Quote:
|
Originally Posted by Mike Betts
Efficiency at Max Power = (1.23 N*m * 280 rad/sec) / (12V * 57A) = 50%
Therefore, the mechanical power at the spindle will be 343W and I would expect the power dissipated in the generator’s brake resistor to be about 172W or so…
|
Mike,
What is your source for these numbers? The only source I have seen for the CIM motor specs is from the FIRST web site but some of the numbers don't quite match:
http://www2.usfirst.org/2005comp/Specs/CIM.pdf
I am by no means trying to nit pick and I understand your numbers were just an example of the calculation but I am just wondering if there is another better source for the CIM motor specs that I should be using that I don't know about.
Using your example but taking the numbers for max power directly from the FIRST PDF spec sheet above I get that the efficiency is even worse (41%).
Torque = 171.7 Oz-In = 1.21 N*m
Speed = 2655 RPM = 278 rad/sec
Current = 67.9 A
Power = 337 W
Efficiency = 41% = (1.21 N*m * 278 rad/sec) / (12V * 67.9A)
The should be at least 337 W * 41 % = 138 W.
Thanks,
Chuck