Compressible flow - help setting up the problem

[mkenwort]

I appreciate any help I could get with this problem. Basically I am trying to analyze the flow of air through 4 sharp-edged entrance orifices into an annulus then out 12 sharp-edged exit orifices. So it flows in, takes a right turn then travels about an inch then out the bottom orifice. It'd be alot easier to describe if I could just draw a picture, but it's not too unusual a case so perhaps that will suffice.

I.D. of the outer cylinder = .413 in.
O.D. of the inner cylinder = .338 in.
(Gives Deff = Do - Di = .075 in.)
The entrance holes are .188 in. dia., exit holes are .046 in. dia. Giving a 5.6:1 area ratio for flow.

Now, the problem is I want to determine the effect of changing the .188 holes to .181. My suspicion is that the effect will be negligible since the exit area will be the limiting factor. However, I am unsure as to the best way to prove this.

I considered just calculating the pressure drop across the entrance then doing the same for the exit (and dividing by 3?) to match them up and show the outlet pressure differential is much greater. But now I'm not so sure what that is even accomplishing what I want to find out or what sort of data I'd need to do that.

Please ask any questions you have that would help you to help me better. I will do my best to answer them, but bear with me as I'm not really a fluids guy so this thing is a real curve ball for me with the orifices and compressible flow.

 

[zdas04]

You gotta know pressures and temperatures upstream and downstream of the contraption. It's a straight forward calculation as long as you know dP.

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[katmar]

mkenwort,

Your suspicion is correct - the effect is negligible. You could do a detailed calculation, but the following "logic" may prove it for you.

As a first approximation the principle parameters for flow through an orifice are related as
Pressure drop = Constant x (Flow / Area)^2

Taking your annular area as an orifice of equivalent area allows you to consider the problem as three orifices in series. The area ratios for Outer Holes to Annular Area to Inner Holes are 1 : 2.2 : 5.6

Because the pressure drop is inversely proportional to the square of the area the pressure drop ratios will be 1/1 : 1/4.9 : 1/31.3

This means the pressure drop through the inner holes is approximately 2.5% of the overall pressure drop. Changing the size of the inner holes from 0.188" to 0.181" decreases the area by about 7%. The pressure drop would increase by the square, or about 16%. So the increase in pressure drop would be 16% of 2.5% or about 0.4% of the overall pressure drop.

A rigorous calculation will probably show that the change in pressure drop will be somewhere between 0% and 1%, which is probably well within thae accuracy with which you can measure the other parameters any way.

Hope this helps.

regards
Katmar