Gravity flow through pipe
Gravity flow through pipe
(OP)
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 of head pressure.
It is very similar to this type of set up: http ://www.str ongwell.co m/selected _markets/e lectrical_ substation /images/ma in/contain ment_0.jpg
My goal is to find the amount of water that will flow through each size of valve so we can install valves that are more of a proper size.
It is very similar to this type of set up: http
My goal is to find the amount of water that will flow through each size of valve so we can install valves that are more of a proper size.





RE: Gravity flow through pipe
RE: Gravity flow through pipe
Peter Smart
HydroCAD Software
www.hydrocad.net
RE: Gravity flow through pipe
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.
RE: Gravity flow through pipe
If you assume a 1' water level:
A 3" valve will do approximately 70 gpm.
A 2" valve will do approximately 14 gpm.
A 1" valve will do approximately 1 gpm.
The equation is Q = 19.65 * d*d sq rt (h/k)
Most people would install a pump because the water flow rate will drop very low when the tank is near empty.
RE: Gravity flow through pipe
Peter Smart
HydroCAD Software
www.hydrocad.net
RE: Gravity flow through pipe
RE: Gravity flow through pipe
I'll take a look through the text and the formula and see if I can figure it out. My Hydro class in college was a number of years ago.
RE: Gravity flow through pipe
On a containment that's 12'W X 40'L X .8'H, that is a lot of water.
RE: Gravity flow through pipe
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 (http
Did I totally FUBAR my math or does this look about right?
RE: Gravity flow through pipe
RE: Gravity flow through pipe
The example containment unit I'm using is 40'L X 12'W X .7'H.
2At(sqrtH1-sqrtH2)/CdAo(sqrt2G)
At= 8.4 ft^2
H1 = .25' (3" of water)
H2 = .08' (1" of water)
Cd = .82
Ao = 3.14 in^2 (.021 ft^2)
(2*8.4)((sqrt.25)-(sqrt.08))/.82*.021(sqrt(2.32.2))
My result is 88s.
This seems way too short of a time span and doesn't exactly mesh with the gpm result I got from the previous equation.
RE: Gravity flow through pipe
RE: Gravity flow through pipe
= 2 * (40*12) * sq rt (3/12)/(0.82 * 0.021 *sq rt (2*32.2)
= 3,473 seconds or 58 minutes
The area of the tank should be 40 x 12.
RE: Gravity flow through pipe
RE: Gravity flow through pipe
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 the top of the valve opening. I am working these equations out one by. The next thing I would like to do, though, is work this on a modified weir formula.
Thanks for all the assistance so far!
RE: Gravity flow through pipe
This is the difference between starting at level H1 and ending at level H2.
For your application, level H1 = 3-Inches, Level H2 = 0-Inches
RE: Gravity flow through pipe