# Chapter 9: Mass Transfer¶

## Example 9.1, Page no:349¶

In :
import math
from __future__ import division

#Variable declaration
wa = 0.76 ;
wb = 0.24 ;
ma = 28 ; #kg/kg mole
mb = 32 ; #kg/kg mole

#calculations
xa = ( wa /ma)/( wa /ma+ wb /mb);
xb = ( wb /mb)/( wa /ma+ wb /mb);

#result
print"The molar fractions are given by";
print"xa =",round(xa,5);
print"xb =",round(xb,5);

The molar fractions are given by
xa = 0.78351
xb = 0.21649


## Example 9.2, Page no:350¶

In :
import math
from __future__ import division

#calculations
#From Table 9.1 at 1 atm and 25 C
Dab = 0.62*10**-5 ; #m^2/s
#Therefore at 2 atm and 50 C
Dab2 = Dab*(1/2)*(323/298)**1.5 ;

#result
print"Dab at 2 atm & 50 C =",'%.4E'%Dab2,"m^2/s";

Dab at 2 atm & 50 C = 3.4982E-06 m^2/s


## Example 9.3 (a), Page no:352¶

In :
import math
from __future__ import division

#Variable declaration
t = 0.04 ; #m
A = 2 ; #m^ 2
rho1 = 0.10 ;
rho2 = 0.01 ;

#calculations
D400 = 1.6*10**-11 ; #at 400K [m^2/ s ]
#Mass Diffusion in solid solution, assuming Ficks law is valid & steady state and one dimensional diffusion
#Subtituting the values in eqn 9.3.3 , At 400 K
m400 = A* D400 *( rho1 - rho2 )/t; #kg / s

#result
print"Rate of diffusion of Hydrogen at 400 K =",m400,"kg/s";

Rate of diffusion of Hydrogen at 400 K = 7.2e-11 kg/s


## Example 9.3 (b), Page no:352¶

In :
import math
from __future__ import division

#Variable declaration
t = 0.04 ; #m
A = 2 ; #m^2
rho1 = 0.10 ;
rho2 = 0.01 ;

#calculations
D1200 = 3.5*10**-8 ; #at 1200k [m^2/s]
#Mass Diffusion in solid solution, assuming Ficks law is valid & steady state and one dimensional diffusion
#At 1200 K
#From eqn 9.3.3
m1200 = A* D1200 *( rho1 - rho2 )/t ; #At 1200 K

#result
print"b) Rate of diffusion of Hydrogen at 1200 K =",m1200,"kg/s";

b) Rate of diffusion of Hydrogen at 1200 K = 1.575e-07 kg/s


## Example 9.4 (a), Page no:356¶

In :
import math
from __future__ import division

#Variable declaration
L = 1 ; #m
D = 0.005 ; #m
Pa1 = 1 ; #atm
Pa2 = 0 ;
R = 8314 ;
T = 298 ; #K

#calculations
#Assuming Equimolal counter diffusion
#From Table 9.1
Dab = 2.80*10**-5 ; #m^2/s
#Substituing in eqn 9.4.12
Na = -( Dab /(R*T)*( Pa2 - Pa1 ) *(1.014*10**5) /L )*(3.14*(D/2)**2) ;
RNH3 = Na *17 ; #kg/s

#result
print"Na = -Nb =",'%.4E'%Na,"(kg mole)/m^2 s";
print"Rate at which ammonia is lost through the tube =",'%.3E'%RNH3,"kg/s";

Na = -Nb = 2.2489E-11 (kg mole)/m^2 s
Rate at which ammonia is lost through the tube = 3.823E-10 kg/s


## Example 9.4 (b), Page no:356¶

In :
import math
from __future__ import division

#Variable declaration
L = 1 ; #m
D = 0.005 ; #m
Pa1 = 1 ; #atm
Pa2 = 0 ;
R = 8314 ;
T = 298 ; #K

