I have read and heard many different numbers about it and after owning my Tesla for more than 2 years I wanted to test it out. Here is what I found. I hope this video helps understanding regen a little.
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How much energy can 'regen' capture?
- Thread starter David99
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- Driving Dynamics Model S regen
FlatSix911
Porsche 918 Hybrid
I would say those 30% are on the higher side. I think in normal every day driving it's less. When going faster, the aerodynamic losses become much bigger so percentage is probably also lower.
An owner here took his car up to the Saddle Road, marked a spot and then drove 10 rated mi. Up. Then he returned to his starting place and recovered 6 mi. When I drive to Waimea, the first 6mi. uses almost 600 watts/ mi.. the round trip( 80 mi. ) averages about 260 watts/mi. Actually this is a little less than the same distance on a relatively level road not a 2500 ft gain/ loss. Nothing scientific about this, but I tend to ignore altitude change if it begins and ends at the same height.
Short tests don't give you a reliable measurement. For no reason that I can understand, the display gives back regen miles in chunks of 3 (occasionally 4) miles at a time. Drive to the top of Pikes Peak from the Colorado Springs SpC (30 miles), it takes 79 rated miles. Coming back down, I got 22 miles of range back, but it displayed unchanged range for a while then went up by 3 miles, again by 3 miles, 4 miles, then 3 mile chunks for the rest of the drive. (There's only about a mile or two at the end of the drive back to the springs that's not down hill.)
Yeah I just didn't have the time to drive to Colorado from LA and back for this video LOL
The rated range display seems to only add miles in 1 kWh chunks. 1 kWh is a little over 3 miles, hence it only updates in 3 miles chunks. That's why I didn't use rated miles but looked at the kWh.
Thanks for the video. By the way, the energy estimates in V8.0 of the Model S/X firmware seem to be pretty accurate, even when climbing/descending mountains.
The way I like calculate the efficiency of regen, which the OP discussed in the video, is to estimate the gravitational potential energy, then subtract the energy generally required to drive at the expected speed on flat terrain, thus yielding us the "available" energy. The efficiency is thus the recaptured energy divided by the available energy.
Personally, I live on a mountain, and the descent down to the "flatlands" involves about 1500m of net elevation loss over 30km. Let's assume the total mass of my Model S is 2250kg (4960 pounds), including a driver and a passenger. Let's also assume a constant speed of 70 km/h (43.5 mph), which isn't quite accurate but shouldn't be far off. Also, from Model S Efficiency and Range we have an approximate energy consumption of 180 Wh/mile, or 112 Wh/km, at that speed.
PE = (2250 kg) * (9.8 m/s/s) * (1500m) = 33,000,000 Joules = 9.2 kWh (divide Joules by 3.6*10^6 to get kWh)
Flat terrain energy required = (30 km) * (112 Wh/km) = 3350 Wh = 3.4 kWh
Available energy = 9.2 kWh - 3.4 kWh = 5.8 kWh
The last time we did that particular descent, we departed with 80% SOC and ended up with 87% SOC. I should have thought to check the kWh, but didn't. From Realistic usable energy in 85kWh pack, I'm assuming that a new "85" pack had 77.8 kWh of usable energy, and that my 2012 Model S "A" pack has about 5% capacity loss, or 73.9 kWh usable. 7% of that would be 5.2 kWh.
So, if I gained about 5.2 kWh, out of 5.8 kWh "available", that's a regen efficiency of somewhere around 90%. If I'm right, this is quite good!
The way I like calculate the efficiency of regen, which the OP discussed in the video, is to estimate the gravitational potential energy, then subtract the energy generally required to drive at the expected speed on flat terrain, thus yielding us the "available" energy. The efficiency is thus the recaptured energy divided by the available energy.
Personally, I live on a mountain, and the descent down to the "flatlands" involves about 1500m of net elevation loss over 30km. Let's assume the total mass of my Model S is 2250kg (4960 pounds), including a driver and a passenger. Let's also assume a constant speed of 70 km/h (43.5 mph), which isn't quite accurate but shouldn't be far off. Also, from Model S Efficiency and Range we have an approximate energy consumption of 180 Wh/mile, or 112 Wh/km, at that speed.
