Water Supply Query

Hello,

I've got a problem I'm not sure how to solve.

A water tower is supplying a public works. The tower's dimensions, capacity and the approx. elevation change between it and site using the supply and the length of pipe and fittings between the two are known.

I'm not sure what formula to use to determine flow at the end of the pipe. The static pressure is known however. Should I assign a 'K' factor to the flowing pipe and/or how does one do that?

MGX said:

Hello,
I've got a problem I'm not sure how to solve.

A water tower is supplying a public works. The tower's dimensions, capacity and the approx. elevation change between it and site using the supply and the length of pipe and fittings between the two are known.

I'm not sure what formula to use to determine flow at the end of the pipe. The static pressure is known however. Should I assign a 'K' factor to the flowing pipe and/or how does one do that?

Click to expand...

PM me. I did a lot of these type problems during studying so I should be able to help.

The short answer is the energy equation:



It should be very straightforward from there... You can look at PE Notes - Hydraulics for more explanation.

sehad: It would be better for everyone if you tried to help in the forums instead of by PM... for every question that's asked there are probably ten others that could learn something useful for the exam.

When all else fails, use Bernoulli's equation!

While you are at it, you can the P/gamma terms since this is a gravity flow system, and simplify the equation.

Good Lord, I must be an imbecile. Maybe its because I've yet to take fluid mechanics.

I'm going to run this by my engineering prof and see if he can give me some pointers.

The tower is about 100' tall and around 15' in diameter.

I think I should be flowing 7050 GPM with a 6" line.

MGX said:

Good Lord, I must be an imbecile. Maybe its because I've yet to take fluid mechanics.
I'm going to run this by my engineering prof and see if he can give me some pointers.

The tower is about 100' tall and around 15' in diameter.

I think I should be flowing 7050 GPM with a 6" line.

Click to expand...

What are the specifics of the problem? I'm certain we can help you understand the problem/answer?

IlPadrino said:

What are the specifics of the problem? I'm certain we can help you understand the problem/answer?

Click to expand...

That's it! I may not have enough info to figure it out, but I used the velocity(2gh)^1/2 with 100'=h then multiplied by the area of the 6" outlet. With the cubic feet per second I multiplied by 7.48 gal/cubic feet and 60seconds/minute to get a GPM.

MGX said:

That's it!

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I thought you said the length of pipe and fittings between the two are known... that let's you calculate the hL. Otherwise, you can assume hL=0. If there's no pump, then hp=0. If both of these losses are zero, you do what VTEnviro said... apply Bernoilli's



Can you take it from there? Essentially, you converting potential head (Z) into kinetic head (v2/2g).

That's it! I may not have enough info to figure it out, but I used the velocity(2gh)^1/2 with 100'=h then multiplied by the area of the 6" outlet. With the cubic feet per second I multiplied by 7.48 gal/cubic feet and 60seconds/minute to get a GPM.

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I think you are onto it. On a second look you really can simplify it down to a few terms. Assume condition '1' is in the tank and condition '2' is at the discharge.

- The Hp term disappears since there is no pump - a gravity fed water tower.

- P1/Y disappears becuse there is no pressure in the storage tank, again it's working by gravity

- P2/Y disappears because it is free discharge - end of the pipe

- V1^2/2g is generally assumed to be negligible - the water in the tank isn't really moving in comparison to the discharge.

This leaves you with v2^2/2g = Z2-Z1.

Subtract the Z's, multiply by 2g, take the square root of V2^2 --> you get V.

Since Flow = Area x Velocity multiply by the area of the 6" pipe (assumed to be flowing full) to get Q.

IlPadrino said:

I thought you said the length of pipe and fittings between the two are known... that let's you calculate the hL. Otherwise, you can assume hL=0. If there's no pump, then hp=0. If both of these losses are zero, you do what VTEnviro said... apply Bernoilli's
View attachment 1679

Can you take it from there? Essentially, you converting potential head (Z) into kinetic head (v2/2g).

Click to expand...

Aye, I think I've got it now. The city has a gridded water supply so now I'm back in familiar territory. This is only for some preliminary stuff so it need not be exact. Typically I'll use hazen-williams to solve for friction losses.

MGX said:

Aye, I think I've got it now. The city has a gridded water supply so now I'm back in familiar territory. This is only for some preliminary stuff so it need not be exact. Typically I'll use hazen-williams to solve for friction losses.

Click to expand...

BOO! Manning (Peyton)!