B Maintaining the same flow rate while subdivding a tube

[Pyper]

I have a 5/8" tube that air is blown through. I want to subdivide the center section using .1" tubes so that the same volume of air can be blown through it. It will start as 5/8", but immediately be divided into the separate .1" tubes, then end as a single 5/8" tube. How many .1" tubes would I need to maintain the flow of the single 5/8" tube?

 

[olivermsun]

Ignoring friction with the pipe walls for the moment... mass conservation ensures that the same air that goes through the 5/8" section has to also pass through the tiny follow-on tubes, right?
Is the question instead that you want to maintain the same flow at the same pressure even though you are splitting the tube? If so, don't you want to maintain the same total pipe cross-section, ignoring friction?

 

[jbriggs444]

maintain the same flow at the same pressure

What does it mean to maintain the same flow at the same pressure? The same pressure gradient? Then Poiseuille's equation should help.

 

[Pyper]

Is the question instead that you want to maintain the same flow at the same pressure even though you are splitting the tube?

yes!

If so, don't you want to maintain the same total pipe cross-section, ignoring friction?

yes!

Basically I want to blow through the tube and it be just as easy to blow through when diverting to smaller tubes as it is with a straight 5/8" tube.

I did these simple calculations just based on area of a circle:

5/8" = 15.875mm = Area of 791.73
.1" = 2.54mm = Area of 20.268
791.73/20.268=39.063
So I would need 40 small .1" tubes to have the same flow rate and pressure of the single larger 5/8" tube.
Does that sound about right?
That seems like a lot!
5/8" is only .625, it seems like 7 smaller tubes would be enough..
My thinking may be flawed here...

 

[jbriggs444]

Basically I want to blow through the tube and it be just as easy to blow through when diverting to smaller tubes as it is with a straight 5/8" tube.

So the total length is unchanged, but we want a bunch of 1mm capillaries to replace a single 5/8 inch tube and still present the same pressure drop at the same total volumetric flow rate.

I did these simple calculations just based on area of a circle:

5/8" = 15.875mm = Area of 791.73
.1" = 2.54mm = Area of 20.268
791.73/20.268=39.063
So I would need 40 small .1" tubes to have the same flow rate and pressure of the single larger 5/8" tube.
Does that sound about right?

It is far far worse than that. The flow rate through a tube goes as the fourth power of the radial dimension. The ratio of your radial dimensions here is 8.75 to 1. Raise that to the fourth power and you will need a bit under six thousand small tubes to get the same flow rate as that 5/8 inch tube at the same pressure drop.

Edit: corrected my figures to use 0.1 inch small tube rather than 1 mm small tube and added...

There is a reason for this seemingly extreme behavior. In a tube with laminar flow, friction means that the fluid near the tube walls is moving slowly while fluid near the center is moving more rapidly. The closer the walls are to the center, the slower the fluid at the center will move -- for a fixed pressure gradient.

Here is another link to the resulting flow equation: https://www.fxsolver.com/browse/formulas/Hagen-Poiseuille+Equation

The situation is worse yet if the flow is not laminar.

 

[olivermsun]

What does it mean to maintain the same flow at the same pressure? The same pressure gradient?]

Yes.

Then Poiseuille's equation should help.

Yes, I agree that's what the OP should use if this is a practical problem where the pressure gradient is held more or less constant.

 

[Pyper]

This is blowing my mind, 6K tubes?
When bundled together they would be MANY times larger than the single 5/8" tube.
All this because of friction, even for air?

 

[jbriggs444]

This is blowing my mind, 6K tubes?
When bundled together they would be MANY times larger than the single 5/8" tube.
All this because of friction, even for air?

Yes.

[Back of the envelope, since your scale ratio is 8.75 to one you'd need something 8.75 times fatter. A bundle of tubes about six inches in diameter and maybe a bit more by the time you account for wall thickness and the hexagonal packing arrangement]

If you want to change the scenario so that instead of tubes we are talking about a thin plate with holes then friction stops being as relevant. We'd need a real fluid dynamics person to talk about the ins and outs of choked flow in that case.

 

[Pyper]

Wow.
No, it is definitely tubes, I can't get around that..
Thank you for the info..
Lots of trial-and-error ahead.