# Homework Help: Flow in anistropic soil

1. Oct 18, 2017

### tzx9633

1. The problem statement, all variables and given/known data
In this question , it's stated that the horizontal scale is sqrt(kx / kz) X vertical scale ... i think the author's is wrong for horizontal scale and vertical scale in example 8.3...

2. Relevant equations

3. The attempt at a solution

I think the vertcal scale should be 6.1m , as shown in the picture , the horizontal scale , which is sqrt(kx / kz) X vertical scale should be sqrt(1/2) X 6.1 = 4.3m

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2. Oct 18, 2017

### tzx9633

notes here :

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3. Oct 18, 2017

### tzx9633

I also noticed that the example in the notes are also wrong ...
How could it be ?
Or i misunderstood something ?
Anyone can help ?

4. Oct 18, 2017

### Staff: Mentor

What equation is being solved?

5. Oct 18, 2017

### tzx9633

What do you mean ?
I mean in the first question :
I think the vertcal scale should be 6.1m , as shown in the picture , the horizontal scale , which is sqrt(kx / kz) X vertical scale should be sqrt(1/2) X 6.1 = 4.3m

6. Oct 19, 2017

### Staff: Mentor

kx/kz = 2, not 1/2. And the original horizontal distance scale before change of variables was 7.6 m. So the new horizontal distance scale after change of variables is 7.6√2.

7. Oct 20, 2017

### tzx9633

Pls refer to this
one , in the notes , it's stated that horizontal scale = sqrt (kz /kx) x vertical scale ...

so , at the example in the first post , it's clear that the horizontal scale = 7.6m , but the vertical scale isn't stated .
I agree that the horizontal scale = sqrt(1/2) x vertical scale

How could horizontal scale = 7.6 x sqrt(2) ??

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8. Oct 20, 2017

### Staff: Mentor

I see what you are saying. It looks like you are correct in your criticism.

9. Oct 20, 2017

### tzx9633

Then, what should the vertical scale be? It's not provided, or I miss out something?

10. Oct 20, 2017

### tzx9633

11. Oct 20, 2017

### Staff: Mentor

I don't like anything about what your book is doing here. There is really no characteristic length scale in the x direction, because the system is infinite in that direction. In the y direction, a logical characteristic length scale would be the height of the permeable formation, which is not specified in the figure. So talking about length scales here makes no sense to me.

What they are really doing here is using a mapping of the independent variable to transform the differential equation into a mathematically more tractable form. This is not the same as specifying a length scale for the system. I have never seen anyone call it this (before now).

Also, I don't like the transformation they have used. I would have done it much differently. I would have written the following:
$$x=x' \left(\frac{k_x}{k_z}\right)^{1/4}$$
$$z=z' \left(\frac{k_z}{k_x}\right)^{1/4}$$
That would transform the differential equation (more symmetrically) into:
$$\sqrt{k_xk_z}\left(\frac{\partial ^2 H}{\partial z'^2}+\frac{\partial ^2 H}{\partial x'^2}\right)=0$$
The original flow equations, in terms of the stream function and the head are:
$$\frac{\partial \psi}{\partial z}=-k_x\frac{\partial H}{\partial x}$$
$$\frac{\partial \psi}{\partial x}=+k_z\frac{\partial H}{\partial z}$$
Applying our coordinate mapping to these equations yields:
$$\frac{\partial \psi}{\partial z'}=-\sqrt{k_xk_z}\frac{\partial H}{\partial x'}$$
$$\frac{\partial \psi}{\partial x'}=+\sqrt{k_xk_z}\frac{\partial H}{\partial z'}$$
So, by applying this transformation of the independent variables, the equations transform into those for an isotropic formation, with a hydraulic conductivity equal to the geometric mean of the vertical- and horizontal hydraulic conductivities of the anisotropic formation.

Anyway, that's what I would do (and also what I actually have done in solving real-world groundwater problems).

12. Nov 11, 2017

### tzx9633

Hi , can you help me in the following thread ? Thanks in advance !