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

(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.

e^{st}f′(t) + se^{st}f(t) = 0
e^{st}f(t) = K
f(t) = Ke^{-st}

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

f(t) = se^{-st}

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

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

e^{at^{2}/2}=K^{\prime}

and:  f(t) = K^{\prime}e^{at^{2}/2}

For unit total surface: 1 = 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 \right)

=K^{\prime}\sqrt{\frac{2}{a} } \frac{\sqrt{\pi } }{2}

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

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

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

where ΔC_{A} is the concentration driving force in moles per unit volume Putting st = x², then: s dt = 2x dx

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

 =\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}

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

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

t=\frac{1}{\pi s} =\frac{100}{\pi } =31.8 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:  \Delta C_{A}2k\sqrt{s} \int_{0}^{k10^{-t}}{} e^{-x^{2}}dx

when t = 10 s: x = \sqrt{st} =\sqrt{0.01\times 10} =0.316

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

 \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} \times 0.345

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.