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Re: The 40KWH Battery Topic

Posted: Thu May 09, 2019 9:30 am
by SageBrush
lorenfb wrote:
SageBrush wrote:
lorenfb wrote:
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 ?

Re: The 40KWH Battery Topic

Posted: Thu May 09, 2019 11:17 am
by danrjones
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.

Re: The 40KWH Battery Topic

Posted: Thu May 09, 2019 11:41 am
by Lothsahn
danrjones wrote: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 wrote: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.

Re: The 40KWH Battery Topic

Posted: Thu May 09, 2019 7:20 pm
by lorenfb
Lothsahn wrote: To answer SageBrush's original question:
SageBrush wrote: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.

Re: The 40KWH Battery Topic

Posted: Fri May 10, 2019 5:25 am
by SageBrush
Lothsahn wrote: To answer SageBrush's original question:
SageBrush wrote: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%

Re: The 40KWH Battery Topic

Posted: Fri May 10, 2019 5:55 am
by lorenfb
SageBrush wrote:
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 wrote: 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.

Re: The 40KWH Battery Topic

Posted: Fri May 10, 2019 6:08 am
by SageBrush
lorenfb wrote:
SageBrush wrote:
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 wrote: 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) ?

Re: The 40KWH Battery Topic

Posted: Fri May 10, 2019 6:25 am
by Nubo
danrjones wrote:...
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.

Re: The 40KWH Battery Topic

Posted: Fri May 10, 2019 6:27 am
by lorenfb
SageBrush wrote:
lorenfb wrote:
SageBrush wrote:
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 wrote: 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) ?
Yes, for a short circuit test of the battery, i.e. you connect the two terminals of the battery together (no load) and the battery resistance
increased. Typically the resistance would decrease as it heated. Or you used to different external resistors and did that test.

Re: The 40KWH Battery Topic

Posted: Fri May 10, 2019 9:11 am
by SageBrush
lorenfb wrote:
SageBrush wrote:
lorenfb wrote:
above is battery power loss - from an increase in battery resistance only



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) ?
Yes, for a short circuit test of the battery, i.e. you connect the two terminals of the battery together (no load) and the battery resistance
increased. Typically the resistance would decrease as it heated. Or you used to different external resistors and did that test.
:: light bulb ignites ::
I think I finally get it

(R1 - R) acts as a additional serial resistor in my simple circuit