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Relation between viscosity and pressure drop through filter element

Relation between viscosity and pressure drop through filter element

Relation between viscosity and pressure drop through filter element

(OP)
Hello all,

I am currently designing a hydraulic system. We have very little margin for pressure drop and we just can't afford to find out the pressure is insufficient once the system is installed on location (can't be tested before). I am confident with most of the dP values of different components of the system, but I am very puzzled by the filtering unit.

For example, I am using Parker model 80CN-2 filter housing with 2micron filtering element: http://www.parker.com/literature/Hydraulic%20Filter/CN%20Series.pdf
On page 76, the lower right hand graph shows the dP vs flow relation. Taking different points on the curve, I get roughly the same Cv factors. This makes sense because the graph shoes a second order curve and Cv = Q*sqrt(SG/dP). So here I have no problem using conversion factors to find the pressure drop caused by the housing from 32cSt (Parker reference fluid) to 8cSt (the fluid in our system). However, when looking at the lower left hand graph, this reasoning no longer works: the relation between flow and pressure loss in the filtering element is linear. Not second order. Which means the Cv factor of the element varies with flow. And I am a bit uncomfortable using conversion factors for a component that has a "variable" Cv.

When I asked Parker's customer support, they told me the relation with viscosity was linear. Basicaly, if the Parker chart shows 6PSID for a 42GPM @ 32cSt (150SSU) and my oil is at 8cSt SG=0,94 (50SSU), it means the pressure drop will be one third of the charted pressure drop (50/150 = 1/3) at 2PSID. As much as I like this explanation because of its simplicity and the pressure drop is where I want it to be, I am having doubts. I would normally trust 100% Parker on this, but the consequences of having a higher pressure drop could be quite frustrating and I really want to understand what is hapening here.

The way I see it, let's say a valve with a Cv factor of 20 in water will have a corrected Cv factor of 16 in 10cSt oil (Cv_water * Fv(10cSt) = 20 * 0,8 = 16). So for a flow of 50GPM, the pressure drop through this valve with water will be 6.25PSI. In oil, the same flow would yield a 9,75PSI pressure drop. Now for a filter, according to Parker, this same flow through an element dropping 2 PSI in water would rise to 20 psi with oil? This sounds strange to me. Plus, this source: http://www.oxfordfiltration.com/selfCleaningFilters_spec.asp shows that the filter elements are not linearly dependant on the flow (scales are logaritmic) and viscosity ( multiplying the viscosity by four anywhere in the second table does not equate to a 4x factor on pressure drop).

If we consider a filter element to be millions of little valves and elbows with a certain Cv factor packed together, how come it doesn't react as such when the flow/viscosity changes? Or could it be that the theory is very different for micro meshes (few microns) and the bigger the mesh holes are, the closer the behavior to a "constant Cv" hydraulic component the strainer/filter becomes?

Thank you for input,
Marc

RE: Relation between viscosity and pressure drop through filter element

Looks like the "Empty Housing" is making the difference. Assuming no filter material inside that one, flow is primarily turbulent and effective diameters of 1.5 and 2" are relatively a higher proportion to flow path length, so it's turbulent flow through an orifice where the fluid's density effects are prevalent. With filter material, effective diameter being, in microns, is much, much less than length, so I think it should be modeled as laminar flow through a tube, for which viscosity effects are far more prevalent.

Independent events are seldomly independent.

RE: Relation between viscosity and pressure drop through filter element

(OP)
Thank you BigInch, your explanation makes much sense.
Guess I'll have to dust off these Fluid Mechanics books of mine.

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