Does Chill mode increase efficiency? Surprisingly, Tesla says yes

Does Chill mode increase efficiency?

This question seems to come up in Tesla circles every so often, and the conventional answer is "no, if you drive exactly the same speeds then acceleration mode does not impact efficiency".

But Tesla has some interesting notes in the manual:

If your vehicle is equipped with a heat pump (to determine if your vehicle has a heat pump, touch Controls > Software > Additional Vehicle Information), you can improve the efficiency of the cabin heating by reducing your selected acceleration mode. This allows the heat pump system to take more heat from the Battery to efficiently heat the cabin, instead of maintaining the Battery's ability to provide peak acceleration performance.

So in weather cold enough to use cabin heating apparently yes it can improve efficiency.

Link to the section for Model 3 (it also exists for S and presumably any heat bump vehicle).

"Tesla Model 3 | E-Cannonball 2018" by JayUny is licensed under CC BY-SA 2.0.
Admin note: Image added for Blog Feed thumbnail

 

[gsmith123]

That's why some EVs have 800 Volt batteries. It results in half the Ampere which 1/4 of the losses.

Yes and no. A higher voltage lower amperage powertrain can also use smaller conductors (for cost and weight savings), which counteracts that somewhat. I'm not sure which factor wins out in practice

800v is I believe more for the reduction in copper usage and for charger benefits (needs less current from the charger, so lower cable heating)

 

[sleepydoc]

Early Teslas had induction motors that have slippage. They are not as efficient, but can roll free without any power. All dual motor Teslas today are running on a permanent magnet reluctance motor in the back, which has no slippage. The rotor follows the magnetic field 1:1. The front motor is an induction motor which most of the time runs free without power. Only when accelerating hard and a little bit when regenerating is the front motor used.

I think the main reason the EV isn't doing so well when pushing it hard is the ohmic losses. Because the voltage is 'constant', the way to increase power is using more Amps. Double the power result in 4 times the losses in all current carrying wires and inside the battery. 4 times the power means 8 times the ohmic losses. That's why some EVs have 800 Volt batteries. It results in half the Ampere which 1/4 of the losses.

Assuming constant resistance and voltage, ohmic losses are constant for a given amount total power delivered. As conductors heat up, the resistance will increase a bit which can have an effect and the internal resistance of the batteries/voltage drop at higher currents can also have an effect.

Also, the power curve for Tesla's motors is relatively flat but they're still not perfect so there are some increased losses at higher torques. It also depends on which motors are being used - if the front induction motor is used (likely with hard acceleration,) there will be slippage.

 

[David99]

Assuming constant resistance and voltage, ohmic losses are constant for a given amount total power delivered.

Resistance is constant but not the losses. P(loss) = R * I²

I was responding to the question why EV do worse when requiring higher power.
During normal cruising you will draw approx. 40 Amp from the battery. If you push the pedal, you will draw over 800 Amp. While that's 20 times the power, the losses are 400 times higher.

 

[sleepydoc]

Resistance is constant but not the losses. P(loss) = R * I²

yes, but do the math. At higher currents, losses increase but for a given total energy delivered (i.e. speed obtained) the total energy loss is constant, regardless of the current. This is because lower currents also deliver less energy so even though the resistive losses are lower, they continue for a longer period.

 

[David99]

yes, but do the math. At higher currents, losses increase but for a given total energy delivered (i.e. speed obtained) the total energy loss is constant, regardless of the current. This is because lower currents also deliver less energy so even though the resistive losses are lower, they continue for a longer period.

I'm not sure how else to explain it. The increase in losses is not in proportion with the increased power but squared. This a well known issue.

 

[gsmith123]

Lets actually do the math. The kinetic energy of a Model 3 at 70 mph is about 1000kJ (calculation on Wolframalpha).

Lets compare accelerating 0-70 mph in 3 seconds vs 30 seconds.

3 second 0-70 mph case
- Average power = 1000kJ / 3 seconds = 333kW
- Average current (assuming 400V pack) = P / V = 333kW / 400V = 832A
- Average Ohmic loss (assuming 0.01 ohm resistance from battery to motor) = I^2 * R = 6920W
- Total Ohmic loss = 3 seconds * 6920W = 20kJ

30 second 0-70 mph case
- Average power = 1000kJ / 30 seconds = 33kW
- Average current = P / V = 33kW / 400V = 83A
- Average Ohmic loss = I^2 * R = 69W
- Total Ohmic loss = 30 seconds * 69W = 2kJ

Conclusion: accelerating 10x faster results in 10x higher total losses from electrical resistance.

Note:
- I'm making rough assumptions about both the pack voltage and resistance from pack to motor, but the outcome doesn't depend on these being accurate since we're using the same estimates for both scenarios and comparing relative outcomes. That said, these are pretty close. At very high SOC some Tesla's are a little below 400V, some a bit over. The resistance was a bit more of a guess
- There's a bunch of a factors I'm not consider here, of course, like air drag, voltage sag, tire deformation, rolling resistance, and many others.
- I'm assuming constant power across the 0-70 mph acceleration, but in practice that's probably not possible for the 3 second case where at low speeds acceleration (and therefore power) is traction limited

 

[Max Spaghetti]

- There's a bunch of a factors I'm not consider here, of course, like air drag, voltage sag, tire deformation, rolling resistance, and many others.

