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So, I have 72 hydraulic McKibben actuators with 30cm in length and 14cm in width under 400 kPa (or 4 bar) that lifts a 5 ton weight that I want to actuate in a third of a second (continuously, actuating and un-actuating).


The explanation below is about McKibben actuators, if you have no clue about how these work, just imagine that these are hydraulic cylinders that work at 4 bar, have 30cm in length and 40cm of diameter and just move by 6cm.


McKibben muscles normally contract 20% of its length and increase 30-40% of its diameter.

So this means that a single one of these muscles have an uncontracted volume of 4.618 liters inside of it.

Taking into consideration the change of length in 20% and diameter in 40%, the contracted McKibben muscle would have a volume of 6.244 liters inside of it.

The difference between these two would be 1.626 liters, which means that if I wanted to actuate all of the muscles in a third of a second, I would need 578.88 liters per second, so 34732,8 liters per minute.


Most commercial hydraulic pumps have a certain limit of RPM, liters per minute and pressure. I don't think that even if the torque of the pump motor was super low, it would be able to pump as much fluid as this.

So, regardless of making a custom pump or not, I believe that this is a really inefficient way of actuating these.

One would need a lot of energy to pump 34000 liters of hydraulic oil, and it would also take a lot of energy to actuate cylinders that work around 12000 PSI, but need a few ml.


So, the question:

Is there a way of achieving some kind of ideal balance between working pressure to fluid flow? Or I will need to test/calculate option to option until I find a certain balance?

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  • $\begingroup$ There are pumps that deliver 1 litre per revolution and can do more than 3500 rpm. Can't remember the max pressure offhand but that is just simple research. $\endgroup$
    – Solar Mike
    Feb 22 at 20:53
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    $\begingroup$ @Fulano if the actutation is continuos then a pump with the flowrate for all actutators would be called for, if I understand you corretly this is not the case - can you explain? $\endgroup$
    – mart
    Feb 23 at 14:49
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    $\begingroup$ Alternativly, use pneumatic actuators, storing of compressed air is a bit easier $\endgroup$
    – mart
    Feb 23 at 14:51
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    $\begingroup$ ideal gas law, approximately. 34k l/min at 4 are roundabout 140k l/min at normal pressure. to provide air at 4 bar, you probably need at least 5 bar in the tank. $\endgroup$
    – mart
    Feb 24 at 8:31
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    $\begingroup$ That said, something is missing from your question. talk to a supplier of hydraulic power supply, normally a sales engineer will help you. Same for compressed air. $\endgroup$
    – mart
    Feb 24 at 8:36

1 Answer 1

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What you are describing is referred to in dynamical systems analysis as an impedance matching problem.

For optimum transfer of power (effort x flow), the impedance of the source must equal the impedance of the load.

See Karnopp & Rosenberg, System Dynamics: A Unified Approach to learn how to solve for the impedances.

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  • $\begingroup$ I found this online version of the book, and searched for the word "impedance", it showed only results related to radio and electronic systems, not hydraulics. Link: archive.org/details/systemdynamicsun0000karn/page/290/mode/… $\endgroup$
    – Fulano
    Feb 23 at 12:56
  • $\begingroup$ impedance is the ratio of effort to flow in any dynamical system. Karnopp provides the tools to represent hydraulic systems as if they were electrical circuits. Have a look at his treatment of hydraulics. I'll see if I can find any other references which might be helpful. $\endgroup$ Feb 23 at 17:26
  • $\begingroup$ See "hydraulic analogy" on wikipedia. hydraulic impedance is system pressure divided by volume flow rate. $\endgroup$ Feb 23 at 17:42

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