1. Diffusion
2. Fick’s law
5. Diffusion Between phases – Phase equilibria
6. Rate of diffusion between phases – Mass transfer
7. Concentration Driving Force
8. Problems
Diffusion is a result of random molecular motion in a fluid, a consequence of which is to tend to reduce differences in concentration. This is because there will be more molecules moving into an area of low concentration than moving out.
Diffusion may be illustrated by reference to the following diagrams (Fig 1.1)
Fig 1.1
The rate of diffusion is governed by Fick’s law. This states that the rate of diffusion is proportional to the concentration gradient and to ease with which the molecules will move. The latter property is called the Diffusivity. Fick’s law states that

_{ } 
1.1 
Where: 
J_{A} 
= net rate of diffusion or molar flux of component A 
kmol m^{2} s^{1} 

D_{AB} 
= diffusivity of A through B 
m^{2} s^{1} 

_{ } 
= concentration gradient along the axis of diffusion 
(kmol m^{3}) m^{1} 
This equation is similar to the Fourier equation for heat transfer and is form of the general transfer equation which states that
In this case, the resistance is 1/D_{AB}
Notes
1. The rate of diffusion is expressed in terms of flow/per unit area. This is known as the Molar flux and is obtained by dividing the rate of diffusion by the cross sectional area.
2. The units above are all expressed in molar quantities. However it is also possible and, in some cases, more convenient to express the units in mass quantities by substituting "kg" for "kmol".
3. In the case of gases, the concentration may be expressed in terms of partial pressures. This gives a modified form of the Fick’s law equation

_{ } 
1.2 
Where: 
P_{A} 
= Partial pressure of component A 
Pa 

T 
= Temperature 
K 

R 
= Universal Gas Constant 
J mol^{1} K^{1} 
Two special cases of molecular diffusion will be mentioned here. They are equimolar counterdiffusion and Stefan diffusion.
In the case of equimolar counterdiffusion, the two components, A & B are diffusing in opposite directions at equal rates.

_{ } 
1.3 
Stefan diffusion is, in effect, the opposite of equimolar counterdiffusion. One component diffuses, while the other remains stagnant
Fig 1.2
In Fig 1.2, point 0 is saturated with component A while a stream of pure B flows past the end of the tube removing any A that has reached point z. In this situation, the concentration gradient along the tube is exponential and the rate of diffusion is given by (in partial pressure terms)

_{ } 
1.4 
In most practical situations, diffusion does not occur only as a result of molecular motion but is aided by the bulk motion of the fluid. For example animals, when hunting, try to remain downwind of their prey so that the odours from the prey are carried towards them and their own odours are carried away from their prey. This is an example of how bulk movement of the air assists the process of molecular diffusion in a particular direction.
In order to account for the effect of bulk flow of the fluid, it is necessary to extend the basic Fick's law equation.
Fig 1.3
Consider a fluid flowing with Volume flow V along a pipe past a point X in the pipe. The fluid comprises two components A & B and the system is at steady state.
The molar flux of component A past X will be due to molecular diffusion plus the bulk flow of Component A. This molar flux is referred to as the mass transfer rate. The rate of transfer is now a combination of bulk flow and diffusivity. This combined term is called the mass transfer coefficient. Thus the basic Fick’s law equation must be modified to give

_{ } 
1.5 
N_{A} is the rate of mass transfer and is equal to J_{A} ´ A, where A is the cross sectional area.
The quantity k is called the mass transfer coefficient and is a function of the diffusivity, the fluid properties and the physical geometry of the system.
When two immiscible phases are in contact, eg. gasliquid, two immiscible liquids, it is possible for diffusing molecules to pass from one phase to the other across the interphase boundary.
When diffusion between two phases occurs, there are two factors to take into account;
1. equilibrium relationships between the two phases and
2. the rate at which the diffusion takes place
Fig 1.4
Referring to Fig 1.4 if a component, A is passing between phases 1 & 2, there will ultimately be a dynamic equilibrium established where the rate of diffusion of A is the same in both directions. Depending on the type of system (liquid – vapour, liquid – liquid etc.), the equilibrium concentration may often be expressed by simple relationships. Two such are mentioned below.
These most are often found in liquidliquid systems and in vapourliquid systems. The simplest form of law is a linear one of the form

_{ } 
1.6 
Where: 
X_{A} 
= concentration of the component A in Phase 1 


Y_{A} 
= concentration of the component A in Phase 2 


K 
= constant 

In more complex systems K is not a constant but varies with concentration.
Henry's law is a special case of the distribution coefficient for gasliquid systems. In the case of systems where one of the phases is a gas or vapour it is common to express the concentration in the gas phase terms of partial pressure. Thus Henry's Law states that

P_{A} = H C_{A} 
1.7 
Where: 
P_{A} 
= Partial pressure of A in the gas phase 

C_{A} 
= Concentration of A in the liquid phase 

H 
= constant  called Henry's law constant 
The theory discussed so far enables us to calculate the rates of diffusion within a single phase and to calculate the concentrations of a component in two phases when the system is in a state of equilibrium. However, many practical problems concern the rate of diffusion between two phases when the two phases are not in equilibrium.
The rate of mass transfer between two phases is dependant on a number of factors including
1. The diffusivity of the diffusing component in the two phases
2. How far the system is from equilibrium
3. The resistance to transfer across the interface between the two phases
There are a number of theories of mass transfer. Three widely used theories are the Whitman two film theory, the Penetration theory of Higbie and the surface renewal theory of Dankwaerts. Each of the theories has its strengths and weakness but ultimately all come up with an equation of the general type

