The 40KWH Battery Topic

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danrjones said:
I also noticed today my Hx jumped up by a full 0.1. SOH stayed the same so far today.
Reading through the threads here I have discovered Hx was the old battery health meter before SOH was created / discovered.
There didn't seem to be a consensus for what it mean now, especially for the 2018's.

The most logical explanation for Hx is that it's the ratio of the present battery conductance (inverse of resistance) to the
the original battery conductance when new. Like SOH, it declines with battery age and degradation.
 
Interesting

It seems reading through here that many notice their hx increase at first when new. It will be interesting to see how long it does.
 
jdcbomb said:
2018 SV - 40 kWh - 2018/09 Manufacture Date - In Service 2019 January.

AHr 111.47 - SOH 96.56 - Hx 115.05 - ODO 6,176 mi - QC 57 - L2 155

Thoughts on how this compares to other 2018 units?

Your car was built a month earlier than mine, and after a year or so we both have about the same SOH. Good so far.
 
danrjones said:
Interesting

It seems resding through here that many notice their hx increase at first when new. It will be interesting to see how long it does.

That implies that the battery has become more efficient in the short term (a few months to a year) after being placed in service.
In the long run, it correlates with SOH.
 
Yep, it drops more or less constantly. Now aftet 13000km I have 96.93% SOH, it was 99.36% when I got the car from the dealer. It dropped 0.01% every day in autumn, stopped to drop in winter (like only 0.01% per week or so), then in february there was 1 bigger drop, like 0.5% and in march agai . Now it is like in winter, dropping 0.01% every few days/week .
When Hx goes upwards, it drops slower, when going down, SOH drops faster... This is my experience.
 
jdcbomb said:
2018 SV - 40 kWh - 2018/09 Manufacture Date - In Service 2019 January.

AHr 111.47 - SOH 96.56 - Hx 115.05 - ODO 6,176 mi - QC 57 - L2 155

Thoughts on how this compares to other 2018 units?

Mine is really early in its "in service" life, but a fact I didn't report before is that the build date was 3/18 according to the door.
So it was sitting for a full year, through a hot Palmdale summer, before I got it a week ago.

As of this AM:

AHr 112.68 SOH 97.61 Hx 98.16 QC 3 L1/L2 16 REAL ODO 228
 
LeftieBiker said:
As far as I can tell, the Hx reading is no longer any help at all, starting with the 40kwh packs. It makes no obvious sense as it once did...

To most before, it didn't make much sense. Never really could understand the preoccupation some have/had. SOH or the ratio of present
Ahrs to max Ahrs are more than adequate to assess the battery's condition.
 
lorenfb said:
LeftieBiker said:
As far as I can tell, the Hx reading is no longer any help at all, starting with the 40kwh packs. It makes no obvious sense as it once did...

To most before, it didn't make much sense. Never really could understand the preoccupation some have/had. SOH or the ratio of present
Ahrs to max Ahrs are more than adequate to assess the battery's condition.

That is mostly my opinion too although I thought Hx was a marker of battery resistance.
 
SageBrush said:
lorenfb said:
LeftieBiker said:
As far as I can tell, the Hx reading is no longer any help at all, starting with the 40kwh packs. It makes no obvious sense as it once did...

To most before, it didn't make much sense. Never really could understand the preoccupation some have/had. SOH or the ratio of present
Ahrs to max Ahrs are more than adequate to assess the battery's condition.

That is mostly my opinion too although I thought Hx was a marker of battery resistance.

Hx declines as the battery ages, as has been noted on MNL. Battery resistance increases with battery age. Given that, a logical
conclusion is that Hx represents the ratio of the new/original battery resistance to the battery resistance as it ages (present),
or the ratio of conductances (present/new).

With regard to the 40kWh battery and Hx, it's possible that the original BMS firmware for Hx wasn't tweaked for compatibility.
 
lorenfb said:
SageBrush said:
lorenfb said:
To most before, it didn't make much sense. Never really could understand the preoccupation some have/had. SOH or the ratio of present
Ahrs to max Ahrs are more than adequate to assess the battery's condition.

That is mostly my opinion too although I thought Hx was a marker of battery resistance.

Hx declines as the battery ages, as has been noted on MNL. Battery resistance increases with battery age. Given that, a logical
conclusion is that Hx represents the ratio of the new/original battery resistance to the battery resistance as it ages (present),
or the ratio of conductances (present/new).
Is it correct to say that battery (A) with the same capacity as battery (B) but 25% higher R would have 25% lower range if we presume identical driving ?
 
I'm not a battery expert but it seems to me resistance by itself would not determine battery capacity, but might impact available power at any given time or maximum charge speed.

There was a good video - I'm not sure but it may have been transport evolved channel on you-tube - where she talked about the oxidation (?) layer that built up on the original Gen 1 batteries.

It appeared resistance was coupled to loss of spaces for the ions to go TO, but I don't think higher resistance in "principle" means a loss of battery capacity. In terms of reality, the oxide layer or whatever built up seems to both add resistance (makes sense, harder for the ions to get through) and also prevented ions from getting into their spaces.

