao = 5.0;# [g/100 cubic cm]
Ca_thetha = 1.0;# [g/100 cubic cm]
Ca_Inf = 0.0;# [g/100 cubic cm]
#*******#

E = (Ca_thetha-Ca_Inf)/(Cao-Ca_Inf);
# E = 0.2;
# From Fig. 4.2: For diffusion from two opposite exposed surface D*thetha/(a**2) = 0.568
D = 4.70*10**(-6);# From Illusration 4.2(a) [square cm/s]
thetha = 0.568*(a**2)/D;# [s]
thetha = thetha/3600.0;# [h]
print'The time taken for the avg. conc. to fall to 1g/100 cubic cm when two faces opposed is:',int(thetha),' hours'
# the solution in the textbook is wrong due to approximation

Illustration 4.2 - Page: 92

Illustration 4.2 (a)

Diffusivity of urea in gel from only one exposed durface: 4.71e-10 square m/s

Illustration 4.2 (b)

The time taken for the avg. conc. to fall to 1g/100 cubic cm is: 302.0  hours
Illustration 4.2 (c)

The time taken for the avg. conc. to fall to 1g/100 cubic cm when two faces opposed is: 75  hours


## Ex4.3: Page 94¶

In :
# Illustration 4.3
# Page: 94

print'Illustration 4.3 - Page: 94\n\n'

# solution

#****Data****#
z = 0.1;# [cm]
pa1 = 1;# [cmHg]
pa2 = 0;# [cmHg]
Da = 1.1*10**(-10)*10**4;# [square cm/s]
#***********#

# Solubility coeffecient in terms of Hg
Sa = 0.90/76;# [cubic cm gas (STP)/cubic cm.cmHg]
# Using Eqn. 4.15
Va = (Da*Sa*(pa1-pa2))/z;# [cubic cm(STP)/square cm.s]
# Using Eqn. 4.16
P = Da*Sa;# [cubic cm gas (STP)/square cm.s.(cmHg/cm)]
print'The rate of diffusion of CO is:',round(Va,8),'cubic cm(STP)/square cm.s'
print'The permeability of the membrane is',round(P,9),'cubic cm gas (STP)/square cm.s.(cmHg/cm)'

Illustration 4.3 - Page: 94

The rate of diffusion of CO is: 1.3e-07 cubic cm(STP)/square cm.s
The permeability of the membrane is 1.3e-08 cubic cm gas (STP)/square cm.s.(cmHg/cm)


## Ex4.4: Page 96¶

In :


# Illustration 4.4
# Page: 96

print'Illustration 4.4 - Page: 96\n\n'

# solution

#****Data****#
a = 0.005;# [m]
# For the KCl diffusion
Dab1 = 1.84*10**(-9);# [square m/s]
thetha = 4.75*3600;# [s]
Ca_Inf = 0;
# For K2CrO4 diffusion
Cao = 0.28;# [g/cubic cm]
Ca_Inf = 0.002;# [g/cubic cm]
Dab2 = 1.14*10**(-9);# [square m/s]
#*******#

E = 0.1;# For 90% removal of KCl
# From Fig. 4.2 (Pg 91): Deff*thetha/a^2 = 0.18
Deff = 0.18*a**2/thetha;# [square m/s]
Dab_by_Deff = Dab1/Deff;
Ca_thetha = 0.1*0.28;# [g/cubic cm]
Es = (Ca_thetha-Ca_Inf)/(Cao-Ca_Inf);
# From Fig. 4.2 (Pg 91): Deff*thetha/a^2 = 0.30
Deff = Dab2/Dab_by_Deff;# [square m/s]
thetha = 0.3*a**2/Deff;# [s]
thetha = thetha/3600;# [h]
print'The time reqd. is:',round(thetha,3),'hours'

Illustration 4.4 - Page: 96

The time reqd. is: 12.778 hours


## Ex4.5: Page 98¶

In :
# Illustration 4.5
# Page: 98
import math

print'Illustration 4.5 - Page: 98\n\n'
print'Illustration 4.5 (a)\n\n'

# solution (a)

#****Data****#
# a = H2 b = N2
Dab_eff = 5.3*10**(-6);# [square m/s]
Dkb_eff = 1.17*10**(-5);# [square m/s]
Dab = 7.63*10**(-5);# [square m/s]
#*******#

