Flow rates division need help

[mech212]

On the left we have a system with a pump and two branches , a pump produces head of 10 bars and Q of 5 m3/h according to the system requirments. On the left the pressure came uot of the pump 10 bars it drops to reach 9 bars at point 3 then the 9 bars drops in branch 1 to reach 1 bar at its end and the 9 bars drops in branch 2 to reach 8 bars at the end , while the flow rate is divided to be Q=2 in branch 1 and Q=3 in branch 2 according to pipe sizing and calculations . Now on the right picture the same system but we adds a resistance ( restricted orfice ) to drop the pressure 2 bars on branch 2 ,, now the pump curve point is the same and the pressure is the same at branch 1 but it drops to 6 bars at the end of branch 2 ... Now the question is the pump still produces the Q=5 as the point of operation doesn't change but at the same time there is much resistance in branch 2 than the previous case , so what will be the valve the flow rate decreases in branch 2 and increases in branch 2 ??
I need to understand this point

 

[LittleInch]

Please proof read your post. The diagram and the text don't match. On the LH diag the end of branch 2 is shown as ending at 8 bar, but the text says 2 bars. Which is it??

Assuming it's 8 bar the flows should be the same because you have been able to lower the end point by the same amount as your artificial pressure resistance (2 bar). Normally that doesn't happen..

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[mech212]

I edit the question sorry for that its 8 bars ,, The artificial resistance drops the pressure 2 bars and branch 2 resistance is 1 bar so pressure drops from 9 bars at point 3 to 6 bars at the end of branch 2 ,, so i don't get "Normally that doesn't happen" ? in my previous thread someone told me when there is a pressure drop or change in pressure like introducing an orfice , this means more resistance so flowrate changes

 

[LittleInch]

Yes but how does the end pressure magically change from 8 bar to 6 bar? That's what I mean by doesn't usually happen. Yes the system resistance has increased but the pressure drop from the branch point 3 to point has also increased by the same amount. So no change to flow.

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[mech212]

The pressure decreases from 8 bars to 6 bars by using an orfice for example , the pressure drop in branch 2 due to friction and static head is 1 bar and we added an orfice of pressure drop 2 so the fluid pressure decreases from 9 bars to 6 bars at the end user , why it is not possible ?

 

[LittleInch]

By using a larger orifice? Only if the ultimate downstream pressure is lower and the 6 or 8 bar is being kept artificially high.

Is there a practical application to this or is it some sort of course work?

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Also: If you get a response it's polite to respond to it.

 

[mech212]

It is a practical solution that when we have a branch of high pressure and we need to decrease the pressure so the pressure reaches the end user at the pressure needed so we use a restricted orfice in that branch , so my question was since the Q from pump is the same , does adding this orfice will affect the flow rate in this branch

 

[LittleInch]

Not as long as the end pressure decreases at exactly the same as the orifice pressure drop.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[mech212]

Ok i got it but it there is equation or rule that i can read to further understand

 

[LittleInch]

Try drawing it as a hydraulic diagram with axes as length and head of your liquid.

If the angle of the hydraulic line stays the same for the same size pipe then you have the same flow rate.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.