Volume Flow Rates in and out of Double Acting Cylinder

[Munro85]

Hi there,

My mind is completely scrambled on this one and really need some help getting to the bottom of it.

If you have a double acting cylinder which is connected to a fixed displacement pump, say 20 litre/min, will the 20 litres/min be the same entering (rod end/bore end) as exiting (bore end/rod end)?

My instinct is that with the conservation of mass, the flow rates should be the same. But doing a very simple sum where the area of the bore is double the area at the rod end, if pump, say 10 litres into the bore end and work out the short distance the cylinder moves in a given time, then the volume displaced on the rod end where area is half, the volume displaced by that cylinder stroke is only 5 litres.

Can someone please shed some light onto which is correct, and if the volume flow rates are the same, what factor am I not taking into consideration with my simple sum?

Thanks.

 

[LittleInch]

My instinct also was of course its the same, but its not.

The difference between the "flat end" being pressurised and the end with the piston in it is that one will generate more force as there is more area than the other, but flow per unit length is different on both sides.

So you can't simply run one side into the other side of the pump. You will need to go via a reservoir.

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

 

[Stick.]

The volume flow rates would only be the same if you had a double rod cylinder (with equal rod diameters). For a single rod cylinder, the flow rates will be related by the area ratio. As a thought experiment, consider what happens as the diameter of the rod approaches the diameter of the bore and oil flows into the cap end of the cylinder. When the rod diameter increases to the point that it equals the cylinder bore, you end up with a single acting cylinder, and zero oil will leave out of the cap end port. So in that case, you'd have 20 l/min going into the cap end, but 0 l/min leaving the rod end.

To LittleInch's point about reservoirs, if you watch a system operate where the cylinder is large enough in relation the the tank size, the oil level in the tank will go up and down very noticeably as the cylinder retracts and extends.

 

[Compositepro]

If the flow rate is constant, then the retraction velocity of the cylinder will be twice the extension velocity of the cylinder.

 

[akkamaan]

If you have a double action DIFFERENTIAL cylinder and supply 10 L/min to the capped side it will come out 10 L/min from the rod side. But the 10 L/min from the rod side is the sum of oil volume and rod volume. Accordingly, if we reverse the direction of travel and supply 10 L/min to the rod side PLUS the rod, 10 L/min plus the volume of the rod will come out as flow from the capped side.
On a differential cylinder, the size of the rod is the crucial point. Look at both sides of the pistons as displacement volumes. The piston rod enters and exits the hydraulic system and the rod volume/volumes have to be accounted for.

Mobile Hydraulics

 

[Munro85]

Thanks for your responses everyone.

So what I'm getting with regards to my confusion with the conservation of mass, the fact that the reservoir essentially makes the system appear to be more of an open loop system rather than being fed back into the pump which would make it a closed loop. I realise I may be missing the ball with my understanding of the conservation of mass and how and when it applies in certain scenarios and when it doesn't. I do realise that the flow entering a pump doesn't necessarily have to match the flow exiting (or does it).

I understanding I'm perhaps veering from my initial question which I can at least accept, the idea of why the flow rates are different - and that is quite obviously attributed to the same stroke translating to either side of the pistons but due to the different volumes of oil being displaced results in different flow rates exiting one chamber compared to that entering the other.

Thanks again.

 

[LittleInch]

Liquid flow into a PD pump needs to equal liquid flow out.

Not enough going in - creates vapour to make up the difference.
More going out than going in - you're creating mass which isn't possible.... Or you're Jesus re-incarnated.

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

 

[Compositepro]

Your system requires an oil reservoir. The volume of oil in the cylinder is not constant. It contains more oil extended than retracted.

 

[zeusfaber]

Double acting cylinders are full of surprises:

Oil is stored in a double acting cylinder as you extend it. The total volume behind the rod seal remains constant but, as you extend the actuator, less of that volume is occupied by rod so more of it has to be occupied by oil. When you pump oil into the annulus to retract the cylinder, the net effect is to reduce the total amount of oil stored in the cylinder (replacing it with rod), so oil has to come out faster than you put in.

If you feed a double acting cylinder with a constant flow pump, it will retract twice as fast as it extends, but will be capable of exerting only half the force on the retract stroke for the same pressure in the system (assuming annulus area is half of full bore).

If you feed oil at pressure into the bore and restrict the outlet from the annulus, you can see a doubling of pressure at the annulus port.

If you have a decent rod seal, a double acting cylinder will hold a load surprisingly effectively even if the piston seal is completely shot.

If you want double acting cylinders to work at the same speed in each direction for the same oil flow, consider connecting a pair of them working in opposition, with the bore of each paralleled hydraulically with the annulus of the other. In this configuration, you won't benefit from the same load holding if you lose a piston seal.

 

[Jacc]

Hoses dimensioned for the flow going into the cylinder on the cap side is a common design mistake.
The flow going out is obviously much larger and this causes excessive flow speeds in hoses and extra back pressure and heat generation.

 

[hydtools]

Start with G= V×A. Q is the pump output. A is the effective piston area. V is the piston velocity.
Make Ah the head end piston area and Ar the rod end piston area. The extending velocity would be Q/Ah = Vext. The flow out of the rod end would be Vext × Ar = (Q/Ah) × Ar = Q × (Ar/Ah) = Qr.
The retraction velocity would be Q/Ar = Vr. The flow out of the head end would be Vr × Ah = (Q/Ar) × Ah = Q × (Ah/Ar) = Qh.

Ted

 

[akkamaan]

This flow intensification/FI Jacc mentions happens on differential cylinders capped side when retracting the cylinder. That means that we get x% higher flow returning from the capped side than the pump flow entering the rod side.
A normally designed hydraulic differential cylinder has a capped side vs. rod side piston area ratio of 1:0.75 (or 1.33:1) which means the rod diameter is 50% of the cylinder diameter. So when retracting the cylinder the meter-out flow from the capped side will be 33% higher than the meter-in flow. Accordingly, when extending the cylinder the meter-out flow from the rod side will be 25% less than the meter-in flow.

If using a regenerative function on a 1:0.75((1.33:1) cylinder during extension the meter-in flow will be 100%+75%=175% of the pump flow or almost double the pump flow.
With a 2:1 piston area ratio we can make a differential cylinder travel with the same speed (at least theoretically) when both extending and retracting. In such an application the hydraulic lines on the capped side need to be 75% larger than the rod side (which only handles the pump flow.

I always think about the piston rod on a differential cylinder as a part of the flow. To the meter-in flow from the pump flow on the capped or rod side, the "steel" flow from the piston rod can be added to make the rod side flow equal to the flow on the capped side. The "piston rod flow" has an equal effect on the oil volume (level) in the reservoir.

On the flip side of the "flow intensification coin", we have pressure intensification/PI that can cause even bigger problems than FI, but that's another story.


Mobile Hydraulics