Diffusion Mass Transfer

Example 14.1 Page 884

In [1]:
import math

T = 293.  	    				;#[K] Temperature
Ma = 2       					;#[kg/kmol] Molecular Mass
#Table A.8 Hydrogen-Air Properties at 298 K
Dab1 = .41*math.pow(10,-4);     #[m^2/s] diffusion coefficient
#Table A.8 Hydrogen-Water Properties at 298 K
Dab2 = .63*math.pow(10,-8);     #[m^2/s] diffusion coefficient
#Table A.8 Hydrogen-iron Properties at 293 K
Dab3 = .26*math.pow(10,-12);    #[m^2/s] diffusion coefficient
#Table A.4 Air properties at 293 K
a1 = 21.6*math.pow(10,-6);      #[m^2/s] Thermal Diffusivity
#Table A.6 Water properties at 293 K
k = .603        				;#[W/m.K] conductivity
rho = 998       				;#[kg/m^3] Density
cp = 4182       				;#[J/kg] specific Heat
#Table A.1 Iron Properties at 300 K
a3 = 23.1 * math.pow(10,-6);    #[m^2/s]
#calculations

#Equation 14.14
#Hydrogen-air Mixture
DabT1 = Dab1*math.pow(T/298.,1.5);# [m^2/s] mass diffusivity
J1 = -DabT1*1;             		#[kmol/s.m^2]  Total molar concentration
j1 = Ma*J1;               		#[kg/s.m^2] mass Flux of Hydrogen
Le1 = a1/DabT1;             	# Lewis Number Equation 6.50

#Hydrogen-water Mixture
DabT2 = Dab2*math.pow(T/298.,1.5);# [m^2/s] mass diffusivity
a2 = k/(rho*cp)              	;#[m^2/s] thermal diffusivity  
J2 = -DabT2*1             	;#[kmol/s.m^2]  Total molar concentration
j2 = Ma*J2               	;#[kg/s.m^2] mass Flux of Hydrogen
Le2 = a2/DabT2             	;# Lewis Number Equation 6.50

#Hydrogen-iron Mixture
DabT3 = Dab3*math.pow(T/298.,1.5);# [m^2/s] mass diffusivity
J3 = -DabT3*1;             	#[kmol/s.m^2]  Total molar concentration
j3 = Ma*J3;               	#[kg/s.m^2] mass Flux of Hydrogen
Le3 = a3/DabT3             	;# Lewis Number Equation 6.50
#results

print '%s %.1e' %('a (m^2/s) in 1 = ',a1)
print '%s %.1e' %('\n a (m^2/s) in 2 = ',a2)
print '%s %.1e' %('\na (m^2/s) in 3 = ',a3)
print '%s %.1e' %('\nDab (m^2/s) in 1 = ',DabT1)
print '%s %.1e' %('\n Dab (m^2/s) in 2 = ',DabT2)
print '%s %.1e' %('\n Dab (m^2/s) in 3 = ',DabT3)
print '%s %.1e' %('\n Le  in 1 = ',Le1)
print '%s %.1e' %('\n Le  in 2 = ',Le2)
print '%s %.1e' %('\n Le  in 3 = ',Le3)
print '%s %.1e' %('\n ja (kg/s.m^2) in 1 = ',j1)
print '%s %.1e' %('\n ja (kg/s.m^2) in 2 = ',j2)
print '%s %.1e' %('\n ja (kg/s.m^2) in 3 = ',j3)

a (m^2/s) in 1 =  2.2e-05

 a (m^2/s) in 2 =  1.4e-07

a (m^2/s) in 3 =  2.3e-05

Dab (m^2/s) in 1 =  4.0e-05

 Dab (m^2/s) in 2 =  6.1e-09

 Dab (m^2/s) in 3 =  2.5e-13

 Le  in 1 =  5.4e-01

 Le  in 2 =  2.4e+01

 Le  in 3 =  9.1e+07

 ja (kg/s.m^2) in 1 =  -8.0e-05

 ja (kg/s.m^2) in 2 =  -1.2e-08

 ja (kg/s.m^2) in 3 =  -5.1e-13

Example 14.2 Page 898

In [5]:
#Variable Initialization

import math
import numpy
from numpy import linalg
import matplotlib
from matplotlib import pyplot

T = 298          			;#[K] Temperature
D = 10*math.pow(10,-6)     	;#[m]
L = 100*math.pow(10,-6);    #[m]
H = .5            			;# Moist Air Humidity
p = 1.01325      			;#[bar]
#Table A.6 Saturated Water vapor Properties at 298 K
psat = .03165;            	#[bar] saturated Pressure
#Table A.8 Water vapor-air Properties at 298 K
Dab = .26*math.pow(10,-4);  #[m^2/s] diffusion coefficient
#calculations

C = p/(8.314/100. *298)     ;#Total Concentration
#From section 6.7.2, the mole fraction at x = 0 is
xa0 = psat/p;
#the mole fraction at x = L is
xaL = H*psat/p;

#Evaporation rate per pore Using Equation 14.41 with advection
N = (math.pi*D*D)*C*Dab/(4*L)*2.303*math.log10((1-xaL)/(1-xa0))    ;#[kmol/s]

#Neglecting effects of molar averaged velocity Equation 14.32
#Species transfer rate per pore
Nh = (math.pi*D*D)*C*Dab/(4*L)*(xa0-xaL)        ;#[kmol/s]
#results

print '%s %.2e %s' %('\n Evaporation rate per pore Without advection effects',Nh,'kmol/s')
print '%s %.2e %s' %('and With Advection effects',N,'kmol/s')

