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Hydraulic Power

2009.07.25





Hydraulic power is defined as Flow x Pressure. The hydraulic power supplied by a pump: P in [bar] and Q in [lit/min] => (P x Q)÷ 600 [kW]. Ex. Pump delivers 180 [lit/min] and the P equals 250 [bar] => Pump power output = (180 x 250)÷600 = 75 [kW].

When calculating the power input to the pump, the total pump efficiency ηtotal must be included. This efficiency is the product of volumetric efficiency, ηvol and the hydromechanical efficiency, ηhm. Power input = Power output ÷ ηtotal. The average for axial piston pumps, ηtotal = 0.87. In the example the power source, for example a diesel engine or an electric motor, must be capable of delivering at least 75 ÷ 0.87 = 86 [kW]. The hydraulic motors and cylinders that the pump supplies with hydraulic power also have efficiencies and the total system efficiency (without including the pressure drop in the hydraulic pipes and valves) will end up at approx. 0.75. Cylinders normally have a total efficiency around 0.95 while hydraulic axial piston motors 0.87, the same as the pump. In general the power loss in a hydraulic energy transmission is thus around 25% or more at ideal viscosity range 25-35 [cSt].

Calculation of the required max. power output for the diesel engine, rough estimation:

(1) Check the max. powerpoint, i.e. the point where pressure times flow reach the max. value.

(2) Ediesel = (Pmax·Qtot)÷η.

Qtot = calculate with the theoretical pump flow for the consumers not including leakages at max. power point.

Pmax = actual pump pressure at max. power point.

Note: η is the total efficiency = (output mechanical power ÷ input mechanical power). For rough estimations, η = 0.75. Add 10-20% (depends on the application) to this power value.

(3) Calculate the required pumpdisplacement from required max. sum of flow for the consumers in worst case and the diesel engine rpm in this point. The max. flow can differ from the flow used for calculation of the diesel engine power. Pump volumetric efficiency average, piston pumps: ηvol= 0.93.

Pumpdisplacement Vpump= Qtot ÷ ndiesel ÷ 0.93.

(4) Calculation of prel. cooler capacity: Heat dissipation from hydraulic oil tanks, valves, pipes and hydraulic components is less than a few percent in standard mobile equipment and the cooler capacity must include some margins. Minimum cooler capacity, Ecooler = 0.25Ediesel

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Complicated...huh?
12 Comments
lifeisgood Complicated, but beautiful.
lifeisgood · 2009-07-25: 17:08
lookagain Very cool captures....I love the splash of light on the water in #3!!
lookagain · 2009-07-25: 20:18
eternity65 sure is, but great shots!
eternity65 · 2009-07-26: 02:01
neusnet beautiful shots! #3 is my fav.
neusnet · 2009-07-26: 05:21
dontblink You made my head hurt with the math:) Nice water shots!
dontblink · 2009-07-26: 05:25
tedbarlow Great shots, Pi, showing the power and potential of water
(Are you sure the math is right...there seem to be some discrepencies!)
tedbarlow · 2009-07-26: 07:35
nunuinpictures OH wow!
Very interesting post.
:O
nunuinpictures · 2009-07-26: 11:42
shutterjitter wonderful pix..but I could not beyond the 2nd line of the text
shutterjitter · 2009-07-27: 04:44
firehorsecri Ehi, nice shots... but I can't follow you in maths... :-(
firehorsecri · 2009-07-27: 14:19
Widcat I love #3!
Widcat · 2009-07-28: 00:20
Ryana Wow, cool (and I think, real cool to .. ;o) set!! xxx
Ryana · 2009-08-01: 14:33
gwena nice set!
gwena · 2009-08-07: 11:50
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