Airflow in a long, cylindrical, perforated pipe 1

[rfinch]

Hello -

First time poster, several time referencer.

My question:
I have air coming off a 3hp regenerative blower at 0.433psig(12 in H2Og) on the outlet of the blower. Flowrate = 300cfm.

This flows through a 6" pipe with 3 90's, then forks into 2 L=50' ID=4" cylindrical segments.

The purpose: provide evenly distributed airflow for a composting operation.

The question: what size hole (probably need to assume), # of holes, and distribution of holes needed to ideally achieve Q = uniform through each airhole.

What has been observed in the field: the purpose of the air is to cool the self-heating pile. With evenly longitudinally-spaced 2" holes the end nearest the blower has lower temps than the end further away. Makes sense.

My thought: have larger spacing between the air holes nearer the blower, tighter spacing further away.

I don't have any CFD resources, which make my life alot easier. I could do it experimentally, but if there is a convenient approximation.. well that's why I'm posting.

Sorry for the long post.

CN:
How do you calculate air flow distribution through holes spaced long-ways on the cylindrical pipe fed by a small regenerative blower such that volumetric airflow is approximately equal through each hole.

Thanks.

 

[MikeHalloran]

Assume an arbitrary flow per hole, and start at the last hole.
Then stage the spacing or diameter of the penultimate hole, and recurse until you run out of flow or pressure at the blower.
Put it all in a spreadsheet to make modifications easy.


Mike Halloran
Pembroke Pines, FL, USA

 

[katmar]

There are two ways to design a distributor (or sparger) like this. One is to recalculate the pressure in the pipe after every hole and adjust the spacing or sizing of the subsequent holes to suit - this seems to be the route you are thinking of and it is certainly a valid way of doing it.

A second method, which is much easier to calculate but which results in larger pipes is to size the pipe such that there is a negligible pressure drop along the pipe. In this way every hole is exposed to virtually the same pressure so they all have the same flowrate. This means you can use the same size hole evenly distributed along the length of the pipe. Easier to calculate and easier to fabricate.

You are in the fortunate position that your 4" pipe falls into the second category. But 2" holes seem way too big to me. My estimates for the number of holes per leg (assuming 2 legs) is
1/2" dia 15 holes
3/8" dia 26 holes
1/4" dia 58 holes
These holes should be evenly distributed along the part of the leg that is in the compost.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[ione]

The second method described by noticeable member katmar is the one I usually adopt when I have to size steam spargers, but it is undoubtedly applicable to your case. Since you do not have any further pressure drop downstream the holes as air directly discharge to atmosphere, ensuring all holes are “exposed” to the same upstream pressure, will lead to an evenly distributed air flow throughout each hole.
You can assume each hole behaves as an orifice (at least this is what I do) and evaluate the pressure drop provoked by the hole. This involves a reiterating procedure until you find the size that meets your requirements.

 

[Subystud]

Thanks for all of your inputs.

Katmar -

I understand this reasoning. In fact, one point I left out was that in the field the piping was littered with 2" holes, certainly causing a significant pressure.

My question is: how did you come to this result? I see that the total cross-sectional area is equal for these results, but how did you come to determine this area?

By my calcs: at 150cfm (per leg) & CS Area = 0.0873 ft2 my velocity equals 1718.87 ft/min. For 150 cfm to go through your area your velocity equals 7352.94 ft/min through each 1/2" hole. Is this correct? Or am I forgetting something like compressibility?

I appreciate all the help.

 

[katmar]

Are Subystud and rfinch one and the same???

The holes are calculated as simple orifices with a fixed average discharge coefficient. It is buried deep in a program I wrote a long time ago and I think I took something like 0.61 for C_D. I made no attempt to correct C_D for different pipe and orifice sizes so the total area should be the same regardless of the hole size.

The logic is to determine the flow through 1 hole with the given pressure drop and then use that flow rate to determine the number of holes required.

I agree with your velocity calcs. At these pressures you can safely ignore compressibility.

I took the basic method from an article by KS Knaebel in Chemical Engineering, March 9, 1981 p118 but Perry describes much the same procedure.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[Subystud]

lol. yeah

 

[Latexman]

The design guidelines I've seen uses the area ratio (total perforation or orifice area to pipe cross sectional area) versus % maldistribution. Little maldistribution (under 5%) is present for Ar < 0.5, and significant maldistribution (over 10%) is present for Ar > 1. There is a weaker dependency on the distributor length to diameter ratio (L/D). I think Katmar's #'s fall around the Ar = 0.5 area.

Good luck,
Latexman

 

[ione]

You can take a glance para "Flow of gases through an orifice" in the link below:

http://en.wikipedia.org/wiki/Orifice_plate

Or look for Darcy formula (not Darcy-Weisbach) in Crane TP-410.

 

[Subystud]

ione-

I've checked this wiki page before and my question is: is it applicable to perpendicular flow, which would be the case in question.

 

[zdas04]

Perpendicular to what? At velocities as high as you're talking about, the flow is quite turbulent (even at these pressures) and it is going from side to side in the pipe as much as it is going down the pipe. Treat it as flow from a pressure vessel.

David

 

[ione]

You’ve already got an answer from zdas04. The key point is to keep pressure practically constant along the length of the 4” pipe (negligible pressure drop), then the pipe will behave as a pressurized vessel with multiple exit points with equal sizes and consequently with an even flow distribution.

 

[Subystud]

Katmar -

Thanks for all your help. I have another question. The blowers will be VFD driven and operate at range of flow rates.

The highest flowrate and static pressure was what you previously calculated. I'm wondering how numbers change at our minimum flowrate.

That being: 100 CFM & 1.2" H2O.

I'd imagine that the hole size decreases... but now I'm wondering (if it's even possible) to have even air flow through the holes at a range of backpressures and blower flow rates.

Thanks --

Rob

 

[chicopee]

If I remember properly, the Chemical Engineering Handbook has a section on spargers.

 

[Latexman]

In the "second method", even distribution results from low pressure drop in the header compared to high pressure drop across the orifices. dP is proportional to flow² both in the header and through the orifices. If the flow is reduced, the pressure drop is still going to be high through the orifices compared to the header, and the flow should still be even. The maldistribution is mostly a fuction of the area ratio (total perforation or orifice area to pipe cross sectional area).

Good luck,
Latexman

 

[Subystud]

I got Perry's open, but am finding it difficult to locate the sparger section... What section is it in?

I about to get a section of pipe, plug 5 pressure gauges in it. And start drillin'.

 

[ione]

Section 6-32 "Fluid distribution" (7th edition)

 

[Latexman]

Section 5-48 (6th Edition)

Good luck,
Latexman

 

[katmar]

There is more good news with a distributor designed this way - as opposed to the alternate method with varying spacing and hole sizes - and that is that it is fairly tolerant of changes in flowrate. The reason is as already described by Latexman. Since the pressure drop is proportional to flow² and the flow has decreased by a factor of 3 (i.e. from 300 cfm to 100 cfm) we would expect the pressure drop to decrease by a factor of 9, i.e. from 12"WC to 12/9= 1.33"WC. You say that your blower delivers 100 cfm at 1.2"WC so it will be very nearly at equilibrium with the system curve.

Distributors are not quite as tolerant on the high flow side because at high velocities the pressure recovery down the distributor can become significant when the decreasing flow towards the end causes the velocity to decrease and the static pressure to increase (Bernoulli strikes again). But you have to over-range your distributor quite severely before you get noticeable maldistribution. In this counter-intuitive situation you can get higher flows out of the holes further from the blower.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[Subystud]

Thanks for all your help. I'll update this thread with my results for the sake of science