If we look at equation (4), which states that our boundary condition is defined at infinity, we quickly realize that it is difficult to come up with a numerical solution that includes an infinite number of grid points. Since we know that there is a characteristic length associated with this problem, λ, we can assume that most of what happens takes place within one characteristic length. Even though we just said that all the fun most likely happens within a time frame of one λ, we want a dense grid over this area, where we can see how the solution varies here.
For our boundary condition at infinity, we know that the value of φ = 0 at grid point N+1. This problem is very similar to problem P6B4, except that we now change the boundary condition at the wall to a fixed charge density. So it looks like we just have a fixed Neumann state, in which we can vary σ0.
Since we have convection in our system, we need to solve for the flow profile. Therefore, if we discretize the y axis at Ny grid points, we should be able to calculate a local velocity at each grid point in the y direction. Moreover, we assume that we have laminar flow and that there is no change in the film thickness so that.
Starting with Navier-Stokes in 2D, which we can get from Deen p.229, we can eliminate the terms to have:.
2 cos
We label n as follows to decrease the bandwidth of the Jacobian, keep the matrix sparse, and improve the computation time. Now that we have a way to get between real space and vector space, in a method we call stacking and unstacking a vector, let's look at the equations we're trying to solve. But the equations for the internal lattice points are mostly of the same form, so in this region we look exclusively at species A, and the same equations can be treated similarly for species B and AB.
By substituting these representations into the terms in the PDEs, we arrive at nonlinear algebraic equations in the form of the following. These coefficients are easy to calculate if we passed the value of the indices m and n, given the vectors z and y and we write a function to take care of this. We must remember to pass values of the Jacobian as this will greatly reduce the amount of computation time.
In our problem here we have boundary conditions for each species at each edge of the domain. We simply replace the value of the concentration at the imaginary grid point to the right of our domain, with the specified concentration. In practice, we can invoke Danckwert's BCs and let the parameters of the problem determine the behavior at the edge.
We can rearrange again to get a value for the imaginary grid point on the left right of the grid. So the effect of the Neumann BC treatment is to select a value of CA(m,Nz+1) and replace this value, which depends only on the values of CA(m,Nz) and cA(m,Nz - 1), in the equation for that grid point. Dirichlet conditions are easy to implement, in that for the imaginary mesh point on the mesh, you replace the concentration with the constant value specified in the boundary condition.
Once you place them, you will define all the interior points of the mesh along with all the edge points. We immediately notice that most of the action takes place very close to the surface, which makes sense since we have a fairly fast velocity with relatively slow diffusion. If you could do that and look at the simulation just near the surface, you might get graphs like these (blown up near the area of interest).
Since the aim of the problem was to increase the amount of absorption of A from the gas stream, through the addition of a reactive falling film, we want to determine how much additional A enters the stream. For further consideration of the Neumann condition, you may be able to get numbers in the order of this (note that the numbers here are calculated for different grid spacings, resulting in different absolute numbers).
Neumann boundary conditions (no flux at surface for AB, B)
By changing the upper boundary condition to a no-flow condition, we can improve the absorption. To achieve this boundary condition, you just need to use our standard 4/3 – 1/3 processing as before (the code is very similar, but included at the end so you can read it if you want). Therefore, it is acceptable for the Dirichlet boundary condition that there is no relative increase in extraction.
We note that it is small, although this is to be expected because the reaction is relatively slow (less than 0.1 M/s) and the fluid flows fast (0.8 m/s) over a short span (0.5 m). If we increased the value of the reaction constant k, we would see a significant improvement in extraction for reactive extraction.
