Hydromechdude
Aerospace
- Jun 3, 2008
- 44
I'm trying to determine how long it would take for a hydraulic accumulator to bleed down from full working pressure to empty (no fluid remaining).
The equation that I have been using is Q = C*(deltaP)^(1/2). In order to determine the constant C, I'm using the initial conditions for Q (leakage rate through the isolation valve) at the deltaP of 3000 PSI.
In order to determine how much fluid is initially in the accumulator, I used isothermal compression/expansion P1*V1=P2*V2. I know pre-charge pressure, initial volume, and then working pressure. Solve for V2 of air. That will then mean we have Vaccumulator total - V2 for the fluid volume, Vfinitial.
Now to solve for the leakage. I did iterations of one minute in length. So, for the first minute I assumed that the leakage flow was constant at the leakage rate (say 4 cc/min). Thus, after one minute we'll have Vfinitial - 4cc of fluid and Vnewofair = V2 + 4 cc.
At this stage, I now calculate P2*V2 (working pressure start conditions) = Pnew*Vnewofair to solve for the new pressure.
Next is taking the new pressure, setting it as the deltaP in Q = C*(deltaP)^(1/2) to solve for the new leakage flow. I've iterated this for several hours to determine how long it will take to drop to initial pressure.
How does this look? I don't have a program to model this with and so I have nothing to bounce my theory off of. Testing this with a real accumulator will not be possible. I realize I'm modeling the ideal maximum as most valves will fall under the maximum 4 cc/min that I've used, however, having this time would be helpful.
Thanks.
The equation that I have been using is Q = C*(deltaP)^(1/2). In order to determine the constant C, I'm using the initial conditions for Q (leakage rate through the isolation valve) at the deltaP of 3000 PSI.
In order to determine how much fluid is initially in the accumulator, I used isothermal compression/expansion P1*V1=P2*V2. I know pre-charge pressure, initial volume, and then working pressure. Solve for V2 of air. That will then mean we have Vaccumulator total - V2 for the fluid volume, Vfinitial.
Now to solve for the leakage. I did iterations of one minute in length. So, for the first minute I assumed that the leakage flow was constant at the leakage rate (say 4 cc/min). Thus, after one minute we'll have Vfinitial - 4cc of fluid and Vnewofair = V2 + 4 cc.
At this stage, I now calculate P2*V2 (working pressure start conditions) = Pnew*Vnewofair to solve for the new pressure.
Next is taking the new pressure, setting it as the deltaP in Q = C*(deltaP)^(1/2) to solve for the new leakage flow. I've iterated this for several hours to determine how long it will take to drop to initial pressure.
How does this look? I don't have a program to model this with and so I have nothing to bounce my theory off of. Testing this with a real accumulator will not be possible. I realize I'm modeling the ideal maximum as most valves will fall under the maximum 4 cc/min that I've used, however, having this time would be helpful.
Thanks.
