17th November 2009 - 07:59 PM
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 Horsepowerhttp://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
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.