Cross-connected cylinder pressures

I can't see a practical purpose for the two cylinders. Someone is likely to think that your post is a student homework post and red flag it for deletion. If it isn't a student post it would be useful for you to explain what you are trying to accomplish.

In case it doesn't get deleted ... without the applied force, the only pressure source is the weight of the pistons which are the same in the two cylinders so the pistons should both fall to the bottom of the cylinder.

I'm assuming the right-hand piston is not constrained from up and down movement.

Applying the blue force downward on the left will increase the pressure below the piston and reduce the pressure above the piston. This causes the pressure above the right-hand piston to increase while the pressure below the right-hand piston to decrease--lowering the right-hand piston at about the same speed as the left-hand piston falls. Of course you cannot reach equilibrium in this system since there is no way to balance the applied load, so the left hand piston will bottom out.

Pressures are indeterminant until the piston bottoms out.

David

Thanks for taking the time to answer David. It's not a homework or assignment question, the numbers I gave are just as an example.

I understand what you are saying but I don't believe it's that simple; I'll try and explain the reasons for my confusion again..

The rods and pistons cannot both move down (into the cylinders) because there is a fixed volume of fluid in the system that cannot escape from the fixed volumes of the cylinders (and pipes). By pushing down on the rods, you are trying to add in the volume of the metal rods. The fluid is incompressible so this cannot work. Likewise if you try to extract the two rods from the volume, without the fluid becoming less dense and expanding this cannot work either.

The way I imagine applying a force to one side is that the large volume of fluid from the lower left chamber is trying to occupy the smaller top right chamber (as they are connected). This will mean the stroke of the right rod would consequently need to be larger than the left rod. Except when the right piston moves down, the lower right chamber volume must occupy the smaller top left chamber volume and essentially cause the left piston to move down more than the right rod - and more than it actually can. This can't physically happen which is why the system must be locked, the rods can't move independent of each other or together.

So when a force is applied to one rod, there is no movement, but there is still a pressure applied to the fluid inside as it is the fluid trying to compress (or expand?) that is stopping the pistons moving. I'm trying to calculate this pressure, or pressures, as there are two circuits.

Hopefully this makes more sense, what do you think?

You are correct, because of the volume differences the pistons will not move.
For pressure, start with applied force, F1, divided by the piston area. That pressure will be the same at the rod end of the righthand cylinder. That pressure times the cylinder area minus rod area will be the force, F2, applied to the piston. The pressure reacting over the piston area will be F2 divided by the piston area. That pressure will be the same at the rod end of the lefthand cylinder.

Ted

Let's take it slow. If the pipes between the cylinders did not exist, what whould happen? The force would raise the pressure in the left-hand bottom while lowering the pressure in the left hand top (no appreciable movement because the fluid is incompressible on this scale).

Now add the pipes back in.

The top of the left hand is sucking on the bottom of the right, tending to pull the right-hand piston down. The bottom of the left is pushing on the top of the right trying to push the right-hand piston down.

I don't know how fast the pistons would fall, but I'm pretty sure that with the unbalanced forces the left hand piston will go to the bottom stop. The right hand pistion will also go down until the unbalanced force is balanced (against the stop), but I'm not sure if it ends up on the bottom or not.

I see the right-hand piston as a force multipier, with all the forces in the same direction. I can't make the math work for locking it up.

David

David, thanks again for the reply, I'm not quite sure if you understood my description though.

Ted, this is how I thought it would be worked out. What confused me were the resultant forces created by these two different pressures on the pistons. I've included the calculations in the new attached drawing below, giving 1MPa in the red circuit and 0.8MPa in the blue (according to my initial conditions specified in post 1).

As far as I can tell, the reason the pistons do not move is because the resultant force of both pressures acting above and below each piston amounts to zero. For the right piston, 800N against 800N cancels out, but the left piston has three forces that don't seem to equalise at all. Which forces should I be considering to act on the left piston which balance to ensure there is no resultant?

The cylinders are locked. As drawn, the the volume of red fluid and the volume of blue fluid and the total volume of both fluids cannot change. There is only one position of both pistons where this can be true. If either or both pistons move, these volumes must change.

If the blue and red were connected then only the total volume would have to be constant and the piston rods would act as the pistons.

If we assume that the fluid is completely incompressible then the pistons cannot move - but it would be wrong to think that they cannot transmit any force simply because they cannot move.

