Question 10.35: For the diffusion of carbon dioxide at atmospheric pressure ...
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?
(b) At the depth of 1 mm?
The diffusivity of carbon dioxide in water may be taken as 1.5 × 10^{-9} m²s^{-1}. 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.
\frac{∂C_{A}}{∂t} =D\frac{∂^{2}C_{A}}{∂y^{2}}
where C_{A}is concentration of solvent undergoing mass transfer.
The boundary conditions are:
\begin{matrix} y=0 \ \text{(interface)} & C_A=C_{As} \ \text{(solution value))}& t\gt 0 & \\ y=\infty & C_A=0 \\ t=0 & C_A=0 & 0\lt y\lt \infty \end{matrix}
Taking Laplace transforms then:
\frac{\overline{∂C_{A}} }{∂t} =\int_{0}^{\infty }\left(\frac{∂C_{A}}{∂t} \right) e^{-pt}dt
=[C_{A}e^{-pt}]_{0}^{\infty }-\int_{0}^{\infty }(-pe^{pt})C_{A}dt=0+p\overline{C} _{A}
\frac{\overline{∂^{2}C_{A}} }{∂y^{2}} =\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}
Thus: p\overline{C} _{A}=D\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}
\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}-\frac{p}{D} \overline{C} _{A}=0
\overline{C}_{A}=Ae^{\sqrt{p/D}y }+Be^{-\sqrt{p/D}y }
For t > 0;
Thus: \frac{C_{As}}{p} =B.1
and: \overline{C} _{A}=\frac{C_{As}}{p} e^{-\sqrt{p/D} y}
Inverting: \frac{C_{A}}{C_{As}} =\text{erfc}\frac{y}{2\sqrt{Dt} } (See Table in Volume 1, Appendix)
Differentiating with respect to y:
\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} } \right) \right.
=\frac{2}{\sqrt{\pi } } .\frac{1}{2\sqrt{D}t } (e^{-y^{2}/4Dt})=-\frac{1}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt}
The mass transfer rate at t, y, N = -D\left\{\frac{-1}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt}\right\} C_{As}
when: t > 0, then: \left(-D\frac{∂C_{A}}{∂y}\right) _{y=y}=C_{As}\sqrt{\frac{D}{\pi Dt} } e^{-y^{2}/4Dt} (i)
At t > 0 and y = 0, then: N_{0}=C_{As}\sqrt{\frac{D}{\pi t} } (ii)
For a concentrated 1% of surface value at y = 1 mm, C_{A}/C_{As} = 0.01 and:
0.01 = \text{erfc}\left\{\frac{10^{-3}}{2\sqrt{1.5\times 10^{-9}t} } \right\}
Writing erf x = 1 – erfc x, then:
0.99 = erf(12.91t^{-1/2})
From tables 1.82 = 12.91t^{-1/2}
t = 50.3 s
The mass transfer rate at the interface at t = 50.3 s is given by equation (ii) as:
N_{0}=C_{As}\sqrt{\frac{D}{\pi t} } =C_{As}\sqrt{\frac{1.5\times 10^{-9}}{\pi \times 50.3} }
= 3.08 × 10^{-6} C_{As} kmol/m² s
The mass transfer rate at y = 1 mm. and t = 50.3 s is given by equation (i) as:
N=N_{0}e^{-y^{2}/4Dt}=3.08\times 10^{-6}e^{-10^{-6}/(4\times 1.5\times 10^{-9}\times 50.3)}C_{As}
= 1.121 × 10^{-7} C_{As} where C_{As} 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 × 10^{-4} kmol CO₂ kmol solution (≈ per kmol water)
Water molar density = (1000/18) kmol/m³
and: C_{As}=7.037\times 10^{-4}\left(\frac{1000}{18} \right) =0.0391 kmol/m³
When y = 0, then: N_{A0} = 1.204 × 10^{-7} kmol/m² s.
When y = 1 mm, then: N_{A} = 4.38 × 10^{-9} kmol/m² s.