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By pass pressure drop calculation

By pass pressure drop calculation

By pass pressure drop calculation

(OP)
Hi,
I am trying to calculate the pressure drop through a by pass valve (simple stage valve).
I am not sure to understand what is the behaviour of the valve when the flow goes through it. For instance, in the by pass I am studying, the cracking is 12 bar but the pressure drop measured for low flow rate is about 8.5 bar (lower than 12 bar).
How do you explain it ?
Thanks

Eric

RE: By pass pressure drop calculation

Daunay

Are you measuring 8.5 Bar across the valve? Or is the
by-pass pressure 8.5 bar?

If the gauge after the valve shows 8.5 This shows a pressure drop across the valve of 3.5 BAR.

If the pressure gauge before the valve shows 8.5 BAR, then the by-pass valve is spilling too early.

Is the valve too big for the application? When using by-pass valves to limit system pressure you have to add the vales pressure drop to the cracking pressure.

The valve may be sized to give a 12BAR system pressure. 8.5 BAR cracking pressure + 3.5 pressure drop.

Should the valve actually spill @ 12 BAR? Or should it limit the pressure to 12 BAR. There is a big difference.

OR

It could be nothing to do with the above...the valve could be broken???

RE: By pass pressure drop calculation

(OP)
Hydromech,

The 8.5 bar is the pressure drop measured accross the valve.

This valve is designed to begin to spill @ 12 bar + or - 10%

In fact, the valve opens when there is 12 bar (+ - 10%) of pressure difference between its inlet and its outlet.
When the flow rate is established through it, the pressure difference decreases to 8.5 bar.

Daunay
 
PS : the valve is not broken !!


RE: By pass pressure drop calculation

Daunay

Is the valve held closed by a spring?

With any valve that is held closed by a spring the reset pressure after the valve has lifted is generally 30% lower than the cracking pressure.

Once a valve has opened and oil has begun to pass through it, the flow forces of the oil help to hold the valve open.

The spring does not have enough power to close the valve.

What I believe you are seeing is a pressure spike to 12 BAR as the valve opens. Once the valve is open the pressure required to keep it open is only 8.5 BAR.

As long as there is oil flowing the pressure drop will remain at 8.5 BAR. Variations in flow will cause slight changes in the pressure drop.

As flow drops the clearance in a valve close and the pressure drop increases. The pressure rise pushes the valve open again.

A natural balance between the flow forces, pressure drop and  the mechanical action of the valve occurs and the valve maintains the pressure drop.

Has this helped?


 

RE: By pass pressure drop calculation

(OP)
Yes, the valve is held by a spring.

What are the forces which maintain the valve opened ?
It is not possible that it is the same force which has open the valve at 12 bar because the pressure drop is lower than 12 bar (8.5 bar), so the valve should be closed !

I think that it is the force created by the flow going through the restriction created by the valve opening.

Is that correct ?





 

RE: By pass pressure drop calculation

This valve is effectively working as a relief valve. I assume that the low pressure side of the valve is at zero.

Somewhere within the system a resistive force has caused the pressure to rise to 12 BAR.

At this point there is 12 BAR acting against the poppet in the valve.

The oil is not moving at this point there are no flow forces.

The force acting against the poppet and spring is only hydraulic. 12 BAR of hydraulic pressure is required to open the valve.



The valve opens and oil flows through the valve.

To know the flow forces you need to know the fluid properties, the velocity of the fluid and the area of the poppet against which the flow force is acting.

The valve only opens enough to limit the pressure. The pressure and flow forces cannot open the valve fully. If the valve was fully open there would be virtually no pressure drop to keep it open. The forces keeping the valve open are reactive forces against the spring.

The gap through which the oil flows is enough to generate a pressure drop of 8.5 BAR. The rest of the force holding the valve open is a reult of the flow against the poppet.

You know that there is only 8.5 BAR hydraulic pressure in the system. And you know that the spring in the valve has a 12 BAR rating...so what is keeping the valve open? It can only be flow forces.

All of the time that oil is flowing through the valve. The spring is trying to close the valve with an effective force equivelant to 12 BAR.

As the gap begins to close the pressure drop increases again  
and the valve is forced open.

Obviously the valve does not keep opening and closing. The force of the oil flowing through the valve helps to maintain the valve in the open position.

If the resistive force elsewhere in the system is removed, the pressure keeping the valve open will reduce. The oil will then flow to/through a different valve.

The forces that were keeping the valve open have gone...the valve will close.  

Daunay...please understand that I can only only offer you the benefit of my experience. I have tried to explain what I think is happening.

Hydromech.

RE: By pass pressure drop calculation

hydromech is correct, the hydrodymanic/drag forces that the flow of fluid exerts on the poppet plus the pressure just before the poppet can hold the poppet in equillbrium against the spring. If this was not the case the valve had to chatter which is undesireable and will destroy the poppet and the sealing seat. To my opinion this is a good design which is not simple to achieve and avoid chatter.

RE: By pass pressure drop calculation

(OP)
Hydromech
Thanks a lot to give me these information !
It helps me to understand better the valve behaviour.

Daunay

RE: By pass pressure drop calculation

Above posts have done a good job of explaiing flow forces.
I would also add that inmost cases, the area exposed to the poppet is slightly larger once the poppet lifts.
Since the spring is 'force', not psi, 12 bar times a small area is enough force to lift the poppet. Once open, 8.5 times a larger area equals the same force.

If it is a problem, is find a valve with less hysterisis, or use a pilot operated one. (Sun, Sterling, HydrForce/Vickers, adn Sauer are ones we typically use.)
Pilot oper are typically very flat curves, but not as fast response.

kcj

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