# Flow in anistropic soil

tzx9633

## Homework Statement

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...

## 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

#### Attachments

tzx9633
I also noticed that the example in the notes are also wrong ...
How could it be ?
Or i misunderstood something ?
Anyone can help ?

Chestermiller
Mentor
I also noticed that the example in the notes are also wrong ...
How could it be ?
Or i misunderstood something ?
Anyone can help ?
What equation is being solved?

• tzx9633
tzx9633
What equation is being solved?
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

Chestermiller
Mentor
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
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.

• tzx9633
tzx9633
Pls refer to this
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.
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) ??

#### Attachments

Chestermiller
Mentor
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) ??
I see what you are saying. It looks like you are correct in your criticism.

• tzx9633
tzx9633
I see what you are saying. It looks like you are correct in your criticism.
Then, what should the vertical scale be? It's not provided, or I miss out something?

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

• tzx9633
tzx9633
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).
Hi , can you help me in the following thread ? Thanks in advance !