branching pipe downstream of pump 3

[MachinaMan]

Hi all,

I am trying to size a pump for my system.
The system branches out downstream of the pump discharge (3 branches) and recombines before discharging into a tank that is open to atmosphere.
There is only one pump in my system, this is not a parallel pump system.

How do I determine the system curve to size my pump ?

Thanks

 

[STYMIEDPIPER]

If the three branches are nearly identical with regards to routing, number of components, pipe length, etc --- you need to be concerned with one branch to calculate your system friction loss.
If the three branches vary --- you need to select the one branch that has the higher friction/head loss and use during calculation to determine system head loss.
Is the flow the same in all three branches??
If not then you need to be able to reduce/ throttle flow thru branch which will add loss coefficients to your branches.

 

[MachinaMan]

Hi Stymidpiper,

All three braches vary and flow varies as well.
This is an existing system but I need to size a new pump for it.

Thanks

 

[FACS]

The flow will divide between the branches in such a manner as to make the head loss in each branch the same.

The head loss between the junctions is the same as the head loss in each branch.

The total flow rate is the sum of the flow rates in each branch.

Charlie

http://www.facsco.com

 

[LHA]

Headloss is equal across all 3, determines Q for each...homework problem?

Engineering is the practice of the art of science - Steve

 

[MachinaMan]

Hi Guys,

What you are saying is also what I read, but I tought it was only applicable to piping with the exact same caracteristics.

So I will need to throttle to get the correct flow into each branch (compressor)

Thank you all for your help

lha:

this is a real life problem, compressor cooling, not homework, I wish it was only homework !

 

[BigInch]

For pump analysis or system head loss purposes, you can "replace" all parallel pipes with one imaginary "equivalent length" of one pipe, a length of one pipe with some theoretical diameter that yields the same head loss of the multiple pipe configuration at any given total flowrate (pump flowrate). This is much like how, in electrical circuit analysis, two or more parallel resistors can be replaced by 1 resistor with the value Req = 1/R1 + 1/R2 +...1/Rn. The system curve will then follow the head loss characteristics of the equivalent pipe.



BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[25362]

There is a typo since parallel resistors add reciprocally so that 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + ...

 

[MachinaMan]

Hi BigInch and 25362,

Yes this is exactly what I thought about doing last night.
It's just like a parallel electrical circuit, find Req.
If Voltage is Head and Current is Flow, what is Req ?

Thanks

 

[MachinaMan]

I suppose R=H/Q
where H = head loss and Q = flow

 

[wfn217]

You should be able to come up with a system of simultaneous equations the pipe segments: P1-P2 = f(Q). You can use Mathcad to solve them.

 

[25362]

Ohm's law can be compared to a simplified model of laminar flow. Following the Hagen-Poiseuille equation for laminar flow (aka parabolic flow) of Newtonian non-compressible fluids, in pipes of constant cross section (radius =r, length=L), constant viscosity [η], the resistance R, in the formula R.Q = [Δ]P (similar to Ohm's law) would be:

R = 8[η]L[÷]([π] r⁴)​

In turbulent flow, [Δ]P = [ƒ](Q), which is not necessarily a linear relationship. For example, when the
Reynolds number for liquids is in the range 30,000-100,000:,

[Δ]P [≈] (L/5[π] ²)[ρ](Q²/D⁵)​

for gases with Reynolds number in the range 100,000 to 500,000:

[Δ]P [≈] (L/6.25[π] ²)[ρ](Q²/D⁵)​

where [ρ] is the density. Same formulas can be used for ducts and channels of other cross sections if the diameter D is replaced by the hydraulic diameter D_h.

 

[BigInch]

Thanks for the correction to 1/Req.

To be precise, the true water <=> electricity analogy is that head loss corresponds to impedence, since both head loss and impedence are nonlinearly proportional to flow, ie. K * Q^n = Hl, whereas electrical resistance is linear at R * I = E

You can write a system of equations to solve for heads and flows, similar to how Spice solves for electric currents, however, as electrical resistance equations are linearly proportional to current and their simultaneous equations can be solved using linear algebra, but head loss equations are non-linear and the corresponding system of equations must be solved using non-some linear algebraic technique. The non-linear equations are replaced by a linear system, solved and the errors evaluated. Corrections are determined and reapplied to the original linear system and the iteration process continues until (hopefully) it converges to a solution. The Hardy-Cross follows the same principles using a series of step by step iterations, rather than a "simultaneous" solution.

It is possible to program the such a system of equations to approximate a small network (say less than 10 pipes and 2 or 3 loops), even including simplified valves and pumps or compressors, using only Excel.





BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[TedLinsle]

Any/All: Wouldn't the Hardy-Cross method be appropriate for this problem, or is that only for multiple outlets?

 

[BigInch]

You can use the Hardy-Cross method. For now, remove the upstream pipe between pump and first branch and the downstream pipe from the last branch to the tank. Now there are 3 unknown branch flows, making 2 loops, with an assumed known inlet head = head coming from the outlet of the pump discharge pipe, and a last branch exit head equal to the head at the inlet of the pipe going to the tank. Both of those pipes can be solved, if you only know one of the heads, either from the pump, or at the tank (fluid elevation), since I guess you probably have a known flow you want to go into the tank. So, those become the boundry conditions for solving the unknown loop flows using HC.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[dcasto]

What is the maximum flowrate through any one system and what is that systems pressure drop at that rate. That pressure is you head requirement for the pump, done. Next, what is the maximum volume you want to move, done. This should be your pumps maximum design point. To find a best effiency point, do the same step as above but sustitute most probable flow and pressure drop. Find a pump curve that gives highest effiency for most probable and will still allow the maximum.

 

[BigInch]

What is the maximum flowrate through any one system and what is that system's pressure drop at that rate?

Not done! The OP is trying to find the system curve. Flowrates depend on the differential head applied... its unknown. He needs to determine the pressure drop through the 3 pipes at any (or all interesting) flowrates first (determine the system curve), therefore he doesn't yet know the required pumping head ...at any flowrate. That's why the equivalent pipe, or the H-C or some other method must be used.



BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[rcooper]

As it is an existing system, if possible I would try and measure the system curves having calculated them. If the piping is not new, you will not know the pipe roughness or the degree of fouling.

Rather than solving equations, I tend to plot graphs of the three systems, and add the flow rates together. It's a nice visual way of solving the problem.

 

[25362]

This is typical, among others, of cooling water and hot oil systems.