For the diffusion of carbon dioxide at atmospheric pressure and a temperature of 293 K, at what time will the concentration of solute 1 mm below the surface reach 1 per cent of the value at the surface? At that time, what will the mass transfer rate (kmol m^-2 s^-1) be: (a) At the free surface?

Question

For the diffusion of carbon dioxide at atmospheric pressure and a temperature of 293 K, at what time will the concentration of solute 1 mm below the surface reach 1 per cent of the value at the surface? At that time, what will the mass transfer rate (kmol [latex]m^{-2} s^{-1}[/latex]) be:

(a) At the free surface?
(b) At the depth of 1 mm?

The diffusivity of carbon dioxide in water may be taken as 1.5 × [latex]10^{-9} m²s^{-1}[/latex]. In the literature, Henry’s law constant K for carbon dioxide at 293 K is given as 1.08 × 10⁶ where K = P/X, P being the partial pressure of carbon dioxide (mm Hg) and X the corresponding mol fraction in the water.

Accepted Answer

[latex]\frac{∂C_{A}}{∂t} =D\frac{∂^{2}C_{A}}{∂y^{2}}[/latex]

where [latex]C_{A}[/latex]is concentration of solvent undergoing mass transfer.
The boundary conditions are:

[latex]\begin{matrix} y=0 \ ext{(interface)} & C_A=C_{As} \ ext{(solution value))}& t\gt 0 & \ y=\infty & C_A=0 \ t=0 & C_A=0 & 0\lt y\lt \infty \end{matrix}[/latex]

Taking Laplace transforms then:

[latex]\frac{\overline{∂C_{A}} }{∂t} =\int_{0}^{\infty }\left(\frac{∂C_{A}}{∂t} ight) e^{-pt}dt[/latex]

[latex]=[C_{A}e^{-pt}]_{0}^{\infty }-\int_{0}^{\infty }(-pe^{pt})C_{A}dt=0+p\overline{C} _{A}[/latex]

[latex]\frac{\overline{∂^{2}C_{A}} }{∂y^{2}} =\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}[/latex]

Thus: [latex]p\overline{C} _{A}=D\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}[/latex]

[latex]\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}-\frac{p}{D} \overline{C} _{A}=0[/latex]

[latex]\overline{C}_{A}=Ae^{\sqrt{p/D}y }+Be^{-\sqrt{p/D}y }[/latex]

For t > 0;

[latex]\begin{matrix} ext{when} & y=\infty & C_A=0,& \bar{C_A}=0 \enspace herefore A=0.\hspace{100 pt} & \ ext{when} & y=0 & C_A=C_{As}& \bar{C_{As}}=\int_{0}^{\infty }{C_{As}e^{-pt}dt}=C_{As}\left[\frac{e^{-pt}}{-p} ight]=\frac{C_{As}}{p}.\end{matrix}[/latex]

Thus: [latex]\frac{C_{As}}{p} =B.1[/latex]

and: [latex]\overline{C} _{A}=\frac{C_{As}}{p} e^{-\sqrt{p/D} y}[/latex]

Inverting: [latex]\frac{C_{A}}{C_{As}} = ext{erfc}\frac{y}{2\sqrt{Dt} }[/latex] (See Table in Volume 1, Appendix)

Differentiating with respect to y:

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

[latex]=\frac{2}{\sqrt{\pi } } .\frac{1}{2\sqrt{D}t } (e^{-y^{2}/4Dt})=-\frac{1}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt}[/latex]

The mass transfer rate at t, y, N = [latex]-D\left\{\frac{-1}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt} ight\} C_{As}[/latex]

when: t > 0, then: [latex]\left(-D\frac{∂C_{A}}{∂y} ight) _{y=y}=C_{As}\sqrt{\frac{D}{\pi Dt} } e^{-y^{2}/4Dt}[/latex] (i)

At t > 0 and y = 0, then: [latex]N_{0}=C_{As}\sqrt{\frac{D}{\pi t} }[/latex] (ii)

For a concentrated 1% of surface value at y = 1 mm, [latex]C_{A}/C_{As}[/latex] = 0.01 and:

[latex]0.01 = ext{erfc}\left\{\frac{10^{-3}}{2\sqrt{1.5 imes 10^{-9}t} } ight\}[/latex]

Writing erf x = 1 - erfc x, then:
0.99 = erf(12.91[latex]t^{-1/2})[/latex]

From tables 1.82 = 12.91[latex]t^{-1/2}[/latex]
t = 50.3 s

The mass transfer rate at the interface at t = 50.3 s is given by equation (ii) as:

[latex]N_{0}=C_{As}\sqrt{\frac{D}{\pi t} } =C_{As}\sqrt{\frac{1.5 imes 10^{-9}}{\pi imes 50.3} }[/latex]

= 3.08 × [latex]10^{-6} C_{As}[/latex] kmol/m² s

The mass transfer rate at y = 1 mm. and t = 50.3 s is given by equation (i) as:

[latex]N=N_{0}e^{-y^{2}/4Dt}=3.08 imes 10^{-6}e^{-10^{-6}/(4 imes 1.5 imes 10^{-9} imes 50.3)}C_{As}[/latex]

= 1.121 × [latex]10^{-7} C_{As}[/latex] where [latex]C_{As}[/latex] is in kmol/m³.

Henry’s law constant, K = 1.08 × 10⁶, where: K = P/X.
Also: P = 760 mm Hg
X = Mol fraction in liquid
X = 760/(1.08 × 10⁶) = 7.037 × [latex]10^{-4}[/latex] kmol CO₂ kmol solution (≈ per kmol water)
Water molar density = (1000/18) kmol/m³

and: [latex]C_{As}=7.037 imes 10^{-4}\left(\frac{1000}{18} ight) =0.0391[/latex] kmol/m³

When y = 0, then: [latex]N_{A0}[/latex] = 1.204 × [latex]10^{-7}[/latex] kmol/m² s.
When y = 1 mm, then: [latex]N_{A}[/latex] = 4.38 × [latex]10^{-9}[/latex] kmol/m² s.

Question 10.35: For the diffusion of carbon dioxide at atmospheric pressure ...

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