Ah, I thought 'At' was the cross sectional area not just the area of the tank.
In your equation, I see that it's just sqrtH and not sqrtH1 - sqrtH2. How come?
cvg - Yeah, I realize that the discharge area (the cross sectional area of teh valve) will change once the water level drops below...
I am trying the 'time to drain tank' now. This is assuming that the containment is almost level so the equation for varying cross-section doesn't come into play for this. I will try the varying cross-section next, though.
The example containment unit I'm using is 40'L X 12'W X .7'H...
Assuming
2" Valve,
3" Water,
Qo=Cd*Ao(sqrt(2gh))
Qo = .82*(3.14in^2)*(sqrt(2*32.2ft/s^2*.25ft)
Qo = 124 in^3/s or ~32Gal/min
So my flow through the valve would be about 32 gallons per minute when the water level is at 3". I used Wolfram Alpha to verify my equation...
I forgot to mention that the height in the containment areas will never reach 12". The levels are usually between 3-8" high.
On a containment that's 12'W X 40'L X .8'H, that is a lot of water.
bimr - thanks, that's exactly what I was looking for. While we won't be installing any pumps to more quickly drain our containments, this should help us size the valves better to different sized containment areas.
I'll take a look through the text and the formula and see if I can figure it...
My question is: how much water will flow through a valve under the above circumstances?
So if I install a 3" valve, will that valve drain 100 gallons per minute or 25 gallons per minute?
I was unable to find a good equation to use for this purpose.
Basically, we have bermed areas that hold aboveground storage tanks. Most tanks have a concrete berm around them. Each bermed area will capture rainwater and we have to drain it. We have a range of 2", 3" and 4" valves. The rainwater is under atmospheric pressure and there's a minimal amount...
