What is the penetration theory for mass transfer across a phase boundary? Give details of the underlying assumptions. From the penetration theory, the mass transfer rate per unit area NA is given in terms of the concentration difference ΔCA between the interface and the bulk fluid, the molecular

Question

What is the penetration theory for mass transfer across a phase boundary? Give details of the underlying assumptions. From the penetration theory, the mass transfer rate per unit area [latex]N_A[/latex] is given in terms of the concentration difference [latex]ΔC_{A}[/latex] between the interface and the bulk fluid, the molecular diffusivity D and the age t of the surface element by:

[latex]N_{A}=\sqrt{\frac{D}{\pi t} } \Delta C_{A}[/latex] kmol/m² s (in SI units)

What is the mean rate of transfer if all elements of the surface are exposed for the same time [latex]t_{e}[/latex] before being remixed with the bulk?
Danckwerts assumed a random surface renewal process in which the probability of surface renewal is independent of its age. If s is the fraction of the total surface renewed per unit time, obtain the age distribution function for the surface and show that the mean mass transfer rate [latex]N_{A}[/latex] over the whole surface is:

[latex]N_{A}=\sqrt{Ds} \Delta C_{A}[/latex] (kmol/m² s, in SI units)

In a particular application, it is found that the older surface is renewed more rapidly than the recently formed surface, and that after a time [latex]s^{-1}[/latex], the surface renewal rate doubles, that is it increases from s to 2s. Obtain the new age distribution function.

Accepted Answer

Assuming the age spread of the surface ranges for t = 0, to t = ∞, consider the mass
transfer per unit area in each age group is t to t + dt and so on.
Then the mass transfer to surface in age group t to t + dt is:

=[latex]\sqrt{\frac{D}{\pi } } \Delta C_{A}t^{-1/2}dt[/latex]

Thus the total mass transfer per unit area is:

[latex]\sqrt{\frac{D}{\pi } } \Delta C_{A}\int_{0}^{t_{e}}t^{-1/2}dt= \sqrt{\frac{D}{\pi} } \Delta C_{A}\left[\frac{t^{1/2}}{\frac{1}{2} } ight] ^{t_{e}}_{0}=2\sqrt{\frac{D}{\pi } } \Delta C_{A}t_{e}^{1/2}[/latex]

The average mass transfer rate is:

[latex]\frac{1}{t_{e}} \left\{2\sqrt{\frac{D}{\pi } }\Delta C_{A}t_{e}^{1/2} ight\} =2\sqrt{\frac{D}{\pi t_{e}} } \Delta C_{A}[/latex]

In the steady state if f(t) is the age distraction function of the surface, then:
surface in age group t to t + dt = f(t) dt
and: surface in age group t - dt to t = f(t - dt) dt
Surface of age t - dt to t which is destroyed is that not entering the next age group
f(t - dt) dt - f(t) dt = s{f(t - dt) dt} dt

As dt → 0, then: -f′(t)dt dt = sf(t)dt dt

[latex]e^{st}[/latex]f′(t) + [latex]e^{st}s[/latex]f(t) = 0

Integrating gives: [latex]e^{st}[/latex]f(t) = K
then: f(t) = K[latex]e^{-st}[/latex] (i)

As the total surface = 1, [latex]K \int_{0}^{\infty } e^{-st}dt=K\left[\frac{e^{-st}}{-s} ight] ^{\infty }_{0}=\frac{K}{s} [/latex] ∴ K=s

and: f(t) =[latex] se^{-st}[/latex] (ii)

The mass transfer rate into the fraction of surface in the age group t - t dt is:

[latex]\sqrt{\frac{D}{\pi t} } \Delta C_{A}se^{-st}dt[/latex]

The mass transfer rate into the surface over the age span t = 0 to t = ∞ is:

[latex]=\sqrt{\frac{D}{\pi } } \Delta C_{A}s\int_{0}^{\infty }t^{-1/2}e^{-st}dt=\sqrt{\frac{D}{\pi } } \Delta C_{A}[/latex]

Putting = st = β² , then: [latex]t^{-1/2} = \frac{\sqrt{s} }{\beta } [/latex]

and: s dt = 2β dβ.

[latex]I=\int_{0}^{\infty }\frac{\sqrt{s} }{\beta } e^{-\beta ^{2}}\frac{2\beta d\beta }{s} =\frac{2}{\sqrt{s} } \int_{0}^{\infty }e^{-\beta ^{2}}d\beta =\frac{2}{\sqrt{s} } .\frac{\sqrt{\pi } }{2} =\sqrt{\frac{\pi }{s} } [/latex]

Thus the mass transfer rate per unit area for the surface as a whole is:

[latex]\sqrt{\frac{D}{\pi } } \Delta C_{A}s\sqrt{\frac{\pi }{s} } =\sqrt{Ds} \Delta C_{A}[/latex]

With the new age distribution function:
[latex]\left. \begin{matrix} 0

Fraction of surface of age 0 to[latex] \frac{1}{s} =K\int_{0}^{1/s}e^{-st}dt=K\left[\frac{e^{-st}}{-s} ight] ^{1/s }_{0}=\frac{K}{s} (1-e^{-1})[/latex]

Fraction of surface of age[latex] \frac{1}{s}[/latex] to [latex]\infty =K^{\prime}\int_{1/s}^{\infty }e^{-2st}dt=K^{\prime}\left[\frac{e^{-2st}}{-2s} ight] ^{\infty }_{1/s} =\frac{K^{\prime}}{2s} e^{-2}[/latex]

The total surface is unity or: [latex]\frac{K}{s} (1-e^{-1})+\frac{K^{\prime}}{2s} e^{-2}[/latex]

At t = (1/s) , both age distribution functions must apply, and:

[latex]Ke^{-1}=K^{\prime}e^{-2}[/latex] or: K′ = K . e

Thus, for the total surface: 1 = [latex]\frac{K}{s} (1-e^{-1})+\frac{Ke}{2s} e^{-2}[/latex]

[latex]=\frac{K}{s} (1-e^{-1}+\frac{1}{2}e^{-1} )=\frac{K}{s} (1-\frac{1}{2} e^{-1})[/latex]

Thus: [latex]K=s(1-\frac{1}{2}e^{-1} )^{-1}[/latex] K′=[latex]se(1-\frac{1}{2} e^{-1})^{-1}[/latex]

Thus 0

[latex]\frac{1}{s}[/latex]

Question 10.45: What is the penetration theory for mass transfer across a ph...

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