Fluid Mechanics Flowrate Question

[Gigawatts]

I am working on a fluid mechanics problem within a piping system and could use a little help:

There is a piping system that starts as a singular line and then branches out into two lines (equal pipe sizes). I have a known initial flowrate; and I realize that (assuming incompressible flow) the flowrate should remain constant. Additionally, I understand that (given the branch areas are equal), each branch experiences a constant flowrate that is 50% of the initial flow. However, in one of the branches there is a valve that is open only a very small amount, enough to still leak fluid through that valve. If I'm just using flowrates and not using Bernouilli's energy equation, does that branch still experience exactly 50% of the original flowrate? Perhaps there's something I'm overlooking because it doesn't seem as though that is possible. Thank you in advance for any input!

 

[LittleInch]

Of course it doesn't have 50% flow in that instance. Prpbably unlikley to have 0.5% of the flow...

At the point where the pipe branches, the pressure or head is equal. If the end point of the branches is the same height and pressure then the flow will depend on the friction losses in the branches. If one branch has a higher friction loss due to a "valve that is only open a small amount" than the other then it will have less flow than the other. All depends on the friction loss between the two branches. Internal size, length, material, fittings etc all have an impact on the friction losses.

Think electricity as an analaogy. You couple up two identical wires to the same juncton box, but in one you place a large resistor in line and then couple up the wires again. Which has the most current (flow)?? It cetainly isn't equal current lfow...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Latexman]

Breath in and hold it. Push one of your nostrils almost closed with a finger of one hand. Cup your other hand in front of your nose. Now, breathe out. Do you understand now?

Good luck,
Latexman

Need help writing a question or understanding a reply? forum1529

 

[IRstuff]

What's missing in your analysis is the consideration of line resistance and pressure, as the previous posters indicated. What propels the input flow rate is the pressure required to overcome the line resistance. When you close down one leg, if the input flow rate is to remain unchanged, the input pressure must go up to account for the increased resistance. If the input pressure is unchanged, then the input flow rate must drop.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[LittleInch]

Latexman, Nice one, but what happens if you have a cold??

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[hacksaw]

it is only a 50% split in the ideal case with two equal legs,
in reality it is a function of reynolds number, fluid distribution, piping layout, etc., even if you are nursing a head cold, a great essay question for an exam

 

[zdas04]

In real life it is never 50/50. There is always an extra bend, more elevation change, more standing liquid, or just messier walls in one pipe vs. the other. I had a loop line once that ended up 45%/55%, and I was quite impressed. The only other time I've ever been able to measure the relative flow rates in parallel lines were 30%/70%.

Gigawatts,
You make a statement that makes me nervous.

I have a known initial flowrate; and I realize that (assuming incompressible flow) the flowrate should remain constant

.

The continuity equation has no restrictions on incompressible flow, constant density, or any other flow parameter. That equation says that the mass flow rate must be constant within any line that does not add or remove fluid. So if you measured the mass flow rate in your system before the branch and after the lines come back together the mass flow rates MUST be identical. If they are not then you either have a leak, new fluid coming int, or a measurement error.

Also, Bernoulli only applies to pipe flow in cases where the distance traveled is VERY short pipe runs (inches, and not very many inches). It allows the AGA-3 equations to work because the distance between the ports is 1-inch which is too short for friction to be material. Any pressure drop due to friction at all invalidates the "inviscid flow" and the "no work is done on or by the gas" assumptions.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

 

[SNORGY]

@ LittleInch,

Depending on the discharge coefficient, you end up with anything between shut off head and total head loss.

 

[Artisi]

Gigawatts
Is this a hydraulic 101 homework problem?

Q. " does that branch still experience exactly 50% of the original flowrate?"

A. No, how can it see 50% of the total the flow rate when in effect the line is closed off, the flow can only equal the leakage rate.

However, the pump will be trying to pump the total flow (50% + 50%) thru' the branch line still open, at this point neglecting any leakage thru' the closed valve on the other branch line, the result being - it will force the pump left on its operating curve and could result in the pump operating in an unstable point on its curve.

You need to calculate a system curve (head loss) for the normal operating condition, or at least know what flow and head was used for the pump selection -- a second system curve must be calculated for the system when operating with one leg closed - to see what this does to the pump operation in terms of where on its curve it is going to operate.


It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[Latexman]

If you have a cold, two-phase flow!

Good luck,
Latexman

Need help writing a question or understanding a reply? forum1529

 

[Latexman]

If you have a cold? Two-phase flow!

Good luck,
Latexman

Need help writing a question or understanding a reply? forum1529

 

[Gigawatts]

First, thank you all for assisting me with this problem. To clarify, I am not a student in an introductory fluids course, but in my current job I don't see many problems like these and it's been too long since I've been out of school. Anyway, this proposed problem makes much more sense to me now.

I realize that it wasn't logical to believe both branches received 50% of the initial flow rate, but I was being tripped up by thinking that by having a barely open valve in one branch (using Q=VA) this would just mean the velocity increases as the area shrinks to keep the flow rate constant. I see the error in that now since I was thinking of the fluid's initial entry to the branch line as opposed to the steady state where the many head losses would serve as a resistance to that flow.

Moving forward, I suppose the best way to approximate the volumetric flow rate through the branch with the failed valve is to use the electrical circuit analogy and approximate (unfortunately I don't have all the data available for this system) an equivalent line resistance for each of the branches and use the ratio of resistances to determine the percentage of the initial flow rate that each branch will experience. Does this seem sensible? Thanks again for your help.

Also, Zdas04, I realize the continuity equation has no restrictions on incompressible flow, but I am looking at a volumetric flow rate and not a mass flow rate. As such, I believe compressibility would be a factor since it can affect the volume of a fluid, but of course leaves the mass unchanged.

 

[LittleInch]

Whilst the electrical analogy is useful, it isn't exact and you need to be careful in it's use. Resistance or frictional lasses in fluid flow are depressant on many things including the square of the velocity. Therefore different flows have frictional losses which are not in proportion. It sounds like you're looking at gas flow which for a long line has even more complications as it expands as it travels and therefore the actual velocity for the same mass flow keeps changing.

For branches with very different flows, you could probably estimate at the start that the one with very little flow has negligible friction losses in the pipe and all the pressure loss is concentrated across the valve where the upstream preassure is the same as that existing at the branch. As flow increases then this ceases to be valid.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way