Let's call the force F. If we call the full bore area of the cylinder A1 and the pressure in the full bore of the left hand cylinder P1 (the red oil), and if we call the cylinder annulus area A2 and the pressure in the annulus of the left hand cylinder P2 (the blue oil), then for the left hand cylinder we can say:

Equation 1: (P1 x A1) - (P2 x A2) = F

For the right hand cylinder we will have P1 in the annulus and P2 in the full bore but no force on the cylinder rod so we can say:

Equation 2: (P2 x A1) = (P1 x A2)

We can rearrange this equation to give:

Equation 3: P2 = P1 x A2/A1

Then we can put this result back into equation 1 to give us:

Equation 4: (P1 x A1) - (P1 x A2²/A1) = F

Then gather everything together and you will get:

Equation 5: P1 = F/(A1 - A2²/A1)

And once you have calculated P1 put that answer back into equation 3 to get P2.

With the numbers given as an example I reckon the red oil pressure will be 27.8 bar and the blue oil pressure will be 22.2 bar.

DOL

And there’s more…

It’s not fair to assume that the oil is incompressible. The oil will have a typical compressibility of 1/2% per 70 bar and this means the pistons have to move in order for the pressure of the oil volumes to change.

I think the calculation goes like this: Let’s assume that we start off with zero pressure in both of the cylinders, both pistons are at exactly half stroke, the stroke of each cylinder is 100 mm and the volume of oil in the interconnecting pipework is negligible.

The volume of the red oil will be:

[1000 mm² x 50 mm] + [800 mm² x 50 mm] = 90,000 mm³

The volume of the blue oil will be the same.

When the red oil is pressurised to 27.8 bar it will shrink by 27.8 seventieths of ½%. The exact value of this shrinkage will be:

27.8/70 x 0.5/100 x 90,000 = 178.7 mm³ so the new volume of the red oil will be 89,821.3 mm³

Similarly the blue oil in rising to 22.2 bar will shrink by:

22.2/70 x 0.5/100 x 90,000 = 142.7 mm³ and the new volume of the blue oil will be 89,857.3 mm³.

The total oil in the system was 180,000 mm³ (when it was unloaded) but this now has to shrink by the sum of the two individual shrinkages, i.e., by 321.4 mm³. One or other (or both) of the cylinder rods are going to have to retract a little to accommodate the shrinkage in the oil volumes – but we don’t yet know which one will move or, if both, by how much each one will retract.

The area of the cylinder rod is 200 mm² so the ‘total’ retraction of rods will be 321.4/200 = 1.607 mm.

Let’s say the left hand cylinder rod retracts a distance R mm, then the right hand cylinder rod will have to retract a distance of {1.607 – R} mm. The new volume of the red oil is 89,821.3 mm³ and the new configuration of rod positions has to match that, so we can say:

[1000 mm² x (50 – R) mm] + [800 mm² x (50 + {1.607 – R}) mm] = 89,821.3 mm³

This isn’t the place to go through the algebra, but when you reach the end you will get R = 0.814 mm. The left hand cylinder rod will retract 0.814 mm and the right hand cylinder rod will retract the remaining 0.793 mm.

If the volumes were different (different cylinder stroke, non-zero volume in the cylinder interconnections) or if the fluid bulk moduli were different and/or if there were hoses between the cylinders, or if you wanted to take friction into account, or the dilation of the cylinder bodies or anything else under the sun then the result would be a little different. I believe, however, the principle of the calculation would be the same.

Incidentally, it’s not quite right that the pistons cannot move. True enough for pushing: if you could push in the rod on the LHS cylinder then part of the full bore oil fill of the LHS cylinder would try to escape into the annulus of the RHS cylinder. This would push down the RHS piston a greater distance than the movement of the LHS piston (because of the ratio of areas of the LHS cylinder full bore and the RHS cylinder annulus). An even greater volume of oil would want to exit the RHS full bore (because of the ratio of the areas of the RHS annulus and the RHS full bore) and this twice amplified volume would need to get into the annulus of the LHS cylinder where there just isn’t room for it. So the pressure goes up in the full bore of the RHS cylinder – this prevents the RHS piston from moving and this prevents the LHS piston from moving.

But you could still pull out one (or both) of the rods. You would just cavitate the oil fill.

DOL

I think Oldhydroman has it right but I would use a system of differential equations to provide. The system if differential equations may be over kill but I can simulate a system like this easy enough.

BTW, I have seen a similar system but it had double rodded cylinders. The two cylinders were used to control a crusher roll of a huge rock crusher that could crush cars and small trucks easily. However if a big rock went through the rock crusher the two cylinders would retract in synchronously and avoid binding.



Peter Nachtwey
Delta Computer Systems