Question 10.19: A crystal is suspended in fresh solvent and 5% of the crysta...
A crystal is suspended in fresh solvent and 5% of the crystal dissolves in 300 s. How long will it take before 10% of the crystal has dissolved? Assume that the solvent can be regarded as infinite in extent, that the mass transfer in the solvent is governed by Fick’s second law of diffusion and may be represented as a unidirectional process, and that changes in the surface area of the crystal may be neglected. Start your derivations using Fick’s second law.
The mass transfer process is governed by Fick’s second law:
\frac{∂C_{A}}{∂t} =D\frac{∂^{2}C_{A}}{∂y^{2}} (equation 10.66)
and discussed in Section 10.5.2
The boundary conditions for the crystal dissolving are:
when \begin{matrix} t=0 & 0\lt y\lt \infty & C_{A}=0 \\t\gt 0 & y=\infty & C_{A}=0 \\ t\gt 0 & y=0 & C_{A}=C_{As} \text{(the saturation value)}\end{matrix}
hese boundary conditions allow the solution of equation 10.66 using Laplace transforms as the most convenient method:
\frac{\overline{∂C_{A}} }{dt} =\int_{0}^{\infty }e^{-pt}\frac{∂C_{A}}{∂t} dt (equation 10.102)
= \left[e^{-pt}C_{A}\right] ^{\infty }_{0}+p\int_{0}^{\infty }e^{-pt}C_{A}dt= 0 + p\overline{C} _{A} (equation 10.103)
Taking Laplace transforms of both sides of equation 10.66:
p\overline{C} _{A}=D\frac{∂^{2}\overline{C}_{A} }{∂y^{2}}
∴\frac{∂\overline{C}_{A} }{∂y^{2}} -\frac{p}{D} \overline{C} _{A}=0
and: \overline{C}_{A} =Ae^{\sqrt{(p/D)y} }+Be^{-\sqrt{(p/D)y} } (equation 10.105)
When y = ∞, C_{A} = 0 ∴ \overline{C}_{A} = 0 and A = 0
When y = 0, C_{A} = C_{As} and \overline{C}_{A} = C_{As}/p_{o}, B = C_{As}/p
∴ \overline{C}_{A} \frac{C_{As}}{p} e^{-\sqrt{(p/D)y} }
Inverting: C_{A}=C_{As}erfc(y/2\sqrt{Dt} ) (See Volume 1, Appendix Table 13)
Mass transfer rate at the surface = -D\left(\frac{∂C_{A}}{∂y} \right) _{y=0}
\frac{∂C_{A}}{∂y} =C_{As}\frac{∂}{∂y} \left\{\frac{2}{\sqrt{\pi } }\int_{(y/(2\sqrt{Dt} ))}^{\infty }e^{-y^{2}/4Dt}d\left(\frac{y}{2\sqrt{Dt} } \right) \right\}
\frac{∂C_{A}}{∂y} =C_{As}\frac{2}{\sqrt{\pi } } \left(-\frac{1}{2\sqrt{Dt} } \right) e^{-y^{2}/4Dt} (equation 10.111)
∴\left(\frac{∂C_{A}}{dy} \right) _{y=0}=-\frac{C_{As}}{\sqrt{\pi Dt} }
(N_{A})_{t}=-D\left(\frac{∂C_{A}}{∂t} \right) _{y=0}=C_{As}\sqrt{\frac{D}{\pi t} }
The mass transfer in time t=\int_{0}^{t}\sqrt{\frac{D}{\pi t} } dt=2\sqrt{\frac{D}{\pi } } \sqrt{t}
and the mass transfer is proportional to\sqrt{t}
Thus:\frac{M_{1}}{M_{2}} =\sqrt{\frac{t_{1}}{t_{2}} }
M_{1} =5% , M_{2} = 10%, and t_{1} = 300 s
and: 0.5=\sqrt{300/t_{2}} and t₂ = \underline{\underline{1200 s}}