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ingunn (Mechanical)
28 Aug 08 11:38
I'm a young engineer in need of som guidance.
I am working on a pilot operated relief valve. The pilot sensing point is placed in such a position that during a relief scenario, the velocity across the sensing point will be quite high (Mach=>0,3). I am affraid this may cause the relief valve to close due to the venturi effect, although pressure in the tank to be relieved, is still high. Incompressible calculations show that the sensed pressure will be lower than the opening pressure of the relief valve. The sensing point is also very near the nozzle of the tank to be relieved, and entrance effects may be expected.
How will compressibility contribute?
What kind of entrance flow should I expect?
To me, this seems unacceptable. Is it?
btrueblood (Mechanical)
28 Aug 08 12:26
字串9
Think through what happens when the valve starts to open, and the pressure at the pilot sense port falls - the valve closes, causing the velocity to drop, and the pressure at the sense port rises again, starting the whole loop over again. In reality, it won't oscillate (unless the spring has a very low stiffness), but reach an equilibrium, with the valve opening bein a function of tank pressure. At lower pressures, the valve will just crack open, and as tank pressure rises, it will open further.
If you must have a full open condition for any pressure above some value, then you can (a) design the relief valve to open earlier, so that even with reduced pressure at the sense port, the valve still opens to the required Cv value. (b) Make the valve grossly oversized, so that the difference between the cracking pressure and the pressure to achieve the minimum vent flow is smaller. (c) Change the sense port, i.e. make it a pitot tube, or have the sense port linked by seperate tube to the tank or other stagnation zone. (d) use a snap-over (e.g. belleville spring with high h/t ratio) mechanism, although these can be unreliable at re-seating if that is a requirement.
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