Gravity flow through pipe 1

[AlecDeSantis]

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.strongwell.com/selected_markets/electrical_substation/images/main/containment_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.

 

[MarineConstruction]

And your question is? You will have entrance and exit losses through the value and fiction along the pipe. You can use this, along with anticipated flow, to find the most suitable valve.

 

[psmart]

Since the berm is probably high enough to completely retain most rainfall events, the valve sizing would seem to be primarily a matter of how quickly you want to empty the water.



Peter Smart
HydroCAD Software

http://www.hydrocad.net

 

[AlecDeSantis]

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.

 

[bimr]

The flow rate through the pipe will vary depending on what the level in the tank impoundment is.

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.

 

[psmart]

The flow isn't constant. The rate will decrease steadily over time as the water level draws down and the head is reduced. A basic hydrology text will discuss a draw-down scenario with a weir, orifice or tube outlet. You can also simulate this with the right H&H software.

Peter Smart
HydroCAD Software

http://www.hydrocad.net

 

[bimr]

I have attached the formula.

 

[AlecDeSantis]

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 out. My Hydro class in college was a number of years ago.

 

[AlecDeSantis]

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.

 

[AlecDeSantis]

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 (

http://www.wolframalpha.com/input/?i=Q=+.82(3.14in^2)(sqrt(2*32.2ft/s^2*.25ft)))

Did I totally FUBAR my math or does this look about right?

 

[bimr]

The calculation looks right.

 

[AlecDeSantis]

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.

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.

 

[AlecDeSantis]

Sorry, I wrote down the incorrect result. The result for this equation should be 26s.

 

[bimr]

T = 2 * AT*sq rt h/(C *AO*sq rt 2g)

= 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.

 

[cvg]

note that the equation assumes complete submergence of the inlet at all times. If it is not completely submerged then you will not be able to use this equation, you will have to switch to a form of the weir equation instead and your flow rate will drop significantly

 

[AlecDeSantis]

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 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!

 

[bimr]

sqrtH and not sqrtH1 - sqrtH2. How come?

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

 

[AlecDeSantis]

Ah, I see. You just shortened it up instead of writing the full thing.