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Hydraulic restrictors

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Michiel1984

Michiel1984

Aerospace
Joined
Dec 2, 2004
Messages
14
Location
NL
How do I have to calculate the pressure drop over a restrictor? If I calculate it with the formula stated below, the pressure drop will be huge!!!

Q=Cf*A*SQRT((2*differential pressure)/Density)

A=0.7 mm^2
Q=1.66 GPM
Density=0.997 g/ml

The units have to be calculated in cubic meters and liters (liters/second). The Cf is mostly in between 0.6 and 0.9, but I don't know what the right value of it is. If somebody does know how I can determine the value of it, it would be really great!

Thank you in advance

Ps: the system pressure is 3000 PSI (20,7 MPa)


 
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The equation is correct, and at 3000 lbs initial pressure, the drop will definitely be huge.

The coefficient varies due to the type of orifice. A sharper-edged orifice will give closer to 0.6, a well-rounded edge will give as high as 0.98.

It also varies with the length of the the constriction, relative to its diameter. Bump it down roughly 0.1 with L=0.5D, up 0.15 for L=2 or 3 D.

The value of C is really an engineering judgement, there are numerous references to back you up.
 
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