The one factor you could include here to make it a better calculation is the cruise time the first car then spends at 70mph to then be at the same physical point down the road as the second car.

The first car will spend some number of seconds cruising at 70mph consuming approx 15kW. This calculation will actually add further to the ohmic losses of the first car.

The distance travelled by the first car (average speed x time):
35 mph x 3 seconds = 35 x 3 / 3600 = 0.02917 mi

The distance travelled by the second car (average speed x time):
35 mph x 30 seconds = 35 x 30 / 3600 = 0.2917 mi

The distance the first car then needs to cruise is:
0.2917 - 0.02917 = 0.2619 mi

It's doing that at 70 mph, so the time spent in cruise is:
0.2619 / 70 x 3600 = 13.47 seconds.

13.47 seconds at 70 mph
- Average power = 15kW
- Average current (assuming 400V pack) = P / V = 15kW / 400V = 37.5A
- Average Ohmic loss (assuming 0.01 ohm resistance from battery to motor) = I^2 * R = 14W
- Total Ohmic loss = 13.47 seconds * 14W = 0.19kJ

Add that to the 20kJ of the first car gives 20.19kJ.

Not as significant as I was expecting, but there's a slightly more complete answer.

 

[David99]

You are correct, it takes 1/10 of the time at 10 times the power to accelerate to the same speed. But you forgot to factor in the voltage drop losses that are due to the higher current draw. You assumed the same pack voltage in both situations. In reality the pack voltage drops under load.

Based on your numbers, the difference between a power draw of 0.3 C vs 3 C results in approx a 15% lower voltage. Instead of 400 Volt we only have 340 Volt. (In reality the pack voltage is lower for both, but that's not important as the delta is the same).

To get the same power, the system has to pull more Amps. 333 kW / 340 Volt = 980 Amps. Total losses are 9600 Watt.

9600 Watt * 3 seconds = 29 kJ.

We have 29 kJ vs 20 kJ for the slower acceleration. That's an almost 30% higher energy consumption.

Based on my research the 0.01 Ohm is probably a good estimate for the battery. But that's only the battery. The bus bars, wiring, and inverter all suffer the same ohmic losses. So in reality the voltage drop is even more, thus an even higher current (Amps) are necessary to deliver the same power needed. Which makes the difference even worse for the hard acceleration scenario. I can't find any info on the resistance of the above mentioned items, but I let's assume they all account for another 10% difference, bringing the total close to 40%. That's significant.

 

[terranx]

All dual motor Teslas today are running on a permanent magnet reluctance motor in the back, which has no slippage. The rotor follows the magnetic field 1:1. The front motor is an induction motor which most of the time runs free without power. Only when accelerating hard and a little bit when regenerating is the front motor used.

Doesn't really change the overall point of your post, but for what it's worth, S/X have permanent magnet front and rear

 

[sleepydoc]

Lets actually do the math. The kinetic energy of a Model 3 at 70 mph is about 1000kJ (calculation on Wolframalpha).

Lets compare accelerating 0-70 mph in 3 seconds vs 30 seconds.

3 second 0-70 mph case
- Average power = 1000kJ / 3 seconds = 333kW
- Average current (assuming 400V pack) = P / V = 333kW / 400V = 832A
- Average Ohmic loss (assuming 0.01 ohm resistance from battery to motor) = I^2 * R = 6920W
- Total Ohmic loss = 3 seconds * 6920W = 20kJ

30 second 0-70 mph case
- Average power = 1000kJ / 30 seconds = 33kW
- Average current = P / V = 33kW / 400V = 83A
- Average Ohmic loss = I^2 * R = 69W
- Total Ohmic loss = 30 seconds * 69W = 2kJ

Conclusion: accelerating 10x faster results in 10x higher total losses from electrical resistance.

Note:
- I'm making rough assumptions about both the pack voltage and resistance from pack to motor, but the outcome doesn't depend on these being accurate since we're using the same estimates for both scenarios and comparing relative outcomes. That said, these are pretty close. At very high SOC some Tesla's are a little below 400V, some a bit over. The resistance was a bit more of a guess
- There's a bunch of a factors I'm not consider here, of course, like air drag, voltage sag, tire deformation, rolling resistance, and many others.
- I'm assuming constant power across the 0-70 mph acceleration, but in practice that's probably not possible for the 3 second case where at low speeds acceleration (and therefore power) is traction limited

I made a post a few months ago and did the calculations - when I did mine they showed no difference. Of course I can’t find that post now but looking at your calculations I can’t see any mistakes. If your calculations are correct and the answer’s different then I have to assume I had a mistake in mine.