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

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:

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

Thus the total mass transfer per unit area is:

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

The average mass transfer rate is:

$\frac{1}{t_{e}} \left\{2\sqrt{\frac{D}{\pi } }\Delta C_{A}t_{e}^{1/2} \right\} =2\sqrt{\frac{D}{\pi t_{e}} } \Delta C_{A}$

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

$e^{st}$f′(t) + $e^{st}s$f(t) = 0

Integrating gives: $e^{st}$f(t) = K
then: f(t) = K$e^{-st}$ (i)

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

and: f(t) =$se^{-st}$ (ii)

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

$\sqrt{\frac{D}{\pi t} } \Delta C_{A}se^{-st}dt$

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

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

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

and: s dt = 2β dβ.

$I=\int_{0}^{\infty }\frac{\sqrt{s} }{\beta } e^{-\beta ^{2}}\frac{2\beta d\beta }{s} =\frac{2}{\sqrt{s} } \int_{0}^{\infty }e^{-\beta ^{2}}d\beta =\frac{2}{\sqrt{s} } .\frac{\sqrt{\pi } }{2} =\sqrt{\frac{\pi }{s} }$

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

$\sqrt{\frac{D}{\pi } } \Delta C_{A}s\sqrt{\frac{\pi }{s} } =\sqrt{Ds} \Delta C_{A}$

With the new age distribution function:
$\left. \begin{matrix} 0 <t<\frac{1}{s} , \text{surface renewal rate/area = s        f(t)} = Ke^{-st} \\ \frac{1}{s} <t< ∞ , \text{surface renewal rate/area = 2s      f(t)} = K′e^{-2st} \end{matrix} \right\}$from equation (ii)

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

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

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

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

$Ke^{-1}=K^{\prime}e^{-2}$ or: K′ = K . e

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

$=\frac{K}{s} (1-e^{-1}+\frac{1}{2}e^{-1} )=\frac{K}{s} (1-\frac{1}{2} e^{-1})$

Thus: $K=s(1-\frac{1}{2}e^{-1} )^{-1}$  K′=$se(1-\frac{1}{2} e^{-1})^{-1}$

Thus 0 <t<$\frac{1}{s}$  $f(t)=s(1-\frac{1}{2}e^{-1} )^{-1}e^{-st}$

$\frac{1}{s}$<t<∞ f(t) = se$(1-\frac{1}{2}e^{-1} )^{-1}e^{-2st}$=$=s(1-\frac{1}{2}e^{-1} )^{-1}e^{1-2st}$