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If there are two hydraulic cylinders connected in parallel, each with a different load (shown in the picture below), will they start extending at the same time?

I'm having a disagreement with my tutor, he has told me that the cylinder with least resistance will fully extend, and only once it has fully extended will the other cylinder start to extend. I would have thought that both cylinders will start extending at the same time, and the cylinder with lower resistance will just extend faster. Much like parallel resistors in an electrical circuit.

The reason I think this is because there is still a net force on both cylinders. In the picture below, if the pressure of the system is 50000 Pa and the area of each cylinder is 0.03m^2, then there is a force of force of 1500N on the cylinders. Cylinder 1 has a load of 491N, and cylinder 2 has 981N, so cylinder 1 has a net force of 1009N, and cylinder 2 has 519N. So surely both cylinders will accelerate as there is a force on them?

Another reason I believe the tutor is wrong is that if the masses have very similar weights, say 50kg for cylinder 1 and 50.0000000000001Kg for cylinder 2, would they still only extend one at a time? If that is true then surely it is impossible for them ever to extend at the same time as no two weights will be exactly the same.

The only explanation I can think of to them extending one at a time is that the pressure relief valve (shown in the picture below, just to the left of the switch) plays some part. The pressure relief valve works exactly as the picture makes it look like it would. if the pressure of the fluid coming down the dotted pipe is enough to push the arrow so that it connect the above pipe and the reservoir, then water coming from the pump will flow straight to that reservoir. There is a spring that pushes the arrow to the right, so if that spring takes e.g 2000N to be compressed so that the arrow can connect to the reservoir, then this will prevent the system from ever having a higher pressure than 2000N. Everything else in the circuit is pretty self explanatory so I wont bother explaining what they do.

So please can somebody explain to me, in the diagram below If switch A is pressed down, will both cylinders start extending at the same time?

enter image description here

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  • $\begingroup$ Is the pressure in both cylinders the same? $\endgroup$ – Martin Beckett Apr 18 '14 at 22:09
  • $\begingroup$ Both cylinders are connected to the same pipe, so the pressure underneath each cylinder is the same. There is no fluid above the cylinder. Both cylinders have the same diameter so the same force acts on each cylinder. $\endgroup$ – Blue7 Apr 18 '14 at 22:17
  • $\begingroup$ Then of the same force acts on both cylinders at the same time - then they must both move! $\endgroup$ – Martin Beckett Apr 18 '14 at 22:22
  • $\begingroup$ Exactly what I thought. But as I said in a comment on Ross Millikan's answer, this question was on an exam, and I answered with this answer and got it wrong. I think we must be missing something as I doubt the tutor will have got it wrong. $\endgroup$ – Blue7 Apr 18 '14 at 22:27
  • $\begingroup$ I think I've just found the answer. Think about the pressure as a step input, from 0Pa to 50000Pa. This is the correct way to think about it because if you assume the system is already at 50000Pa when you start analysing it, then the cylinders must already be extended. So as the system steps from 0Pa to 50000Pa, the force on the cylinders steps from 0N to 1500N. However as soon as the force becomes larger than 491N the first cylinder will start extending, so the force can no longer carry on increasing. Only once the first cylinder has extended can the force carry on increasing to 981N. $\endgroup$ – Blue7 Apr 18 '14 at 22:33
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Initially, I agreed with Olaf Chujko's answer to this question; however, on further reflection, I think the most accurate answer is 'it depends':

Firstly, from the schematic that is given, when switch A is pressed, the cylinder volumes above the two cylinders will be vented to a reservoir at ambient pressure (trust me, I work in Oil & Gas and I look at schematics like this every day!).

The answer to the question will depend on the particular setup of the apparatus, specifically, it's a balance between how quickly the fluid can flow into the cylinder chambers (i.e. how restrictive the tubes/hoses are) and how fast the pistons can move.

The answer that Olaf gives is the situation where the pipes are very restrictive and the pistons have no friction: flow into the cylinders is small compared to the piston movement. Here, the cylinder pressure will increase relatively slowly. Once the pressure required to lift the smaller piston ($p_1$) is reached, that piston will start to lift and, as more fluid flows into the cylinder, the pressure will stay the same until the smaller piston has been completely lifted. The pressure will then continue to increase until the pressure required to lift the larger piston ($p_2$) is reached. Therefore, the larger piston will not start to move until the smaller one is fully lifted.

