Determining parallel flow off header 4

[dfa1979]

I have a header of 8" line pressurized up to 70 psig with water. The end of the header is capped.

Branching off of the header are a 5" line, 3" line, and 1" line. They are all the same length of 72' and dumping to atmospheric pressure. What I am trying to determine is the flow rate in each line.

I realize that more flow will come out of the 5", then 3", and the smallest flow coming from the 1". The problem I am having is determining the flow rate in each pipe.

Is there an equation to determine this?

 

[dfa1979]

By the way this is schedule 40 pipe. My calculations were the following and wanted to check.

5" - 3400 GPM
3" - 920 GPM
1" - 55 GPM

This is assuming constant 70 psig pressure in the header.

 

[cvg]

are you sure your 8 inch line and the header can sustain this amount of flow without a pressure drop? ie: at 4,375 gpm, will the pressure in the header still be 70 psi? I doubt it since you will have nearly 30 fps velocity in your 8 inch line.

 

[dfa1979]

Let's just assume the header is a large tank instead of a pipe. I'm trying to just get an idea of how the flow changes.


Also how would I account for the pressure drop across each pipe penetration if the header is in fact a 8" line fed from a pump upstream?

 

[BigInch]

Account for the head losses at the joints by using equivalent length of tees with appropriate inlet and outlet dimensions.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[dfa1979]

Wouldn't there be a pressure drop not just because of the friction of T's, but also because there is fluid lost at each penetration?



So the pressure in the header, assuming a pipe, would have a lower pressure after the first 5" penetration than before it? And so on for the other 2 penetrations?

 

[BigInch]

Yes. And the flow is less after each outlet too, so pressure drop per unit length would decrease in each header pipe segment (between the outlets) as flow is removed at each outlet.

A more uniform pressure drop along the whole header might be obtained by placing the inlet near the middle of the header's length and arranging the smallest outlets closer to the header's inlet with larger outlets closest to the end caps.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[Latexman]

Pressure changes in distributor piping occur due to frictional and momentum effects. Of course, friction always works to decrease pressure. Momentum effects in a diverging manifold, like the one described, cause a 1.2 velocity head pressure rise. Conversely, momentum effects in a converging manifold cause a 1.8 velocity head pressure drop.

Good luck,
Latexman

 

[dfa1979]

BigInch do you know of a method or equation to calculate this pressure drop/flow rate in each of the pipes in which I originally described? I can't find anything anywhere about this.




One thing that was really counterintuitive was when I saw a demonstration of a manifold that was not capped at the end and allowed to flow through to a open pot. Each of the 5 pipes off the manifold had closed valves. As the valves were opened, the static pressure in the manifold across each pipe went UP not DOWN. They attributed this to a reduction in velocity as the flow dropped out the first pipe, then the second, and so on.


Check it out here. The demonstration starts about 7 minutes into the video.

http://modular.mit.edu:8080/ramgen/ifluids/Pressure_Fields_and_Fluid_Accel.rm

 

[Latexman]

One thing that was really counterintuitive was when I saw a demonstration of a manifold that was not capped at the end and allowed to flow through to a open pot. Each of the 5 pipes off the manifold had closed valves. As the valves were opened, the static pressure in the manifold across each pipe went UP not DOWN. They attributed this to a reduction in velocity as the flow dropped out the first pipe, then the second, and so on.

Read my post above yours.


Good luck,
Latexman

 

[BigInch]

Yes, total head (unit is feet, but it is a measure of potential energy) is the sum of static head (pressure) and velocity head, of which one can be exchanged for the other by changing the velocity in accordance with Bernoulli's theorem.

If inlet pressure is held constant and equal and outlet pressures are held constant and equal, its equivalent to attaching all the inlets together and then attaching all the outlets together, and the 3 pipe flows can be found by the "equivalent pipe" method, 1 among others.

If you want to not consider the header for a minute and approximate a solution,
You have a pressure loss in each pipe of 70 psi, or 161 feet, if its water. You can make any combination of pipes of any diameter you want (all with length 72 feet) that have a head loss of 161 feet, and the flow in each pipe will be the flow that gives you 161 feet of loss. The only other condition is that the flow into the inlet joint of all pipes must be the sum of the individual pipe flows, which will also be equal to the sum of all outflows from each individual pipe at the outlet joint.