PE = (2250 kg) * (9.8 m/s/s) * (1500m) = 33,000,000 Joules = 9.2 kWh (divide Joules by 3.6*10^6 to get kWh)
Flat terrain energy required = (30 km) * (112 Wh/km) = 3350 Wh = 3.4 kWh
Available energy = 9.2 kWh - 3.4 kWh = 5.8 kWh
The last time we did that particular descent, we departed with 80% SOC and ended up with 87% SOC. I should have thought to check the kWh, but didn't. From Realistic usable energy in 85kWh pack, I'm assuming that a new "85" pack had 77.8 kWh of usable energy, and that my 2012 Model S "A" pack has about 5% capacity loss, or 73.9 kWh usable. 7% of that would be 5.2 kWh.
So, if I gained about 5.2 kWh, out of 5.8 kWh "available", that's a regen efficiency of somewhere around 90%. If I'm right, this is quite good!
That would be good but impossible. Just the fact that the drive train (motor/inverter/battery) is in best case about 80% efficient, should be enough to realize that you can't capture 90% of the energy. As I showed in my video, you always have the losses inherent to driving. Tires, aerodynamics and other things. There is no way you can capture 90%.
Well, there are plenty of assumptions in my calculations, so 90% might not be right. But based on the metrics I'm using, I'm not seeing how that figure could be way, way off, either.
The "losses inherent to driving" are hopefully captured in my post, in the 3.4 kWh "flat terrain energy required" value. If that were removed, and we simply calculated efficiency as a percentage of re-captured gravitational potential energy, then we'd have 5.2 kWh regen divided by 9.2 kWh PE, or 57% efficiency.
Also, those efficiency numbers (both the 90% and 57%) only account for storing re-captured energy back in the battery pack. I'm not sure they're accounting for the losses involved in drawing the energy back out of the battery, though that should be fairly small as I believe Tesla batteries themselves are pretty efficient.
This is from Tesla's blog. For example, look at 40 mph (the speed I was going at for my test) and the energy usage. approx 110 is lost in tires, air drag and ancillary. Total is about 175. Subtract that from the total and you only have 33% is available at all for regen. Sounds strangely similar to what I was getting
<-- not an expert, but this doesn't pass my common-sense test. It seems you're comparing 2 different things and getting the same ratio and saying that based on that ratio your conclusion is valid (I'm not arguing if your conclusion is valid or not)This is from Tesla's blog. For example, look at 40 mph (the speed I was going at for my test) and the energy usage. approx 110 is lost in tires, air drag and ancillary. Total is about 175. Subtract that from the total and you only have 33% is available at all for regen. Sounds strangely similar to what I was getting
When you're accelerating you're fighting air drag and tire friction (so these are losses). When you're decelerating those items are helping you (they wont help you regain kwh to your pack, but they help you slow down), so they're not really loses here.
Am I missing something?
You're making invalid assumptions, which is why your numbers look odd. The part you left out is the energy required to cover the 14 miles (13.4 miles using the efficiency and total numbers you gave) at the speed you drove.
The energy used going up should be of the form (energy to drive that distance at that speed on flat ground + potential energy gained*efficiency) and the energy recovered going down is (energy flat - potential energy recovered by regeneration * efficiency).
You didn't give us enough information to fully model the event. In particular, the weight of the car and driver and/or the average speed would have been very helpful. I ran a 5k car with a spreadsheet and came up with a viable match with 250 Wh/mile flat consumption and 95% efficiency.
To look at sensitivity, I switched to a 4500 pound car. I was able to get it all to fit together if the motor/regeneration efficiency is 82% and your flat land equivalent consumption is 230 Wh/mile.
I'm pretty sure the reality is somewhere between these numbers...
aesculus
Still Trying to Figure This All Out
The best pure regen efficiency I have been able to come up by math is 89%. But as was stated above, that was just the effect of regen. By the time you factor in rolling resistance, wind, air density, cabin control ..., you loose a huge amount of that.
So unless you are in a vacuum sealed lab that is miles long with a fixed incline and no resistance at STP, it's going to be hard to tell what it is.
So unless you are in a vacuum sealed lab that is miles long with a fixed incline and no resistance at STP, it's going to be hard to tell what it is.
No matter how you cut it, it seems that the 'S' (and presumably the 'X') has very good regen.
And it's wonderful to be able to make a big mountain descent without needing to touch the brake pedal!
And it's wonderful to be able to make a big mountain descent without needing to touch the brake pedal!
SageBrush
REJECT Fascism
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%
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%
These losses slow the car down so they are not helping, they are taking energy away from regen. Anything that slows the car down is not available for regen hence it is a loss.
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