(btw, this is all assuming the piston diameters are equal).

The answer that Floris and Ross Millikan are supporting represents the other extreme, where the fluid will flow into the cylinders much more quickly than the pistons can move. For example, consider a high-pressure source with large, relatively smooth hoses and the case where the pistons have a large amount of friction, so their maximum speed is very limited. Here, the pressure in both cylinders will increase to be equal to the supply pressure very quickly, so (assuming $p_s>p_2$) both pistons will lift simultaneously. However, the smaller piston will always start to move a short time before the larger one, because in a real fluid system pressure has to increase at a finite rate - you can't have a step change.

tldr: either answer could be correct, depending on the exact setup of the system.

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Based on your comments (that the exam question said "switch A is pressed"), the question can be answered - and the tutor was correct. The key is to look closely at the diagram, and observe that the lower halves of the compartments are connected together, as are the upper halves.

enter image description here

In this diagram, $p_1$ represents the driving pressure. Now across the piston on the left there is a pressure drop equal to the weight of the 50 kg mass divided by the area of the piston - let's call the pressure above the piston $p_2$. On the right, the same thing would lead to a pressure $p_3<p_2$ if the 100 kg object was getting lifted. But this pressure differential cannot exist while the two compartments are connected, so the right hand weight stays at the bottom - where it exerts a force of $50 g\ N$. This restores the balance of pressure and force.

Once the 50 kg weight reaches the top, the pressure against the end stop will result in a force of $50 g\ N$, and as that force increases, the force of the piston against the bottom of the right hand piston decreases until it, in turn, is lifted up.

All this assumes that the pressures at the top of both pistons are equalized by the tube joining them. It is hard to know from the diagram whether that is the case.

If the exit pipe represents essentially no resistance to fluid flow, then $p_2=p_3=1 atm$, and both pistons could be lifted at the same time.

However I'm pretty sure, given the way the diagram above is drawn, that this is not the case, and that my explanation above is the one your tutor had in mind.

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  • $\begingroup$ As drawn both the bottom and top pipes go to the same reservoir so nothing would move at all - I'm not sure you can trust the diagram $\endgroup$ – Martin Beckett Apr 19 '14 at 2:55
  • $\begingroup$ @MartinBeckett no, when A is pressed down, the inlet pressure goes to the bottom of the piston (through the pipe with the right arrow) and the return pressure goes to the outside (pipe with the left arrow). At least that's how I interpret the drawing - it is a four way, three position valve. The center position ("neutral") is the one where the two pipes connect to the same reservoir. But in position A, pistons are driven up - and in position B they are driven down. Maybe I should have copied that bit in my sketch? $\endgroup$ – Floris Apr 19 '14 at 3:34
  • $\begingroup$ Yeah that is how the drawing is meant to be interpreted. The switch is pushed all the way down so the right arrow connects the pump and the bottom of the pistons, and the left arrow connects the return and reservoir. The fluid in the reservoir can then be pumped around the circuit again. I'm struggling a bit understanding your answer, but thats probably because it 4:20 in the morning, i'll get some sleep then read it again in the morning. I just want to make something in the diagram clear first: Fluid only can return through the return pipe once the piston is fully extended. $\endgroup$ – Blue7 Apr 19 '14 at 3:42
  • $\begingroup$ @blue7 I suspect that there is fluid on both sides of the piston since position B of the switch would drive them the other way. Thus the fluid above has to "go somewhere" even before the piston reaches the top. If anything I think the piston should stop before that point is reached (or it stops when fluid can escape). $\endgroup$ – Floris Apr 19 '14 at 11:54
  • $\begingroup$ @Floris Ah good point, I never thought of it like that, I always assumed there was no fluid above the piston, and the return pipe was just so the fluid under the piston can escape once the piston is high enough. I can see I was wrong now. Thanks for your answer! $\endgroup$ – Blue7 Apr 19 '14 at 17:23
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Probably you have your question answered already, however, let me point out that:
You are incorrect.
You can't think of a hydrostatic system as it was an electric circuit. In an electric circuit what you (or the source) supply(ies) is the voltage and the result of that voltage acting upon the resistor of given resistance is the current. In hydrostatic systems what you (the source) supply(ies) is the fluid flow, not the pressure. If the hydrostatic pump is rated, say 200 bar what it really means that it can work with pressure (pressure difference really) up to 200 bar, it does not in any way mean that it supplies 200 bar of pressure.
As I said, what the pump supplies is flow, not pressure. If you had no cylinders in the system and the fluid would run directly back to the tank, the pressure would be, theoretically, 0 bar.

