State the assumptions made in the penetration theory for the absorption of a pure gas into a liquid. The surface of an initially solute-free liquid is suddenly exposed to a soluble gas and the liquid is sufficiently deep for no solute to have time to reach the far boundary of the liquid. Starting

Question

State the assumptions made in the penetration theory for the absorption of a pure gas into a liquid. The surface of an initially solute-free liquid is suddenly exposed to a soluble gas and the liquid is sufficiently deep for no solute to have time to reach the far boundary of the liquid. Starting with Fick’s second law of diffusion, obtain an expression for (i) the concentration, and (ii) the mass transfer rate at a time t and a depth y below the surface. After 50 s, at what depth y will the concentration have reached one tenth the value at the surface? What is the mass transfer rate (i) at the surface, and (ii) at the depth y, if the surface concentration has a constant value of 0.1 kmol/m³ ?

Accepted Answer

[latex]\frac{C_{A}}{C_{As}} = ext{erfc}\frac{y}{2\sqrt{Dt} }[/latex](equation 10.108)

[latex]\frac{C_A-C_{A0}}{C_{Ai}-C_{A0}}= ext{erfc}\left(\frac{2}{2\sqrt{Dt}} ight)=1- ext{erf}\left(\frac{2}{2\sqrt{Dt}} ight) [/latex] (10.108)

Differentiating with respect to y:

[latex]\frac{1}{C_{As}} \frac{∂C_{A}}{∂y} =\frac{∂}{dy}[/latex] [latex]\left\{\frac{2}{\sqrt{\pi } } ight.\int_{y/2\sqrt{Dt} }^{\infty }e^{-y^{2}/4Dt}d\left(\frac{y}{2\sqrt{Dt} } ight) [/latex]

[latex]=-\frac{1}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt}[/latex]

Thus: [latex](N_{A})_{y,t}=-D\left\{-\frac{1}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt} ight\} C_{AS}=\left(\sqrt{\frac{D}{\pi t} } ight) C_{AS}[/latex]

When t = 50 s and [latex]C_{A}/C_{AS}[/latex] = 0.1, then:

[latex]0.1= ext{erfc}\frac{y}{2\sqrt{10^{-9} imes 50} }[/latex]

or: [latex]0.9 = ext{erf}\frac{y}{2\sqrt{Dt} } [/latex]

From Table 13 in the Appendix of Volume 1, the quantity whose error fraction D 0.9 is:

[latex]\frac{y}{2\sqrt{Dt} }=1.16 [/latex]

At the surface: [latex](N_{A})_{y=0,t}=\sqrt{\frac{D}{\pi t} } C_{AS}[/latex]

[latex]=\sqrt{\frac{10^{-9}}{\pi imes 50} } imes 0.1[/latex]

= 0.252 × [latex]10^{-6}[/latex] kmol/m² s

At a depth y: [latex]N_{A} = 0.252 imes 10^{-6} imes e^{-(1.16)^{2}} = 0.0656 imes 10^{-6}[/latex] kmol/m²s.

Question 10.42: State the assumptions made in the penetration theory for the...

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