Scavenge pump for hydraulic oil.

[Turbo20V]

I need to design an oil scavenging system for the output of a hydraulic system. The flow is not very much perhaps 0.1 GPM but it is hard to establish the flow rate. The source of the oil is the output of two servo hydraulic actuators, and their hydrostatic bearings. The flow of oil is not constant. I plan on switching the pump on and off as needed with an existing power switching system.

I was thinking of going with a Gerotor pump, 120V. I need this pump to produce vacuum to scavenge out the hydraulic oil effectively. However I also need some positive pressure to force the oil against 100g. The pump will be at the end of a large centrifuge, and will pump oil against this large G load and through about 50 feet of tubing before returning to the hydraulic pump reservoir.

Can anyone provide me any expert insight into this task? I'm finding it hard to even find a supplier of gerotor pumps at all. Let alone good quality technical specs such as vacuum at inlet and output pressure.

Any assistance would be much appreciated.

Thank you.

 

[MikeHalloran]

Tuthill provides gerotor assemblies to build into other stuff, along with a bunch of lube oil pumps that are not gerotors.

http://devcocorp.com/tuthill.htm?gclid=CNzpisC54KQCFYpa2godfFzbdw

Mike Halloran
Pembroke Pines, FL, USA

 

[Turbo20V]

Ok so im having a bit of a problem with calculations.

I have a piece of rotating machinery I am putting this pump on and the pump is at the full 3 meter radius.

The pump will be subjected to 100g conditions. Now not all of the 3 meters to the center is subjected to 100g however. The normal acceleration should be (w^2)r. In terms of G's this should be ((w^2)r/9.81)= g's

In a normal system pressure head is =pgh a linear function. i believe in my situation I would have to substitute in the above equatin for "g" to yield:

p=rho
P=pressure

P= p[((w^2)r)/g]r

which simplifies to : P= p[((w^2)(r^2))/g]

Does this seem right? What pressure UNITS does this yield if I use 880kg/m^3 and 3 meters, and 19.02rad/sec?

I need to add this answer to a base return pressure of 200psi but im not sure what the metric units work out to or if my math is even remotley correct.

Thank you.

 

[Turbo20V]

Hello?? Anybody here to confirm?

 

[vpl]

Is "G" different from 'g"? If you are substituting for g, then why does it appear in the last equation?

Also in your first equation, is the 9.81 the gravitational constant (g_c? If so, that means you need to use metric units throughout (or use the english g_c.)


Working out the units should be the easy part, once the equation is correct.

Patricia Lougheed

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