hacer
Active Member
The battery alone has ~ 0.09 Ohms of internal resistance. A steeper grade requires a higher power output to travel at the same speed. For example, if climbing a grade causes your Wh/mi to go from 500 Wh/mi to 1000 Wh/mi at 60 mph, then the power output (from the battery) went from 30kw to 60kw while climbing the grade. At 30kW, a nominal battery current would be ~76.3A. At 60kW, the current is ~155.4A (more than double because the battery voltage has dropped). In the battery there is 524 W lost to heat when delivering 30 kW, but 2.17kW lost to heat when delivering 60kW. Note that this extra 1.6kW increased consumption rate is not even seen by the energy meter because it is internal to the battery. This is for a fully charged battery. It gets worse as the battery discharges. Note that the battery SOC (state of charge) estimator tries to account for these internal losses when estimating SOC which is one reason why energy consumed since last charge can sometimes seem inconsistent with the remaining SOC.
While the power efficiency of an induction motor is highest at heavy load, the total losses are also greatest there! Made up example: If the motor is 95% efficient when delivering 10kW, there is 523W of heat. If it is is 96% efficient delivering 20kW there is 833W of heat. I have no actual numbers for the motor, but its losses at high power are primarily resistive and will go up slightly higher than the square of the power delivered at constant speed.
The majority of loss in the DC/AC convert (motor driver) at high power are resistive losses (at lower power switching losses are more significant). These too are increasing slightly faster than the square of the power because the voltage droops a bit. I don't have any numbers for the converter efficiency but it's certainly going to be worse than the battery. Besides these losses, the cooling system has to work harder (pump running more) to keep the systems cool.
While the rolling resistance losses go down slightly with increased grade, they were always the least significant source of energy loss at highway speeds. If you drive both the steep and shallow grades really, really slow so that the power is always small and the rolling resistance (and time) are more significant then you might win with the steeper grade, but mainly because the total linear distance is less (and therefore the time is less) so the fixed power consumption (computers, etc) are less along with slightly less rolling resistance. But inn any real-world drive it's going to be worse because the power required for the rolling resistance is not significant compared to that required for the steep grade.
