Hydrualic pressure to HP formula

[sdelanty at sonoma.net]

>> therefore Power = Volume Flow * Pressure (drop)

>Wrong.  You must also consider the density of the liquid, and the head
>loss.  Head loss is not pressure drop.

   Well, yeah, technically it's wrong, but by how much?

   Head loss is inconsequential on closed hydraulic systems.
   The fluid is returned to the tank at roughly the same height as the inlet. 
   Any difference between these points is worth less than 0.44psi per foot
with    water.
   Hyd oils usually have a S.G. between 0.6- 0.8 , so a -100- foot head
causes a 
   loss of 35 psi or less. 
   A 3 foot head is about the most You will ever see in closed hyd systems.
  _Maybe_ a 1 psi loss out of 2000 or so....

    Unless You need to know power requirements down to 0.1percent, 
    power= volume of flow x pressure.

    I didn't check the gentlemans metric formula, but in my world 
    of U.S. measurements and fixin lotsa hyd stuff for a living:

   HP = GPM X PSI / 1714  

    This is the formula I use for sizing hyd equipment and it appears in most
    books on hydraulics. I've never seen corrections for "head" pressure or
S.G. 
    in hyd formulas.

    The only other correction factor is for pump or motor efficiency but fluid 
    motors are quite good, usually 95 percent or better....
 
>> Note that this is a derivative of the Work=Force*Distance =>
>> Power=Force*Vel.

>Your derivation is wrong.

   Yeah, but by an inconsequential amount.

  Happy motoring,


 Steve Delanty (sdelanty at sonoma.net)

  1971 F100, FE390, T-18 4-speed shortbox.