Capacity Loss in Non-Concentric Quasi-Annular Compressible Flow

[sloquick]

In other words a circular tube lying bottom center of a circular pipe through which gas is flowing. We've had an entity performing some actual tests for us using air, but we can't believe the results. The capacity loss associated with an area reduction due to the innerduct of less than 3%, is estimated to be almost 20%. It's not intuitive that it could be this high, especially given that the diameter ratios hardly suggest annular flow.

Tube and pipe are smoothwall polyethylene. Pipe nominal diameter is 4". Velocity is about 70 ft/sec.

Does a 20% loss in capacity from inserting a tube in a pipe that resricts flow area less than 3% seem reasonable?

Thanks!

 

[fvincent]

If the area ratio (free pipe / resulting pipe) is 0.97, the pressure drop global coefficient for a turbulent flow would increase circa 8%. As a result, for the same pressure difference, the flow would decrease less than 4%.

I am considering characteristic diameter = D-d, where D is pipe inner diameter and d is tube external diameter,
and [(D-d)/D]^2 = 0.97, as you quoted (area ratio)

Is there any fitting else inside the pipe? Something that could provoke singular losses? I mean some connection, inserts for tube alignment. etc

Last but not the least - you don't have to consider the flow at 70fps as compressible, because Mach is less than 0.07. As a normal rule consider compressible effects only for Mach > 0.2

regards
fvincent


fvincent
Figener S/A

 

[sloquick]

fvincent,

Thanks for your reply. There may indeed be a problem with their measurement or calculations. This is my suspicion bolstered by your response.

Actually d/D = 0.163, so a/A = 2.7%. Does this change your response? I'm interested and would be greatly pleased to understand how you obtain the global coefficient and resulting flow loss.

Thanks again!

Regards

Kevin

 

[fvincent]

I've considered the following:

Delta p = f * L / D * ro * v^2 / 2

which can be written as

Delta p = f * L / D * m ^ 2 / (2* ro * A ^2)

where
A = section area = PI/4 * D ^2
ro = density

Again this can be rewritten as

Delta p = f * L * m ^ 2 / (2* ro * (PI/4)^2 * D^5)

That is: Delta p is proportional to the fifth power of 1/D

Well, D is the characteristic dimension of the flow inside the pipe

When you insert the tube of external diameter d inside the pipe of inner diameter D, the new characteristic dimension changes to (D-d)

You write the new equation for Delta p':

Delta p' = f * L * m' ^ 2 / (2* ro * (PI/4)^2 * (D-d)^5)

( m' and Delta p ' are the new mass flow and pressure drop)

Now you want Delta p = Delta p' to check the new capacity, so:

m'^2/(D-d)^5 = m^2/D^5

or:

m'/m = ((D-d)/D)^(5/2)


Pressure drop coefficient = f * L / (2* ro * (PI/4)^2 * (D-d)^5)
(this is a coefficient related to mass flow not to velocity!)

Hope this helps (and I have made no mistake!)

Regards
Fabio











fvincent
Figener S/A