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Theoretical flow rate?

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RandyP

Computer
Jul 28, 2003
6
Need help in getting a rough idea for the maximum potential flow two openings will permit. The diameter of the two orifices are 0.078" and 0.219". Fluid is motor oil 20W-50. Orifices are fed off the same input line that has a diameter of 0.25". I'm guessing this supply line will probably be the one that dictates the flow to the two openings?

Any help in solving the above problem is greatly appreciated!!!! Thanks Randy
 
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Forgot to add maximum pressure for the input line has already been tested at 60lbs. Randy
 
fluid dont "choke" as gasses do. Therefore its not proper to say that its the line or the orifices that "dictates" the flow.

Use standard Dp calc for the line and the orifices. You also need to know the length of the feed line.

The easiest way i believe would be to consider the opening as one area and then calculate the max flow with your available dP (60 PSI i assume). The distribution will then be equal to the ratios of the two area. (i.e your small orifice gets .078^2/(.078^2+.219^2 == 11% of the total flow). This last part is independat (almost) of the actual max flow

Best regards

Morten
 
W = Cv*SQRT(DP/Gw), Where W is Flow rate in lb/hr, DP = 60 psi.g, Gw = specific gravity of liquid (no units).
Cv is the flow coefficient used in valve design. This is should be tested but in general Cv = 30*Area of orifice if the orifice is long in length i.e.>2*Diameter, Cv=40 if short in length I.e. < 0.5* Diameter. Note this does not consider viscosity either.
0.078&quot; dia 1.11 - 1.48 lb/hr
0.219&quot; dia 8.75 - 11.67 lb/hr
 
BHMech,

Your equation is a handy way to quickly estimate the hydraulic resistance through an orifice. But I have one question. I am used to calculating flow in gpm when using valve flow coefficients as you described. So in your post, did you mean to say lb/hr or gpm?
 
Correct gpm.
Also, I calculated for water. For Oil Gw = 0.8 not 1.0 so my estimates should be multiplied by 0.8 to compensate.
 
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