Quantifying The Effect Of Additional Loads On The Powertrain In Regards To Range??

[RaySuave]

Well, then the output will be more like 7.35 kW.

An alternator is not a free energy machine, you only get out around 90% of the mechanical energy you input as electrical power.

If the alternator is spinning at around 5,000 RPM for the rated output, and what is driving it is spinning at 10,000 RPM, how would the alternator output be 7.35kW?

 

[Yggdrasill]

You have a couple of different possibilities for reconciliating the different equations.

If we assume alternator generates 14.7 kW at 5000 RPM, that means that the torque is:

(14.7 * 9549) / 5000 = 28 Nm (without losses)

((14.7/0.9) * 9549) / 5000 = 31.2 Nm (with losses)

This means that the alternator needs 31.2 Nm of torque on the input shaft at 5000 RPM

Working our way back in the chain to 10 000 RPM means converting with a 1:2 ratio. This gearing halves the torque but doubles the RPM, which means that you have 10 000 RPM and 15.6 Nm.

How many hp would this be?

(15.6 Nm * 10,000 RPM) / 9549 = 16.34 kW = 21.9 hp

The gearing doesn't change the amount of hp, it only changes the RPM and torque.

If you assume you start with 10,000 RPM and 7.8 Nm (10.95 hp) , gearing this by 2:1 means you end up with 5,000 RPM and 15.6 RPM, which is only half the torque the alternator requires to produce 14.7 kW. In other words it would not produce 14.7 kW.

 

[RaySuave]

The gearing doesn't change the amount of hp, it only changes the RPM and torque.

If you assume you start with 10,000 RPM and 7.8 Nm (10.95 hp) , gearing this by 2:1 means you end up with 5,000 RPM and 15.6 RPM, which is only half the torque the alternator requires to produce 14.7 kW. In other words it would not produce 14.7 kW.

You are correct the gearing indeed does not have any effect on HP, my over sight. This is exactly the reason my intent was not to convince anybody of anything because it has just been me looking at what I have done so far and I currently lack the benefits of a 2nd or 3rd person to point out mistakes and inconsistencies. Even using the additional load on the drive train of 30hp (300 mile battery range will reduce to a 110 mile range @ 60mph), I still get an increase of range using the appropriate charge rate and figuring the total drive range. Thanks for explaining your point with actual equations and numbers.

...forgot to add the comment that my initial question should have been how much torque is required at the drive shaft to maintain the speed of 60mph and how additional torque loads would effect the estimated range since Im primarily manipulating RPM.