Determining PSI by Pipe Length and Water Flow

Just an idea...

If it is only water and a reasonably low pressure, and a steel pipe why not;

1. Weld a coupling to the pipe with FNPT.
2. Install a ball valve to the coupling.
3. Use a pneumatic drill to cut a hole.
4. Close the valve.
5. Install a guage.
6. Open valve and view pressure.

Now no calculations are necessary.

That would work but you can't get to the water it is covered . Really a unique situation. Other than where it actually discharges at it is viewable through three small drains that are welded in place with very small holes. Come on you mathematicians.

"water is covered"

But is is vented? There may also be a pressure head in the mystery "water supply". Take that into account if so, but otherwise follow the advise given by Reid.

Each foot of water column will amount to 0.433 psi static pressure in the mystery water supply, irregardless of area dimensions. However, a larger tank "or whatever this thing is" will supply more flow with less loss as it gets bigger since the outlet flow (into a pipe) velocity will be less. Since the velocity head is the square of the pipe diameter, the head loss will depend greatly on the pipe diameter.

As mentioned by Reid, the loss per 100 feet can be picked out of Cameron's Handbook once pipe information becomes available.

I have taken Cameron off the shelf and placed it on my desk in anticipation of the next comment.

It can be estimated based on the elevation head and velocity head.

Convert the PSI to feet of water head. Add or subtract the change in elevation to determine the head at the end of the pipe.

Velocity head loss is proportional to velocity of the water and the total length of pipe. I usually look in a Cameron handbook find the head loss for a velocity in feet of head per 100 feet and end up with a number to add or subtract depending on which point in the pipeline I am at.

Assuming you know the initial pressure or head at the origin, take the initial head, add the head of the elevation change (since you are dropping in elevation), and subtract the velocity head lost from friction factor. If you want PSI again, convert feet head back to PSI.

Thanks Reid but my problem is the origin is not available it is buried under a mountain and inacessible. You can only see the water in three small drains that you can't get into and where it discharges and it discharges 15 feet above the creek so inacessible there to. The water actually comes out in a channel and not a pipe. The channell is 24 " wide andat least 200 feet long I can't really tell the elevation drop either right now the water depth is 13". Suppose the elevation drop is 40 feet and you put an 8 inch pippe into the channel that is 100 feet long and filled it with water with the push of the water behind the full pipe can the psi be closely approximated. I have calculated the flow to be about 6 cfs

Use a topo map or "Goggle Earth" in that case to get an idea about the supply depth.

You may have to make some assumptions.

While it has perhaps not yet been conclusively stated it sounds like you want to somehow replace some length of an existing open channel with a gravity flow pipe. At least if you can make a dependable, at least some minimal pressure-tight connection to whatever is underground, perhaps the information and example on page 17-9 at http://www.acipco.com/adip/products/Sect17.pdf would be of help to you.

If you know the elevation head and flow rate at the exit point into the channel, you can estimate the pressure in the pipe. You must know the elevation difference between where you measured the head and flow rate and the point where you want to know the pipe pressure. You will also need to know the pipe materials, pipe size (ID), condition of pipe lining, oultet and any fittings to estimate head losses between the outflow and the point you want to estimate. It then becomes a case of using equations like Darcy-Weisbach or Hazen-Williams, and Bernoulli to estimate pressures, as you know the flow rates and elevation head difference. Of course these equations are estimates and are dependant upo assumptions about the variable, primarily assumptions about pipe friction losses and minor losses can be quite variable. It is possible that you do not have pressure flows in which case it must be analyzed as a gravity flow open channel type system and water pressures at any point are just the depth of water in the pipe.

RCE

I agree with much of this, but keep in mind that while Randy Conner has speculated that the "mystery water source" is an "open channel", this has not as yet been confirmed and in fact OP has stated that said mystery source is "buried".

We are all on standby with our charts, formula, equations, and in some cases solutions.

But as yet, unfortunately, without the benefit of any real knowledge about the geometry or properties of the mysterious "buried" water source having difficulty rendering real opinion.

