Modeling Advection

 

Let us consider the distribution of a certain constituent in space and time. Suppose we are looking at only one spatial dimension, so this can be a long pipe, or a canal. To account for spatial heterogeneity we will assume that the whole length of the canal can be divided into equal segments, each Dx long. The concentration of the “stuff” in each segment will be then a function of both time and length: C(t, x).

We also assume that there is a certain velocity of flow in the canal, r, and it is constant.

Let us now define the concentration of stuff in any given segment at time t+Dt, assuming that we know the concentration there at time t. Since calculating concentration may be confusing let us write the equation for the total amount of stuff in segment x at time t+Dt:

Rearranging the terms, we get:

C(t+Dt, x).Dx - C(t, x).Dx = C(t, x-Dx).r.Dt - C(t, x).r.Dt

 

Dividing both sides by DxDt:

Or, cancelling Dx on the left hand side and Dt on the right hand side:

Now if we let Dx ➔ 0 and Dt ➔ 0, we get the well know advection equation as a partial differential equation:

In discrete notation, the equation for concentration at the next time step is:

C(t+Dt, x) = C(t, x) - [C(t, x) - C(t, x-Dx)].r.Dt / Dx. (1)

If we know the concentration at the previous time step we can calculate the concentra-
tion at the next time step. To be able to use this equation at any (x,t) we still need to
define two more conditions. First, we need to know where to start, what was the distribu-
tion of stuff along the canal at the beginning, at time t=0. That will be the initial condi-
tion:

C(0,x) = co(x)

Besides, if you look at equation (1) you may notice that to solve it for any t we need to
know what is the concentration at the leftmost cell, where x=0. That is the boundary
condition:

C(t,0) = b(t).

There may be other ways to initialize equation (1) on the boundary. For example,
instead of defining the value on the boundary, we may define the flow, assuming, say,
that

C(t,0) = C(t,1).

This will a condition of no flow across the boundary, and this will also be sufficient to
start the iterative process with equation (1).