1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 '19 at 5:47
  • $\begingroup$ Engine power input $\endgroup$ – Jordan McBain Jan 23 '19 at 15:17
1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.