Parallel flow pipe network help 4

[hydronicengr]

Let's say I have a simple parallel flow problem as shown in the attachment. The flow is pressurized and the downstream of the return goes back to a piping system that is negligible. It does not vent to atmosphere.

I've tried reading up on the subject, but I got stuck at a certain point. At point A, I know that the flow going in must equal the flow going out to the two pipes. Therefore, it will look like Q = Q1+Q2. Expanding Q1 and Q2 gives me V*A for each respective pipe.

I also know that the head loss of Pipe 1 must equal the head loss of Pipe 2. Expanding those equations, I find that I am faced with an unknown, which is the friction factor for each pipe. The friction factor can be explicitly calculated using the Swamee and Jain formula, but is dependent on the Reynolds Number, which is dependent on the flow velocity, which is ultimately dependent on the flow rate of each of the pipes. After getting the friction factor, everything else is easy. Simply plug the velocities from the head loss equations back into the Q formula shown above and solve for the unknown.

I know one can attain the correct friction factor value by many iterations, but is there a way without doing iterations by hand or by Excel and not fancy software?

Also, how would I find the total pressure loss of the system?

Thanks!

 

[hydronicengr]

See picture below.

 

[BigInch]

There is no direct solution.
Sum all pressure drops in any path you can construct from source to sink.
Each path will have an equal pressure loss.

If it ain't broke, don't fix it. If it's not safe ... make it that way.

 

[rutherford703]

Yes. Using Excel goal seeker you can easily solve the problem if you have an Excel spreadsheet

 

[MikeHalloran]

What I usually do is calculate the pressure drop for each pipe for >an arbitrary flow rate<, then use the pressure drop and flow rate to calculate a Cv for that pipe.

Using the Cvs, you can calculate the flow rates that will make both pipes have the same pressure drop.




Mike Halloran
Pembroke Pines, FL, USA

 

[jnich]

We had this argument in my office recently on the total pressure loss in a system with a run of parallel pipes. Since the pressure loss in both pipes has to be equal (assuming steady state conditions), would you analyze the streamline flow path of one particle through the system (traveling through only one of the parallel pipes) and determine the loss through that flow path to be equal to the total pressure loss of the entire system (which includes both parallel pipes)? It's an interesting question yet to be addressed on this thread.

 

[hydronicengr]

MikeHalloran, can you please elaborate?

This problem is driving me crazy.

 

[TD2K]

What I think Mike is suggesting is run an arbitrary fraction (within reason, running 0.1% isn't going to give you an accurate estimate) of the total water flow through each line. That will give you a dP for that flow. Now, use the standard control valve equation Q = Cv (dP/SG)^0.5 you can determine a "Cv" for that line.

I still think you have a assume a split between the two 'valves' until the pressure drop through the first 'valve' is the same as the pressure drop through the second 'valve'. Since Qt = Q1 + Q2 (Qt is the total flow and Q1, Q2 is the flow through each line) and dP is the same for a solved solution, you might be able to substitute into the two equations and come up with a single formula.

Mike, correct me if I'm wrong regarding your suggestion.

One thing this approach does not do is account for the change in friction factor as you adjust the split of the total flow through each 'valve'. That's why I prefer to use Excel to calculate the pressure drop through each line explicitly and then adjust the split until my pressure drops are equal.

 

[hydronicengr]

TD2K: Thank you for the input! I am indeed using Excel.

Do you mean to just treat each of the lines separate from each other first? (as if there was only that line in the system and not the other one). If so, I've already calculated the pressure drop for each of the individual lines in Excel already. I will try what you have mentioned and will report back.

Again, thanks all for the help!

 

[stanier]

The loss of pressure in each pipe will be the ssme the flow will vary if the diameter or the surface roughness is different.

“The beautiful thing about learning is that no one can take it away from you.”
---B.B. King

http://waterhammer.hopout.com.au/

 

[MikeHalloran]

TD2K has it about right.
I also use Excel.
After finding a Cv for each branch, I just use goal seek on the split to get the pressure drops equal.


Mike Halloran
Pembroke Pines, FL, USA

 

[hydronicengr]

Thanks all.

Without using the Cv, can I just keep entering estimates for flow rates in each branch in my spreadsheet until the pressure drop of branch 1 and branch 2 are equal? I tried it just now and although they're not 100% exact, they are quite close. I understand that this takes a lot of trial and error, but with knowledge of these systems one could probably enter in a good estimate.

For the rest of the day, I will try to figure it out with using Cv. I just want some further clarification. Basically, I have two pressure drops (one for each branch) that have been calculated individually. I use the Cv equation by using each of those pressure drops and an estimate of the flow rate for each of the branch. Now I have two Cv values, one for each branch. How would you implement goal seek for this?

 

[BigInch]

Yes. You can do it with your own estimates.
I don't know what your Cv equals, but it must equal some valid expression for head loss as a function of flow.

Something like attached.
Could be better, but its my weekend already.

If it ain't broke, don't fix it. If it's not safe ... make it that way.

 

[hydronicengr]

BigInch, thank you very much! I will definitely check it out and I'll try to improve on it.

 

[hydronicengr]

Last question (which may be quite silly to many experienced engineers), but like the flow, is the inlet pressure also distributed at the first junction? Or does pressure remain constant until there is a head loss due to a fitting etc.?

 

[Latexman]

At the junction (tee or wye) where flows divide or combine, the pressure is the same (constant) for both flows. P = P[sub]1[/sub] = P[sub]2[/sub].

Good luck,
Latexman

 

[davefitz]

Well, in the general case it is more complicated then the above problem. The general case has a pump performance curve to be matched, and also may have heat transfer and bouyancy effects. Also, there could be as many and 900 pipes in parallel( boiler waterwall), and the circuit can be statically unstable .

In any case, for ( n pipes) in parallel, each pipe can have a DP vs W curve drawn , by assuming a cerain flow, calculating the DP and storing the resulting data in a file. For each calculated DP, you can vary the friction coeficient , as well as the heat transferred and the bouyancy or 2 phase flow effects.

Next, you assume a certain pressure drop across the circuit, and for each assumed DP, there is only one flow for each of the tubes. Add up the sumof these flows , for each assumed DP.

If the circuit is thermal-hydraulically stable ( static stability aka Ledinegg stability) then there will be only one unique total flowrate for each assumed DP. For the known total flowrate, select the DP that generates that total flowrate.

The above topic was widely discussed in the 1960's and was titled "multi channel flow instability"

 

[stanier]

I guess then you would invest in AFT's Fathom software or equivalent.

http://www.aft.com/fathom.

This is not something for an Excel solution.

“The beautiful thing about learning is that no one can take it away from you.”
---B.B. King

http://waterhammer.hopout.com.au/

 

[adamlcarp]

Since this thread is still open I'll post this question here:

I'm working on a parallel pipe scenario where there is one inlet and multiple outlets all exiting to free air. Do the constraints of the Hazen-Williams equation still apply (ie all hL values equal)? I've found solutions assuming so but in practice we have found that in a case where all pipes have identical ID, roughness, and length we will consistently see greater dispense volumes from the pipe farthest from the inlet.

a little background: inlet is horizontal, all dispense pipes are vertical exiting the bottom with all dispense inlets at same height. Prior to a timed dispense all pipes and inlet tube (manifold) have been primed and are assumed full of fluid (ie. no air volumes displaced at beginning of dispense).

Thanks in advance for any guidance,
Adam

 

[vpl]

adamlcarp, please post your question as a new question. Thanks

Patricia Lougheed

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