Open Channel - storage volume with longitudinal slope
Open Channel - storage volume with longitudinal slope
(OP)
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.
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.





RE: Open Channel - storage volume with longitudinal slope
On program that you can use is HEC RAS:
http://r
Here is a link to an example problem.
RE: Open Channel - storage volume with longitudinal slope
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.
RE: Open Channel - storage volume with longitudinal slope
RE: Open Channel - storage volume with longitudinal slope
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).
RE: Open Channel - storage volume with longitudinal slope
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)
RE: Open Channel - storage volume with longitudinal slope
d2=do-So*L. Use the area of a trap for each point then by successive approximation or iteration solve for L.
RE: Open Channel - storage volume with longitudinal slope
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.
RE: Open Channel - storage volume with longitudinal slope
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.
RE: Open Channel - storage volume with longitudinal slope