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I am trying to figure out what the efficiency loss of a leaking hydraulic cylinder is. This master's thesis found that a controllable pitch propeller (essentially a hydraulic cylinder inside a rotating shaft to control the pitch of the propeller) had external leakage of 0.5 ltr/min at 40 bar and internal leakage of 3 ltr/min at 100 bar. I was wondering if it would be possible to convert these numbers into a rough efficiency loss?

I have been google searching for days to try to find a reference to indicate how to make this calculation. I also went all out and went to a local university library to use Engineering Village -- hoping to find some paper that did an analogous study but couldn't find much to help.

This seems relevant but not sure how to use it: Losses through holes drilled in a pipe

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  • $\begingroup$ What would you base the efficiency on? engine power input? Air or fluid moved? Blade Positioning error? Peak hydraulic fluid flow? $\endgroup$
    – Solar Mike
    Jan 23, 2019 at 5:47
  • $\begingroup$ Engine power input $\endgroup$
    – Jordan McBain
    Jan 23, 2019 at 15:17

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The efficiency is a dimensionless quantity and is calculated by power output devided by power input.

I hope you have the value of your power input? But you can also calculate it with the following formula. The left one is for your (electro-)motor which is convertible in your hydraulic pump.

$P_{mechanical}=M[Nm]*2*n[1/sec]*\pi=p[bar]*Q[l/min]=P_{hydraulic}$

There are three (main-)possibilities to get some energy loss in your system:

  • the mechanical one caused by friction in your pump, motor, valves … - $P_{loss-mechanical} [W]$
  • the pressure one caused by perfusion-resistance of the elements in your hydraulic system, e.g. in your piping – $P_{loss-\Delta p} [W]$
  • the volumetric one caused by leakage in your hydraulic system – $P_{loss-volumetric} [W]$

The volumetric loss is your issue - mentioned by the two leakage (internal and external).

$P_{l- vol} [W]= \Sigma (p[bar]*Q[l/min])=10^5*40bar*1/60000*0,5l/min + 10^5*100bar*1/60000*3l/min=533,33 W$

Note the $10^5$ and the $1/60000$ are values because of the different units…

Now the equation for efficiency:

$n= P_{output}/P_{input}=(P_{input}-\Sigma(P_{loss}))/P_{input}=(P_{input}-P_{l- vol})/P_{input}=(P_{input}-533,33 W)/P_{input}$

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