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Hydraulic Motor torque 3

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Deaks

Mechanical
Oct 30, 2002
23
AU
HAs anybody got a mathematical proof that [Nm]=[cc/rev]x[bar]/20/Pi? Perhaps with explanation diagrams?
 
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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
 
"Neither irony or Sarcasm is argument" - Samuel Butler
"[Nm]=[m^3] x ([N] / [m^2])" - Ian Deacon
 
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
 
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"

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

Regards
 
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