Calculating Flow into Vertical Pipe 3

[Sigma47]

Hello,
Thanks for taking the time to read... I am currently trying to solve the problem of flow into a 8" vertical pipe draining approx 10 Ha, connected to a horizontal drain in an ag-field.

The overall project is dealing with drained prairie potholes and trying performing a CBA on the practice of draining them. The entrance to the pipe is approx at ground level, and I have continuous level data for ponded water near the pipe. I have been using a weir equation til now and would like to get some input as to what methods others would approach this with.

My concerns are with the levels I observe adjacent to the pipe I know that the weir equation breaks down. I have recorded depths of .75 m. Although I've never been on site to see it full plug flow I might suggest its been there. I have seen the pond at a "slurping and gurgling" vortex stage around the drain, approximately .4 m, and I presume just cm's from becoming full. So, my guess is that this system behaves at 3 interval a weir, vortexing, then full plug. My questions are:

1.) Any thoughts on the threshold heads at these various stages and

2.) what equations/coefficients would be used to model them.

Thanks much for your time,

Jay

 

[Drew08]

There may actually be 4 governing equations, not counting intermediate stages, depending on the exact configuration and connection to the horizontal drain. Option 3 may not apply if a pipe bend connects the vertical to horizontal pipes.

Riser (vertical) Pipe Controls:
1: Weir equation Q = CLH^1.5; C = 1.7 (SI)
2: Orifice equation Q = CA(2gH)^0.5; C <= 0.6

Outlet (horizontal) Pipe Controls:
3: Throat control – Orifice equation applied to the inlet to the horizontal pipe (H = head above centerline of pipe)
4: Friction control (full flow capacity of the horizontal pipe)
Q = A{H2g/[1+K+184 L n^2/D^(4/3)]}^0.5;
where H is the difference between the water surface and the higher of the top of the horizontal pipe at the outlet or tail water elevation.
K = sum of entrance and other minor (bends/valves) loss coefficients = 0.5 (entrance losses)+ 0.2 to 0.4 (Losses in bends if applicable).

The governing equation would be that which results in the least flow. Each of these 4 controlling factors have some intermediate stage (i.e “slurping and gurgling”) where actual flow is less than both the preceding flow and developing flow. These intermediate stages can be reasonable estimated by plotting out the governing equations (Q vs H) and drawing out by hand a smooth connection line at the sometimes abrupt intersections.
Also note that if you are not measuring depth at the outlet, H for weir and orifice control may need to be reduced depending on the surface vegetation. Some frictional losses may occur as the water flows towards the outlet.

 

[Sigma47]

Hey Drew08, thanks much for the response...I will run the constraints, but am wondering if you could clarify that your second orifice coefficient is in [SI] and could you if you have an explanation for the constraint that its less than .6 tell me? I have seen other values that approach 0.8.

Thanks again.

 

[Drew08]

The weir coefficient ‘C’ mathematically equals: C =Fg^0.5/(1+F^2/2)^(3/2), where F is the Froude number for any given energy head ‘H’ and Flow ‘Q’ in a rectangular channel. It has unites related to g^0.5. For F = 1, 'C' differs between 1.7 (SI) and 3.1 (US). Whereas, the ‘C’ in the orifice equation is unit less and is the same for SI and US.

C is theoretically 1, but experimentally determined to be 0.6 in reality for a sharp-crested inlet. C = 0.6 can be thought of as a theoretical maximum, unless you have a specially designed inlet such as an ogee or other specifically rounded edge, which I assume you don’t. If you have a square-edged culvert, or CMP sticking up out of the ground, use 0.6.

Without an anti-vortex hood on the opening, there will be more turbulence and thus more energy losses and you won’t likely reach the full C = 0.6. Since to date you’ve only been using the weir equation, I’m guessing an exact calculation isn’t necessary to prevent catastrophic failure, so 0.6 is a good starting point.

Once you calculate your rating curve you can do a sensitivity analysis on these coefficients. If outlet control flows develop relatively quickly, which they might if the horizontal pipe is also 8”, then the result will have little sensitivity to these parameters.

If you’re at the site often, you can always check your assumptions to measured flow rates and adjust your as necessary.

Also, note that I miss typed in the Option 4 equation above: 184 relates to U.S. units only – it is a combination of parameters: 2g 4^(4/3)/a^2; where g = 9.81 (SI), or 32.2 (US); and a = 1 (SI), or 1.49 (U.S.) – Sorry for the confusion.