Meter-Out Flow Control Set Higher than Would-be Normal Flow

[HydraulicsGuy]

Please see the simplified hydraulic circuit below. Many components not shown for simplicity. For now, this is just a thinking exercise and not an actual situation.

Cylinder retracting. Gear pump putting out 10 GPM pump flow to rod end. 10 GPM is its absolute maximum flowrate at motor absolute maximum RPM. 1.5:1 cylinder ratio. That means flow from cap end would normally be 15 GPM. But pressure-compensated flow control is set to 16 GPM. Overrunning load acting on cylinder attempting to push fluid out of the cap end as cylinder retracts. What happens? I think flow from cap end would be 16 GPM. Pump is not able to fill up the rod end fast enough, which would require it putting out 16 GPM / 1.5 = 10.67 GPM. Pressure at rod end would drop to 0. Pressure at cap end is load-induced pressure. Cylinder retracts at a speed equivalent to 16 GPM. Am I thinking about this correctly?

This is a way to control the speed of the cylinder as it retracts with the overrunning load, while avoiding a pressure intensification situation. Any downsides? If the pump can't fill the rod end fast enough, is that a problem?

 

[hydtools]

With less than 0 gage pressure in the rod end, the rod seal may allow air to enter the system.

Ted

 

[akkamaan]

Air in via rod seal or will vacuum will occur on rod end (if pushing force on rod is big enough). And vacuum will make the rod side cavitate. But if you have a work port relief on the rod side of the directional valve, it should be combined with anti-cavitation check valve. That way the rod side will get additional flow from the return gallery in the directional valve. Actually the same oil as just left the capped side.

 

[73lafuite]

Your system may work if you add a scheck valve between tank and orifice off cylinder. But we never do such a circuit. Today we have cunterbalance valves, an efficient and inexpensive system.

 

[HydraulicsGuy]

As I said, it's just a thinking exercise.

 

[hydtools]

Would tank atm pressure over drive the pump/motor? Just a random thought.

Ted

 

[LittleInch]

Could you have two flow controls?

One set at <10 as shown and the other at >6 but feeding back into the gear pump inlet line with a check valve in place to only allow flow in one direction?

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

 

[HydraulicsGuy]

OP here. To those who say negative pressure and cavitation will result - isn't the fluid in the rod end of the cylinder at atmospheric pressure? The fluid is flowing into the vacant rod end and almost able to keep up, just not at any meaningful hydraulic pressure. Why wouldn't the pressure in the rod end be 0 psig, or 14.7 psia? And the pump discharge pressure is still above atmospheric because of pressure drops from pump to cylinder.

 

[HydraulicsGuy]

Could you have two flow controls?

One set at <10 as shown and the other at >6 but feeding back into the gear pump inlet line with a check valve in place to only allow flow in one direction?

I'm not sure I understand fully. You mean two flow controls in series back-to-back?

 

[LittleInch]

"Almost able to keep up" - Yes, but not able means there is a volume deficit between the cylinder movement and the oil flow in. That needs to be accommodated somehow to preserve mass - result the piston pulls a vacuum.

If mass flow required by the volume created in the piston space isn't filled with oil (almost keeping up) then what is it filled with?

Does that make sense?

NO I mean add a horizontal line between your two vertical lines between the cylinder and the controller and insert a flow controller in there with a NRV (flow left to righ). Make that 16 gpm one 10 gpm and make this new one 6 gpm.

The in effect the balance of the oil (6gpm) is simply being transferred from one side of the piston to the other.

So two flow controllers in parallel, but feeding different end points.

See my terrible mark up below.

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

 

[akkamaan]

To those who say negative pressure and cavitation will result - isn't the fluid in the rod end of the cylinder at atmospheric pressure?

As you drew the opening statement schematic noting indicates that the rod side is directly connected to the atmospheric pressure. You are obviously using e positive displacement pump which disconnects (theoretically no internal leakage) the rod side from the atmosphere.

As I said in a previous comment atmospheric ventilation can be done with an anti-cavitation check valve

it should be combined with anti-cavitation check valve

And the pump discharge pressure is still above atmospheric because of pressure drops from the pump to cylinder.

Nope, if cylinder is retracting faster than the pump can fill on the rod side the pump pressure will be negative vs atmospheric pressure.

 

[akkamaan]

See my terrible mark up below.

AS long as we understand what you mean the "mark up" is OK. It's called "effort"

But what do you do with your additional "flow orifice" when you extend the piston?

 

[LittleInch]

There should be a no return valve added to allow flow from left to right only. I ran out of space...

The issue I see about the nrv as shown is that it leaves the return uncontrolled.

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

 

[HydraulicsGuy]

if cylinder is retracting faster than the pump can fill on the rod side the pump pressure will be negative vs atmospheric pressure.

I'll buy the "vacuum pressure in the rod end of the cylinder" argument for a second. Let's say the rod end is at -5 psi gauge, or 9.7 psi absolute. As I said, some components not shown for simplicity. If there are components and plumbing to the tune of 200 psi worth of pressure drops between pump and cylinder, then pump discharge pressure would be 9.7 psi + 200 psi = 209.7 psi, well above atmospheric pressure. Is that not right?

 

[HydraulicsGuy]

If mass flow required by the volume created in the piston space isn't filled with oil (almost keeping up) then what is it filled with?

Good way to think about it. I'm with you. The answer is air, and it's at less than atmospheric pressure because it's being pulled in (flowing) across the rod seals, as hydtools and akkamaan noted originally. The fact that the air is flowing means there's a pressure drop, meaning it's at less than atmospheric pressure once it gets inside the cylinder. It's good to interact with other engineers about these things and think them through.

 

[LittleInch]

Sounds like you've got there.

One issue here is that your 10gpm is being fed by a gear pump which regardless of the pressure downstream will only let 10pm through the pump. Other pumps might let more through if the pressure downstream went negative to inlet pressure into the pump ( probably atmospheric but then we don't have details.

But follow the mass flow. It cannot be created or destroyed!

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

 

[hydtools]

Why connect the rod end to any hydraulic input? The prevailing load is enough to move the piston to retract all the time. Make the setup single acting.

Ted

 

[akkamaan]

The answer is air

Is "air" and "vacuum" the same thing?
If we theoretically do not let any air in on the rod side while expanding the volume on the rod side there is no air. It must be a vacuum.
That vacuum will be in form of uncompressed oil, extracted gas "bubbles" from the oil, or just "nothing". The process that leads to "cavitation".

So I think it is better to give it the term "vacuum" than calling it "air"...

 

[LittleInch]

Yes but when is a "vacuum" not a "vacuum"? - When it is more than 0 psia. With the gases and oil vapour it would be, as hydraulics guy says, "air but at a pressure lower than atmospheric. The only question is how much lower? Could be 100mbara, could be 900 mbara, but it's still "air".

Semantics and different to what most people understand as "vacuum", but otherwise you need to define quite closely what you mean by "vacuum"

Even a search on definition has both of those - "a space entirely devoid of matter" and "a condition well below atmospheric pressure"


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

 

[akkamaan]

If you start with a 100% sealed container filled with 10cui of atmospheric pressure air and expand the volume we will get "vacuumed" air (with its normal contents of the gases of O, N, CO2, etc) we will get "air" under a certain level of vacuum.
BUT
If we start with a container filled with 10cui liquid at atmospheric pressure and expand the volume we will get "liquid" under a certain level of vacuum.
AND
If we start with a container filled with "nothing" at atmospheric pressure ie the volume 0 (zero) and expand the volume we will get a vacuum with "nothing" in it (if you don't extract molecules from the casing material.