 Evaporation rate per pore Without advection effects 1.30e-14 kmol/s
and With Advection effects 1.34e-14 kmol/s

Example 14.3 Page 898

In [6]:
#Variable Initialization

import math

D = .005     						;#[m] Diameter
L = 50*math.pow(10,-6);          	#[m] Length
h = .003            				;#[m] Depth
Dab = 6*math.pow(10,-14)        	;#[m^2/s] Diffusion coefficient
Cas1 = 4.5*math.pow(10,-3)        	;#[kmol/m^3] Molar concentrations of water vapor at outer surface
Cas2 = 0.5*math.pow(10,-3)        	;#[kmol/m^3] Molar concentrations of water vapor at inner surface
#calculations

#Transfer Rate through cylindrical wall Equation 14.54
Na = Dab/L*(math.pi*D*D/4. + math.pi*D*h)*(Cas1-Cas2);    #[kmol/s]
#results

print '%s %.2e %s' %('\n Rate of water vapor molar diffusive ttansfer through the trough wall ',Na,'kmol/s');
#END

 Rate of water vapor molar diffusive ttansfer through the trough wall  3.20e-16 kmol/s

Example 14.4 Page 902

In [7]:
#Variable Initialization

import math

D = .2             			;#[m] Diameter
L = 2*math.pow(10,-3)       ;#[m] Thickness
p = 4              			;#[bars] Helium Pressure
T = 20+273.        			;#[K] Temperature
#Table A.8 helium-fused silica (293K) Page 952
Dab = .4*math.pow(10,-13)	;#[m^2/s] Diffusion coefficient
#Table A.10 helium-fused silica (293K)
S = .45*math.pow(10,-3)		;#[kmol/m^3.bar] Solubility
#calculations

# By applying the species conservation Equation 14.43 and 14.62
dpt = -6*(.08314)*T*(Dab)*S*p/(L*D);

#results
print '%s %.2e %s' %('\n The rate of change of the helium pressure dp/dt',dpt,' bar/s');
#END

 The rate of change of the helium pressure dp/dt -2.63e-11  bar/s

Example 14.5 Page 904

In [8]:
#Variable Initialization

import math

Dab = 8.7*math.pow(10,-8)       ;#[m^2/s] Diffusion coefficient
Sab = 1.5*math.pow(10,-3)       ;#[kmol/m^3.bar] Solubility
L = .0003              			;#[m] thickness of bar
p1 = 3                			;#[bar] pressure on one side
p2 = 1                			;#[bar] pressure on other side
Ma = 2                			;#[kg/mol] molecular mass of Hydrogen
#calculations

#Surface molar concentrations of hydrogen from Equation 14.62
Ca1 = Sab*p1       				; #[kmol/m^3]
Ca2 = Sab*p2       				; #[kmol/m^3]
#From equation 14.42 to 14.53 for obtaining mass flux
N = Dab/L*(Ca1-Ca2) ;       	#[kmol/s.m^2]
n = Ma*N            ;       	#[kg/s.m^2] on Mass basis
#results

print '%s %.2e %s' %('\n The Hydrogen mass diffusive flux n =',n,' (kg/s.m^2)');
#END

 The Hydrogen mass diffusive flux n = 1.74e-06  (kg/s.m^2)

Example 14.6 Page 909

In [1]:
#Variable Initialization

import math

Dab = 2*math.pow(10,-12)        	;#[m^2/s] Diffusion coefficient
Ca0 = 4*math.pow(10,-3)        		;#[kmol/m^3] Fixed Concentration of medication
Na = -.2*math.pow(10,-3)       		;#[kmol/m^3.s] Minimum consumption rate of antibiotic
k1 = .1            					;#[s^-1] Reaction Coefficient
#calculations

#For firsst order kinetic reaction Equation 14.74
m = math.pow((k1/Dab),.5);
L = math.acosh(-k1*Ca0/Na) /m;
#results

print '%s %.1f %s' %('\n Maximum Thickness of a bacteria laden biofilm, that may be siccessfully treated is ',L*math.pow(10,6), 'mu-m');
#END

 Maximum Thickness of a bacteria laden biofilm, that may be siccessfully treated is  5.9 mu-m

Example 14.7 Page 913

In [10]:
#Variable Initialization

import math

Dap = .1*math.pow(10,-12)       ;#[m^2/s] Diffusion coefficient of medication with patch
Das = .2*math.pow(10,-12)       ;#[m^2/s] Diffusion coefficient of medication with skin
L = .05                			;#[m] patch Length
rhop = 100             			;#[kg/m^3] Density of medication on patch
rho2 = 0               			;#[kg/m^3] Density of medication on skin
K = .5                 			;#Partition Coefficient
t = 3600*24*7          			;#[s] Treatment time
#calculations

#Applying Conservation of species equation 14.47b
#By analogy to equation 5.62, 5.26 and 5.58
D = 2*rhop*L*L/(math.sqrt(math.pi))*math.sqrt(Das*Dap*t)/(math.sqrt(Das)+math.sqrt(Dap)/K);
#results

print '%s %.1f %s' %('\n Total dosage of medicine delivered to the patient over a one-week time period is',D*math.pow(10,6) ,'mg');

 Total dosage of medicine delivered to the patient over a one-week time period is 28.7 mg