How can I determine the power generated by a regenerative braking system?

Question

If I have an electric vehicle on an incline (m=17,300 kg) I want to figure out how much energy it would generate by regenerative braking.

I've used $a= g \cdot \sin({\theta}) $ and $f=m\cdot a$.

at 05 deg incline, acceleration is 0.85 m/s/s; this creates a force of 11114.972 N
at 10 deg incline, acceleration is 1.70 m/s/s; this creates a force of 22145.352 N

So bearing these values in mind if I don't want to accelerate and maintain a set speed I will need to create a braking force of 11114.972 N at 5 deg incline - this in turn will keep the machine at the same speed (ignoring friction, bumpy/uneven surface, etc.).

How can I figure out the Watts generated for the regenerative braking force of 11114.972 N?

The DC motor (I think it's 3 phase) is a 105 kW motor rated at 350 V and 300 A, max torque is 190 Nm @ 285 V drawing 285 A. (That's all the information I have for the motor.) Efficiency of the motor would be 0.85 (Assumed).

Answer 1

If I have an electric vehicle on an incline (m=17,3oo kg) I want to figure out how much energy it would generate by regenerative braking ... and maintain a set speed.

Change in potential energy per time is as good an approach as using kinetic energy.
The answers should be the same :-).

E= mgh

  h = change in vertical axis  
  m = mass
  g = gravitational acceleration.

In one second power = energy numerically.
Say g = 10 (more than good enough for 'playing')

1 kph = 1000 metres in 3600 seconds.
1 kph = 0.278 m/s (or 1 m/s = 3.6 kph)

V_vertical / V_vehicle = sin(slope angle) V_vertical = V_vehicle x sin(slope_angle)

Mass = 17300 kg
First calculate for 1 metre/second vertical rate of change to get a general feel.
Power = energy change in one second at constant V
P = mgh = 17300 x 10 x 1 = 173,000 Joule/1s
= 173 kW at 1 m/s !!!!

At your example 45 degrees and 15 kph

V_vertical = 15 x sin(45) = 10.6 kph
10.6 kph = 10.6/3.6 m/s = 2.95 m/s vertically.
So power = 2.95 x 173 kW = 510 kW. Change g to 9.8 instead of 10 and I get 500 kW
= your result.

This tells you that under the stated conditions your motor would be trying to regenerate at about 5 x it's rated power. It is almost certain that it would not achieve this, and quite likely that it would be damaged if you ran it in that manner. 500 kW (or even 100 kW) of magic smoke would be very impressive indeed!.

It also tells you that the motor could not drive the vehicle up a 45 degrees slope at that rate. Maximum speed would be about 105 kW/500 kW x 15 kph = 3 kph

This is not a vast surprise when you consider what you are asking for.
Braking at constant V at 15 kph is equivalent to lifting a 17 ton vehicle ~= 10+ cars vertically at 3 m/S

________________________

Useful rule of thumb

Energy  ~= kg x metres lifted vertically x 10 x efficiency 

This is good for energy required to lift mass OR power delivered by falling masses.

Power is energy per unit time so power can be used instead of energy is rates are in meters per second rather than total metres moves.

So, for water wheel

Power = litres/second x head in metres x 10 x efficiency 

For pump

Power = litres/second x head in metres x 10 / efficiency 

Note that efficiency Z = power out / power in in both cases, but

Efficiency for water wheel = Power delivered / Power from water flow
Efficiency for pump = Power in outlet water flow / Power used to drive pump

Answer 2

The calculation perfectly right, bu the normal electric vehicle weights just 1/10th on the weight that you have considered. Hence the results are scaled in that ratio (10x more than what the real world EV could generate)... Hope that helps!

Answer 3

The simple answer is based upon your speed. $P = F*v$, so if you're going at 2 m/s, Wolfram gives you 22.4 kW. Quite a bit of energy to start with! Of course, your final design will eat up some of this power due to energy conversion.