to determine the head along a flow path, you need to know both the upstream and downstream head conditions. So, you first define a datum. I'm taking your datum as the bottom of the soil column. So, your upstream head is at El 30M. Now tell us what's the head at the downstream (i.e., the underside) of the sample? In other words, if you were to position a piezometer at the lowest depth of the soil where would you measure the water elevation? Also at 30 meters? Something more like 1 meter? You see it's the conceptual ground water regime that's critical to understanding how to calculate such data.

f-d

¡papá gordo ain’t no madre flaca!

I would measure it at 30 meters with the elevation heads at 0, 10, 20, and 30 meters.

Is water flowing through the column? If water is NOT flowing through the column, then YES, the total head is the same at all elevations. Total head = Elevation head + Pressure head. At El. 0m, Pressure head = 30m, Total head = 0 + 30 = 30m. At El. 15m, Pressure head = 15m, Total head = 15 + 15 = 30m. At El. 30m, Pressure head = 0m, Total head = 30+0 = 30m. The elevation of the ground surface is irrelevant in this case.

There is water flowing (40 cc/s) in the column to maintain water level at the top of column.

giving us volumetric water flow requires that we have the area. Can you please explain the whole story? Is this a field test in a bore hole? Is this a lab test with a mighty tall soil column? Do you have any boundary conditions (i.e., well measurements)?

Fundamentally, this all relates to Darcy's law; v=ki or q=kia

You are ultmately striving to determine the "i" term, or gradient, which is related to the length of flow and the head conditions at the upstream and downstream observation points. You have some sense of "q" (i.e., 40cc/sec). So, now we can't help you with either "i" or "a." You got the "q" though. . .

f-d

¡papá gordo ain’t no madre flaca!

This is a lab test. The column is short and immersed in a water tank. The area is about 78.5 square meters and k is assumed to be 0.34 m/s. No well measurements are known.Thanks.

please make a sketch and add a pdf to your post so we can visualize the problem. If the bottom of the test is submerged in water there must be some tailwater head. Ultimately, we need the tailwater head, the headwater head and the length/area of the soil. Saying the column is short and then using dimensions of 20 and 30 meters (i.e., 60 and 90 ft)is a bit confusing.

f-d

¡papá gordo ain’t no madre flaca!

I have attached a quick sketch in Microsoft Word. I had the units messed up. It is actually cm and not meters.

• http://files.engineering.com/getfile.aspx?folder=f5238aea-06c5-44cb-8ead-26

So total head at the top of the cylinder is 30cm. No headloss from top of water to top of soil, so total head at top of soil = 30 cm. Elevation head = 20 cm, pressure head = 10 cm. Total head drops from 30 cm to 0 cm over 20 cm of soil, so the gradient = i = delta H / L = 30 cm /20 cm = 1.5 cm of head lost per cm depth below soil surface. So the total head at any point in the soil column is 30cm - 1.5 * z where z is depth below the soil surface in cm.

I recommend checking out Lambe and Whitman's Soil Mechanics, page 252 for a refresher/further explanation.

o.k. so now we can run with our first principal - Darcy's Law!

Q=kia, where
Q=40 cc/sec
k=permeability (unknown)
i=hydraulic gradient, DeltaH/DeltaL
DeltaH is the head loss that occurs over the length of the sample
DeltaL is the length of the sample.
A=78.5cm2

Using a bit of algebra, I get k=3.4x10^-1cm/sec

f-d

¡papá gordo ain’t no madre flaca!

Thanks for all of the help. I know have a better understanding of finding the parameters. I do have a second question. If Q is no longer supplied so that the water level is at the soil surface, is the Total H at the surface 20 cm or is it still 30 cm?

What I was thinking about doing was finding how long it will take for the water level to drop to the surface using the velocity of the water through the soil.

there are equations for interpreting the falling head test. Somewhere there's an integral. I'm no longer involved in this thread, 'cause it reads like a homework question and all these equations are spread throughout textbooks and the internet.

f-d

¡papá gordo ain’t no madre flaca!