As always, ready to crunch numbers ASAP after if and when problem is formulated.

I really appreciate ya'll. It is indeed a channel but definitely not open. It is buried and totally inacessible. The channel is 24 to 30 inches deep and 24 inches wide and currently has 13 inches of water flowing through it. I can measure the depth through one of the drain covers that has slots in it. In no way can I get to it to put a pipe in it. I just wondered by mathmatics if I put an imaginary pipe in it if the pressure in an imaginary pipe couold be figured and thus helping me establish a head pressure.

Ya'll deserve a brief history of this mystery water source. It is an abandoned coal mine since the sixties. In the early 1940's they cut into another abandoned mine and was flooded by millions of gallons of water and 30 some men almost drown. It gets more complicated. I have no knowledge of the mine they cut into and don't really need to. The mine abandoned in the sixties ran for 13 mile, big mine. They pumped it for a while then decided to drain it to the outside, now another coal mine enters the picture. They had worked the seam below the mine closed in the sixties so the layed a pipe into the water ran it to where they had drilled bore holes to the lower seam and dumped it in there filled it up and it then flows out the channel. What will be my water source the upper mine or the lower mine. The upper mine is constantly feeding the lower one. This water has ran like this since the 1940's and is not going to dry up. I am hunting some old mine maps but a problem there they were burned in a fire so the one at the time isn't available but the last one is. I will hunt for elevations and see what I can find. Since the water is constant I will look for the highest elevation and use that for my head. I am going to use the upper seam. So there are literally millions and millions of gallons of water pushing on the water being discharged. I will take all of the help I can get. As I told you it is definitely a unique situation.

Robert

I think there is a little confusion about the terminology. The use of the term "open channel flow" mean the water is flowing such that it is at atmospheric pressure at the water surface in the conduit. Open Channel flow occurs in sewers and storm drainage system, as well as canals. Pressurized flows occur typically when the fluid is under pressure such that it fully fills the conduit ans if the conduit were to be penetratred the water would rise some distance equaliing the pressure in head) above the top of the pipe. Currently, it appears that you have open channel flows occuring as the water level in the pipe is only 13 inches in a conduit that is a full depth of 24 to 30 inches. You would use a different set of equations for open channel flow. What you can assume in open channel flow is the water surface pressure is equal to atmospheric or zero gauge pressure. so if you know a point where you want the water to exit to atmospheric pressure, yes you can design a pressure pipe or and open channel system (storm drainage/culvert type conduits) to replace the existing conduit, quite easily. If you keep the flow in the pipe pressurized you can design it using the pressurized flow equations i mentioned earlier. you just need to know something about the conditions you want in the pipe, entrance and exit elevations, length of pipe, types of materials, sizes of pipe, flow velocities, etc.. You could make all of these vary to assess for the most efficient design parameters to use.

My favorite book is "Open Channel Flow" by Henderson, although many hydraulic engineers prefer Chow.

This stuff I know well, but thanks anyway.

We now have more info from the OP on the mystery water source mine so I'm going to go with that.

Can you clarify what you are trying to accomplish?

Are you trying to use the water for power, irrigation, or consumption?

Trying to see if enough for power generation

OK. You assume aa flow of 5 cfs and 15 feet head(vertical drop)

A cubic foot of water weight about 62.4 pounds per cubic foot. 5 cfs = 312 lbs per sec.

312 lb/sec x 15 feet = 4680 foot lbs/sec or approximately 8 HP

Multiply by 40% efficiency, you might get 3 to 3.5 HP.

Your mileage may vary depending on typ of equipment used to harness this "free" source of power.

Humor08, please speak to us. Do you need more information? Are plans moving forward? Or is the cost/benefit ratio too high?

I think i would look for another source than water from an abandoned coal mine, just from a water quality perspective.

There may be another way to get the information you want. But even that information may not be correct.

