clarification on pumps and flow rates (and prevalence)

[Fulvio Romano]

I can change the flow with a valve... but the pressure after the valve is the one that comes from the pump's characteristic curve?

Yes, minus the load losses induced by the valve, if they are not negligible.

You do something like that with the parallel pumps... But I don't remember why, there was a video of a course... If I find the place.

 

[MBT]

I swear I'm making it up! :

Don't swear and tell the truth! ! !
what brand and pump model you are using! hunting out that characteristic curve!!!


:biggrin:

 

[PierArg]

I did not understand this valve:
- or works by putting in by-pass a part of the flow rate (an input and two exits to understand us)
(@fulvio: do you understand what I'm talking about? )

- or it works as an adjustable nozzle. the fall of the nozzle in itself is negligible, but the one due to the strozzatura is function of the dish and flow.
based on how I go to position myself on the characteristic curve of the pump and I find the corresponding flow rate.
in this case by applying bernoulli you have the pressure out of the valve.

 

[mir]

ok we make a small upgrade to the problem:

I have my classic pump (with 1000 lt/h I have 30 mt. of prevalence) with the valve and two gauges.

I switch the pump and direct the valve until I read 3 bar on the pressure gauge 1. from the above (if there was no free discharge) I should read 3 bars also on the two gauge (always with valve without losses).

but now we see that downstream there is a free discharge ... questions:

the pressure gauge 2 always reads 3 bars?
Excluding losses of sort the hspruzzo is 30 mt?

Thank you guys!

 

[PierArg]

no pressure gauge 2 does not always read 3 bars.
But I'm referring to a flow regulator like choke.

your regulator valve is a real nozzle that creates upstream pressure and moves me the pump operating point that will erode the flow according to its characteristic.
downstream you will have water with lower pressure and higher speed.

 

[ale_.93]

I take this old post that I read out of curiosity.
paradoxically, that valve causes the pump to erode a greater flow, but the prevalence (lower jet) is reduced as the pump's operating point slips to the right (because the characteristic of the circuit is crushed). Am I right?
- - - updated - - - -I take this old post that I read out of curiosity.
paradoxically, that valve causes the pump to erode a greater flow, but the prevalence (lower jet) is reduced as the pump's operating point slips to the right (because the characteristic of the circuit is crushed). Am I right?

 

[paulpaul]

I take this old post that I read out of curiosity.
paradoxically, that valve causes the pump to erode a greater flow, but the prevalence (lower jet) is reduced as the pump's operating point slips to the right (because the characteristic of the circuit is crushed). Am I right?
- - - updated - - - -I take this old post that I read out of curiosity.
paradoxically, that valve causes the pump to erode a greater flow, but the prevalence (lower jet) is reduced as the pump's operating point slips to the right (because the characteristic of the circuit is crushed). Am I right?

Why paradoxically? Chopping the valve introduces a greater load loss: the characteristic of the circuit - which is parabolic, as function of the flow to the square - moves to the left, and consequently the intersection with the characteristic of the pump moves towards lower rates and greater prevalences. then on the pressure gauge 1 light a x pressure, on the pressure gauge 2 x-y, where y is the load loss introduced by the valve (for that flow). of course, the jet will rise of z meters, that is the piezometric height corresponding to the pressure y minus the load losses in the nozzle (and in the tube section between valve and nozzle).

 

[paulpaul]

errata corrige:
of course, the jet will rise of z meters, that is, the piezometric height corresponding to the pressure x-y less load losses in the nozzle (and in the tube section between valve and nozzle).

 

[meccanicamg]

ok we make a small upgrade to the problem:

I have my classic pump (with 1000 lt/h I have 30 mt. of prevalence) with the valve and two gauges.

I switch the pump and direct the valve until I read 3 bar on the pressure gauge 1. from the above (if there was no free discharge) I should read 3 bars also on the two gauge (always with valve without losses).

but now we see that downstream there is a free discharge ... questions:

the pressure gauge 2 always reads 3 bars?
Excluding losses of sort the hspruzzo is 30 mt?

Thank you guys!

I feel so much like the jominy hardening test sprayer.
gauge 1 and 2 give different value by force of things.
if you close the pressure gauge valve 1 read 100bar and pressure gauge 2 read 0bar.
if you open the pressure gauge valve 1 read 90bar and pressure gauge 2 read 80 bar.
There are load losses in the middle and you have to assess the balance of pressures....trinomio bernoulli bla bla bla bla bla bla bla bla bla bla