How it works is:
1. The pump pumps fluid into the system.
2. The flow is restricted due to the loads on the pistons.
3. Pressure rises up to the level required to move the piston with lower load up.
4. The first cylinder moves up.
5. The pressure is not enough to move the other piston, but it is not necessary to move at this point, because the first cylinder consumes all the flow that the pump provides.
6. The first cylinder reaches the top and it can move no more.
7. The fluid is still pumped into the system. The flow is restricted, so the pressure rises up to the leved required to move the cylinder with the bigger load.
8. The other cylinder moves up.
9. The other cylinder reaches the top
10. The pump still keeps pumping the fluid.
11. The flow is restricted, the pressure rises even higher up to the point at which...
12. The relief valve opens. The flow is redirected directly into the tank at this point.

In real world it works a bit different. There is no "0 bar" pressure. Everything is rescticting the flow, every piece of pipe or hose, every valve, fitting, etc creates what is called "pressure drop". There is also leakage. There is no cylinder, that will not let the fluid under pressure move from one chamber to another, it will always leak, the better the seal, the slower it will leak, of course, but you can never avoid that. There is also leakage in the pump. And also, the greater the pressure in the hydraulic system, the bigger the strain that is put on the motor that drives the pump and the RPM lowers (and since the flow supplied by the pump is directly proportional to its RPM, the flow lowers as well).

Hope that helped.

P.S. This probably will not help you at all, but if you wanted the pistons to move at once, you'd need what is called a proportional valve with load sensing function.

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  • $\begingroup$ I would like to discuss it with you guys, but I am afraid my reputation is too low to comment on other answers. Sorry. $\endgroup$ – Olaf Chujko May 1 '15 at 12:47
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You are correct. Each cylinder has a hydraulic force applied at the bottom that is sufficient to accelerate its mass upward. The one with the lighter mass will accelerate more rapidly and reach the top of travel more quickly, but the other will already be moving.

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  • $\begingroup$ Thanks for your answer. This exact question, with very similar values (I just rounded some of the numbers for this question), was asked on an exam that counts towards my university degree, and I answered with this answer but apparently it is wrong. I've just emailed the tutor, but I somehow doubt he would be wrong considering he is the tutor. Are you sure we are not missing something? $\endgroup$ – Blue7 Apr 18 '14 at 22:24
  • $\begingroup$ Your logic is fine. Each cylinder doesn't know the other is there. It has pressure (assumed fixed, not reduced by flow rate) at its input, mass on top making a downward force, and you do a force balance as you have done. If the force is net upwards it will move. I'll be interested in your tutor's explanation. $\endgroup$ – Ross Millikan Apr 18 '14 at 22:31
  • $\begingroup$ See my comment under the question for a possible explanation. I'll add my tutors explanation as an answer once he replies. $\endgroup$ – Blue7 Apr 18 '14 at 22:35
  • $\begingroup$ That is a reasonable explanation for one starting before the other, though if it is not specified that the pressure rises in the problem one assumes it is a step function. It still doesn't explain the second cylinder waiting for the first to finish. The second should start to rise as soon as the pressure is high enough. Of course, if the pressure rises slowly enough, the first might be done by the time the second starts. $\endgroup$ – Ross Millikan Apr 18 '14 at 22:38
  • $\begingroup$ If the first cylinder is rising then the pressure cannot increase, as any added fluid into the pipes below the cylinder is counteracted by an increase in volume due to the fact the cylinder is extending. So the new fluid will just take the space of the new volume (so no increase in pressure). Also the question actually asked what happens when switch A is pushed down. I thought this was the same question as I originally asked but it's not. Sorry, I shouldn't have said "This exact question" was asked on the exam. $\endgroup$ – Blue7 Apr 18 '14 at 22:46
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piston with lesser resistance will move first. that is correct. because these two pistons are connected to same pressure source and pressure will build gradually and not spontaneous. i.e. the pressures inside both cylinders gradually increase from 0 to a value where less resistance piston starts move. once this moves pressure will not increase further until that piston reaches dead end. because if there is no resistance, there is no pressure.

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