Hydraulic Motor torque 3

[Deaks]

HAs anybody got a mathematical proof that [Nm]=[cc/rev]x[bar]/20/Pi? Perhaps with explanation diagrams?

 

[GregLocock]

Well as a professional engineer I'm sure your skilled dimensional analysis of that equation will show that it cannot be true, as written. No diagram necessary.
Cheers

Greg Locock

 

[Deaks]

"Neither irony or Sarcasm is argument" - Samuel Butler
"[Nm]=[m^3] x ([N] / [m^2])" - Ian Deacon

 

[GregLocock]

Ouch. You are right.



Cheers

Greg Locock

 

[GregSmith]

The formula is wrong (by a factor of 100)

For a hydraulic, positive displacement motor or pump, torque is the same as work per radian and there are 2.pi radians in a revolution. Thus 2.pi.T = work done per rev, in joules per rev if T is in Nm.

Work done by fluid/rev = P.v where
P = pressure in Pa (note 1 Pa = 1 N/m^2)
v = volume per rev in m^3/rev

Thus T = P.v/2/pi, using the units above.
Thus, Nm = Pa.m^3/rev/2/pi.

Now substitute for the unit conversions
1m^3 = 10^6 cc
1 Pa = 10^-5 Bar
and we get
Nm = 10^-5 Bar.10^6 cc/rev / 2 / pi which becomes:

Nm = 10.Bar.(cc/rev)/2/pi
Which is not quite what you had but is of a similar form.

Regards

 

[Deaks]

Greg,
Thanks for this, but I'm afraid the original formula is correct. Look at any Fluid Power Manufacturers Literature. eg

http://www.sauer-danfoss.com/domdb/...44a5a2c7605c8625658b00027fe4/$FILE/10030c.pdf

Regards

 

[Nick55]

Greg:

I think you applied the "cc to m3" and "Pa to bar" conversions the wrong way round....

Quoting the last part of your argument:

"Now substitute for the unit conversions
1m^3 = 10^6 cc, so if the input value is in cc it must be divided by 10^6
1 Pa = 10^-5 Bar, so if the input value is in Bar it must be multiplied by 10^5
and we get
Nm = (Bar.10^5).((cc/rev)/10^6) / 2 / pi which becomes:

Nm = Bar.(cc/rev)/20/pi"

 

[GregSmith]

Thanks Nick55 for pointing out the mistake in my derivation. Deak's formula is indeed correct.

 

[Deaks]

Thanks for your inputs guys

Regards