Multi-Suction Point Pipe Flow

[ATKrank]

If I a pump with a main suction header and 4 smaller branch connections of difference sizes. Let's say for this example we have (2) 3", (1) 1", and (1) 2". Also let's say our pump is requiring 20 gal/min.

The flow rate of each branch would be proportional to the area of the pipe correct? flow = Vel. * Area. Or would they all have an equal flow rate and just vary in the velocity? How would friction losses play a part in determining the flow rate of each branch? I feel like we need to use Bernoulli's/Darcy-Weisbach equations.

If we close a valve on one of the branches, how does that affect the flow rate of each of the other branches? Would the flow rate increase equally amongst all the other branches? or would it be proportional to their area's? If it is proportional, would we be able to do a sort of weighted average to determine how much more flow one branch would receive vs another? Also again how would viscous/friction losses play a roll in determining this?

 

[JBA]

For a given pressure differential the flow rate will vary proportionally but not linearly with pipe diameter, pipe internal surface finish, length, fittings, inlet elevation, etc; and, the pump suction required to provide sufficient differential pressure across the combination of all inlet branches to provide the desired 20 gpm system flow. One majorly significant, and potentially poorly controllable, additional element in all of this is that any change in the inlet pressure of any individual or a combination of the branch feeders will substantially effect all other branch flows and the total flow rate of the pumping system as well.

A suggested analysis method is to assume a given pump suction pressure and calculate the flow rate (using standard "Bernoulli's/Darcy-Weisbach equations" including inlet pressure, inlet, outlet and fitting loses, branch inlet height, etc) for each individual intake branch to determine the individual branch and total intake header flow at that assumed header suction pressure. Once that is established then I would expect the change in flow for the set should be reasonably (but not directly) proportional to any required changes (within a reasonable pressure range) in header suction pressure required to achieve the needed pressure differential across the branch set to attain the desired 20 gpm flow rate. Note: The "Crane Piping Handbook" is a generally accepted "go to" reference for piping liquid flow loss calculations.

As for closing one of the branches feeds, given the original calculations I would anticipate that the increase in flow between the remaining branches would remain essentially, but not exactly, proportional; but, a reduced suction header pressure (greater pressure differential across the remaining branch set) will be required to maintain the 20 gpm flow through the system.

 

[ATKrank]

A suggested analysis method is to assume a given pump suction pressure and calculate the flow rate

By "assume a pump suction pressure", are you referencing what would be known as the NPSHR for a specific pump? I think this concept of suction pressure is where I am stuck.

Below I am showing two examples:

To evaluate example 1: Let's assume its a 6" pipe and the medium is water. By continuity, my flow rate in is equal to my flow rate out and the velocity can be determined by Q=V*A. So my velocity is 20 gal/min * .13368ft³/gal / .19635 ft² = 13.6165 ft/min. Got it, that's easy. Now to find my pressure at point 2 I need to use Bernoulli's and Darcy-Weisbach equations. So the equation should look like this: p₁+1/2*rho*v₁²+rho*g*h₁=p₂+1/2*rho*v₂²+rho*g*h₂+h_f

but v₁=0 and h₁=0

So it simplifies to:
p₁=p₂+1/2*rho*v₂²+rho*g*h₂+h_f

But h_f is a function of velocity, pipe length, pipe diameter, and a friction factor (estimated by the moody chart or by the Colebrook-White equation which is a function of the Reynolds number, relative roughness, pipe diameter and the Reynolds number is a function of velocity, Pipe length and velocity)

which looks like:
h_f=f(L/D)*(v²/2g)

substituting in:
p₁=p₂+1/2*rho*v₂²+rho*g*h₂+f(L/D)*(v²/2g)

looks like we have everything except for friction factor f, to use the moody diagram we need to Reynolds number
Re=(rho*V*d)/(mu)
So we would now be able to calculate p₂ as long as it is in turbulent flow. Which after calculating the Reynolds number I realized I chose poor numbers but just substitute the laminar head loss equation instead of the turbulent one and we are good to go.

So now if I want to evaluate example 2: let's say our header from points 4-2 is 6" and branches 1-2 and 3-4 are both 3". Now the velocity at point 2 is the same as example 1 since the flow rate and pipe size are still the same. But this is where I get stuck and seems like I just keep going back and forth that I need to calculate the flow rate to get the velocity through each branch. But to calculate the velocity, I need the pressure drop through the leg. But to get the pressure drop I need the velocity. Can you please explain how to evaluate this in a similar manner as I did for example 1? That is the easiest way I can understand these concepts and I think most people feel the same. Thanks!

 

[JBA]

OK, now we come to the point where you see why the majority of engineers hate to do piping flow analysis and E&C companies use piping flow programs for their designs; and, that is the fact that the calculations generally end up requiring multiple trial and error repetitions to achieve their final solution.

At this point, more important than my explaining in more detail how you could proceed with such a calculation process with your multiple branch feeds is that you have not provided the overall function or basic parameters of your pumping system so it is hard to judge the ability of the whole design concept to provide a system meeting your requirements.

For example, the delivery rate of each branch is going to depend upon the combined effect of its inlet head/pressure and flowing pressure drop compared to all of the alternative branches; and, anyone branch with the combined inlet pressure and pressure drop that can deliver 20 gpm to the pump at its suction pressure will be the branch feeding the pump until its inlet pressure falls below that of the next appropriate branch inlet head/diameter at which point that branch will be the primary pump source in a continuous back and forth cycling through the set of headers. In other words, unless each branch inlet maintains exactly the correct inlet pressure and flowing pressure drop combination at the pumps suction pressure then your system will never be able of sustaining a given constant flow ratio across the branches.

In that respect, on your above diagram you show the inlet pressure of the the 2 branches to be atmos (14.7 psia) however I have to assume that the ends of those pipes must be submerged in water to some depth and so the inlet head (ft) or psia will not be atmospheric pressure and the pressure drop from your calculation will determine what that minimum inlet head/pressure must be to provide a 20 gpm delivery at the pump's inlet NPSHR. If your inlet branches are sourcing from some set of water retention tanks that are going to have varying water heights this is going to defeat your efforts to deliver a consistent water delivery rate ratio through the branches to the pump inlet header.It is for this reason that systems for collecting fluids from multiple sources and pumping that fluid away are generally designed to have a common collection tank (or pond) from which the pump draws its fluid; or, alternatively if there is a required delivery rate from multiple sources, each of those sources has its own discharge pump.