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I have a design that would solve a hindering problem if I could simply charge one end of an actuator with about 200 psi of oil and not tie it into a circulating system.  My question is would it heat up too much in doing this?  The other end of the actuator is going to fluctuate from about 200 psi to 1200 psi...so it would spike the trapped end to these pressures as well...in some cases it will see this fluctuation 120 times per second.  How much heat would this generate...and what affect would it have on the piston that separates the two chambers?  Can it be done?


First, how do you intend to trap a volume of oil at 200psi but not allow it to communicate with the rest of the system? If it is trapped in the cylinder it can not move. If it is not trapped oil will be exchanged --- unless it is tied into an accumulator.

Unless there is some type of thermal path besides the oil I would have to say heat will be a problem. Most hydraulic seals are only good to ~225 deg F. It's also hard on the oil.

I can think of two heat sources.
1. fluid compression - I haven't had a fluid dynamics class but I would think that it will heat when compressed but cool equally when allowed to expand - so it should be a net gain of 0.
2. Friction - This will add heat, which must be dealt with.



Thanks ISZ...I was hoping that the one end would act more along the lines of a shock.  The compression of the fluid I wasn't too concerned with but like you stated I was wondering if the friction would be an issue...and it probably would be.  Even if I had a way of cooling the trapped end, I think you are right about it still being to hard on the oil.  Thank you for your input...it is back to the drawing board.


Hydraulic fluid is nearly non-compressible at the pressures most systems work at - therefore you can't use trapped oil as a "spring." You can however compress air.

Also, the seals in almost all hydraulic cylinders leak - its just to what degree. Even if you could pressurize one side of the cylinder it would eventually loose pressure and have to be recharged somehow. The leakage would get worse at higher temps.

Some ideas
1) If the one end is dry - You might be able to apply pressurized air. (Like shop air at 90psi) Might be an explosion hazard though if it gets mixed with an oil mist.
2) Install an orifice on the back side to restrict the flow, which will generate the pressure. You could use a "tee" and check valve to limit the restriction to one direction only if needed. This would not be very effective at low speeds or displacements.



I don't see where you should be concerned about heat. If your using trapped oil, there will be no flow, subsequently no power dissipation.

Trapped oil in itself won't do anything for you, other than take up space. If it's shock absorbtion, then a traditional accumulator will work, these typically being charged with nitrogen.

Your needs sound specialized however in that a 120hz pressure wave requires attenuation. Flexible hosing or coiled steel tubing should buffer this.


My concern for the heat strictly has to do with the seals and what the oil can tolerate.  This may sound a bit out of the ordinary but the more heat that could be generated without destroying the seals and oil, the better.  The one end of the cylinder is "charged" at system start up but is then disconnected from the circut after this point...there is a way for this end to maintain a constant 200psi, however, and the fluid is able to be displaced when subjected to the 1200 psi to keep the constant pressure of 200.  There is no traditional pump so the only heat that would be inducted in the system would be from friction involving the piston, drag in the lines (although there is no hose...all metal tubing), and across the valves/orifice plates, etc.


well, I'd have to say at this point that I have absolutely no idea what your talking about. Sorry


My thoughts...

A volume of oil trapped will absorb little or no shock. The bulk modulus of oil is far less than that of air, it will compress slightly, however, at such low pressures the compression will be almost zero.

Energy that is stored in oil will be transfered to heat if it is not used. When compressing the oil, if it is done isothermally, i,e slowly, there will be almost no heat increase. When pressurised 120 time per second, the oil may tend to warm up. However, the amount of energy required to compress the oil could be very low. By the way, what is the volume of the actuator? The size of the actuator makes a big difference to the oil temperature. Specific heat capacity of oil...the more oil the more energy required to raise the temperature. The bigger the surface area of the actuator, the more heat it will disipate.

If the "other" end of the actuator is to be charged by a pump you may find that there is an inherent ripple that is transmitted from the pump. Almost all hydraulic pumps have a ripple of some sort. In most hydraulic systems the ripple is absorbed by the components. In this application there would be no way to absorb the ripple. Depending on the rotational speed and the type of pump, the frequncy of the ripple could be as high a 500Hz. If the pump is working upto 1200 PSI, that is the pressure that will be seen in the actuator.

As I amagine the system to work, there will no oil flowing at the actuator and the volume of oil is such the low power required to raise the pressure will limit the heat input.

There are many other hidden factors that can effect that heat input, can you provide more details about the pump and the pressure controls etc...?


Adrianwinky smile



Thanks for the great info, Adrian.  At the moment, the actuator's volume is a rough design at 9.75in3...surface area of 3.25in2.  There is no rotary pump in the system at all...so no ripple and no heat inducted.  After the initial startup, there are no valves, etc. involved with the hydraulic portion of the design.  What I should include is that the actuator side that maintains the constant 1200psi is open at the one end (not fed by valves) and that there is an identical actuator attached directly across from it with only a few inches of tubing connecting the two.  However, the tubing has an identical ID of the two cylinders ID.  The oil sort of yo-yo's, if you will, back and forth between these two.  The "trapped" oil portion of 200 psi that I was talking about, there is one each of these at both ends of the 1200psi actuators.  As weight of the machine shifts back and forth, it applies force to the the 200psi section bumping it up to 1200psi, which in theory will apply this 1200 to the middle section of actuators that should maintain this 1200.  The other end, however will be at 200psi (due to some weight still applied to this end) and the 1200psi will travel this direction due to the weaker pressure and in turn shift the weight of the machine back in the other direction (giving us a -200psi as sort of braking pressure/shock absorbtion).  I hope I explained this well enough!  I do not want heat to be generated to a point that it will damage the seals in the pistons or destroy the oil.  As far as I could tell, the only heat inducted will be from friction of the pistons pretty much but will that be enough to generate 225F or more?  I'm estimating that it is not.     


absorbtion qualities are based on the recession of the supportive oil mass. A high pressure gas is commonly used to allow this recession. The very small amount of compression within the oil itself does not allow flow of consequence to produce heat. However, flexural characteristics of seals, and metalic containers, will produce heat within thier structures.

Your heat problem if it occurs, will likely be developed with the sealing elements, or within the flexation of metalic components.


Thanks automatic2, that helps me quite a bit.  I assume there is no way to determine what kind of heat will be generated from the seals and flexation...thus only by trial and error.  Thank you again for the help!

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