Pump outlet pressure in function of circuit pressure when OFF

[mfqd13]

Hi,

I would like to hear some opinions about this situation.

Gived data
I have a pump with max head of 50mH2O (S.I.units) and in it's working point in this specific circuit is of aprox. 35mH2O.

Question
What happens to the pump outlet pressure when the system is turned ON, if i have a 2bar pressure when the systen is OFF? And the difference for example if it is at 4bar (also when it is OFF)?

Thanks

 

[BigInch]

Assuming you mean you have a maximum pump DIFFERENTIAL head at shutoff of 50m, starting the pump with a 2 bar suction pressure (say 10m suction head), the initial differential would be 50 m (say 10 bar) and the discharge head would be 60 m (say 12 bar).

As flow builds up in your loop, your suction head would tend to fall as friction losses increase, thereby reducing your suction pressure to S₂ when full flow is finally stablished. At that time the total discharge head would fall accordingly to S₂ + 35m (say 7 bar) - loop loss.

So that says, if loop loss is greater than pump differential, you must begin with a suction pressure (a surcharge pressure, if you will) equal to at least the absolute value of the difference, otherwise your pump will cavitate at normal flowrate. Alternatively, if your suction pressure is controllable, you would have to increase it to that value as you came up to normal flowrate. As I say, your pump would starve for flow and cavitate otherwise.

If loop loss is less than pump differential at normal flowrate, which is how it should be, then your suction pressure will increase until it finally reaches a number equal to the difference of 35m - loop loss. IN other words, if you do not use all the pump differential head in friction loss to your loop, the extra head delivered by the pump will (in effect) increase your system surcharge level whenever it is running.





Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[mfqd13]

Dear BigInch,

I understood perfectly your explanation, but i have 1 simple question to make, regarding your pressure units:

-You mentioned pressure units in "bar" and in "m". 1bar = 10m, so i see with some surprise tour convertions;


THANKS!!

 

[BigInch]

I'm multinational. Roughly speaking ... assuming water .. etc. etc. etc. Normally I don't like to use equal signs between bars and meters, or psi and ft, but we as hydraulic engineers should be allowed to knowingly use that convention amongst ourselves without any further concern.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[mfqd13]

Dear BigInch,

Now that i'm applying this to a real case, i found a strange thing related to your explanation.
I have a case where i have the following values when the system is OFF and ON:

OFF
-Pump inlet pressure = 2 bar
-Pump outlet pressure = 2,1 bar

ON
-Pump inlet pressure = 5 bar
-Pump outlet pressure = 9 bar

So, in your 1st answer you said 2 things:

1) "total discharge head would fall accordingly to S2 + 35m (say 7 bar) - loop loss", wich i can convert to my case in: 9 = 5 + A - B -> A - B = 4 bar

Where: A = Pump nominal DP
B = Circuit pressure drop

2) "your suction pressure will increase until it finally reaches a number equal to the difference of 35m - loop loss", wich i can convert to my case in: 5 = A - B

So, as you can see the 2 statements mean 2 different values for A-B. One is 4 bar and the other 5 bar.

I'm a bit confused....can you help me?

 

[BigInch]

Suction pressure changed from 2 to 5, so there is 3 bars of excess pump differential head that is being converted to a surcharge pressure, giving a total 2 + 3 = 5 bars as a suction pressure when running.

Discharge pressure when running you say is 9 bars.
Minus the 4 bar loop loss = 5 bars suction... so that's OK.

According to my theory, the pump must be making 7 bars total diff pressure during operating conditions, of which 7 - 4 loop loss = 3 bars that is an excess being converted into elevating the suction pressure, 2 initial suction + 3 converted = 5 suction when operating, so that seems to be correct.

5 bar suction when operating -3 raise in suction is 2 bar initial suction pressure, correct.

2 bar initial suction + 7 bar total differential head = 9 bars discharge, correct. -4 for loop loss = 5 suction, correct.

2 bar + 7 bars total pump differential = 9 discharge, correct.

Seems OK.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[mfqd13]

Ok. Good.
I understood perfectly!

So, how can i reduce the outlet pressure to protect some components that are not suitable for this pressure values?
there is an "over-head" that i may cut and i could put a pressure reducing valve, but it isn't energy efficient.

In this case i have also a VFD. Can i obtain the reduction in the pressure, using the VFD??? I think that no, but can you give me you opinion?

Thanks!

 

[BigInch]

Differential head reduces by square of speed. Try your vfd, or perhaps trim the impeller, if you want a permanent solution.
With the VFD, you should provide pressure control and relief in order to protect those weak components from full speed/overspeed pressures.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso