flow direction

[SloFlo]

Hopefully this will be an easy question for most of you.

80Psi(alpha)
_____\_
/ |__/_____20Psi(beta)
| \
\|/
|
ATM(charlie)

If I connect a 1/4" pipe @ 20psi (alpha) to an 8" pipe @ 80psi (beta) will both fluids exit at "charlie"? Or will there be backflow into the "beta" stream. Please help. thanx

 

[vpl]

my bet's on backflow into beta

Patricia Lougheed

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[LHA]

You need to know losses through the exit section. Also, the relative elevations of all three nodes.

Engineering is the practice of the art of science - Steve

 

[SloFlo]

Alpha stream high point is 20 ft. Beta would be injected at 18 ft and the discharge would be at 17ft. I'm not sure about the exit losses.

 

[vpl]

Just noticed there is a dichotomy between your drawing (which shows alpha at 80 psia) and your message which states that alpha is a 1/4" tube at 20 psia. I was going off the drawing.

Patricia Lougheed

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[SloFlo]

My apologies. Alpha is the 80psig stream and Beta is the 20psig stream. But I think you are absolutely right. I would have backflow into the beta stream because the 20psig is obviously insuffiecient to overcome the 80psig pressure.
Thanks.

 

[iainuts]

Hi SloFlo,
Just to make sure I understand you right, there is a 1/4" pipe (alpha) at 80 psi dumping into an 8" verticle header (charlie). 2 feet below that is another pipe (beta, unknown diameter) at 20 psi, also dumping into this 8" verticle header. 1 foot below beta is the 8" pipe opening to atmosphere.

Because of the relatively large size (8") of your header, I'd bet there is a good chance that both pipes, alpha and beta, will both discharge perfectly well, and there will be no backflow into beta. Unfortunately, the problem isn't sufficiently defined. The pipe ID's are unknown, and the fluid is unknown. Answer those and you can calculate losses more accurately, though I doubt the single restriction (1/4" pipe rapid expansion for alpha) is sufficient to properly define the flow from either pipe.

If you could better define the piping systems, say if there is a valve on alpha or beta near the header for example, and what that valve Cv and upstream pressure is, that would help tremendously in more accurately calculating the flow rates and thus the pressure drops inside the header.

 

[SloFlo]

Overall System for clarification...
An 8" line is coming from a pump at ground level and going up and into a tank which is 20' in heigh. The pressure in this pipe is 80psig. This is transporting water. I would like to inject NaOCL into the 8" pipe just before it dumps into the tank using a 1/4" line having a pressure of 20psig.

 

[iainuts]

The system you're describing now sounds very different than the OP. Sounds like water is coming out of a pump, going horizontal through an 8" pipe and into the bottom of a 20 foot high tank. You have a 1/4" pipe at the junction just before the tank entrance you wish to inject NaOCL which is at 20 psi. Is that configuration correct? Please correct as necessary.

Couple questions:

Do you know the pump flow rate?

Is the 20 foot high tank at atmospheric pressure at the top (20 ft up)?

Where is the 80 psi measured (ie: how far from the tank/how long is the pipe before entering the tank)?

Are there any valves or other restrictions such as bends in the 8" pipe between where you're measuring pressure and the entrance to the tank?

 

[SloFlo]

Not quite...The 8" line comes out of the pump which is at ground level and then rises above the top of the tank, approximately 23'. It then turns horizontally toward the center of the tank and then drops 3' and dumps into the top of the tank. There are no other bends in that line besides those just mentioned. There is however a control valve in the line between the discharge of the pump and the 90 degree turn at 23'.

 

[iainuts]

Ok, good. I take it the 1/4" pipe you're trying to inject the NaOCL is on the verticle section where it dumps into the tank and about 2 feet from the pipe opening. Is that right? I'm assuming also the pipe opening is at atmospheric pressure.

I'm also assuming the control valve is in the 8" pipe between the pump outlet and before the 1/4" pipe. Do you know the Cv of that valve? Should I also assume the point at which you're measuring the 80 psi is before the valve?

Do you know the pump flow rate?

 

[Ashereng]

If the pressure in the 8" line is greater than 20 psi at the point of injection, then you will not be able to inject. With regards to backflow, this can be prevented with a check valve (various sorts).

You need to determine the pressure in the 8" line where you are injecting at 20 psi via the 1/47" line.



"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
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[Artisi]

Without reading all the in's and out's of the system, (please excuse the pun)- why not inject into the inlet side of the pump - surely much easier.

Naresuan University
Phitsanulok
Thailand

 

[katmar]

Artisi's solution will not only overcome the pressure problem, it will also give you better mixing. This is a standard setup for this type of blending operation.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[25362]

That would be right if the pump is only dedicated to this service. Is it ?

 

[Scotsinst]

Your pressure is not 80psi after the control valve. It is atmosphreric pressure plus the hydraulic losses between the valve and the outlet.
You can therefore add in the 20 psi line. It is however standard practice to include a check valve to prevent backflow in the line.
If you add the small line just downstream of the valve you'll also get good mixing.

 

[legs]

the basics of this problem is that the flow of the 80psi will pull the flow in the 1/4" 20psi into the flow of the 8" 80psi line. Unless there are other conditions that make this unique, stay with the basics and build on them.

 

[BigInch]

Can't forget the "ATM" part. Why is it all not just blowing out through the atmospheric pressure nozzle? I don't see where any backpressure caused by a nozzle restriction at "charlie" has been defined. It is defined at "ATM", which I assume to be atmospheric 15 psiA and all the rest are psiG. If the configuration is anywhere near realistic, the 20 psi(a or g? take your pick) will be reduced at the connection point and the water evacuated from that line by the even lower pressure created by the stream from the 80 psi(a or g) point as it passes beta and all will blow out through "charlie", until somebody turns charlie off, then flow will start into the 20 psi(a or g), probably increase the pressure there while full flow begins into the previously 20 psi line.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[johnp]

There appears to be some confusion between compressible and incompressible flow.

Your 1/4" (6 mm) 80 psi nozzle flow has to expand into an 8" (200 mm) volume, so your Pressue Volume values have to change. You will probably get critical flow at the nozzle, a shock wave front, cooling and condensation, if not freezing in a two-phase fluid.

This is different to single phase liquid flow.

Johnp.Rz

http://www.mets.net.au

 

[BigInch]

Then you're saying it's 3-phase?

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com