Water Flow PSI

The structure of the question screams HOMEWORK. Or some type of examination.

If you had put it in the first person and pretended that it was you having the problem, maybe someone would have helped you.

Good luck.

Hmmm, I am 55 years old, finished school long ago and am working on a hydroponic project. Your assumption is incorrect.

OK,

That's a horse of a different color. 2,000 GHP is a lot of water though. Is this constant flow?

Sorry I don't have more time tonight. _{(Veterans Day Parade today)} It's like you need to overcome some friction losses and 2 feet of head. 1 foot = .43 PSI.

Unless it will incriminate you, what are you growing?

Yes its a closed, recirculating system. I am not growing anything. Trying to develop a commercial product to sell through a local hydroponic store.

For your 33.3 gpm, the friction loss in the 2 ft of 1.25" pipe is about 0.5 ft.
For 6.7 gpm in each 16-ft 0.5" pipe, the friction loss is about 17 ft.
Other piping and fittings guess 4 ft of friction loss.

This comes to 21.5 ft of head (≈ 9.3 psi) at 33.3 gpm; I would guess about a 1/3 hp pump.

The 17 ft of loss in the 0.5" pipes is rather high (about 7.4 psi difference). Thus the upstream holes will pass more water than the downstream ones. You might compensate by drilling the downstream holes a bit larger, maybe the middle third of them 9/64" and the final third 5/32".

Thanks

To clarify

The 16' x 1/2 " lateral have 33 nozzles. (one on each end)

There are 5 laterals ==> 165 nozzles

Each nozzle must deliver 12.12 gph.

the max pressure is 24"

How deep below the surface of the water are the nozzles / emitters placed?

Uniformity of discharge would not be that important because the water inn the tank will find its own level. Right?

How critical is the 2000 gph?

Actually there are no nozzles, water will just be coming out of holes which will be drilled with a .125" bit.

165 holes is correct, and you are right, each hole will be emitting 12.12 gallons per hour according to my original specs. However, I just realized I made an error in the total gallons which should be 1,000 per hour, not 2000. So the correct answer to the flow from each hole is approximately 6 gallons per hour.

The water will be sprayed 2 inches above the surface.

You are correct about uniformity not being critical, main reason for the number of holes is to maximize aeration of the water. A drain will keep the water height at 3 inches. The high exchange rate of water in the tank is also crucial to keeping the water well oxygenated.

At the new flow of 1/2 previous, the various pipe friction losses will become 1/4 of before. The uniform hole size is now probably okay, and a smaller pump will work.

So about 17 gpm at about 10 psi?

And the 1.25 inch inlet will be ok?

Yes, at this reduced flow rate, your pipe sizes are all fine. (The velocity was about 7 ft/s before, which was rather high though not prohibitive; 3.5 ft/s is nice and gentle.)

Thank You

I woke up this morning with a couple of questions relating to water flow calculations.

It seems that no consideration was given to the number and size of holes that are in the .5 inch pipes when calculating water flow.

This seems counter intuitive to me.

For example what if the pipe had only 1 pinhole in it as opposed to 32, .125 inch holes. Would one not expect much less flow of water through the pipe with the pinhole?

Perhaps I should have mentioned that the ends of the pipes are sealed so that the water can only leave the holes through the drilled holes. Would that make a difference in calculating pump psi?

In the original description, the total cross-sectional areas of the (165) holes, the (5) 0.5" pipes, and the 1.25" pipe were all about equal. So that part was pretty good already.

As a refinement, the flow capacity of a pipe varies approximately as the 2.5 power (rather than square) of the pipe diameter. That's why the friction loss in the 0.5" pipes was so much more than in the 1.25" pipe (with length also being a factor).

Excellent, a light bulb went off! So if I want to make maximum use of the flow rate of any pipe using .125 inch holes, I can simply divide the cross sectional area of the main pipe by the cross sectional area of .125 inch holes and drill that many holes into the pipe. Thus,

With a .75 inch pipe, which has a cross section of .4417, I divide by the area of a .125 hole, which is .0123 which gives me about 36 holes.

Interesting how I just happened to choose the right sizes of pipe and holes from the get go for my project. Sometimes luck does work in our favor.

Thanks for the explanation.

Friction in a pipe depend on the area and the length of the wetted perimeter. that is why the power of 2.5 is used.

The hole will act like a nozzle and discharge approx K x D² x √P

For USGPM the constant is (i think) 30 where D in inches and P in PSI

Your K may differ because the hole may not be perfect,

The pressure losses can be calculated by doing a point to point calculations but that is hard work. if I remember correctly (40years) 60% of the friction of total volume in the pipe may be a good guess.