"They had worked the seam below the mine closed in the sixties so the layed a pipe into the water ran it to where they had drilled bore holes to the lower seam and dumped it in there filled it up and it then flows out the channel"

What size is the pipe they used? Based on approximate water drop hight, you should get close. What size are the drilled bore holes? Same procedure. Which is higher? You would have the higher figure.

"it discharges 15 feet above the creek "

Is it possible to catch the falling water into a large homemade funnel on a barge in the creek? Allow the funnelled water to spill back to the creek. Start with a funnel opening that is too big, retaining nothing in the funnel. Slowly make the opening in the small end of the funnel smaller until the water starts to build some in the funnel. at that point you have a round pipe(funnel small end) read that with a pitot gauge. The gauge reading will convert to GPM. The utility that services your fire hydrants will have a pitot gauge. The funnel could be made out of plywood if you reinforce it well, and mount it strongly to the barge. Be careful with it there is a lot of energy in that falling water.

How come everybody has a Cameron Diaz manual but me? Any good pictures?

I think you are asking a question for which there is no adequate answer. The question asks the pressure of the water in the pipe at the flow of 5 cfs. Therein lies the problem: 5 cfs is equal to about 19,200 GPM. Three is no way you can push 19,200 GPM through a 4", 6" or 8" pipe. The 12" pipe would require about 157 feet of water (about 68 psi) above the pipe to get that amount of flow through a 100' section of 12" pipe. If the pipe is any longer than that, the friction losses would render it impossible to accomplish with reasonably available equipment. Therefore, the question is unanswerable as stated. Try again.

You are wrong.

5 cfs equates to 2,244 gpm.

Thank you so much for correcting me. I was in outer space when I was answering that thread. Should have paid more attention to what I was saying.

To All: I stand corrected. Please disregard that answer. I would try again, but time doesn't permit me to make an intelligent reply, as I have just shown.

Thanks again.

Let me say you are not the first to be caught in a mistake on this site and you won't be the last. Welcome to the club!

Anyway, my Cameron Handbook is missing (no doubt some one is looking at the pictures!) But my Fairbanks-Morse Hydraulic Handbook show an 8" pipe with a water flowrate of 2200 GPM will have a velocity around 14 fps, and a head loss of 7 feet per 100 feet of pipe. If the intent is to generate power from the flow, a lot of energy will be lost in friction losses with 8" pipe.

In that case, go to 12-inch to reduce velocity to 6.3 fps resulting in a velocity head loss of 0.9 ft/100 feet (sch 40 new steel pipe asphalt lined) according to Darcy's equation (source Cameron Diaz, page 3-25).

Since the head loss is a function of velocity squared and since velocity is a function of area which is a function of diameter, you need a BIGGER pipe (assuming that you cannot afford to burn up this much energy).

Head loss is also a function of pipe materials, not just pipe diameter.

Wouldn't a composite pipe of some sort be better? Fiberglass, etc.

Yup. I reverse engineered a solution offered by Ried varying only the diameter, but holding the material constant.

To compare a smooth pipe to a rough pipe (of equal diameters) would be like comparing apples to oranges.

Darcy's formula is:

hf = fl/D v**2/2g

where f is the friction factor.

A plastic pipe will have a lower coefficient of friction than a steel pipe for example.

All of these formulas depend on the Reynolds number and Darcy's f can be related to Chezy's (French engineer) C.

Typically nomographs are used to facilitate solutions to these equations (or the Tables given in handbooks).

Do you need a pipe? The water is already being delivered. Or do you need to relocate the water energy?

Good point. Use the natural channel by all means if available.

In post #10 you mentioned that the water flow was started when the minors drilled some holes, and that the minors ran a pipe to the older mine.

If you were to either drill more holes near the original ones, or use an additional pipe to drain the mine, you would be able to increase water flow available for your project.

The elevation of the water might be able to be determined by sending a long length of clear plastic tubing down the water passage from the old mine to the water drop. If the mine end is submerged under water, and the other end is pulled up above the water drop point till the water stops coming out of the tube, the water elevation can be determined.