But I can imagine a condition where there is more resistance getting ions into a position but all the positions are still available.
Perhaps though without applying a higher voltage to charge as the battery ages, you can't get all ions to penetrate that oxide (?) layer.

An interesting experiment would be to take a highly degraded battery and try and charge it at higher voltages.
 
danrjones said:
I'm not a battery expert but it seems to me resistance by itself would not determine battery capacity, but might impact available power at any given time or maximum charge speed.

A higher internal resistance causes more heat production, especially at higher current. In particular, I noticed that when I replaced the battery pack on my 2011, the car had significantly more power than before. In addition, accelerating rapidly had a massive impact on the "fuel economy" with my old battery, but much less so with the new battery.

Also, you'd want to avoid rapid charging at high internal resistances as well, due to heat production and possibly other factors. Nissan resolves this issue by basically disabling regen braking (limited to <10 kW & only below 20 mph) as the battery begins to degrade. My old pack had 0 regen braking at speeds above 35 mph.

The effect of this is that you're burning up the friction brakes below 8 bars or so, and your range is significantly less as a result of all the above.

To answer SageBrush's original question:
SageBrush said:
Is it correct to say that battery (A) with the same capacity as battery (B) but 25% higher R would have 25% lower range if we presume identical driving ?

No, it is not. The vast majority of energy discharged by the battery goes into the motor and wires, not into resistance in the battery. So yes, you would have 25% higher resistive losses in the battery, but these account for a small fraction of the total energy used. However, at higher power outputs, the battery internal resistance losses are amplified.

Practically speaking, with a Leaf, your range may still decline significantly from all factors, but primarily due to the loss of regen braking. My new battery took me from SOH 55%->87%, but more than doubled the range.
 
Lothsahn said:
To answer SageBrush's original question:
SageBrush said:
Is it correct to say that battery (A) with the same capacity as battery (B) but 25% higher R would have 25% lower range if we presume identical driving ?

No, it is not. The vast majority of energy discharged by the battery goes into the motor and wires, not into resistance in the battery. So yes, you would have 25% higher resistive losses in the battery, but these account for a small fraction of the total energy used. However, at higher power outputs, the battery internal resistance losses are amplified.

Good answer. Actually it depends on the relative size of the battery resistance losses compared to the other key losses,
i.e. motor/controller, rolling resistance, and drag. So for all practical purposes, the answer is no.
 
Lothsahn said:
To answer SageBrush's original question:
SageBrush said:
Is it correct to say that battery (A) with the same capacity as battery (B) but 25% higher R would have 25% lower range if we presume identical driving ?

No, it is not. The vast majority of energy discharged by the battery goes into the motor and wires, not into resistance in the battery. So yes, you would have 25% higher resistive losses in the battery, but these account for a small fraction of the total energy used.
This is one of those problems that I get different answers depending on how I think about it.

What you say makes sense: if baseline heat generation in the battery is say 2% of the power draw then a 25% increase in resistance would presumably (?) increase heat generation up to 2*1.25 = 2.5% of the power draw. No big deal.

But then I start from
P = I*I*R and I conclude that an increase in R to R1 would lead to a drop in I = sqrt (R/R1)
Then e.g., an increase in R of 25% would lead to a drop in I to sqrt(0.75) = 86.6%
 
SageBrush said:
What you say makes sense: if baseline heat generation in the battery is say 2% of the power draw then a 25% increase in resistance would presumably (?) increase heat generation up to 2*1.25 = 2.5% of the power draw. No big deal.

above is battery power loss - from an increase in battery resistance only

SageBrush said:
But then I start from
P = I*I*R and I conclude that an increase in R to R1 would lead to a drop in I = sqrt (R/R1)
Then e.g., an increase in R of 25% would lead to a drop in I to sqrt(0.75) = 86.6%

above is total power loss - from a total load resistance increase

You're confusing the two different situations.
 
lorenfb said:
SageBrush said:
What you say makes sense: if baseline heat generation in the battery is say 2% of the power draw then a 25% increase in resistance would presumably (?) increase heat generation up to 2*1.25 = 2.5% of the power draw. No big deal.

above is battery power loss - from an increase in battery resistance only

SageBrush said:
But then I start from
P = I*I*R and I conclude that an increase in R to R1 would lead to a drop in I = sqrt (R/R1)
Then e.g., an increase in R of 25% would lead to a drop in I to sqrt(0.75) = 86.6%

above is total power loss - from a total load resistance increase

You're confusing the two different situations.
I'm obviously being dense about this, thanks for the patience.

If I just consider the battery in isolation, is my use of I*I*R correct ?
That is, does an increase of R to R1 reduce the current to sqrt(R/R1) ?
 
danrjones said:
...
There was a good video - I'm not sure but it may have been transport evolved channel on you-tube - where she talked about the oxidation (?) layer that built up on the original Gen 1 batteries.

It appeared resistance was coupled to loss of spaces for the ions to go TO, but I don't think higher resistance in "principle" means a loss of battery capacity. In terms of reality, the oxide layer or whatever built up seems to both add resistance (makes sense, harder for the ions to get through) and also prevented ions from getting into their spaces. ...

Sounds like you're describing the SEI (Solid electrolyte Interphase) layer.
 
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