## REGEN Question

joeriv
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### REGEN Question

I am not an engineer so perhaps some others can answer this:

Assume you're on the highway straight and level 60 mph. You accelerate to 65 mph, then ease off and regen to 60, then repeat the same cycle. Will the energy you recapture be the same, less than or more than what was used to accelerate to 65 mph, ie is it possible to have a net gain? My intuition says there's no such thing as a free lunch, so I would guess there would be no net gain, although 3000+ pounds of car at 65 mph is a lot of energy.
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gshepherd
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### Re: REGEN Question

Short answer is no. As Homer Simpson once said: "Lisa, the laws of physics will not be violated in this household!" It is often written here that regen only captures about 40% of the energy originally used to accelerate up to speed or climb a hill. Better than nothing (friction brakes). Every time energy is changed from one form to another, some is lost. Add it up each step of he way and it is not insignificant.

You can check it out for yourself, though. I would map out a safe course and conduct a controlled experiment: constant speed for one run, and accel/regen for a second run, all else being equal. Best to make round trips to account for wind and grade. A newer LEAF that can display SOC% should have enough instrumentation to give you evidence (while interesting, don't rely on the Miles to Empty gauge since it has its own fuzzy math to extrapolate the future from the recent past). Rinse, repeat. Report back what you learn.
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IssacZachary
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### Re: REGEN Question

I've only read that electric regen is about 25-30% efficient (70-75% energy loss). Maybe it could be more efficient (less loss) in a larger battery powered EV, such as the Leaf, than a typical electric hybrid because a larger battery will have less internal resistance. This is why commercial hybrids normally use hydraulic regen instead of electric regen, which is around 70% efficient (30% loss).

The most efficient way to use your kinetic (moving) energy is to "shift" into neutral and coast to a stop. The less you use your braking, be that regen or discs, the more efficient you'll be driving. Of course if you need to slow down faster then using regen braking is better than using the disc brakes. But still, the longer you take to come to a stop the more efficient you'll be driving. That's because if you take a long time to stop, even if you're using a little bit of regen braking, part of your slowing down is from regen (30% efficient) and part is from coasting (100% efficient). So the total efficiency ends up somewhere in between.

The efficiency losses are rotating to electrical in the motor/generator, AC to DC in the inverter, then electric to chemical in the battery, then chemical back to electrical, DC to AC, electric to rotating. With each change in energy there is a loss. All these losses add up to as much as 75% efficiency loss.
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NeilBlanchard
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### Re: REGEN Question

There are ALWAYS be more losses than gains - otherwise EV's would be perpetual motion machines.

Regen is better than friction brakes - which are a 100% loss - but coasting back down to 60MPH (especially if you can utilize a slope down) will be much more efficient IF what you need to do is move the car forward.

alozzy
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### Re: REGEN Question

One thing I've often wondered is, at what speed does coasting in neutral results in higher losses than using regenerative braking at a lower speed?

The reason I suspect that happens eventually is that wind resistance is proportional to the square of the speed at which you are travelling. So, it seems to make logical sense that, at some point, if you keep picking up speed in neutral down a steep hill, the losses due to wind resistance will negate any efficiency gains due to coasting and will only get worse the faster you go. Stated a different way, since you would be travelling at a considerably lower speed when in B mode (vs coasting) going down the same steep hill, then the efficiency losses due to regen braking would be more than offset by the lower losses due to wind resistance.

I'm guessing that you'd have to be coasting for a long time on a steep grade though before you would hit the speed at which coasting becomes less efficient than regen braking.

Is there a physicist in the house?
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LeftieBiker
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### Re: REGEN Question

The regen efficiency figure I've read is "up to 39%." It could well be 25% under many circumstances.
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IssacZachary
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### Re: REGEN Question

alozzy wrote:One thing I've often wondered is, at what speed does coasting in neutral results in higher losses than using regenerative braking at a lower speed?

The reason I suspect that happens eventually is that wind resistance is proportional to the square of the speed at which you are travelling. So, it seems to make logical sense that, at some point, if you keep picking up speed in neutral down a steep hill, the losses due to wind resistance will negate any efficiency gains due to coasting and will only get worse the faster you go. Stated a different way, since you would be travelling at a considerably lower speed when in B mode (vs coasting) going down the same steep hill, then the efficiency losses due to regen braking would be more than offset by the lower losses due to wind resistance.

