equation for fluid flow in pipes with multible inlets. 1

[RuneRRA]

Hi I would like to calculate the flow and pressure distribution in a pipe with multible inlets.

The pipe is basically a 25 m long straight pipe, Ø500mm, with 10 gaps (Ø110) drilled in to its side. at the end of the pipe a ventilator is sucking the air through the pipe at a flow of 2000 m3/h.

I would like to calculate the corresponding flow through each gap and the pressure below each gap. I know i need to use some kind of iteration process (proably based on Bernouilli's equation and the minor losss coeficient for the gaps and the loss coeficient for the tube) but it would be nice to se some one who has descriebed how this is done in practice.

Can anyone refer me to where i can see such description? a book/ homepage or alike.

Thanks in advance

 

[bimr]

Maybe this will assist:

http://www.pdhcenter.com/courses/m246/m246content.pdf

 

[RuneRRA]

Thank you for your answer but unfortunately not. Im looking for at set of equations and the method to use for using them as to iterate the answer.

 

[goodguy1405]

What is thr I.D. Of the pipe?

 

[RuneRRA]

Im not sure what ID is but if it is inner diameter this is roughly 478 mm.

 

[bimr]

Not sure that there are any equations for what you are trying to accomplish.

With air systems, the ports should be sized to obtain a desired velocity. It is usually around 2-4 meters/sec. An addition, if you have particulate present, your piping should also be sized for a velocity that will carry the particulate.

The friction loss through the ports will be negligible. Size the fan for the loss in the straight pipe which appears to be oversized and will cause an expansion loss.

 

[Latexman]

Use Search (upper left between Forum and FAQs) on the word maldistribution, or the phrase "flow distribution" and you will get a number of old threads that will be very helpful to you.

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[katmar]

You are correct that this has to be solved iteratively. However, if you set it up correctly in a spreadsheet you can use the Goal Seek function to handle the mechanics of the iteration for you.

For ease of reference let us number the inlets from 1 to 10, with #1 being the furthest from the fan and #10 being the closest. Let the absolute pressures inside the duct in line with each inlet be numbered P₁ to P₁₀, and the flows through each of the inlets be numbered F₁ to F₁₀.

The only number you have to guess is the pressure P₁. Everything else after that is easily calculated. Once you have specified P₁ you can calculate F₁ using a simple orifice equation. The flow along the duct towards the fan between inlet 1 and inlet 2 is the F₁ you have just calculated. From the flow F₁, plus the dimensions of the duct, you can calculate the pressure drop due to friction between inlets 1 and 2. Call this ΔP_1-2. Now P₂ = P₁ - ΔP_1-2.

This puts you in a position to repeat everything you did for inlet 1. You have P₂ so you can calculate F₂. The flow between inlets 2 and 3 is F₁ + F₂ so you can calculate the pressure drop between inlets 2 and 3 - and therefore you can calculate the pressure P₃.

You keep stepping along the duct like this until you have solved for each of the flows F₁ to F₁₀. If you add them all up you should get 2000 m³/h. If the total flow is not 2000 m³/h adjust P₁ and repeat. Goal Seek takes all the pain out of this.

You will find that for a given number and diameter of inlets, as the diameter of the duct increases the flows into each inlet become more uniform. But of course at the cost of a larger duct.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[RuneRRA]

Thank you all so much. Unfortunately om attending meetings all day. I vant wait to get started making this spreadsheet based on your comments.

Right now things look simple I'll let you know if things work out as I expect.

 

[bimr]

There is no reason to guess at pressures. The total pressure at the fan inlet point is equal to the fan VP plus the total duct resistance. You start the spreadsheet at the suction side and calculate from there.

These problems are usually calculated using the velocity pressure method. Here is a link to a spreadsheet:

http://www.google.com/url?sa=t&rct=...fogZAJ&usg=AFQjCNGUNEsooer8paMWo9XkAxjSK4r3Hw

The background for the calculations can be downloaded at:

https://law.resource.org/pub/us/cfr/ibr/001/acgih.manual.1998.pdf

 

[RuneRRA]

Once Again thank you all. Unfortunately i wont be back at the office before NeXT week. Ill let you know how it Works out.

