Steady-State Molecular Diffusion in a Binary Mixture Through a Constant Area

In many important practical cases, it often becomes necessary to measure the rate of molecular diffusion of a component from one point to another. But a flux equation like Eq. (2.18) cannot be directly used for the purpose since it involves the measurement of concentration gradient at a point which is extremely difficult. On the other hand, the concentration of a component at any two points at a known distance can be measured much easily and accurately. Suitable working equations can conveniently be developed by integrating any of these Eqs. (2.18), (2.19) or (2.20). While developing the working equation, it has been assumed that

(i) diffusion occurs in steady-state*, (ii) the gas mixture behaves as ideal gas, (iii) the temperature is constant throughout, (iv) diffusion occurs through constant area.

The following two situations which are commonly encountered in practice have been considered here:

(a) diffusion of component A through a stagnant layer of component B, and (b) equal molal counter-diffusion of components A and B.

Diffusion of component A through a stagnant layer of component B

Let us consider a pool of water placed in a tray in contact with a stream of unsaturated air. So long as the air remains unsaturated, water molecules will diffuse into the air. The bulk of air is in motion, but a thin layer of air in contact with water will be stagnant and then moving in laminar motion in a direction normal to the direction of diffusion. Water vapour will diffuse through this layer by molecular diffusion before being carried away by the moving air.

In this case, since the component B is not diffusing NB = 0 and Eq. (2.20) becomes

NA = NA (2.22)

or, NA dz = - dp A (2.23)

Here, NA is constant since diffusion is taking place at steady-state through constant area.

Integrating between limits 1 and 2, we get

or, (2.24) since the total pressure is uniform, P = p A1 + p B1 = p A2 + p B2

or, p B2 - p B1 = p A1 - p A2.

Since, p BM =

so that, we get

(2.25)

The molar flux of component A in a binary mixture of A and B where B is nondiffusing can be calculated from any of these Eqs. (2.24) or (2.25).

After determining the flux NA , the partial pressure of A at any point within the diffusion path can be calculated from Eq. (2.25). The partial pressure of component B can then be evaluated from the relation p B = (P - p A). A typical distribution of partial pressure of components A and B during diffusion is shown in Figure 2.3. Component A diffuses due to the gradient (-dp A/dz). Component B also diffuses relative to the average molar velocity having a flux which depends on (-dp B/dz). But it has been assumed that B is nondiffusing. This apparent anomaly can be explained with the help of the following example:

Let us consider a man swimming in a river against the current, i.e the velocity of the flowing water.

The absolute velocities of the man and the current are equal in magnitude but opposite in direction. To a stationary observer on the river bank, the man appears to be stationary. But an observer moving with the velocity of the current will see the man moving in a direction opposite to that of the current.

Similarly, during diffusion of component A through nondiffusing component B, the flux of B to a stationary observer appears to be zero. But an observer moving with the average velocity of A will find a flux of B in the opposite direction.

Figure 2.3 Distribution of partial pressure for diffusion of A through stagnant B.

EXAMPLE 2.2 (Diffusion of one component through the stagnant layer of another component):

Oxygen is diffusing through a stagnant layer of methane 5 mm thick. The temperature is 20°C and the pressure 100 kN/m2. The concentrations of oxygen on the two sides of the film are 15% and 5% by volume. The diffusivity of oxygen in methane at 20°C and 100 kN/m2 is 2.046 10-5 m2/s.

(a) Calculate the rate of diffusion of oxygen in kmol/ m2$s

(b) What will be the rate of diffusion if the total pressure is raised to 200 kN/m2, other conditions remaining unaltered?

Solution: This being a case of diffusion of one gas through the stagnant layer of another gas, Eq.

(2.25) is applicable.

T = 293 K, P = 100 kN/m2, z = 5 mm = 0.005 m.

p A1 = 100 0.15 = 15 kN/m2, p B1 = (100 - 15) = 85 kN/m2 p A² = 100 0.05 = 5 kN/m2, p B² = (100 - 5) = 95 kN/m2 p B^M = (95 + 85)/2 = 90 kN/m2;

Since the values of p B¹ and p B² are very close, arithmetic average is quite justified.

Substituting the values in Eq. (2.25), we obtain

NA =

The rate of diffusion of oxygen = 1.87 10-8 kmol/s m2 At 200 kN/m2 pressure:

P = 200 kN/m2, DAB = (2.046 10-5) (100/200) = 1.023 10-5 m2/s as the diffusivity is inversely proportional to the pressure (Section 2.6).

p A1 = 200 # 0.15 = 30 kN/m2, p B1 = (200 - 30) = 170 kN/m2 p A² = 200 # 0.05 = 10 kN/m2, p B² = (200 - 10) = 190 kN/m2

p BM = = 180 kN/m2.

From Eq. (2.25),

NA = = 1.87 # 10-8 kmol/ s$m2

The rate of diffusion of oxygen at 200 kN/m2 pressure = 1.87 # 10-8 kmol/s$m2. Thus, the rate of diffusion remains unchanged.

