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How effective is regenerative braking in recharging?

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I expect there must be a thread about this but I've not been able to find it. Apologies if this is old stuff.

Can someone tell me how effective/efficient regen braking actually is? By which I mean at recharging the batteries, not slowing the car. It's a feature I love, and I try hard to use the brakes as little as possible, it feels pretty good being able to slow down in this economic way and it's good for one's driving too, making one think ahead much better. and drive smoother

But how much current does it actually produce? . Of course I understand that with the battery full there's nowhere for generated charge to go, but for example, slowing down over say 500ft from say 60mph with a battery half full: how much use is that in terms of mileage gained? Is this even a reasonable question? But I can't find any evaluation of just how beneficial regen actually is.

For the record, I drive a RHD M3P- stealth, new last August.
 
I expect there must be a thread about this but I've not been able to find it. Apologies if this is old stuff.

Can someone tell me how effective/efficient regen braking actually is? By which I mean at recharging the batteries, not slowing the car. It's a feature I love, and I try hard to use the brakes as little as possible, it feels pretty good being able to slow down in this economic way and it's good for one's driving too, making one think ahead much better. and drive smoother

But how much current does it actually produce? . Of course I understand that with the battery full there's nowhere for generated charge to go, but for example, slowing down over say 500ft from say 60mph with a battery half full: how much use is that in terms of mileage gained? Is this even a reasonable question? But I can't find any evaluation of just how beneficial regen actually is.

For the record, I drive a RHD M3P- stealth, new last August.
The full cycle of wheel-battery-wheel is +/-70% efficient. In other words of the kinetic energy you'd normally expend using friction brakes about 70% of that is regained when you speed up again, and 30% is lost to heat in the battery, heat in the motor (AKA generator), imperfect tire grip, and so on.

That's pretty good for an energy round trip.

Of course ultimately the most efficient regen is the regen you don't use, in that if you always free-roll coast down to a stop without using friction brakes (or in the case of ICE, engine retarding) that will always be the more efficient path from energy use POV. But for times when it isn't practical to not use some sort of braking, and the lower your rolling resistance and air resistance is the more often that'll be the case, the Model 3 regen is really good.
 
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I expect there must be a thread about this but I've not been able to find it. Apologies if this is old stuff.

Can someone tell me how effective/efficient regen braking actually is? By which I mean at recharging the batteries, not slowing the car. It's a feature I love, and I try hard to use the brakes as little as possible, it feels pretty good being able to slow down in this economic way and it's good for one's driving too, making one think ahead much better. and drive smoother

But how much current does it actually produce? . Of course I understand that with the battery full there's nowhere for generated charge to go, but for example, slowing down over say 500ft from say 60mph with a battery half full: how much use is that in terms of mileage gained? Is this even a reasonable question? But I can't find any evaluation of just how beneficial regen actually is.

For the record, I drive a RHD M3P- stealth, new last August.

Regeneration on flat ground is converting the car’s kinetic energy into electric form, so the math is fairly simple, especially if you go metric.

60 mph = 96 km/h =~ 27 m/s.

With an 1800 kg car, your kinetic energy is 0.5*1800*(27)^2 = 656100 Joules.

3.6 MJ are one kWh, so 0.656 MJ is about 0.18 kWh.

So the total energy available to be recaptured from a 3 going 60 mph is a little less than you need to cover an EPA mile - and as noted above, you won’t get all of it back - in the best conditions the motor is 94% efficient and the pack is something similar.

Note the squared term, though. The energy available increases rapidly as speeds rise.
 
So the total energy available to be recaptured from a 3 going 60 mph is a little less than you need to cover an EPA mile - and as noted above, you won’t get all of it back - in the best conditions the motor is 94% efficient and the pack is something similar.
This is where the roughly 70% I mentioned above comes from. An estimate of the round trip by .92 wheels to motor * .94 on the motor * .95 battery (round trip in and out) * .94 motor * .92 motor to wheels = estimate of 71% efficiency. Those are guesses, the motor to wheels could be lower even though the drive train is fantastically simplistic, so it being around 65% wouldn't be surprising at all. Assuming 10% loss in the drivetrain instead of 8%, but it won't be much above that, or you'd never see the sort of Wh/mile or acceleration performance the Model 3 sees.
 
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Another way to see it is to look at the odometer. If you reset your odometer to zero right at the top of the incline, at the bottom, you will see the distance, and a negative value for wh/mile or km. Multiply them, and you should have a sense of exactly how much energy was reclaimed.
 
Everyone's hills, mountains, roads, etc. are going to vary. I've been meaning to test the efficiency at different speeds, will probably have data a couple months from now.

