Energy Storage, Hydraulic Accumulator

Hey there, I need help sizing a bladder type hydraulic accumulator. Could someone please tell me how i size an accumulator to produce enough power to power a 1 watt L.E.D continuously for 24 hours. Also how do I determine the flow rate? I am finding it quite difficult to source information on this topic. I have already tried downloading the InPhorm software package from the Parker website but for some reason it won't download to my computer. The maximum system pressure is 210 bar. Any help on this will be greatly appreciated.

Thanks.

You have difficulty in part because a hydraulic bladder accumulator is completely inadequate for you needs. They serve to provide a huge power (MW+ range) for a short time (10ms to a few seconds).

To run your LED for 24 hours, use an chemical accumulator, like Li-ion, or NiMH. Simpler for electric energy, better adapted to your power.

Or even a supercapacitor. Something like 0.1F *2V. Should last more cycles than a chemical accumulator. From Maxwell and several others.

I strongly doubt you'll find a pump nor a hydraulic motor for the ridiculous power you mention. They're rather for kW to MW+. With 210e5 Pa, 2.1W (with losses) would need 100mm3/s or 0.0001L/s or 0.006L/min: check that such a motor isn't available.

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Hydraulic accumulators are often buffers (serge protectors) in hydraulics systems. Now that is not to say that they can't be used to provide flow in place of a pump.

1 watt is equal to 0.00134102209 horsepower (google) so you need a .002 horsepower motor (compensating for energy losses).

Now a quick look at Horsepower

http://en.wikipedia.org/wiki/Horsepower#Hydraulic_horsepower

Has hydraulic hp as..

hp = flowrate * pressure * 7/12000

So we have the max system pressure at 210 bar

1 bar = 100000 N/m^2 so 210 bar = 21 000 000 N/m^2

.002 = m^3/s * 21 000 000 N/m^2 * 7/12 000
.002 = m^3/s * 7000
.002/7000 = m^3/s
2.857e-7 m^3/s

There are 86 400 seconds in a day so multiply the flow rate by that number and you get...0.0247 cubic meters which is equal to about 24.7 litres.

Of course an accumulator loses pressure as the hydraulic fluid flows out. So you would have to use a pressure regulating valve set to 210 bar and your accumulator will have to have a precharge set to at least that pressure when empty.

Now you say that the 210 is max system pressure, is there a minimum? If there is I would suggest using that instead in the above equations. The result will be a larger accumulator.

However now that I see it...if you consider the minimum 0 bar and max 210 then the mean would be 105.

.002 = FR * 10 500 000 * 7/12 000
.002 = FR * 6125
.002/6125 = FR
FR = 3.265e-7 m^3/s
3.265e-7 m^3/s * 86 400 = .0282 m^3 = 28.2 litres

If anyone sees any mistakes please correct them. I hope I did it all right and I hope that answers your question.

At least one mistake: hydraulic accumulators don't start at zero bar, but at an important pressure.

Next: I can't imagine a hydraulic motor of 1 watt, so take a battery.

Typically yes, accumulators are precharged to a designated pressure. I can't think of an example in which this is not the case. However that is not to say that an accumulator can't start at 0.