How much energy can 'regen' capture?

[Saghost]

175 Wh/mile on flat terrain at 40 mph strikes me as too low a number, but that is easily checked with another test. Assuming it is correct and that no energy was used going down,

175 Wh/mile * 14 miles = 2.45 kWh of consumption expected without the hill. 1.225 kWh each way
Without regen, the trip would have cost (5.7+1.225) = 6.925 kWh total, of which 2.45 kWh is expected from flat terrain over 14 miles,
And the height gained consumed (5.7 - 1.225) = 4.475 kWh

With regen, the total trip consumed (5.7-1.7) = 4.0 kWh
Since 2.45 kWh is flat terrain driving, the height up and down cost (4.0 - 2.45) = 1.55 kWh

So.... Addition of the hill up and down cost 1.55 instead of 4.475, or 34.6%. Regen then was (100-34.6) = 65.4%

Color me confused. The video said 5.2, not 5.7, and I have no idea where the 175 is coming from (the 40 mph for the Roadster from the blog post?)

 

[sorka]

We know from CANBUS logging paired with vbox power at the wheel measurements that we lose about 22% from the battery to the wheels. Lithium Ion batteries charge at nearly 100% efficiency, but once you throw drivetrain and inverter costs into it, if it's the same 22% going the other way, you'll won't recapture more than 78% percent of the cars kenetic energy to the battery. If you're talking a round trip, after removing air resistance, you can't capture more than 60%. This all assumes the battery itself is 100% efficient at capturing the electrons put into it. If less than 100% efficient, then that 78% one way and 60% round trip will be even less.

 

[SageBrush]

Color me confused. The video said 5.2, not 5.7, and I have no idea where the 175 is coming from (the 40 mph for the Roadster from the blog post?)

Right you are -- 5.2 kWh up, not 5.7
The flat terrain consumption is from post #18. If that is the Roadster then another test on flat terrain at 40 mph is in order for the Model S
I made a mistake in my prior post, so try #2

In the meantime, continuing to use 175 Wh/mile for flat terrain consumption at 40 mph, the corrected numbers are then:

Non-regen car:
5.2 kWh total, of which 2.45 kWh spent on flat terrain so up and down cost 2.75 kWh

With regen:
5.2-1.7 = 3.5 kWh spent on 14 mile trip
Since 2.45 spent on flat terrain, up and down cost was 1.05 kWh

Regen efficiency then is 100*(1-1.05/2.75) = 61.8%

 

[David99]

The Magic of Tesla Roadster Regenerative Braking

Quote from their blog post.

How much energy does it recover?


Unfortunately, the adage “your mileage may vary” applies to regen as well. The amount of energy you can recover depends on how and where you drive. From the powertrain point of view it looks pretty good. The energy conversion efficiencies from chemical to electrical (battery), DC current to AC current (inverter), electrical to mechanical (motor), and torque to force (transmission and wheels) are all quite high and work just as efficiently returning energy into the battery. The bigger problem is aerodynamic losses and higher speeds and rolling friction of the tires. These both act to slow the car, but the energy dissipated cannot be recovered. We must also remember that, even though the battery-to-wheel conversion efficiency is pretty good (up to 80% or so), the energy makes a full circle back into the battery and it gets converted twice for a net efficiency of at most 80% * 80% = 64%.

 

[sorka]

The Magic of Tesla Roadster Regenerative Braking

Quote from their blog post.

How much energy does it recover?


Unfortunately, the adage “your mileage may vary” applies to regen as well. The amount of energy you can recover depends on how and where you drive. From the powertrain point of view it looks pretty good. The energy conversion efficiencies from chemical to electrical (battery), DC current to AC current (inverter), electrical to mechanical (motor), and torque to force (transmission and wheels) are all quite high and work just as efficiently returning energy into the battery. The bigger problem is aerodynamic losses and higher speeds and rolling friction of the tires. These both act to slow the car, but the energy dissipated cannot be recovered. We must also remember that, even though the battery-to-wheel conversion efficiency is pretty good (up to 80% or so), the energy makes a full circle back into the battery and it gets converted twice for a net efficiency of at most 80% * 80% = 64%.

Amazingly close to my 78% one way and (.78 * .78) 60.8% round trip two posts above for a P85D based on KW out of the battery to power at the wheels.

 

[TIppy]

We know from CANBUS logging paired with vbox power at the wheel measurements that we lose about 22% from the battery to the wheels. Lithium Ion batteries charge at nearly 100% efficiency, but once you throw drivetrain and inverter costs into it, if it's the same 22% going the other way, you'll won't recapture more than 78% percent of the cars kenetic energy to the battery. If you're talking a round trip, after removing air resistance, you can't capture more than 60%. This all assumes the battery itself is 100% efficient at capturing the electrons put into it. If less than 100% efficient, then that 78% one way and 60% round trip will be even less.

Does the vbox power at the wheels subtract out the drag terms, or are they just looking at the acceleration of the mass of the vehicle?

 

[sorka]

Does the vbox power at the wheels subtract out the drag terms, or are they just looking at the acceleration of the mass of the vehicle?

No. That's why I did the calculation with the coast down curve to adjust power. Without that, it looks like the power decreases as speed goes up much more than it actually does but that calculation was done over two narrow speed ranges. But after you apply the coast down, which has greater and greater power drag as the speed increases, it flattens out the power curve at the wheels to almost perfectly flat at least up to 90 MPH.

Last weekend was a bust for me getting the more complete data because of the storm and then I traveled to the Bay Area for work like I usually do. This weekend will be fine though as there's no weather except warm and calm.

 

[TIppy]

There are canbus messages that give the instantaneous mechanical power and dissipation of the drive units. During acceleration the drive units are as high as 93% efficient while during regen they are about 86% efficient. This is just for the motor/inverters themselves. Under heavy acceleration (1600 amps) there are significant losses from the internal resistance of the battery (0.057 ohms) and the 0.013 ohms of resistance between the battery and the inverters. During regen (140 amps) these losses are pretty small.