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Capillary Flow

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blckwtr

Mechanical
Oct 30, 2006
204
Does it exist a formula for calculation pressure drop and/or time for given volume through a capillary tube with given viscosity, diameter, length? We are interested in various time delays for a hydraulic chamber...
 
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I beleive most people on this forum have an internet connection, and therefore when searching on capillarity get up Wikipedia on the first hit.

However, the reason for asking on this forum is to gain access to experts on this area, not the height of a liquid column (see thread starting question)
 
blckwtr,
No need to get huffy, sometimes people focus on a particular word ("capillary" in this case) and jump to an answer.

Very small tubes are often called "capillary tubes" because the ratio of the surface area to flow area is supposedly so large that surface tension and adhesion effects will dominate other flow parameters. This isn't actually the case and the flow through these small tubes can be modeled with pipe-flow equations.

I'm pretty reluctant to use the "simplified" equations like Weymouth or Panhandle A & B with very small or very short (feet instead of miles) tubes. It is probably just my bias, but I use the AGA Fluid Flow equation for small or short tubes and have gotten results in the same universe as measured data (never +/-10%, but I have regularly gotten within 25%).

David
 
blackwater,

If you ask a better question (more detail about what you're really interested in) and you might get a better answer. You'd be surprized at the number of people that haven't figured out where Google is.

Time delay and capilliary and hydraulic chambers, don't usually go together, unless you've got a lot of time on your hands.



**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
I would use standard pipe pressure drop calc methods; but with a capilliary, you may find that the resulting Reynolds Number is low. So you will have to use a laminar flow curve to get the friction factor. If so, the friction factor will be a lot higher that you get in a normal pipe.
 
From Yeaple's 1966 collection of articles capillary flow through small tubing is given as:

Qin = d^4 * deltap * .0245/(mu * l)

Qin = in^3/sec
d = tube ID, in.
deltap = delta pressure, psi
mu = mean absolute viscosity, lb-sec/in^2
l = tubing length, in.

Rearrange the equation if you know flow rate and want to solve for pressure drop.

Ted
 
Thanks hydtools...

This is actually comparable with

headloss = 32 * nu * L * V / ( g * D^2 )

- nu = kinematic viscossity
- L = tube length
- V = velocity
- g = constant
- D = tube diameter

I think this is "Hagen-Poiseulle law" for laminar flow in tubes, so the assumption that you can use a laminar flow curve is correct, to all you other guys...

at least I get the time delay approximately the same with both formulas.;)
 
Hydtools, is "deltap" the same as the hydrostatic head pressure?
 
As one of my old bosses used to say, "if you want the right answer, you have to ask the right question".

Good luck,
Latexman
 
And incidentally my Late Father(basically a Teacher by profession)

Always said,

'Half of the answer is hidden in the question itself'


Best Regards
Qalander(Chem)
 
Chicopee;
Head loss is not the same, but is easily found by:

deltap = rho * g * h ,

where rho is density,
g is constant,
h is "head loss" or hydrostatic pressure.
 
I'd like to take an opportunity for suggesting review of the Process TGML link when posting formulae. For example, Greek letters are easily created by use of brackets "[" and "]" surrounding "&Greekletter;" (Unfortunately, the brackets don't show up if I just put them the phrase.) So if I use brackets instead of parentheses, (Δ) becomes [Δ].

Also blkwtr, what does "g = constant" mean? Are you talking about gc, the gravitational constant? ...one of the reasons I'm asking is that Crane Technical Paper 410, "Flow of Fluids through Valves, Fittings, and Pipe," define g as the acceleration of gravity in terms of distance/time squared (i.e. "ft/sec2"). Obviously, this doesn't work out dimensionally in your formulae.

Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.
 
If it's obvious, then the probability for this to be the right assumption is close to 1, also if the units match, then this strengthen your assumption. However, if I would choose another gravity constant than the most likely to be used, and which are universally known, I would have specified this.

I will take into consideration to use greek letters the next time, but my impression is that the experts on the area don't mind, if it's a known expression, also it is easier to write "nu" than brackets-what-ever...
 
chicopee,
No deltap is not hydrostatic head pressure. It is the pressure differential across the capillary tube related to the flow through the tube. This is a dynamic condition.

Ted
 
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