I'm guessing that you'd have to be coasting for a long time on a steep grade though before you would hit the speed at which coasting becomes less efficient than regen braking.

Is there a physicist in the house?

Going downhill, yes. If the slope is exactly the right grade to maintain 100kph, then by going 50kph will only be the result of regen. Actually, the only way to slow down from 100kph in that situation is to use regen. But let's you could change the grade of the road, then things would be different since regen would not be the only factor.

Actually, the most efficient road would be one that has a grade just steep enough to maintain the exact speed you're after. Such a road would not cause any regen losses at all.

But what about a flat road? Say you're traveling along at 200kph or around 125mph. That's about 55.556m/s. 55.556^2 x 1500kg = 4,629,703.704J, or around 1.29kWh of kinetic energy. Now if you could slam on your regen and instantaneously convert all that into electric energy at 39% then you'd get around 1/2 kWh hour back. At 6km or 4 miles per kWh you could go 3km or 2 miles. In reality you'd probably get a bit less than 1/2kWh back since that would require you to come to a stop instantaneously from 200kph to 0kph. So you won't be able to go even 3km or 2 miles on regen energy from slowing down from 200kph. But how far could you coast from 200kph on a flat road? My guess is that you'd be able to coast farther than 3km or 2 miles. But that's just my guess.
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GerryAZ
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### Re: REGEN Question

It really depends how you measure or define regeneration efficiency. The electric motor and inverter/converter system are quite efficient (probably over 90%) in both directions (the only difference between power generated by the motor and power consumed by the motor is direction of current flow). Since mechanical friction (bearings, tires, etc.) and aerodynamic drag are present at a given speed, the power required by the motor is the amount required to overcome those losses plus the amount of power required to move the car up the hill. Conversely, the power generated by the motor going down the same hill is the amount of power generated by the car going down hill minus the sum of mechanical and aerodynamic losses. The mechanical and aerodynamic losses make it appear that regeneration is not efficient.

Example:
10 kW for mechanical and aerodynamic losses at given speed
20 kW for energy to move car up hill and recover going down hill at given speed

Assume 100% efficiency in motor and inverter/converter system:
Net power from battery going up hill is 10 kW losses + 20 kW hill = 30 kW
Net power to battery going down hill is 20 kW hill - 10 kW losses = 10 kW

Assume 90% efficiency in motor and inverter/converter system:
Net power from battery going up hill is (10 kW losses + 20 kW hill) / 0.9 = 33.3 kW
Net power to battery going down hill is (20 kW hill - 10 kW losses) * 0.9 = 9.0 kW

This is why many people claim that only 1/3 of power is recovered by regeneration when regeneration is probably about 90% efficient.
Gerry
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LeftieBiker
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### Re: REGEN Question

I think that most people asking about regen efficiency are talking about 'efficiency in being able to use regenerated power to move the car' (like the OP here) as opposed to 'efficiency in recovering power from the road to the motor/generator output'. Just storing that power in the pack and then drawing it back out ruins the 90% estimate.
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alozzy
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### Re: REGEN Question

Awesome explanation, thanks @GerryAZ! Hadn't thought about it that way, makes total sense.

Dug into this a bit, out of curiosity...

Since drivetrain inefficiencies and rolling resistance are linearly related to speed, total resistance is dominated by air resistance at even moderate speeds, because wind resistance varies as the cube of the speed. So aero power (power needed to overcome wind resistance) at 100 mph is roughly 3 times the power requirement at 70 mph!

For the Nissan Leaf, aero power at 70 mph in a wind tunnel was 13.4 kw and at 100 mph was 40 kw, according to this source:

So, maintaining a speed of 70 mph, down a steep hill, while in B or ECO mode (i.e. using more regen) would reduce aero power by 26.6 kw versus coasting in neutral while maintaining an average speed of 100 mph.

If I've understood the physics properly, it seems clear that at higher downhill speeds it makes more sense to reduce speed and regen than it does to coast in neutral.

When traveling at lower speeds, I think it still makes sense to occasionally coast in neutral because reducing drive-train inefficiency will have a greater impact on reducing total resistance at those speeds.
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