**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[vzeos]

dfa1979
Your calculated flows seem to be inline. You can assume the header to be a big tank if the take-offs are close to each other, then you only need to account for the equivalent length of the tees. If the take-offs are far apart, then you will need to calculate the pressure at each take-off at points B, C & D. The pressure at A is 70 psi. See diagram below (assuming the diagram compiled correctly), as follows:


Since you don't know the pressures at B, C & D, you will need to do a trial and error solution. The flow rates you calculated were based on 70 psi and can be used for your first approximation.

5" - 3400 GPM
3" - 920 GPM
1" - 55 GPM

You will need to calculate the pressure at B using the flow in the 8" line from A to B of Q=(3400 GPM+920 GPM+55 GPM).

You will calculate the pressure at C using the flow in the 8" line from B to C of Q=(920 GPM+55 GPM).

And you will calculate the pressure at D using the flow in the 8" line from C to D of Q=(55 GPM).

Once you obtain the pressures at B, C & D you will need to re-calculate the 5", 3", and 1" flow rates. Iterate this process until the change in pressure at points B, C & D is trivial.


[tt]
A 8" B 8" C 8" D
??????????????????????????????????????????????????????????
? ? ?
? ? ?
? ? ?
5" 3" 1"

[/tt]

 

[BigInch]

Good work vezos. I was just going to tell him how to add the header into the equivalent pipe system in a second post, but I think that'll do the trick.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[dfa1979]

Thanks all. Latexman, can I ask where you got the 1.2 and 1.8 velocity head pressure rise and drop from?

 

[Latexman]

Those particular interpretations are from a company design manual. The source data is very scattered. Perry's Chemical Engineers' Handbook has a reference to the original journal articles, and says, "Experimental data (Van der Hegge Zijnen, Appl. Sci. Res., A3, 144–162 [1951–1953]; and Bailey, J. Mech. Eng. Sci., 17, 338–347 [1975]), while scattered", show that K is probably close to 1 velocity head rise for a diverging header and a 2 velocity head fall for a diverging header. I had to adjust the text somewhat because Perry's actually uses 2K in their equations (+0.5 and -1).

Good luck,
Latexman

 

[dfa1979]

ok sounds good. The spreadsheet attached is my calculation for 3 iterations. I think it is correct. It would be greatly appreciated if someone could confirm the accuracy. The cells labeled "A-B in 8"' and so on use Bernoulli's equation to determine pressure rise due to the decrease in velocity. This pressure was used in the next interation as the motive force pushing the fluid through the 3" and 1" branches.

 

[vzeos]

dfa1979,

You need to do a frictional flow calculation taking into account the 8" pipe lengths A to B, B to C and C to D. Your pressures should not increase in the direction of flow. The pressure at A is 70 psi the pressure at B will be less due to frictional loss. The pressure at C will be less than the pressure at B, etc.

Try using a pipe flow calculator such as the one on this website:

http://kirkmansoftware.com/

or similar.

 

[Latexman]

vzeos,

Pressure changes in distributor piping occur due to frictional and momentum effects. Of course, friction always works to decrease pressure. Momentum effects in a diverging manifold, like the one described, cause a 1.2 velocity head pressure rise. Conversely, momentum effects in a converging manifold cause a 1.8 velocity head pressure drop.

In a short diverging header, momentum effects rule and the pressure increases. In a long diverging header, frictional effects rule and the pressure decreases.

See and read the parts on Figure 3 at:

http://web.mit.edu/hml/ncfmf/06PFFA.pdf

Good luck,
Latexman

 

[Latexman]

dfa1979,

Both friction and momentum should be accounted for.

Good luck,
Latexman

 

[vzeos]

Latexman,

Thank you for the article. It looks good. I printed it out and am looking forward to reading it.

I do recall a little about the De Laval nozzle from my college days. As I recall, for M<1, as the fluid enters the convergent section of the nozzle the fluid pressure reduces from the initial P1 to the throat pressure PT while the fluid speed increases. As the fluid enters the divergent section of the nozzle the fluid pressure increases from PT to P2 while the fluid velocity decreases. At no time is P2 ever greater than P1 as you seem to imply; this would violate the first law of thermodynamics.

Looking at dfa1979's worksheet, he is clearly confused by this because he has the fluid pressure increasing as the fluid flows through the drainage distribution system. It bears repeating; this is a violation of the first law of thermodynamics. You cannot have the system pressure higher than the source pressure without introducing some shaft work.

I agree with you that nozzle effects can be very important in fluidic circuits. However, in distribution system design it is a very small effect and in 30 years of design I have never had occasion to account for nozzle effects other than the use of K factors. The pressures throughout a distribution are always lower than the source pressure because of fluid friction.