major said:
How do you explain lower power output with reduced back-EMF?
No answer to my question huh. Okay, let's see what we can do with your made up numbers.
Anomaly 1: 100VDC, 10A draw, and you cite a resistance of 0.5Ω. But R = V/I, so R should be 10Ω.
So, I'll work with 10Ω. And since you haven't supplied it, to keep the numbers simple, I'll assume a conductor cross-sectional area of 1 square millimetre or 1e-6 m
Now R = rho * l / A:: 10.0 = 1.68e-8 * l / 1e-6 := l = 10.0 * 1e-6 / 1.68e-8 = 595.23 metres of 1mm squared conductor. (@ 20ºC) (Sounds like a big damn motor, but c'est la vie.)
At 80ºC, the resistance of copper rises to 2.07312e-8. So, R = 2.07312e-8 * 595.2 / 1e-6 = 12.34Ω
As you say all else is equal, at 80ºC the current flowing will drop to: I = V/R; 100 / 12.34 = 8.1A.
Now, given the current flow has dropped, the magnetic field in the coils will have dropped -- I cannot calculate the exact drop because you haven't supplied sufficient information for that -- but as field strength is proportional to current; and force is proportional to field strength; and torque is proportional to force, the torque has dropped. Inevitably, the speed will drop. And if the speed drops, the back-EMF reduces.
But, motor peak efficiency tends to be (ignoring losses) ~ the no-load maximum speed / 2, and it falls either side of that. Assuming (since you didn't supply it) that your 95% efficiency was measured at that peak efficiency, then that "slight reduction in speed" you mentioned, is all that is required to cause the efficient to drop to the 93.82% at that lower speed.
And so to anomaly ... (I lost count):
"a slight reduction in RPM due to the lower generated voltage"
YOU ARE TAKING THE SYMPTOMS OF THE DISEASE AS ITS CAUSE!
The back-EMF drops, because the motor slows down and back-EMF is proportional to speed. And the motor slowed down because the temperature of the coils rose; and with it the resistance. And when the resistance rises, the current flow reduces; and when that happens, there is less torque being applied to the (unchanged load) and so it slows down
And here I suspect is the crux: The reason that (for the second time) I've asked you how the back-EMF was being measured, is because it isn't. Directly anyway.
For real motors, that motor velocity constant (and the back-EMF) is determined empirically. You take the motor, with no load attached, and apply voltage to it whilst measuring the rpm. Slowly increasing the voltage until the rpm fails to rise further. At that point, the back-EMF must equal the EMF being supplied, and Kv is thus determined, Simply divide the final maximum, no-load speed by the input voltage required to achieve it and there's your motor velocity constant.
Now the physical layout of the motor, all its losses -- core, copper, hysteresis skin effect, proximity, et. al.
and the back-EMF -- and thus its efficiency, are all ready set; and Kv simply records that; it does not determine it.
More simply stated: back-EMF is a symptom of the motor design, not a determining factor.