Explain the basis of the penetration theory for mass transfer across a phase boundary. What are the assumptions in the theory which lead to the result that the mass transfer rate is inversely proportional to the square root of the time for which a surface element has been expressed d?(Do

Question

Explain the basis of the penetration theory for mass transfer across a phase boundary. What are the assumptions in the theory which lead to the result that the mass transfer rate is inversely proportional to the square root of the time for which a surface element has been expressed? (Do not present a solution of the differential equation.) Obtain the age distribution function for the surface:
(a) On the basis of the Danckwerts’ assumption that the probability of surface renewal is independent of its age.
(b) On the assumption that the probability of surface renewal increases linearly with the age of the surface.
Using the Danckwerts surface renewal model, estimate:
(c) At what age of a surface element is the mass transfer rate equal to the mean value for the whole surface for a surface renewal rate (s) of 0.01 m²/m²s?
(d) For what proportion of the total mass transfer is surface of an age exceeding 10 seconds responsible?

Accepted Answer

(a) Danckwerts age distribution function
Dividing the total unit surface into elements each of duration dt, then:

If the fraction of surface in age band t to t + dt is f(t)dt, then:
the fraction of surface in age band t - dt to t will be f(t - dt) dt.

The surface not going from t - dt/t to t/t + dt = f(t - dt)dt - f(t)dt

= -f′(t - dt)dt dt

This will be surface destroyed in time dt
= (destruction rate × area of surface × time interval)

= s[f(t - dt)dt]dt

Thus: -f′(t - dt)dt dt = s[f(t - dt)dt]dt

As dt → 0 then: f′(t) + sf(t) = 0

Using the Danckwerts model: s = const.

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

There is no upper age limit to surface
Thus total surface = 1 = [latex]K\int_{0}^{\infty } e^{-st}dt =K\left[ \frac{e^{-st}}{-s} ight] _{0}^{\infty } =K/s[/latex] or : K = s

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

(b) s = at where a is a constant.

Thus: [latex]e^{at^{2}/2}f^{\prime}(t)+ate^{at^{2}/2}f(t)=0[/latex]

[latex]e^{at^{2}/2}=K^{\prime}[/latex]

and: f(t) = [latex]K^{\prime}e^{at^{2}/2}[/latex]

For unit total surface: 1 = [latex]K^{\prime}\int_{0}^{\infty }e^{-at^{2/2}}dt =K^{\prime}\sqrt{\frac{2}{a} } \int_{0}^{\infty }e^{-at^{2/2}}d\left(\sqrt{\frac{a}{2} }t ight) [/latex]

=[latex]K^{\prime}\sqrt{\frac{2}{a} } \frac{\sqrt{\pi } }{2} [/latex]

and: [latex]K^{\prime}=\sqrt{\frac{2a}{\pi } } [/latex] and [latex]f(t) = \sqrt{\frac{2a}{\pi } }e^{-at^{2/2}}[/latex]

(c) regarding surface renewal as random
For unit total surface, mass transfer (mol/area time) = [latex]\int_{0}^{\infty }{} kt^{-1/2}(se^{-st})dt\Delta C_{A}[/latex]

=[latex]\Delta C_{A}ks\int_{0}^{\infty }t^{-1/2}e^{-st}dt[/latex] .

where Δ[latex]C_{A}[/latex] is the concentration driving force in moles per unit volume Putting st = x², then: s dt = 2x dx

The mass transfer rate = [latex]ks\Delta C_{A}\int_{0}^{\infty }x^{-1}\sqrt{s} e^{-x^{2}}2xdx.\frac{1}{s} [/latex]

[latex]=\Delta C_{A}2k\sqrt{s} \int_{0}^{\infty }e^{-x^{2}}dx=\Delta C_{A}2k\sqrt{s} \frac{\sqrt{\pi } }{2} =\Delta C_{A}k\sqrt{\pi s}[/latex]

The mass transfer rate at time t is: Δ[latex]C_{A}kt^{-1/2}[/latex]
Thus the age of surface at which rate = average is given by or:

[latex]kt^{-1/2} = k \sqrt{\pi s}[/latex]

[latex]t=\frac{1}{\pi s} =\frac{100}{\pi } =31.8[/latex] s

(d) Surface of age less than 10 seconds
The mass transfer taking place into surface of age up to 10 s is given by the same
expression as for the whole surface but with upper limit of 10 s instead of infinity

or: [latex]\Delta C_{A}2k\sqrt{s} \int_{0}^{k10^{-t}}{} e^{-x^{2}}dx[/latex]

when t = 10 s: [latex]x = \sqrt{st} =\sqrt{0.01 imes 10} =0.316[/latex]

The mass transfer into surface up to 10 s age is then:

[latex]\Delta C_{A}2k\sqrt{s} \int_{0}^{0.316} e^{-x^{2}}dx=\Delta C_{A}2k\sqrt{s}\frac{\sqrt{\pi } }{2} erf 0.316 = \Delta C_{A}k\sqrt{\pi s} imes 0.345[/latex]

Thus a fraction: 0.345 is contributed by surface of age 0 - 10 s and:
a fraction: 0.655 by surface of age 10 s to infinity.

Question 10.33: Explain the basis of the penetration theory for mass transfe...

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