I'm out after 6 months

[edcoble]

Thanks, Dan.

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[edcoble]

Dan, another question: did you drive the other direction, from Amarillo to Trinidad? If so, were you able to drive it without charging?

 

[Grease Lightning]

The safest speed to drive is that of surrounding traffic regardless of the posted speed limit. Differential speed (higher or lower) is less safe.

I would disagree on that one 1000%. Just because so 🫏 is speeding and driving erratically doesn’t mean i should match their driving style.

Rather on the hwy it is safer to move over and let them pass.

Equal the other way. If someone is impeding traffic, you safely pass them where you legally can and drive the speed limit.

 

[Texas Dan]

Dan, another question: did you drive the other direction, from Amarillo to Trinidad? If so, were you able to drive it without charging?

No, there is a 3K foot elevation change going the other way and there is usually a strong headwind. I usually charge in Wagon Mound but of course that adds 60 miles. When the Superchargers open up I will be able to charge in Clayton and drive straight through.

There is also a couple of new CCS chargers going up in Des Moines. You should be able to make to Des Moines from Amarillo if you start with a full charge. Des Moines to Trinidad would be a piece of cake.

 

[edcoble]

Thanks! I was afraid of that. I've been watching for that promised Des Moines installation but haven't seen any actual work occurring yet.

 

[Texas Dan]

Thanks! I was afraid of that. I've been watching for that promised Des Moines installation but haven't seen any actual work occurring yet.

There are two new Tritium chargers at the Phillips 66 station, these are not the Sierra Grande chargers that have been promised for so long. The chargers look ready go but I don’t know what network they will be on and I don’t know what the power rate will be. I will be going through Des Moines this weekend and I will take a closer look at the chargers.

 

[edcoble]

Ah, thanks. We'll be going through there the day after Christmas . A working charging station there would be great.

 

[wighty]

My limited understanding (I did not stay at a Holiday Inn Express last night) is that the available power output might be less when super cold, but the stored energy is not affected.

The kWh’s are all there, they’re just less effective/effecient.

Bit of a bump here but I wanted to expand on this point here with some actual data and the temps have been too warm... the "stored energy"/kWh is not the same when the temperature of the pack is lowered... temperature lowers the voltage of a battery, which directly reduces the amount of energy available. Heat is energy, so if you reduce the temperature of the battery you are removing energy from the electrons (demonstrated via voltage drop, since voltage is basically the 'potential energy' of the electrons).

I left my truck out overnight last night and looked at the sensor data via OBDII and car scanner (my battery state of health is 98.5% after 16k miles)...
Truck dash Reporting state of charge: 86% (SR so 98 kWh available to me)
HVB State of charge: 76.61% (presumably this is the "true" SOC including the buffer, so 107 kWh battery)
HVB Energy to empty: 64.47 kWh
HVB temp: 26.6*F

After a short drive, then charge up to 90%:
Truck reporting: 90% charge
HVB SOC: 82.9%
HVB Energy to Empty: 80.06 kWh
HVB temp: 55.4*F

So for the cold state (26.6*F), you would expect your available energy to be 98 kWh * 98.5% health * 86% SOC = 83 kWh... but the truck was reporting available to empty as 64.5 kWh (-18.5 kWh)

Warmer state (55.4*F)... 98 kWh * 98.5% * 90% SOC = 86.9 kWh. Truck is reporting 80 kWh, much closer (-6.9 kWh). I have not seen anyone do some detailed analysis of what the "energy to empty" sensor is reporting, but at the very least you can see that the delta between the cold and warmer states are at least different as you would expect.

Unless you are using an external source of energy (charging/grid power, or weather temp increases, I would include heat losses from driving/regenerative braking I guess too but no idea how strong of an effect that is) you are not going to magically 'unlock' more energy out of the pack... if the pack uses its own energy with the conditioning system to heat the battery up it is going to consume an equivalent amount from the battery to do that. This part will likely change with a heat pump system, but cold temp COP isn't going to be the greatest so it may not make that big of a difference overall.

 

[TaxmanHog]

I've been watching my SOC & projected range over the weekend ever since charging the truck Saturday evening to 85%, while charging the heat created helped to improve the range, as the truck cooled overnight and into all day Sunday, the expected range diminished as the truck lost all the pack heat.

What I found peculiar is the [Distance added] when colder is actually the [Distance LOST] due to the cold battery pack, if I had used departure timer, and warmed up the pack it would have improved.
Saturday 1/6--------------------Sunday 1/7---------

 

[Zprime29]

Bit of a bump here but I wanted to expand on this point here with some actual data and the temps have been too warm... the "stored energy"/kWh is not the same when the temperature of the pack is lowered... temperature lowers the voltage of a battery, which directly reduces the amount of energy available. Heat is energy, so if you reduce the temperature of the battery you are removing energy from the electrons (demonstrated via voltage drop, since voltage is basically the 'potential energy' of the electrons).

