Flow rate inside concentric tubes

[KevinH673]

I have a problem dealing with concentric tubes. Technically, the internal one is a solid rod, and the flow is flowing between the inside of the large diameter tube and the outside of the rod (of course). Current flow rate is 9 gal/min. The inside rod is 10 mm, and outside tube is 15 mm (internal dia). We are going to a 6 mm inside rod, and want to keep flow rate at 9 gal/min. Now, I know the flow rate equation is

Q=A*Velocity

So I solved for velocity, with Q/A.

Q = 9 gal/min = 567,823.3 mm^3/s
A = pi/2*(15^2-10^2) = 196.35 mm^2

So, V = 2891.9 mm/s

Now, I don't think I want to change the velocity (unless anyone knows of a reason that I don't), so basically I just need the area's to remain the same...

pi/2*(15^2-10^2) = pi/2(x^2 - 6^2)
x = 9.43 mm

Correct? Seems simple enough, but just looking for a confirmation that I dont need to change the velocity for any reason....Just wanted to check, though.

 

[CdnMechEng]

I'm not huge into piping but my work has recently had me going through piping systems. So I can't speak with 100% authority on the subject, but I can give you my opinion on what I would do.

The only concern you should have with the velocity is pressure drop. Obviously higher velocities lead to higher pressure losses. Now the flow, flow area and the velocity may be the same but from what I remember your hydraulic diameter will be different (your new concentric tube has a smaller hydraulic diameter). So you will want to look at the frictional losses to make sure that they haven't increased.

 

[BigInch]

Not exactly. See Chapter 5 of this doc,

http://txspace.tamu.edu/bitstream/handle/1969.1/4334/etd-tamu-2006B-PETE-viloria.pdf?sequence=1

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"Pumping systems account for nearly 20% of the world’s energy used by electric motors and 25% to 50% of the total electrical energy usage in certain industrial facilities." - DOE statistic (Note: Make that 99.99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[hydtools]

In the calculation of circular area you need to use pi/4 when using diameters^2. r = d/2 r^2 = (d^2)/4

A = pi*(D^2-d^2)/4 = pi*(R^2-r^2)

Ted