Steady-state momentum balances have been made in shell momentum balances in Chapter 1. Also, we have made shell energy balances for heat conduction in Chapter 6. After formulation of the steady-state differential equations, these equations have been solved by setting boundary conditions.
Similarly, we shall see that the steady-state diffusion problems may be formulated by shell mass balances. The steps involved are as follows:
1. Select the coordinate system of the problem.
2. Select the shell, where mass balance has to be applied.
3. Under steady-state conditions, apply the shell balance as:
Rate of “mass in” of A – Rate of “mass out” of A
+ Rate of mass of A produced by chemical reaction = 0 (11.13)
4. Express the terms on the basis of mass, if mass balance is to be applied. Otherwise, express the terms in molar form for molar balance.
5. We have selected the fluxes as molecular, convective or combined as the case may be. Generally, we select the combined flux and simplify the other terms.
6. Formulate the differential equation in terms of fluxes.
7. Integrate the differential equation formed and either keep the constant of integration or solve its value from the physical concepts.
8. Apply the Fick’s law of diffusion and formulate the differential equation for concentration.
9. Integrate the differential equation formed to get the concentration equation.
10. Apply the boundary conditions to solve for the constants of integration.
11. Get the concentration profile and solve other important items.
Before discussing the boundary conditions, let us discuss the molar flux and mass flux produced by chemical reactions. Let NA be the number of moles of A diffusing per unit area per unit time in a binary system. The molar flux of A in the z-direction can be written as:
. . . (11.14)
We may come across the problem of solving for NBz. This can be solved by physical or chemical reasoning. One may say that in a system, A is diffusing but B is not diffusing. In such a case, NBz = 0.
In other cases, NAz/NBz may be known from the physical or chemical concepts.
The reactions may take place by two mechanisms:
1. Homogeneous reactions 2. Heterogeneous reactions
In homogeneous reactions, the source term may appear in the differential equations. This is similar to the heat source problem for energy shell balances.
For heterogeneous reactions, there is no source term in the differential equations in the shell mass balance. These differential equations can be solved with the help of boundary conditions. Let the reaction be taking place as:
A products
The reaction rate equations can be written for homogeneous and heterogeneous reactions.
Homogeneous reaction:
cA = concentration of A, in mole of A/cm3 n = order of reaction, for first order n = 1 NAz = combined molar flux, in mol/cm2·s
kn ≤ = reaction rate constant based on the surface area Boundary conditions:
1. The concentration at the surface may be specified, e.g. xA = . 2. The mass flux at the surface may be specified, e.g.
NAz = or = given
or NBz = 0 for B is not diffusing.
3. If the solid substance A is lost to the surroundings, then we can write:
molar flux concentration gradient
or molar flux = kc(concentration gradient)
or . . . (11.17)
where
= molar flux of A at the surface = concentration of A at the surface
cAb = concentration of A in the bulk fluid stream
kc = mass transfer coefficient, similar to heat transfer coefficient 4. The rate of chemical reaction may be specified, e.g.
. . . (11.18)
where k1 ≤ = rate constant for the first-order reaction based on the surface area.
SOLVED EXAMPLES
EXAMPLE 11.1 CO2 gas is diffusing through N2 in one direction at atmospheric pressure and temperature 15°C. The mole fraction of CO2 at point A is 0.2, at point B, 3 m away in the direction of diffusion, it is 0.0195. The concentration gradient is constant over this distance and diffusivity is 1.5 10–5 m2/s. The gas phase as a whole is stationary, i.e. N2 is diffusing at the same rate as the CO2, but in the opposite direction.
(a) What is the molar flux of CO2 in kmol/m2·s?
(b) What is the net mass flux in kg/m2·s?
Solution:
Let A be the CO2 species and B be the N2 species.
DAB = 1.5 10–5 m2/s
Figure 11.3 Example 11.1.
Mole fraction of A (CO2) at z2 = 0.0195 Mole fraction of A (CO2) at z1 = 0.2 Total pressure = 1 atm
= 1.0132 105 Pa
Temperature, T = 15°C = 273.1 + 15 = 288.1 K z2 – z1 = 3 m
Let us write the Fick’s law:
. . . (i)
where c = concentration of the CO2–N2 system.
or
Now the molar flux of A becomes
. . . (ii) where xA = mole fraction of A.
Integrating Eq. (ii), we have
. . . (iii)
or
or . . . (iv)
R = 8314
Substituting the above values in Eq. (iv), we get
= 0.38 10–7 kg mol of A/m2·s For the component B, i.e. N2,
mole fraction of B, = 1 – 0.2 = 0.8
mole fraction of B, = 1 – 0.0195 = 0.9805 The mole flux for B, , can be written as
. . . (v) Substituting the values in Eq. (v), we get
= –0.38 10–7 kmol of B/m2·s
EXAMPLE 11.2 Ammonia gas (A) and nitrogen gas (B) are diffusing in counter diffusion through a straight glass tube 0.61 m long with an inside diameter 24.4 mm at 298 K and 101.32 kPa. Both ends of the tube are connected to a large mixed chamber at 101.32 kPa. The partial pressure of NH3 is constant at 20.0 kPa in one chamber and 6.666 kPa in another. The diffusivity at 298 K and 101.32 kPa is 2.3 10–5 m2/s. Calculate the diffusion of NH3 in kg mol/s and also the diffusion of N2.
Solution:
Let A(NH3) be diffusing in the chamber as shown in Figure 11.4 and B(N2) be diffusing in the opposite direction.
Figure 11.4 Example 11.2.
DAB = 2.3 10–5 m2/s = 20 103 Pa = 6.666 103 Pa z2 – z1 = 0.61 m
Total pressure, P = 101.32 103 Pa Temperature, T = 298 K
Diameter of the pipe, D = 24.4 mm
= 0.0244 m
Let us write the molar flux of A:
. . . (i) Substituting the values in Eq. (i), we get
= 2.03 10–5 kmol of A/m2·s Diffusion of A (NH3):
= 9.49 10–11 kmol of A/s Diffusion of B (N2):
= (101.32 103 – 20 103)
= 81.32 103 Pa
= (101.32 103 – 6.666 103)
= 94.654 103 Pa
. . . (ii) Substituting the values in Eq. (ii), we have
= –2.03 10–5 kmol of B/m2·s
Diffusion of B = –2.03 10–5 (0.0244)2
= –9.49 10–11 kmol of B/s
PROBLEMS
1. Prove that DAB is equal to DBA for an ideal gas. Is this relation true for a diffusion in a binary liquid system?
2. A mixture of He and N2 gases is contained in a pipe at 298 K and 1 atm total pressure which is constant. At one end of the pipe at point A (1), the partial pressure of He is 0.6 atm and at the other end at 0.2 m, = 0.2 atm. Calculate the flux of He at steady state if DAB of theHe–
N2 mixture is 0.687 10–4 m2/s.
[Ans. 5.63 10–6 kg mol A/m2·s]
3. Calculate the steady-state mass flux jAy of helium for the fusedsilica–helium system at 500 K.
The partial pressure of helium is 1 atm at y = 0 and zero at the upper surface of the plate. The data given is:
Thickness of the silica plate = 10–2 mm Density of silica = 2.6 g/cm3
Solubility of helium in silica = 0.0084 m3/m3
Diffusivity of helium–silica, DAB = 0.2 10–7 cm2/s Make suitable assumptions.