Parallel Pumps Head Loss

[SMontgomery119]

Hello,

I have two parallel pumps: pump-1 has a 32-gpm flowrate and pump-2 has a 96-gpm flowrate. These two pumps feed into a loop that has a 128-gpm flowrate and 58-ft of head loss.

How much headloss do I use from the 128-gpm loop for each of these pumps for pump sizing?

--Steve M.

 

[Artisi]

I assume you are asking what is the head on each of the pumps - it is 58ft.

 

[dcasto]

You need to get out both pump curves and look at the 58 ft head point on each curve. Then come back with what the pump curv says. It it says 44 gpm an 102 gpm, then you'll get 146 gpm, not a drop more, not a drop less.

Then look at 60 feet, if it says 32 and 96, then thats 128 gpm. This means you'll need to add 60 - 58 ft = 2 feet of added head loss to each pump.

 

[SMontgomery119]

Yes, I want to know the head loss for each pump.

What's the logic to your answer?

--Steve

 

[hydtools]

If the parallel pumps are connected to the loop at approximately the same point and the pumps have little or no plumbing between each of them to the loop, then each pump must deliver against the same pressure or head. 58 feet, the pressure it takes to move 128 gpm through the loop.
Delivery pressure is the same for each pump and flows are additive in a parallel connection.

Ted

 

[dcasto]

thanks hdtools. The pumps see the exact same head, less any piping head loss differences. If you want to force pump 1 to have exactly 32 gpm, then you must adjust the head on that pump per the pump curve. If the pump curve for 58 feet says it'll do 500 gpm, then thats what it will do. Follow the pump curve to 32 gpm, then look at the required feet.

 

[SMontgomery119]

It turns out that the fitting losses associated with each individual pump are significantly different: pump-1 32-ft and pump-2 2-ft

Do I simply add each pumps fitting loss to the main section loss of 58-ft? to compute pump-1 90-ft loss and pump-2 34-ft loss?

Comparing piping to circuit analogy, the piping head loss or pressure loss is equivalent to the voltage decrease across a resister.

But, whereas in circuits a resistance value is normally provided for the resister in piping systems a pressure drop (equiv. voltage drop) is normally provided for differing flow rates.

Any thougts?

--Steve M.

 

[Artisi]

Lets go back a step or two- there are you measuring the 58m head loss - is this the head loss in the pipe system only?

If this is the case then pump 1 will have a total head across the pump of 32 + 58 = 90ft, pump 2 will have 2 + 58= 60ft

 

[SMontgomery119]

That's right! My mistake!

The bypass pump is 2-ft head piping at the pump plus 58-ft in the common pipe system that equals 60-ft of head total for the bypass pump.

The 58-ft head section connects to the input/output of the two pumps in parallel.

Thanks!

--Steve M.

 

[Artisi]

No problem - pleased to help out.

 

[dcasto]

Now, you need to look at the pump curves where each pumps individual head is. 90 ft and 60 ft, to get their individual flows. Then add the two.

The head we are talking includes physical distance plus loss for fittings and pipe friction?