# Regenerative Braking Calculation Help!

1. Feb 21, 2012

### Ksterr

Hey everyone,

I've been a long time lurker on this forum and finally just made an account.
My final year project is in the design of a regenerative braking system for a EV. I'm reading a Modern electric vehicle design Book and came across an equation that I don't quite understand.

This equation finds power at the drive wheels:

Pd = V/1000 (Mgfr + 1/2*ρa*CD*A*V2 +M*δ*dV/dt) (kW),

The book says to integrate this equation w.r.t driving time over a given driving cycle. This will than yield energy in traction and energy in braking.

What I don't get is, aren't all these constants except for dV/dt

M = mass
g = 9.81
fr = tire rolling resistance
ρa = air density
Cd = coeffificent of drag
A = frontal area
v = speed
δ = rotational inertia factor

2. Feb 21, 2012

### OldEngr63

If the vehicle is accelerating, that is, dV/dt is not zero, then V^2 is not constant and the air drag term is thus a variable term as well.

I have some questions about the rest of the equation; I'm not quite sure where all of it comes from. I'd need to spend a while to become convinced it was all correct (if it is).

3. Feb 21, 2012

### Ksterr

I got it straight from a EV Design book.

I think it basically calculate the total tractive power?

4. Feb 23, 2012

### Mech_Engineer

It looks like the equation is calculating the minimum required power for a given speed. This equation is basically assuming the maximum power required by a vehicle is when it is at cruising speed and accelerating. When you split the equation up:

m*g*f*r - Calculation of rolling friction (Force)

1/2*ρa*CD*A*V2 - Drag force exerted on the vehicle by air at a certain speed (Force)

M*δ*dV/dt - Required acceleration the vehicle is capable of at cruising speed. (Force)

Force multiplied by Velocity results in power (W), divide by 1000 to convert from watts (W) to kilowatts (kW).

5. Feb 24, 2012

### Ksterr

This equation from what I see equates to the total tractive force for the vehicle.

Which ends up being Power in KW. If I brake that down into positive and negative power, does that mean all the negative power would be what's available for regen braking?

6. Feb 24, 2012

### OldEngr63

In a word, NO. The air drag is still going to consume power that will be unavailable for regen, as will the rolling friction. The only power available for regen is that coming from the mass*accel term, diminished by the other (parasitic) terms that are always present taking power away.

7. Feb 24, 2012

### Mech_Engineer

NO, the equation calculates the total required power for a certain cruising speed based on air drag, rolling drag, and a required amount of acceleration overhead.

You can caluclate the total energy available for regenerative braking by simply calculating the vehicle's total kinetic energy; for regenerative purposes I don't think you need to capture it with a deceleration any faster than say 0.5G. So given this acceleration and a vehicle weight THAT can be used to find your required deceleration power.