dkntec
Civil/Environmental
- Aug 13, 2011
- 3
How do I calculate the ‘discharge distance’ from a storm water Outfall pipe for a certain elevation drop?
I’m not sure what the correct term is for this - have heard it referred to as ‘waterfall’, outflow jet’, ‘outflow projection’, launch, or ‘throw’ distance.
The application is outflow discharge approximately 3’ above a creek water level.
Due to adjacent factors, a riprap slope below the Outfall is essentially fixed at 1.5H:1V slope (66.66…%).
I need to calculate at what distance from the pipe outlet the waterfall ‘curve’ will contact or land on this slope, or as I suspect, beyond this slope.
I am in Canada & normally use SI (metric) units, but no problem in work in Imperial (feet) and simply convert.
As I assume most of the members on this site are in the USA, I will post the information in feet;
- 24” (2’) diameter storm pipe
- Gravity flow with pipe slope of 2.00% (0.02 feet/foot)
- Manning N=0.012 (using ADS N-12 HDPE smooth interior pipe)
Assume ½ full flow in pipe.
Using Manning’s equation at ½ diameter flow, I calculate;
Q (flow) = 17.33 cfs
V (velocity) = 11.03 fps
I anticipate the distance I am trying to determine will be of significance.
This ‘discharge distance’ is needed to establish an appropriate set back distance to stay clear of the creek
Review of alternate pipe slopes, riprap slope angles, and discharge heights above creek water level may be needed as a result of these calculations to find a solution.
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I found the following post in searching the forum
“Waterfall laminar flow to mist” started 21 Aug 02
Is the formula in this link appropriate in my application?
I assume NOT, and assume this is for ‘level water’ (i.e. behind a weir or reservoir spillway), and NOT for flowing water discharging from a pipe slope with velocity, as in my case.
Also, is this formula in feet or metric units?
Thank you for your advice.
This is my first post here.
DKN
I’m not sure what the correct term is for this - have heard it referred to as ‘waterfall’, outflow jet’, ‘outflow projection’, launch, or ‘throw’ distance.
The application is outflow discharge approximately 3’ above a creek water level.
Due to adjacent factors, a riprap slope below the Outfall is essentially fixed at 1.5H:1V slope (66.66…%).
I need to calculate at what distance from the pipe outlet the waterfall ‘curve’ will contact or land on this slope, or as I suspect, beyond this slope.
I am in Canada & normally use SI (metric) units, but no problem in work in Imperial (feet) and simply convert.
As I assume most of the members on this site are in the USA, I will post the information in feet;
- 24” (2’) diameter storm pipe
- Gravity flow with pipe slope of 2.00% (0.02 feet/foot)
- Manning N=0.012 (using ADS N-12 HDPE smooth interior pipe)
Assume ½ full flow in pipe.
Using Manning’s equation at ½ diameter flow, I calculate;
Q (flow) = 17.33 cfs
V (velocity) = 11.03 fps
I anticipate the distance I am trying to determine will be of significance.
This ‘discharge distance’ is needed to establish an appropriate set back distance to stay clear of the creek
Review of alternate pipe slopes, riprap slope angles, and discharge heights above creek water level may be needed as a result of these calculations to find a solution.
-----------
I found the following post in searching the forum
“Waterfall laminar flow to mist” started 21 Aug 02
Is the formula in this link appropriate in my application?
I assume NOT, and assume this is for ‘level water’ (i.e. behind a weir or reservoir spillway), and NOT for flowing water discharging from a pipe slope with velocity, as in my case.
Also, is this formula in feet or metric units?
Thank you for your advice.
This is my first post here.
DKN