Predict Fluid Flow 1

DonU · Sep 10, 2002

I’ve always worked Bernoulli’s equation from point 1 to point 2 using a known source of water usually a pond (point 1) running through a pump to the discharge typically a nozzle (point 2). How would one use Bernoulli’s equation from say a fire hydrant to predict the flow from a nozzle downstream before spending all of the money for plumbing?

Would point 1 be merely the pressure head as read from a gauge on the hydrant (before turning it on) with the velocity head of zero, potential head of zero and of course head added from outside sources as zero? And then could I solve for velocity and head losses at point 2 (outside of nozzle) by considering the pressure to be zero and the potential head of my elevation change.

I would think that the pressure at point 1 would represent the total amount of energy available at this point, regardless of what’s happening upstream with the city’s plumbing. I would also expect that when I do plumb from the hydrant that I would see a slight decrease in pressure from the gauge where the energy is being converted to velocity head, head loss, and potential head.

Am I screwed up? If so, can someone please help? This problem seams to come up for me quite often.

KernOily · Sep 10, 2002

Bernoulli's is just an energy balance:

what goes in = what goes out, minus losses.

That's all there is to it. Just set your elevation datum and get after it. The rest is making sure you are straight on your units and the two energy terms: potential (pressure head + elevation head) and kinetic (velocity head).

I used to get all screwed up on this too. Then one day it hit me that this is nothing more than energy conservation. All I do is set my elevation datum, which is purely arbitrary, and then calc out the various terms: elevation head, pressure head, velocity head, and friction losses.

Remmeber that the energy at any point is independent of what's going on elsewhere (loose paraphrase, there...). Yes, your inlet condition at the hydrant exit at zero flowrate is independent of what's going on upstream in the city plumbing.

Note that this analysis only holds for the steady-state case. Transient events (e.g. waterhammer) are something else entirely, although you do still have the energy balance - with transients though you do have a pressure signal communication both upstream and downstream.

Hope I answered your question?
Thanks!
Pete

poetix99 · Sep 10, 2002

Well, not "screwed up", but not properly framing your problem.

The pressure just inside of the hydrant will drop after turning it on. This is due to pressure drops through the piping upstream, which you seem to have wanted to neglect. The dead-headed supply pressure will, of course, give you (via Bernoulli's equation) the MAXIMUM velocity that you might get at the hydrant if there were no upstream drops.

For simplicity's sake, you would work your way upstream to piping that is at some point large enough that you could (as a first approx.) assume that the velocity is zero. But you must account for the losses in the smaller piping from that "supply plenum" forward.

If you are talking about a small take-off at the hydrant, then the "zero velocity" assumption might be accurate enough right at the hydrant.

Account for the change in elevation if you want; for the first cut calculation that I've described, I don't think that 10 - 15 feet of head will matter much vs. 100 - 200 psi.

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Predict Fluid Flow 1

DonU

Mechanical

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KernOily

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poetix99

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