Thanks

 

[katmar]

You could start at the fan end of the duct if you know the head that the fan will generate at 2000 m³/h and you know the pressure drop in any ducting downstream of the fan (so that you can get the pressure at the fan suction). You would do the reverse of what I suggested earlier, calculating the pressure drop in each segment between the inlets to give you the pressure in line with the inlet. As before, knowing these pressures would allow you to calculate the flow through each inlet and, by subtraction, the total flow from all the upstream inlets. If the pressure you used as the fan's suction pressure was correct you would find that the flow in through inlet #1 exactly matched the flow between inlets 1 and 2.

But it is unlikely that these flows will match exactly because underlying this whole problem is the assumption that the given fan will move 2000 m³/h of air through the given duct. In reality, you need to find the point on the fan curve that also satisfies the system (duct flow) curve. This means that the method proposed by bimr is also a trial and error method because when you get to the end of the duct and find that the flows do not match you will have to re-run the calculation with a new combination of pressure and flow rate selected from the fan curve.

The method I proposed suffers from the same problem. After performing the iterative solution I outlined earlier you would obtain a pressure at the suction of the fan and if this did not match the fan curve you would be forced to select a new flow rate. In the end, both my and bimr's methods would be iterative trial and error solutions. It does not matter which end you start from - if the pressures and flows all balance in the end you should get the same answer from both methods. Maybe do it both ways and tell us how close the 2 answers came. That would be very interesting and would certainly increase your confidence that you have found the correct solution.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[bimr]

Would have to disagree.

The original post states the air is "sucking the air through the pipe at a flow of 2000 m3/h", which gives you the static head (velocity head) at the outlet.

One works through the exhaust system adding up the head losses until you determine how much air flow is being pulled through the last opening. The static head remaining at the last opening will determine how much air is being pulled in at the last opening.

Conservation of energy means that all energy changes must be accounted for as air flows from one point to another.

The pipe velocity is low in this problem, so there does not seem to be any reason that the pipe would not be able to easily pass the volume.

This method is supposed to be accurate to 5%. No guesswork is required. Refer to Figure 5-7 for an example.

 

[katmar]

If the air flow is fixed at 2000 m³/h and we know the ID of the pipe then I agree that we can calculate the velocity pressure at the downstream end of the pipe (fan suction). But I don't see how this gives us the static head at the outlet - unless we take it from the fan curve.

I also agree with bimr's statement "The static head remaining at the last opening will determine how much air is being pulled in at the last opening." My concern is that this quantity of air is not guaranteed to balance with the assumed total of 2000 m³/h unless the static pressure used for the outlet end was correct.

It seems to me that Example 1 in Fig 5-7 in bimr's attachment starts its calculation from the end away from the fan and that the static pressure at the fan suction is the result of the calculation rather than an input to the calculation. BTW, this attachment is an excellent resource - thanks very much for that.

So the essence of our disagreement is how we determine the static pressure at the outlet. I am saying that it can only be determined by trial and error, but bimr is saying that it can be calculated from the known information. Hopefully RuneRRA will run both versions of the calculation and give us a report back.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[bimr]

In looking at this again, the orifices are significantly smaller in area than the straight pipe.

The orifices will have a pressure drop of 0.178 inches of water at 200 m3h. The head loss for 2000 m3/h through 2.5 m of Ø500mm straight pipe to the first orifice is only 0.0019 inches of water. The total headloss through the straight pipe will be around 1% of an orifice headloss.

You should have equal distribution through the orifices since the headloss through the straight pipe is minimal.

You can calculate it with the spreadsheet. Unfortunately, it is in American units.

 

[katmar]

Good calculation bimr. Sometimes it pays to take a practical approach and not get bogged down in theory. I agree that within the limits of our ability to calculate, and well beyond our limits to measure, the flow through the 10 orifices will be equally distributed. A star for you for saving the OP a good few hours work.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"