Diffusion across an interface in .NET Creator gs1 datamatrix barcode in .NET Diffusion across an interface

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6.2.3 Diffusion across an interface generate, create gs1 datamatrix barcode none for .net projects GS1 DataBar Family When solutions wit visual .net ECC200 h different concentrations are in contact, molecules will diffuse across the interface. One can imagine a tube or pipe filled from one side with a solution and from the other side with just water.

We will assume that the pipe is very long so that we do not need to worry about what is happening at the ends. If the tube is filled carefully so that there is no mixing, then immediately. 6.2 S O L V I N G T H E D I F F U S I O N E Q U A T I O N C= 0. C = C0 x= 0 C( x). Fig: 6:2: A pipe ( above) is filled from x 0 to the right with a solution at concentration C0 and to the left with concentration 0. The concentration profile (below) of the diffusing solute (solid curve) is represented as infinitely many closely spaced Gaussian functions (dotted curves)..

after this delicat gs1 datamatrix barcode for .NET e operation is complete we have C C0 for x > 0 and C 0 for x < 0 (the interface is at x 0). We need to find a solution to the one-dimensional diffusion equation that satisfies this initial condition.

This can be developed by using the Gaussian solution obtained above for the initial condition of a point source of diffusing particles. We consider the region to the right of the interface as an infinite row of infinitesimal point sources, each spreading according to Eq. (6.

8). If these point sources are close enough together their sum will approach a constant function (Fig. 6.

2). The solution to this problem is then the superposition of all these point source solutions (Fig. 6.

1). This depends on the assumption that solute starting in one of the point sources diffuses independently of the solute from all the other point sources. This turns out not to be a new physical assumption because if there were interference between solute molecules starting at different places then we would not have a simple linear relation between force and flow, and Eq.

(6.1) would no longer hold. Because linear relations add, superimposability is automatic.

With this view of diffusion across an interface we can sum up the contributions from each point source. Using x0 as the center of a point source, and dx0 as the distance between adjacent point sources, we obtain C(x) as an integral in the limit of small dx0. C0 C x p 4pDt Z1 x0 0. e x x = 4Dt (6:12). This integral cann DataMatrix for .NET ot be solved analytically, but it can be converted to a common special function called the complementary error function, defined as. 2 erfc x p p Z1 2 e y dy (6:13). x p Now with a change in variable to z x , Eq. (6.12) becomes 4Dt DIFFUSION AND BROWNIAN MOTION Fig: 6:3: Concentr ation profiles around an interface between a solution with C0 and pure water. Equation (6.14) is plotted for a sequence of times, starting with a step function at t 0.

. t = 0. t = 1 t = 4 t = 40 0 10 5 0 x 5 10. C0 C x p p x= 4Dt Z1 p C0 2 e z dz erfc x= 4Dt 2 p (6:14). This result is plo tted in Fig. 6.3.

We see that with time the interface spreads. The sharp gradient at time t 0 becomes less and less steep as time passes. This behavior has much in common with spread from a point source.

If we note that the derivative of Eq. (6.14) with respect to x gives a Gaussian function, then we realize that with diffusion at an interface the concentration gradient looks just like the concentration for diffusion from a point source.

The method used to solve this problem can readily be generalized to an arbitrary initial concentration distribution, C0(x). Again, we treat each point as a spreading Gaussian, and integrate over the distribution, to give. 1 C x p 4Dt Z1 x0 0. C0 x0 e x x 2 =4Dt (6:15). Compare this resul t with Eq. (6.12).

. 6.2.4 Diffusion with boundary conditions The problems treat ed so far were for infinitely large regions in space, so the boundary conditions did not require any special attention. But in a finite region of space the shape has to be taken into account by specifying what happens at the boundaries. To illustrate this, we look at a pipe that is open at both ends and filled with solution.

The pipe is placed in water so that solute diffuses out the ends into a large surrounding reservoir (Fig. 6.4).

There is an enormous excess of water surrounding the pipe. When a molecule reaches an open end at x 1 or 1 it is immediately diluted and effectively disappears. The boundary is therefore called absorbing, and the mathematical representation of this condition is C 0 at x 1 and 1.

This means that the solution to Eq. (6.4) must have a form which obeys this condition for all t ! 0.

It must also satisfy the initial distribution C C0 for 1 < x < 1 at t 0..
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