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
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.
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:
