The system consists of 2 immiscible phases, I (-inf%26lt;z%26lt;0) and II (0%26lt;z%26lt;+inf), separated by a partition. Initially, the solute conc. in phase I is c(I,0) and C(II,0) in phase II. At time t=0 the partition at z=0 is removed and diffusion is allowed to take place. Assuming that the diffusion can be described by Fick's second law in both phases, and there is equilibrium at the interface (C(II)=mC(I)), derive the concentration profile. The mathemeatical statement of the problem is:
dC(I)/dt=D(I)d^2C(I)/dz^2, with similar for C(II)
Boundary conditions:
At z=0, C(II)=mC(I)
At z=0, D(I)(dC(I)/dz)=D(II)(dC(II)/dz)
At z=-inf, dC(I)/dz=0
At z=inf, dC(II)/dz=0
Initial conditions:
At t=0, C(I)=C(I,0) and C(II)=C(II,0)
This should be done using Laplace Transform. I am so lost on this. Please help and try to explain it out.
Derive the conc. profiile for each phase in a two-phase system in terms of their respective initial conc.?
Go to a library and find a copy of "The Mathematics of Diffusion", by John Crank. The derivation you need is given in section 2.4.2, which discusses the use of the Laplace transform to solve the diffusion equation in a semi-infinite medium. (Your problem is slightly different, in that the boundary conditions are different, but you should be able to adapt Crank's solution to your particular case.)
Alternatively, find a copy of "Conduction of Heat in Solids", by H.S. Carslaw and J.C. Jaeger, and look at chapter XII, where the application of the Laplace transform to linear heat flow problems is discussed. The mathematics of heat conduction and material diffusion are identical, with concentration being analogous to temperature, and the material diffusion coefficient being analogous to the thermal diffusion coefficient.
In all cases, application of the Laplace transform to the differential equation governing time-dependent diffusion (heat flow) results in the conversion of the partial differential equation to an ordinary differential equation, which can then be solved using conventional means.
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