Regen Calculation Formula

[fangas]

Hi,

I'm in need of a formula to calculate the speed of a hydraulic cylinder during regenerative extension. It's probably something as simple as adding the rod end volume to the delivery rate of the pump. However, I'm unsure due to the apparent paradox of the speed being partially dependant on the flow rate from the rod end, which, is itself, dependent on the speed of extension.

The system in question is 6" X 5.5" X 18" stroke w/ a 17 GPM pump.

Also, it appears that a 1:2 ratio of rod end to cap end volume would be optimum. Is this so? If not, what is?

Thanks
Ed

 

[hydromech]

Fangas/Ed

I have written an Excel spreadsheet to help work out just what you need. Trouble is that its in metric units and I have no idea how to get it to you.

A 2:1 volumetric ratio between the bore and the annulus( the bit with the rod)would be ideal. Unfortunately it is very rare that this is the case as there are many features on a cylinder that can effect the internal volumes. Stroke, cushions and stop tubes to name a few.

With your cylinder the bore volume is 509 cubic inches (2.2 U.S Gal). With a pump delivery of 17 GPM the cylinder will extend in 7.77 seconds.

The volume of the cylinder annulus is 427.7 cubic inches (1.85 U.S Gal).

We know that the cylinder will be fully stroked in 7.77 seconds. In 7.77 seconds the cylinder will displace 1.85 gallons. That equals 14.28 gallons per minute.

The volumetric ratio is 1.2:1. 509 cubic inches in the bore. 427.7 cubic inches in the annulus.

Obviously the cylinder will retract quicker because of the lower volume in the annulus. But the cylinder will displace 20% more oil. If you put 427 cubic inches into the annulus you will get 509 cubic inches out of the bore.

The volumes in the cylinder are fixed and the speed of operation of the cylinder depends on the displacement of the pump. The more flow the faster the cylinder will move.

I think these volumes are right...i'm not used to working in imperial units.

Hope it helps.

Hydromech

 

[Icanfix]

If you are feeding back all the displaced oil from the front end of the cylinder it seems simple to me. Subtract the area of the front end from the area of the back of the cylinder. That gives you how much oil is needed to move the cylinder forward for each unit of stroke after taking into account the oil that is displaced from the front of the cylinder.

From the example hydromech. 2.2 gallons - 1.85 gallons = .35 gallons. That is what is needed to stroke out the cylinder. At 17 gpm it would take 1.23 seconds.

Or am I missing something really basic?

 

[kcj]

Icanfix is correct, use the rod area only for the calcs. The oil coming back from the rod side annular space goes into closed end, and exactly balances out that area, leaving only the rod area.

To stir up this topic, a cylinder does not always retract faster than extend.

When the circuit is FLOW limited, by pump, it will retract faster (with less force) because there is less area on that side, Flow divided by area is speed. This situation applies when the valve is large in relation to flow, and pressure drops in the valve are low (5 or 10% of load pressure.) This is typical of bang/bang or manual spool valves running a cylinder, say the common log splitter or backhoe. It is also what most people are familiar with, and why they so vehemently disagree with the next statement.

When a circuit is PRESSURE limited, it will retract slower than it extends. This situation applies in servo and proportional valve circuits where valve pressure drop is high in relation to load pressure (say 25 to 50% of load pressure.) This is done by intent, and is how servos obtain control of the circuit. Essentially, the pressure in is reduced through the valve inlet, and applied to a smaller area on the rod side. Backpressure fom flow out of the valve is applied to the much larger closed side of piston. The resulting force balance means the cylinder retracts SLOWER than it extends.

Whether a circuit is pressure or flow limited depends on valve sizing, load, cylinder, etc. It is easy enough to calculate, but beyond the scope of a quick response.

This topic has been discussed at length at:

http://patchn.com/forum1/viewtopic.php?t=353&postdays=0&postorder=asc&start=50

and

http://admin.ifps.org/phpBB2/viewforum.php?f=2&sid=d11a2fdd687b8e3e2fefad60578c97fd

but not resolved to understanding in either case. I am working on a larger explanation to post ‘some day’, but this will at least make readers aware of the issues.


k

 

[fangas]

Hi,

Thank you all for the input.

I think I got it.

Ed

 

[danhelgerson]

I know I am a little late jumping in here but I hope this will be helpful. A simple but accurate formula for cylinder speed in Imperial units is Q=.204 Vdsqr where Q= gpm, V=velocity in inches per second and d = the diameter of the piston. In your application the formula would be Q/.204dsqr = V and use the rod diameter as the piston size.