_{ } 
1.5 
Where: 
N_{A} 
= mass transfer rate of A across the phase boundary 


K 
= mass transfer coefficient 


DC 
= concentration driving force 

7. Concentration driving force
The concentration driving force is a measure of how far the system is from equilibrium and may be developed with reference to the diagram below. (Fig 1.5)
Fig 1.5
Consider a component, such as oxygen, diffusing from air to water. Assume the system is at steady state but not at equilibrium.
· P_{A} is the partial pressure of oxygen in the air and C_{A} is the concentration of oxygen in the water.
· C_{A}^{*} is the concentration of oxygen in water that will be in equilibrium with P_{A}
· P_{A*} is the partial pressure of oxygen in air that will be in equilibrium with C_{A}
The rate of oxygen mass transfer between air and water may be expressed in two ways. One based on partial pressure of oxygen in air and one based on the concentration of oxygen in water.

_{ } 
1.6 

_{ } 
1.7 
Where: 
P_{A} –
P_{A}^{*} 
= partial pressure driving force 

C_{A} –
C_{A}^{*} 
= concentration driving force 

K_{L} 
= mass transfer coefficient based on the liquid phase. 

K_{G} 
= mass transfer coefficient based on the gas phase. 
The approriate equilibrium value P_{A}* or C_{A}* may be calculated using Henry's law

P_{A}* = H C_{A} 
1.8 
or 
P_{A} = H C_{A}* 
1.9 
It is usual in fermentation to use concentration driving force and liquid phase mass transfer coefficients (equation 1.7)
In practical mass transfer systems, there is a usually a problem of determining the interfacial area. Most practical mass transfer involves either passing the two phases over some form of packing or bubbling one phase through another. Either way the aim is to maximise the interfacial area. In such circumstances, it is not practically possible to determine the actual interfacial area.
To overcome this problem, a quantity called specific surface area is defined as surface area divided by volume. Specific surface area is usually denoted by a, hence a = A/V. Specific surface area replaces a in equation 1.7 to give.

_{ } 
1.10 
Even in such circumstances, the specific surface area is not separately determined. Rather the total quantity K_{L}a is measured and this value is quoted in most sources.
1. 
A tube 300 mm long and 6 mm diameter contains a mixture of ammonia and air at 20^{o}C. The partial pressure of ammonia at one end of the tube is maintained at 10·000 Pa and at the other end the ammonia is removed at by a current of air so that its partial pressure is negligible. How long will it take to transfer 0.1mg of ammonia along the tube? 
Data 




Diffusivity, 
PD_{AB} 
= 2.33 Pa m^{2} s^{1} 

Total Pressure, 
P 
= 100 kPa 

Universal Gas constant, 
R 
= 8.314 J mol^{1} K^{1} 

RMM ammonia, 
NH_{3} 
= 17 
2. 
A tube 60 mm high having a cross sectional area of 3 mm^{2} contains 3 mg water at 30^{o}C. The partial pressure of water vapour above the water surface is 7370 Pa and at the top of the tube is 2950 Pa. Assuming only molecular diffusion, How long will it take to evaporate the water in the tube? 
Data 




Total Pressure, 
P 
= 10^{5} Pa 

Diffusivity 
PD_{AB} 
= 2.76 Pa m^{2} s^{1} 

Universal Gas constant 
R 
= 8.314 J mol^{1} K^{1} 
3. 
Oxygen is diffusing from air to water under conditions such that oxygen concentration in water is constant at 2.0 mg l^{1} and the oxygen concentration in air is constant at 21% vol. The temperature is 20^{o}C and the pressure atmospheric (= 10^{5} Pa). Calculate; 

a. 
The rate of diffusion of air into water given an gas phase mass transfer coefficient K_{G} of 1.05 x 10^{7} mg m s^{1} Pa^{1} 

b. 
The liquid phase mass transfer coefficient K_{L}. 


Values of Henry's law constant for oxygenwater are given in Fig 1.8 
4. 
Oxygen gas is diffusing from air to water. Conditions are such that the oxygen concentration in the water is maintained at 10% of saturation. Under these conditions Henry's constant for oxygenwater is 2350 Pa l mg^{1} and the liquid phase mass transfer coefficient, K_{L} is 1.7 x 10^{3} m s^{1}. The concentration of oxygen in air is 21% vol and the atmospheric pressure is 10^{5 }Pa. Calculate; 

a. 
the rate of diffusion of the oxygen from air to water, 

b. 
the gas phase mass transfer coefficient Kg, 

c. 
The rate of diffussion if the oxygen concentration in the air is enriched to 30% v/v. The mass transfer coefficient remains constant and the oxygen content of the water is 10% of the saturation value. 
Fig 1.8 Henry's Law constant for
water/oxygen
(Based on data in Sander,
1999)
Sander R. (1999)
Compilation of Henry's Law Constants for Inorganic and Organic Species of
Potential Importance in Environmental Chemistry (Version 3) [Online
at]:
http://www.mpchmainz.mpg.de/~sander/res/henry.html
(Accessed 19/11/03)
Produced by Geoff Walker
Last Modified 21 November 2003