But when the battery warms up it somehow gains that energy back? The physics of this is a little out of my wheel house but something that does interest me. While I was on vacation over the holidays, I was plugged into a 120V outdoor outlet to top up. I was at 100% indicated SOC one morning when we left to go to lunch, about 10 miles away. It was in the 30's overnight and almost 60 after lunch. Despite driving 10 miles (50-60mph, slight elevation gain), I had 100% SOC upon turning the truck on to go home. Where did that extra energy come from? Does the battery absorb electrons from the radiant heat of the parking lot? Or is it from the energy that was already stored in the battery but didn't become available until the battery warmed up? The latter makes more sense but seems to contradict the claim that batteries don't store as much energy when cold.

 

[wighty]

But when the battery warms up it somehow gains that energy back?

The voltage of the pack will increase, so the capacity will increase. You have the same number of electrons, but they have more energy. This is only a net positive for you if you have an external source of energy for the heat, though.

In your case, the weather changed, so the higher air temp warmed the pack up... That is where the energy came from... My original entire point was whether or not Ford had the truck take this into consideration on the battery gauge, and I think the answer is "no", or at least not substantially. I think they would be mostly worried about owners charging to 90 or 100% and unplugging, and then in the morning seeing their charge drop to like 70-80% without any driving.

 

[Jim Lewis]

Heat is energy, so if you reduce the temperature of the battery you are removing energy from the electrons (demonstrated via voltage drop, since voltage is basically the 'potential energy' of the electrons).

Not quite the right wording. The energy isn't removed directly from the electrons but from the effect of temperature on the chemical reaction(s) that produces the electrons.

Here's what some battery chemistry professionals say:

Introduction

Lithium-ion batteries (LIBs) operating at a low temperature are highly wanted in the cold seasons or locations for different applications such as electric vehicles, submarines, and airplanes. Anxiety rises for the reduced battery capacity or drive range but the increased safety issues at the low temperature, which dampens the enthusiasm for widespread usage of LIBs. This is largely due to the slowed-down kinetics of Li ions associated with movement and reactions, resulting in increased resistance, reduced power capability, and even dendritic Li plating1,2,3,4. Thoroughly understanding the influence of temperature on the underlying microstructure and performance of the battery is essential to solving the kinetic bottlenecks and achieving high performance at a low temperature.

Source: Temperature-dependent interphase formation and Li+ transport in lithium metal batteries | Nature Communications

(A 2023 research paper. Nature is one of the preeminent scientific publications in the world. The wording of the paper introduction is likely that the authors are non-native English speakers)

 

[wighty]

Not quite the right wording. The energy isn't removed directly from the electrons but from the effect of temperature on the chemical reaction(s) that produces the electrons.

Fair point... My post was already longer than I meant

 

[Jim Lewis]

Not quite the right wording. The energy isn't removed directly from the electrons but from the effect of temperature on the chemical reaction(s) that produces the electrons.

Here's a good explanation of why, when low temperature causes the internal resistance of a Li-ion battery to increase, the voltage across the load decreases. Basically, because the voltage the chemical reaction can deliver will be distributed as the voltage internal to the battery plus the voltage across the load. When the voltage drop internal to the battery rises because of increased resistance, the voltage available to the load has to drop.

Consider the actual circuit of the above battery feeding a load resistance RL.

By Ohm’s Law, Is = Voc/Rt = Voc/(Rint + RL)

The voltage across the load, VL = Is ٠ RL = Voc ٠ RL / (Rint + RL)

The voltage “drop”, is the voltage lost across the internal resistance of the battery.:

Vdrop = Voc - VL = Is ٠ Rint = Voc - [Voc ٠ RL/(Rint + RL)]

or Vdrop = Voc [ 1 - RL/(Rint + RL)]

As you can see, the higher the internal resistance of the battery (Rint), the greater the voltage drop will occur across Rint for a given load current Is, and the less voltage you’ll have available across the load (VL). Similarly, the smaller the load resistance RL, the larger effect the internal voltage drop across Rint will have on the voltage available across the load (VL).

I was tempted to show you how to solve the above relationship for Rint, the internal resistance of a battery by (a) measuring its open circuit voltage Voc, and then (b) its terminal voltage VL when you connect a load resistance RL across the battery, but I suspect that if you want that information, you can readily do the algebra yourself.

Source: https://qr.ae/pKx0I5

The guy who contributed the Quora post says his credentials are: " MSEE & BSEE, 50+yrs NW R&D EE, Power/Control/Comm Systems."

 

[MickeyAO]

Ohms law is in full effect! As the temperature rises (to a certain point) the resistance goes down, thus the voltage goes up. Since the GOM is based on the voltage of the HVB, the range and SOC will go up.

If you want to see something wild, when I plot voltage/discharge capacity on a cell that is at -30C, I can see the voltage drop like a rock and then suddenly climb (while still discharging) as the internal temperature of the cell climbs, and then go back to Vmin.. Typically happens at higher C rates during out static capacity tests.

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