In this article we will discuss about the general equation for mass transfer. Also learn about equimolal diffusion.
General Equation of Mass Diffusion:
Consider mass balance of species B diffusing through a control volume (dx × dy × dz) in a solid or stationary fluid medium C.
Accumulation of mass of species B in the elemental volume due to its mass diffusion in the x-direction is given by the difference between the mass influx and the mass efflux. Therefore mass accumulation of species B due to its diffusion in x-direction is-
If qb is the mass of species B generated per unit volume, then the total mass of species B generated in the control volume equals
= qb dx dy dz … (14.8)
The total mass of species B accumulated in the control volume due to mass diffusion along the coordinate axis (Eq. 14.7) and the mass generated within the control volume (Eq. 14.8) serves to increase the mass concentration of species B.
This increase is reflected by the time rate of change in mass concentration of species B in the control volume and is given by which is the general mass diffusion equation. This equation is similar to heat conduction equation and its solution can be obtained by applying the relevant boundary conditions.
ADVERTISEMENTS:
A few typical boundary conditions are listed below:
(i) Specified boundary concentration-
Cb = Cbo at x = 0
ADVERTISEMENTS:
(ii) Impermeable surface at boundary-
∂Cb/∂x = 0 at x = 0
(iii) Specified mass flux at a surface-
mb/A = – D ∂Cb/dx at x = 0
ADVERTISEMENTS:
(iv) Specified mass transfer coefficient (convection) at a surface-
Nb = hm (Cbs – Cb∞)
where hm is the convective mass transfer coefficient, Cbs is concentration in the fluid adjacent to the surface and Cb∞ is the bulk concentration in the fluid stream.
Upon separating the variables of Ficks relation (Eq. 14.1) and subsequently integrating, we obtain.
This is upon the presumption that diffusivity is independent of concentration and that the rate of diffusion is constant.
The above relation does suggest that the conduction heat transfer and diffusion mass transfer are analogous.
The convective film resistance on a solid surface in terms of convective mass transfer coefficient will be
These expressions for diffusion resistance and the convective film resistance can be conveniently used to solve problems on composite planes.
Quite often, the species concentration at the gas-solid interface is prescribed in terms of the partial pressure of the gas adjoining the interface and a solubility factors S.
Concentration Cb = S pa
Data is available for values of solubility factor S for several gas-solid combinations.
For steady state conditions with no homogeneous reactions and one-dimensional diffusion the diffusion rate in the radial direction of a cylindrical system of inner and outer radii of r1 and r2 respectively and length I is-
Example 1:
A rectangular container having steel walls of 8 mm thickness stores gaseous hydrogen at elevated pressure. The molar concentration of hydrogen in the steel at the inner and outer surfaces of the wall are approximated to be 1.0 kg-mol/m3 and 0.0 kg-mol/m3 respectively.
Presuming that the binary diffusion coefficient for hydrogen in steel is 0.24 × 10-12 m2/s, workout the diffusion flux for hydrogen through the steel wall. Point out the assumptions made in the derivation of the relation used by you.
Solution:
The molar diffusion flux of hydrogen (h) through the steel wall (s) is prescribed by the Fick’s law-
Which has been worked out with the following assumptions:
(i) Steady-state conditions
(ii) One-dimensional species diffusion through a plane wall which is approximated as a stationary medium
(iii) No chemical reaction of the diffusing substance in the solid wall
Inserting the appropriate data in consistent units:
Since the molecular weight of hydrogen is 2 kg/kg-mol, the mass flux of hydrogen is-
= 2 × 3.0 × 10-11 = 6.0 kg/s-m2
Equimolal Diffusion:
Equimolal diffusion between species B and C of a binary gas system implies that each molecule of component B is replaced by each molecule of constituent C and vice-versa.
Invoking Fick’s law, molar diffusion rates of species B and C are given by-
where pb and pc are the partial pressures, Nb and Nc are the molar diffusion rates.
Reference Fig. 14.3, the component B is shown to be diffusing in the direction of its drop in concentration and the constituent C is diffusing in the opposite direction. Distillation operations form good example of this process and the concentration at any point in the gas mixture remains constant with time.
From Dalton’s law of partial pressures, the total pressure pt of the system equals the sum of the partial pressures of the constituents.
Pt = Pb + Pc
Differentiating with respect to x,
dpt/dx = dpb/dx + dpc/dx
Under steady conditions, the total pressure of the system remains constant. Therefore-
dpt/dx = 0 and so dpb/dx = -dpc/dx … (i)
Further, when the two species diffuse simultaneously in opposite directions but at a constant rate of given number of molecules per unit time, then both Nb and Nc are numerically equal.
Thus for equimolal diffusion, the diffusion coefficient Dbc for diffusion of gas component B into gas constituent C is equal to the diffusion coefficient Dcb for diffusion of gas constituent C into gas component B. The value of this diffusion coefficient for a binary mixture of gases can be calculated by using the equation 14.6.
Thus if the diffusion coefficient Dbc is assumed constant, equation 14.18 may be integrated between any two planes to give:
where pb1 and pb2 are partial pressures of gas B at locations x1 and x2 of the system.
Equation 14.15 is valid only for equimolal counter diffusion.
Example 2:
A distillation column containing a mixture of benzene and toluene is at a pressure of 1 atmosphere and 100°C temperature. The liquid and vapour phases contain 30 mol % and 45 mol % of benzene. At 100°C temperature, the vapour pressure of toluene is 70 kN/m2 and the diffusivity is 5 × 10-6m2/s. Workout the rate of interchange of benzene and toluene between the liquid and vapour phases if resistance to mass transfer lies in a film 0.25 mm thick.
Take atmospheric pressure = 101 kN/m2 and Universal gas constant G = 8.314 kJ/kg-mol K.
Solution:
Let subscripts b and t refer to benzene and toluene respectively. At the liquid plane 1, the partial pressure of toluene is
The negative sign indicates that transfer of toluene is from vapour to liquid. Benzene will diffuse in the opposite direction at the same rate.