EXAMPLE 2.3 (Calculation of rate of diffusion of one component from data on diffusion of another component): The rate of evaporation of water from a surface maintained at a temperature of 60°C is 2.71 # 10-4 kg/s$m2. What will be the rate of evaporation of benzene from a similar surface but maintained at 26°C if the effective film thicknesses are the same in both the cases?

Given

Vapour pressure of water at 60°C = 149 mm Hg.

Vapour pressure of benzene at 26°C = 99.5 mm Hg.

Diffusivity of air-water vapour at 60°C = 2.6 # 10-5 m2/s.

Diffusivity of air-benzene vapour at 26°C = 0.98 # 10-5 m2/s.

Atmospheric pressure in both the cases = 1.013 # 105 N/m2 Solution: For evaporation of water

Rate of evaporation = 2.71 # 10-4 kg/m2 s = 1.505 # 10-5 kmol/s m2 T = 333 K, p A1 = 149 mm Hg = 0.1986 # 105 N/m2

p B1 = 1.013 # 105 - 0.1986 # 105 = 0.8144 # 105 N/m2 p A2 = 0, p B2 = 1.013 # 105 N/m2

p BM = = 0.910 # 105 N/m2

From Eq. (2.25), we have z2 - z1 = z =

Substituting the values, we obtain

z = = 0.0138 m For evaporation of benzene

T = 299 K, p A¹ = 99.5 mm Hg = 0.1326 # 105 N/m2

p B1 = 0.8804 # 105 N/m2, p A2 = 0, p B2 = 1.013 # 105 N/m2 p BM = 0.9451 # 105 N/m2

Substituting the values in Eq. (2.25), we get

= 4.06 # 10-6

The rate of evaporation of benzene = 4.06 # 10-6 kmol/s$m2 Equimolal counter-diffusion of two components

When two components of a mixture of A and B diffuse at the same rate but in opposite directions, the phenomenon is known as equimolal counter-diffusion. In practice, there are numerous examples of equimolar counter-diffusion. Distillation provides a good example of this type of diffusion. Let us consider a mixture of benzene and toluene being rectified. The vapour rising through the column is brought into intimate contact with the liquid flowing down so that mass transfer takes place between the two streams. The less volatile component (toluene) in the vapour condenses and passes on to the liquid phase while the more volatile component (benzene) in the liquid vaporises by utilizing the latent heat of condensation released by toluene. Thus, the vapour phase gets more and more enriched in benzene while the liquid phase gets more and more enriched in toluene. Since the molar latent heats of vaporisation of benzene and toluene are almost equal, one mole of toluene gets condensed while one mole of benzene gets vaporised provided there is no heat loss from the column.

The burning of carbon in air is another example of equimolar counter-diffusion. Assuming complete combustion, carbon dioxide is the only product of combustion and for one mole of oxygen diffusing to the carbon surface through the surrounding air film, one mole of carbon dioxide diffuses out.

For steady-state equimolar counter-diffusion of two components A and B, NA = -NB, or, NA + NB = 0

Substituting the values of (NA + NB) in Eq. (2.20), we have

NA = - (2.26)

For steady-state diffusion through constant area, NA is constant.

Integrating Eq. (2.26) between the limits z1 and z2, and p′A¹ and p′A2 yields

NA dz = - dp A

or, NA = (2.27)

where, z = z2 - z1.

The distribution of partial pressure is shown in Figure 2.4.

Figure 2.4 Distribution of partial pressure for equimolal counter-diffusion of A and B.

EXAMPLE 2.4 (Equimolal counter-diffusion of two components): In a rectification column, toluene is diffusing from gas to liquid and benzene from liquid to gas at 30°C and 101.3 kN/m2 pressure under conditions of equal molal counter diffusion. At one point in the column, the molal concentrations of toluene on the two sides of a gas film 0.5 mm thick are 70% and 20%, respectively.

Assuming the diffusivity of benzene-toluene vapour under the operating conditions to be 0.38 # 10-5 m2/s, estimate the rate of diffusion of toluene and benzene in kg/hr across an area of 0.01 m2.

Solution: This is a case of equal molal counter-diffusion, Eq. (2.27) is applicable.

T = 303 K, P = 101.3 kN/m2, DAB = 0.38 # 10-5 m2/s, z = 0.0005 m.

p A¹ = 1.013 # 105 # 0.70 = 7.091 # 104 N/m2, p A² = 1.013 # 105 # 0.20 = 2.026 # 104 N/m2, Substituting the values in Eq. (2.27), we get

NA = # (7.091 - 2.026) # 104

= 1.528 # 10-4 kmol/s$m2

The rate of diffusion of toluene per 0.01 m2 surface

= 1.528 # 10-4 # 92 # 3600 # 0.01 = 0.506 kg/hr The rate of diffusion of benzene per 0.01 m2 surface

= 1.528 # 10-4 # 78 # 3600 # 0.01 = 0.429 kg/hr

Other diffusional processes sometimes practised in the industry are discussed here.