R = 8314;#[Nm/kmol]
Mb = 2.02;# [kg/kmol]
T = 293;# [K]
Dtrue_by_Deff = Dab/Dab_eff;
# Since the ratio is strictly a matter of the geometry of the solid.
Dkb = Dkb_eff*Dtrue_by_Deff;# [square m/s]
# From Eqn. 4.20
d = 3*Dkb*((math.pi*Mb)/(8*R*T))**0.5;# [m]
print'The equivalent pore diameter is: ',round(d,9),' m\n\n'

print'Illustration 4.5 (b)\n\n'

# Solution (b)

#****Data*****#
# a = O2 b = N2 c = H2
Ya1 = 0.8;
Ya2 = 0.2;
Pt = 10133;# [N/square m]
z = 0.002;# [m]
T = 293;# [K]
#*******#

# From Table 2.1 (Pg 31):
Dab = 1.81*10**(-5);# [square m/s] at STP
Dkc = 1.684*10**(-4);# [square m/s] From Illustration 4.5(a)
Mc = 2.02;# [kg/kmol]
Ma = 32;# [kg/kmol]
Mb = 28.02;# [kg/kmol]
Dab = Dab*(1/0.1)*((293/273)**1.5);# [square m/s] at 0.1 atm & 20 C
DabEff = Dab/14.4;# [square m/s] From Illustration 4.5(a)
Dka = Dkc*((Mc/Ma)**0.5);# [square m/s]
DkaEff = Dka/14.4;# [square m/s]
Nb_by_Na = -(Ma/Mb)**0.5;
# Na/(Na+Nb) = 1.0/(1+(Nb/Na))
Na_by_NaSumNb = 1.0/(1+(Nb_by_Na));
DabEff_by_DkaEff = DabEff/DkaEff;
# By Eqn. 4.23
Na = (Na_by_NaSumNb)*(DabEff*Pt/(R*T*z))*log((((Na_by_NaSumNb)*(1+DabEff_by_DkaEff))-Ya2)/(((Na_by_NaSumNb)*(1+DabEff_by_DkaEff))-Ya1));# [kmol/square m.s]
Nb = Na*(Nb_by_Na);# [kmol/square m.s]
print"Diffusion flux of O2 is ",round(Na,8)," kmol/square m.s\n"
print"Diffusion flux of N2 is ",round(Nb,8)," kmol/square m.s\n"
#the answer in textbook is slightly different due to approximation while here calculation is precise

Illustration 4.5 - Page: 98

Illustration 4.5 (a)

The equivalent pore diameter is:  2.88e-07  m

Illustration 4.5 (b)

Diffusion flux of O2 is  2.95e-06  kmol/square m.s

Diffusion flux of N2 is  -3.16e-06  kmol/square m.s



## Ex4.6: Page 100¶

In :
# Illustration 4.6
# Page: 100

import math
print'Illustration 4.6 - Page: 100\n\n'

# solution

#***Data***#
# a = N2
# For N2 at 300K
viscosity1 = 1.8*10**(-5);# [kg/m.s]
Pt1 = 10133.0;# [N/square m.sec]
T = 300;# [K]
z = 0.0254;# [m]
T2 = 393.0;# [K]
#***********#

Ma = 28.02;# [kg/kmol]
R = 8314.0;# [J/K.kgmol]
#From Eqn 4.22
Lambda = (3.2*viscosity1/Pt1)*(R*T/(2*(math.pi)*Ma))**0.5;
d = 10**(-4);# [m]
d_by_lambda = d/Lambda;
# Kundsen flow will not occur
# N2 flow corresponding to 9 cubic ft/square ft.min at 300K & 1 std atm = 0.0457 cubic m/square m.min
Na1 = 0.0457*(273.0/T)*(1/22.41);# [kmol/square m.s]
Pt1_diff_Pt2 = 2*3386/13.6;# [N/square m]
Ptav = Pt1+(Pt1_diff_Pt2/2.0);# [N/square m]
# From Eqn. 4.26
k1 = Na1*R*T*z/(Ptav*(Pt1_diff_Pt2));# [m**4/N.s]

#For N2 at 393K
viscosity2 = 2.2*10**(-5);# [kg/m.s]
k2 = (k1*viscosity1)/(viscosity2);# [m^4/N.s]
# From Eqn 4.26
Na = (k2*Ptav*Pt1_diff_Pt2)/(R*T2*z);# [kmol/square m.s]
print"Flow rate to be expected is",round(Na,6)," kmol/square m.s"

Illustration 4.6 - Page: 100

Flow rate to be expected is 0.001159  kmol/square m.s