Open Channel - storage volume with longitudinal slope

[Ekenny_G]

Hi,
I am trying to find the storage volume behind a check dam on an open channel. The channel has a longitudinal slope and depth of 6-inches.
If I need to store say 100 cf of water, with a 6-in check dam, and a longitudinal slope of 0.01 ft/ft. what is the required length?


Many approaches make assumptions that I am trying to avoid.
(1)One of the assumptions is that the channel bottom is flat and depth is uniform along the lenght of channel. However, there are variations to the depth as the storage propagates in the channel. The maximum depth in this case being 6-in.
(2)Manning equation can be considered, but thats just to compute flowrate. Can you convert manning's eqntn to find the volume?
(3)The other method is using the Average-End-Area method. However, at stepper channel slopes it doesn't seem valid.

 

[bimr]

Backwater profiles, upstream from a reservoir, are routinely modeled using steady flow profile calculations.

On program that you can use is HEC RAS:

http://rpitt.eng.ua.edu/Class/Water Resources Engineering/HEC-RAS Introduction.pdf

Here is a link to an example problem.

 

[jgailla]

If I was reviewing this, with no other information, I would assume that the storage area is the triangle with the level water surface on one side, the 6 inch check dam on one side, and the slope on the third. The available storage volume would be this area multiplied by the channel width, assuming vertical sides.
So at 0.01 ft/ft, you would have a length of 50 feet for the water level.
The triangle area would be 50*0.5*0.5=12.5 cf of storage per foot width of channel, so the channel width required would be 8 feet.
You could get a lot more complicated than this, but this calculation would work where I am in North Florida.

 

[gbam]

Ekane, a sketch would be helpful.

 

[Ekenny_G]

Thanks for the comments.

Per gbam comment, I am attaching a sketch here for clarification.

It can be modeled as a backwater Profile upstream of a reservior, in this case a "dam".

Like "jgailla" stated, it can be looked at as a triangle with level water surface on one side (see attached sketch). However, this is still an approximation and would work if their is a way to integrate along the length of the channel. The integral of the Area is the volume. However, the depth of flow varies from 6-inches to 0-inch (zero).

 

[Drew08]

This method is exact assuming the water surface is level, and the face of the dam is vertical, neither of which would strictly be true if the channel is flowing.

The longitudinal section through the flat channel bottom is the area of the triangle x the width:

Volume = base width x 0.5 x depth^2 / Slope

The volume along each side slope can be assumed a pyramid, where the base is the triangular area adjacent to the face of the dam. The Volume of pyramid is V = AH/3 or:

Volume = (depth^3 x sideslope) / (6 x slope)

Add the base section to the two side slopes and the total volume is:

Volume = (base width x 0.5 x depth^2) / Slope + (depth^3 x sideslope) / (3 x slope)

 

[gbam]

Okay looking at your problem, one can determine the volume using average end area (A1+A2)/2*L and upstream depth in terms of do by
d2=do-So*L. Use the area of a trap for each point then by successive approximation or iteration solve for L.

 

[bimr]

This seems to be a simple 8th grade math problem. Determine the volume of a Triangular Prism. There is no complex math required.

Volume of a Triangular Prism = 1/2*length*width*height

If you assume vertical channel walls, the answer is an 8 ft channel width as shown by jgailla above.

L = 50 ft derived from 0.5 ft divided by the slope of 0.01 ft/ft.

The channel width volume is 1/2( db1L). (Volume of a Triangular Prism)

To calculate the side slopes, assume a 1:3 channel wall slope.

The channel side volumes are 2 sides * 1/2 (d *1.5* L). The width of each side is 0.5 divided by 3 or 1.5 ft.

100 cu ft = 1/2( db1L) + 2 sides * 1/2 (d *1.5* L)

where L = 50 ft
d = 0.5 ft

100 = 1/2( 0.5* b1* 50) + 2 (1/2 *0.5 *1.5* 50)

b1 channel width is 5 feet.

b2 channel width is 8 feet.

S = 1.58 feet.


The volume on the face of the upstream side of the dam is not included, but can be determined with the same method.

Note that if water is flowing over the dam, that the depth of water held back by the 6-Inch dam will be greater than 6-Inches.

 

[jgailla]

Drew08,
While you are correct in that the water will be moving, every time I've had to satisfy a "storage volume" for an agency, the static approximation I've described above has fulfilled the requirements.
My experience is limited to a few agencies in Florida and Georgia, so it could be different elsewhere.

bimr,
If I tried to get the agency to accept the height of water is greater than the dam, as it actually is, there would probably be some glazed eyes on the reviewer's part and the permit would be held up. It's usually best not to get all cutesy with the calculations unless there is a good economic reason for doing so.

 

[eea]

I think you've got your answer on the volume side. While it should be minimal, also make sure that what every open channel hydraulics you need during the design event aren't impacted by the presence of the check dam or its backwater. I've done these with a "low flow" base pipe to help drain the ditch rather than to rely ground infiltration. Essentially a two stage detention pond, something like a Q(2) release pipe and Q(25) storage and overflow weir.