For now, I know in around-town driving (there are some very mild hills), if I use 12kWh I get back around 4kWh. Depending how you look at that, I recovered about 33% (total at end == 8kWh used, just to be clear). I actually got about the same in a completely flat area as well, so this seems to be a good very-rough estimate for in-town driving.

For highway driving it varies mostly depending on elevation change. If there's no elevation change, you get nothing significant back. With elevation changes, you can get a decent amount if the hills are steep enough (otherwise you just consume less since you're still fighting aerodynamics).
 
Thanks to all for really helpful responses, how nice to find a forum where people don’t just make sarky remarks!

Julien wrote “I remember reading the max on dual motor is about 75kW to 100kW.”

I take this to imply that if one were (improbably) to run a fully discharged vehicle downhill continuously for an hour it would completely charge the battery, ie, taking the lower efficiency figure, 75Kw/hr.

In terms of mechanics, KE=mV^2/2, so distance travelled seems irrelevant, it’s just the change in velocity? So in the example I gave it’s the energy recovered as the car comes to rest from the given speed (60 mph here) rather than the distance over which the regen operates?

However, more simply and perhaps more practically, taking SammichLover’s point about c.70% overall efficiency, using the regen would recover 70% of the distance travelled.

Or am I being over simplistic here? I think I’m understanding, just want to be clear!
 
Thanks to all for really helpful responses, how nice to find a forum where people don’t just make sarky remarks!

Julien wrote “I remember reading the max on dual motor is about 75kW to 100kW.”

I take this to imply that if one were (improbably) to run a fully discharged vehicle downhill continuously for an hour it would completely charge the battery, ie, taking the lower efficiency figure, 75Kw/hr.

In terms of mechanics, KE=mV^2/2, so distance travelled seems irrelevant, it’s just the change in velocity? So in the example I gave it’s the energy recovered as the car comes to rest from the given speed (60 mph here) rather than the distance over which the regen operates?

However, more simply and perhaps more practically, taking SammichLover’s point about c.70% overall efficiency, using the regen would recover 70% of the distance travelled.

Or am I being over simplistic here? I think I’m understanding, just want to be clear!

If the car was moving in a vacuum with no rolling resistance and the motor and pack efficiency were identical at all power levels, then the distance wouldn’t matter at all for the regeneration on a flat surface to a stop case.

I’m the real world, it doesn’t matter much, but results likely will change slightly.

Note that you’re also seeing a lot of discussion that’s assuming you’re talking about regeneration while descending hills instead of regeneration to stop the car on the flat.

The car’s efficiency is basically the same in both cases, but the energy source is different. As the car descends, it releases potential energy proportional to the weight of the vehicle, strength of the gravitational field, and altitude change.
 
This is where the roughly 70% I mentioned above comes from. An estimate of the round trip by .92 wheels to motor * .94 on the motor * .95 battery (round trip in and out) * .94 motor * .92 motor to wheels = estimate of 71% efficiency. Those are guesses, the motor to wheels could be lower even though the drive train is fantastically simplistic, so it being around 65% wouldn't be surprising at all. Assuming 10% loss in the drivetrain instead of 8%, but it won't be much above that, or you'd never see the sort of Wh/mile or acceleration performance the Model 3 sees.

I don’t think that the three spur gears, differential, and four CV joints between the motor and wheels are eating anywhere close to 8% of the power, so my guess is the car might return slightly above 80% wheel to wheel, but I haven’t seen anyone test it on a modern Tesla and I’m not sure the exact numbers matter that much right now anyway.
 
I would take issue with how the question is phrased in the first place.

"Can someone tell me how effective/efficient regen braking actually is? By which I mean at recharging the batteries, not slowing the car."

You shouldn't think of it as recharging the car because in doing so I think you are missing the point that the car had to expend more energy to be able to try and recapture this energy.
Even in the case of a hill where folks have seen range increase on decent it still isn't near the power it took to climb said hill.

Don't think of it as recharging think of it as not wasting energy as heat from the friction brakes.
 
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A lot of great answers. The question is about regen efficiency though, not round-trip. I would thus only calculate the pieces that take that energy from the wheel back to the battery. From what someone else presented, that would be .92 wheels to motor * .94 on the motor * .95 battery = 82%. Spending the energy in the battery to get the car to move is a separate discussion and applies whether you charged from the wall or regen'ed to get that energy in the battery.

TeslaFi indicates the max regen for an AWD model 3 at 85kW, when the battery is warm enough.
 
Thanks to all for really helpful responses, how nice to find a forum where people don’t just make sarky remarks!

Julien wrote “I remember reading the max on dual motor is about 75kW to 100kW.”

