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Old 03-18-2017, 08:56 AM
GBA GBA is offline
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Question How to determine how much energy recovered from regenerative braking

Hi,

I am considering a Regenerative Braking system implemented with a AC permeant magnet motor with an efficiently of ~96%. It is easy to determine via energy equations the kinetic energy of the car and therefore how much energy is possible at max efficiently but regenerative systems also rely on the brakes of the car to slow it down to a stop. Is there a particular rule or reference to how much conventional braking is required for a given scenario? I wonder if it is possible to determine how much the motor would actually decelerate the car and then work out energies from there. Could anyone offer some advice?

Thanks very much
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  #2  
Old 04-13-2017, 03:56 PM
brian_ brian_ is offline
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Default Re: How to determine how much energy recovered from regenerative braking

It sounds like you could use some basic physics education. There should be lots of good material on websites.

The kinetic energy of a moving body is
E = m * v
where
E is the energy (in joules if using SI units; a joule is a watt-second)
m is the mass (in kilograms if using SI units)
v is the velocity (in metres per second if using SI units)
The energy which could possibly be recovered in Regenerative Braking is the difference between the kinetic energy of the moving vehicle at the beginning (v1), minus the kinetic energy of the moving vehicle at the end (v2), so
(m * v1) - (m * v2)
or
m (v1 - v2)

You can also use your suggested power approach. The power dissipated by brakes or absorbed by Regenerative Braking is the force multiplied by the speed:
P = F * v
P is the power (in watts if using SI units)
v is the velocity (in metres per second if using SI units)
... but you need the force. To decelerate a mass, the force is the rate of deceleration multiplied by the mass
F = m * a
F is the force (in newtons if using SI units)
m is the mass (in kilograms if using SI units)
a is the acceleration (in metres per second squared if using SI units)
Example: if braking a one-ton (1000 kg or 2200 pound) mass at half "g" (half of the acceleration of gravity, so 5 m/s), the force would be 1000 kg * 5 m/s = 5000 N (or 5 kN), and if the vehicle is moving at 20 m/s (72 km/h or 45 mph) the power would be 5000 N * 20 m/s = 100,000 N (or 100 kW). That's about as hard as anyone brakes on the street - normal would be less.

Any energy used by drag (rolling drag and aerodynamic drag) will not be available to be recovered, and of what is available there are the losses in the generator (motor) and battery.

Last edited by brian_; 04-13-2017 at 04:11 PM.
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