Principles of Mass Transfer: (Chapter 3) Molecular Diffusion in Liquids

Principles of Mass
Transfer
(CHAPTER 3)
Molecular Diffusion in Liquids
Where solute A is diffusing and solvent B is stagnant or nondiffusing.
Example: Dilute solution of propionic Acid (A) in water (B) being contact
with toluene.
EXAMPLE 6.3-1
An ethanol (A)-water (B) solution in the form of
a stagnant film 2.0 mm thick at 293 K is in
constant at one surface with an organic solvent
in which ethanol is soluble and water is insoluble.
Hence, NB = 0. At point 1 the concentration of
ethanol is 16.8 wt % and the solution density is
ρ1 = 972.8 kg/m3. At point 2 the concentration
of ethanol is 6.8 wt % and ρ2 = 988.1 kg/m3.
The diffusivity of ethanol is 0.740 x 10-9 m2/s.
Calculate the steady-state flux NA.
Solution
The diffusivity of ethanol is 0.740 x 10-9 m2/s. The
molecular weights of A and B are MA= 46.05 and
MB=18.02. for a wt % of 6.8, the mole fraction of
ethanol (A) is as follows when using 10 kg solution:
Then XB2 = 1-0.0277 = 0.9723. Calculating XA1 =

0.0732 and XB1 = 1-0.0732 = 0.9268. To calculate the
molecular weight M2 at point 2,
Similarly, M1 = 20.07. From Eq. (6.3-2),
To calculate XBM from Eq. (6.3-4), we can use the linear

mean since XB1 and XB2 are close to each other:
Substituting into Eq. (6.303) and solving,

Principles of Mass
Transfer
(CHAPTER 4)
Molecular Diffusion in Solids
EXAMPLE 6.5-1
The gas hydrogen at 17 °C and 0.010 atm partial
pressure is diffusing through a membrane of
vulcanized neoprene rubber 0.5 mm thick. The
pressure of H2 on the other side of the neoprene
is zero. Calculate the steady-state flux, assuming
that the only resistance to diffusion is in the
membrane. The solubility S of H2 gas in
neoprene at 17 °C is 0.0151 m3 (at STP of 0 °C
and 1 atm)/m3 solid.atm and the diffusivity DAB
is 1.03 x 10-10 m2/s at 17 °C.
Solution
A sketch showing the concentration is shown in Fig.
6.5-1. The equilibrium concentration cA1 at the inside
surface of the rubber is, from Eq. (6.5-5),
Since pA2 at the other side is 0, cA2 = 0. Substituting

into Eq. (6.5-2) and solving,
FIGURE 6.5-1 concentrations for Example 6.5-1

EXAMPLE 6.5-2
A polyethylene film 0.00015 m (0.15 mm) thick is being
considered for use in packaging a pharmaceutical product
at 30 °C. If the partial pressure of O2 outside the package
is 0.21 atm and inside it is 0.01 atm, calculate the
diffusion flux of O2 at steady state. Use permeability data
from Table 6.5-1. Assume that the resistances to diffusion
outside the film and inside are negligible compared to the
resistance of the film.
Solution
From Table 6.5-1, PM = 4.17 (10-12) m3 solute
(STP)/(s.m2.atm/m). Substituting into Eq. (6.5-8)
EXAMPLE 6.5-3
A sintered solid of silica 2.0 mm thick is porous,
with a void friction ε of 0.30 and a tortuosity τ of
4.0. The pores are filled with water at 298 K. At
one face the concentration of KCl held at 0.10 g
mol/liter, and fresh water flow rapidly past the
other face. Neglecting any other resistance but
that in the porous solid, calculate the diffusion of
KCl at steady state.
Solution
The diffusivity of KCL in water from Table 6.3-1 is DAB =
1.87 x 10-9 m2/s. Also, cA1 = 0.10/1000 = 1.0 x 10-4 g
mol/m3, and cA2 = 0. Substituting into Eq. (6.5-13),
Molecular Diffusion in Biological
Solutions and Gels
Diffusion of Biological Solutes in
Liquids
 Important: The diffusion of solute molecules
(macromolecules)
 Food processing: drying of fruit juice, coffee, tea,
water, volatile flavor
 Fermentation process: diffuse through microorganisms

like nutrient, sugar, oxygen
• Table 6.4-1 shows Diffusion Coefficient for Dilute

Biological Solutes in Aqueous Solution.
Prediction of Diffusivities for Biological
Solutes
 Small solutes in aqueous solution with molecular
weight less than 1000 or solute molar less than about
0.500 m3/kgmol
T
DAB  1.173  10 16
(M B )
12
 BVA
0.6
 For larger solutes, an approximation the Stokes-

Einstein
equation can be used
9.96 x 10-16 T
DAB 
VA
13
Prediction of Diffusivities for Biological
Solutes
For a molecular weight above 1000
9.40 x 10 -15 T
DAB 
 ( M A )1 3
EXAMPLE 6.4-1
Predict the diffusivity of bovine serum albumin at
298 K in water as a dilute solution using the
modified Polson equation and compare with the
experimental value in Table 6.4-1.
Solution
The molecular weight of bovine serum albumin (A) from
Table 6.4-1 is MA= 67500 kg/kg mol. The viscosity of
water at 25 °C is 0.8937 x 10-3 Pa.s at T = 298 K.
Substituting into Eq. (6.4-1),
This value is 11% higher than the experimental value of

6.81 x 10-11 m2/s
Diffusion in Biological Gels
 Gel can be looked as semisolid materials which are

porous.
 Example: agarose, agar, gelatin.
 The pores or open spaces in the gel structure are
filled with water.
 The rates of diffusion of small solutes in the gels are
somewhat less than in aqueous solution.
 A few typical values of diffusivities of some solutes
in various gel are given in Table 6.4.2.
EXAMPLE 6.4-2
A tube or bridge of a gel solution of 1.05 wt %
agar in water at 278 K is 0.04 m long and
connects two agitated solutions of urea in water.
The urea concentration in the first solution is
0.2 g mol urea per liter solution and is 0 in the
other. Calculate the flux of urea in kg mol/s.m2
at steady state.
Solution
From Table 6.4-2 for the solute urea at 278 K, DAB =
0.727 x 10-9 m2/s. For urea diffusing through stagnant
water in the gel, Eq. (6.3-3) can be used. However,
since the value of xA1 is less than about 0.01, the
solution is quite dilute and xBM 1.00. Hence Eq. (6.3-
5) can be used. The concentrations are cA1= 0.20/1000
= 0.0002 g mol/cm3 = 0.20 kg mol/m3 and cA2= 0.
Substituting into Eq. (6.3-5),

Principles of Mass Transfer: (Chapter 3) Molecular Diffusion in Liquids

Uploaded by

Copyright:

Available Formats

Principles of Mass Transfer: (Chapter 3) Molecular Diffusion in Liquids

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Principles of Mass Transfer: (Chapter 3) Molecular Diffusion in Liquids

Uploaded by

Copyright:

Available Formats

Principles of Mass

Then XB2 = 1-0.0277 = 0.9723. Calculating XA1 =

To calculate XBM from Eq. (6.3-4), we can use the linear

Substituting into Eq. (6.303) and solving,

Since pA2 at the other side is 0, cA2 = 0. Substituting

FIGURE 6.5-1 concentrations for Example 6.5-1

 Fermentation process: diffuse through microorganisms

• Table 6.4-1 shows Diffusion Coefficient for Dilute

 For larger solutes, an approximation the Stokes-

For a molecular weight above 1000

This value is 11% higher than the experimental value of

 Gel can be looked as semisolid materials which are

You might also like