#calculations
#Since the tank is large and the pressure and temperature at the two ends of the same tube are same,
#we are assuming Equimolal counter diffusion
Dab = 2.80*10**-5 ; #m^2/s
#Substituing in eqn 9.4.12
Na = -(Dab /(R*T)*( Pa2 - Pa1 ) *(1.014*10**5) /L )*(3.14*(D/2)**2) ;
#Since equimolal counter diffusion is taking place
Nb = - Na ;
#therefore rate at which air enters the tank
Rair = abs (Nb)*29 ; #kg/s

#result
print"Rate at which air enters the tank =",'%.4E'%Rair,"kg/s";

Rate at which air enters the tank = 6.5219E-10 kg/s


## Example 9.5 , Page no:359¶

In :
import math
from __future__ import division

#Variable declaration
D = 0.04 ; #m
h = 0.20 ; #m
hw = 0.03 ; #m
R = 8314 ; #J/kg mole K
Psat = 0.02339 ; #bar
xa2 = 0 ; #mole fraction at open top

#calculations
#Evaporation of water, one dimensional
Tw = 20+273 ; #K
P = 1.014*10**5; #Pa
xa1 = Psat /1.014 ; #mole fraction at liq-vap interface
c = P/(R* Tw );
#From Table 9.2
Dab = 2.422*10**-5 ; #m^2/s
#Substituting above values in eqn 9.4.18
flux = 0.041626* Dab /0.17* math.log ((1 -0) /(1 - xa1 )); #kg mole/m^2 s
rate = flux *18*(3.14/4) *(D**2) ;

#result
print"Rate of evaporation of water =",'%.4E'%rate,"kg/s";

Rate of evaporation of water = 3.1290E-09 kg/s


## Example 9.6 , Page no:364¶

In :
import math
from __future__ import division

#Variable declaration
l = 1; #length, m
w = 0.25; #width, m
T = 293 ; #Temperature, K
rhoinfinity = 0; #kg/m^3
R = 8314; #J/ kg K
paw = 2339; #Saturation pressure of water at 20 degree C. [N/m^2]
rhoainf = 0 ; #since air in the free stream is dry

#calculations
#From Table A.2
v = 15.06*10**-6; #m^2/s
# From Table 9.2
Dab = 2.4224*10**-5; #m^2/s
Re = 2.5/ v;
Sc = v/ Dab ;
#Since Re > 3*10^5, we may assume laminar boundary layer
Sh = 0.664* Sc**(1/3) *Re**(1/2) ; #Sherwood number
h = Sh*Dab;
rhoaw = paw /(R /18* T); #kg/m^3
mh = h *(2* l*w)*( rhoaw - rhoinfinity );

#result
print"Rate of evaporation from plate =",'%.4E'%mh,"kg/s";

Rate of evaporation from plate = 4.8335E-05 kg/s


## Example 9.7 (a) , Page no:366¶

In :
import math
from __future__ import division

#Variable declaration
D = 0.04 ; #m
V = 1.9 ; #m/s
rho = 1.177 ; #kg/m^3
Pr = 0.7015 ;
Cp = 1005 ; #J/kg K
k = 0.02646 ; #W/m K

#calculations
#a) Colburn anology and Gnielinski equation
#Properties of air at 27 degree C
v = 15.718*10**-6 ; #m^2/s
# From Table 9.2
Dab = 2.54 * 10**-5 ; #m^2/s
Sc = v/ Dab ;
Re = V*D/v;
#The flow is turbulent and eqn 9.6.5 may be applied
#let r = h/hm
r = rho*Cp *(( Sc/Pr)**(2/3) ); #let r = h/hm
#From Blasius equation 4.6.4a
f = 0.079* Re**( -0.25) ;
#Substituting this value into Gnielinski
Nu = ((f /2) *(Re -1000) *Pr )/(1+12.7*(( f/2)**(1/2) )*(( Pr**(2/3) ) -1));
h = Nu*k/D;
hm = h/r; #m/s