I take this to imply that if one were (improbably) to run a fully discharged vehicle downhill continuously for an hour it would completely charge the battery, ie, taking the lower efficiency figure, 75Kw/hr.

In terms of mechanics, KE=mV^2/2, so distance travelled seems irrelevant, it’s just the change in velocity? So in the example I gave it’s the energy recovered as the car comes to rest from the given speed (60 mph here) rather than the distance over which the regen operates?

However, more simply and perhaps more practically, taking SammichLover’s point about c.70% overall efficiency, using the regen would recover 70% of the distance travelled.

Or am I being over simplistic here? I think I’m understanding, just want to be clear!

If you found such a long hill, you'd eventually lose a significant amount of regen. Peaking out around ~80kW (I can't recall the exact I've seen) is pretty much Supercharging, but without the battery being preconditioned (heated) to be able to take more power. Without a hot battery, the charge rate a Model 3 can take at a Supercharger drops off quickly as the battery gets a higher state of charge. So likewise, barrelling down a steep hill for a while, you'd eventually start to get significant reductions in regen ability.

This can actually be seen in practice even with realistic hills. Coming down off a mountain highway, you'll often see some regen limitation increase. This goes away quickly with a bit of non-regen driving. They seem to be limiting how long you can pull max regen for, even more than what I outlined above.

You may be oversimplifying, but this is how we understand... basically everything in life. Little thought experiments are fun and can lead to learning things (when you see where you've oversimplified usually). I sure didn't realise that regen was variable and would decrease as it was used when I first got the car!

I don’t think that the three spur gears, differential, and four CV joints between the motor and wheels are eating anywhere close to 8% of the power, so my guess is the car might return slightly above 80% wheel to wheel, but I haven’t seen anyone test it on a modern Tesla and I’m not sure the exact numbers matter that much right now anyway.

Same. I hear people like sources, so I found this while looking (it just happens to be on the subject of EVs oddly enough): Electric Vehicles: Prospects and Challenges | ScienceDirect

"The latter study claims that 21.5% of fuel energy in an ICEV is used in the traction of the vehicle. Fuel energy is lost through [...] mechanical losses (38%) as illustrated in Fig. 2.7.1. These mechanical losses can be further broken down into 5% for air drag and 33% to overcome friction. The figures mentioned apply to an average-sized ICE passenger vehicle (year 2000 model). The friction losses of an average-sized passenger vehicle can be further subdivided into 35% to overcome tyre's rolling friction, 35% to overcome friction of the moving parts in the engine, 15% to overcome friction in transmission, and 15% to overcome friction created during brake contact."
I'm assuming their 15% for the transmission includes all mechanical losses from engine output to the wheel hub (minus the called-out brake friction). This is 15% of the 33% total mechanical losses, so actually only 5% or so. I'd be very surprised if the limited mechanical stuff in the Model 3 adds up to 8% as well in this context, since that would imply it has more transmission losses and a year 2000 ICE vehicle.

Looking this up is confusing, since a lot of sources are considering the losses inside the engine as well, or heat losses/inefficiencies (e.g. exhaust). This is the only one I skimmed that made it clear which part is purely the loss in mechanical transmission.

Would regen be different for RWD owners since it's a single motor?

Not really. From zero to full regen, the AWD version only uses the rear motor for regen. There are a couple exceptions to this (usually seen when navigating a parking lot, or when in reverse) but it's not meaningful in terms of power recovered.
 
I don’t think that the three spur gears, differential, and four CV joints between the motor and wheels are eating anywhere close to 8% of the power, so my guess is the car might return slightly above 80% wheel to wheel, but I haven’t seen anyone test it on a modern Tesla and I’m not sure the exact numbers matter that much right now anyway.
Bjorn Nyland monitored the OBD2 output while performing repetitive rabbit starts. The regen is easy to calculate. ~ 70 - 80% IIRC

I calculated the same one day after I drove up and down a mountain near my home.

The Model 3 is just amazing in this regard. It is also why extra car weight has nowhere near the energy consumption penalty in city driving we have come to expect from gassers
 
on the qualitative side, I've observed many instances during local driving where my roundtrip Wh/mi values are better when, avg speed being similar, the terrain is rolling hills vs flatter ground. I would have expected the opposite. It leads me to think that if my average speed is higher, then coasting becomes less efficient than regen due to aerodynamic losses.

I have often wondered whether it's best to float the accelerator pedal at the 'neutral' point to coast down the hill to let momentum carry me up the following uphill, or regen down the hill to maintain constant speed, and then throttle up the hill at the same speed. The answer seems to depend on the speed and aero drag forces. in my experience, regen is more efficient going down a hill if coasting would make me exceed 45mph.