#result
print"hm using Colburn anology and Gnielinski equation =",round(hm,6);

hm using Colburn anology and Gnielinski equation = 0.009495


## Example 9.7 (b) , Page no:366¶

In :
import math
from __future__ import division

#Variable declaration
D = 0.04 ; #m
V = 1.9 ; #m/s
rho = 1.177 ; #kg/m^3
Pr = 0.7015 ;
Cp = 1005 ; #J/kg K
k = 0.02646 ; #W/m K

#calculations
#(b) mess transfer correlation equivalent to the Gleilinski equation
#Properties of air at 27 degree C
v = 15.718*10**-6 ; #m^2/s
#From Table 9.2
Dab = 2.54 * 10**-5 ; #m^2/s
Sc = v/ Dab ;
Re = V*D/v;
#From Blasius equation 4.6.4a
f = 0.079* Re**( -0.25) ;
#Substituting in eqn 9.6.7
ShD = ((f /2) *(Re -1000) *Sc )/(1+12.7*(( f/2)) *(( Sc**(2/3) ) -1));
hm1 = ShD * Dab /D;

#result
print"(b) hm =",round(hm1,6);

(b) hm = 0.007258


## Example 9.7 (c) , Page no:366¶

In :
import math
from __future__ import division

#Variable declaration
D = 0.04 ; #m
V = 1.9 ; #m/s
rho = 1.177 ; #kg/m^3
Pr = 0.7015 ;
Cp = 1005 ; #J/kg K
k = 0.02646 ; #W/m K

#calculations
#c) To show that mass flux of water is very small compared to the mass flux of air flowing in the pipe
#Properties of air at 27 degree C
v = 15.718*10**-6 ; #m^2/s
#From Table 9.2
Dab = 2.54 * 10**-5 ; #m^2/s
Sc = v/ Dab ;
Re = V*D/v;
#The flow is turbulent and eqn 9.6.5 may be applied
#let r = h/hm
r = rho*Cp *(( Sc/Pr)**(2/3) ); #let r = h/hm
#From Blasius equation 4.6.4a
f = 0.079* Re**( -0.25) ;
#From steam table
rhoaw = 1/38.77 ; #kg/m^3
#let X = (m_a/A)_max
X = f* rhoaw ; #kg/m^2 s
#let Y = mass flux of air in pipe = (m/A)
Y = rho*V ; #kg/m^2 s
ratio = X/Y ;
percent = ratio *100;

#result
print"(c) (ma/A)max/(ma/A) =",round(percent,6),"percent Thus,\nmass flux of water is very small compared to the mass flux of air flowing in the pipe.";

(c) (ma/A)max/(ma/A) = 0.010927 percent Thus,
mass flux of water is very small compared to the mass flux of air flowing in the pipe.


## Example 9.8 , Page no:369¶

In :
import math
from __future__ import division

#Variable declaration
V = 0.5 ; #m/s
Th = 30 ; #C
Tc = 26 ; #C
#From table A.2
rho = 1.173 ; #kg/m^3
Cp = 1005 ; #J/kg K
k = 0.02654 ; #W/m K
Psat = 3363; #From steam table
PP30 = 4246 ; #From steam table partial pressure of water vapour at 30 C, N/m^2

#calculations
Tm = (Th+ Tc )/2;
#From Table 9.2 at 301 K
alpha = k/( rho *Cp); #m^2/s
Dab = 2.5584*10**-5 ; #m^2/s
hfg = 2439.2*10**3 ; #J/kg
#Substituting in equation 9.7.5
#let difference = rho_aw-rho_a infinity
difference = rho *Cp *(( alpha /Dab)**(2/3) )*( Th - Tc )/hfg ;
#From steam table
rhoaw = Psat /(8314/18*299) ;
rhoinf = rhoaw - difference ;
x = rhoinf / rho ; #mole fraction of water vapour in air stream
PP = rhoinf *8314/18*303; #Partial pressure of water vapour in air stream
relH = PP/ PP30 ;
percent = relH *100;

#result
print"Relative humidity =",round(relH,4),"i.e.",round(percent,4),"percent";

Relative humidity = 0.7441 i.e. 74.4122 percent