eKrom wrote: ↑
Tue Aug 04, 2020 7:12 am
Although 30% lose of conductance (Hx=70%) would mean the 30% more energy is lost to heat
R = 1/0.7 in the battery. And those heat losses are proportional to I^2*R where I is current, itself proportional to power
But for most typical driving battery losses are only a small fraction of energy use. Somewhere along the line in the life of the LEAF an owner might decide to only fast charge if no L2 option, and perhaps not stomp the accelerator to the floor at every opportunity. If that is your behavior anyway then the low(er) Hx has little daily effect on energy consumption. Someone who runs up and down the hills of San Francisco might say otherwise.
I think the arithmetic goes like this, but I'll rely on loren to correct any errors:
Say a new cell is 50 milliOhm
The pack is 192 cells arranged as 96S2P. The 2P parallelism reduces the resistance of the pack by 50% compared to a 96S pack.
We'll presume a 360 volt pack
Using I = W/V, If you are driving at 3*10*360 watts then the pack is pushing 30 Amps. That would be city driving at ~ 50 kph
Using I^2*R and taking account of the parallelism, Battery power losses are then 1/2 of 30*30*0.05 = 22 watts
Now say internal resistance has increased 50%. Losses in the battery will increase from 22 to 33 watts, and overall energy consumption will increase from 30*360 to 30*360+11.
Call it 0.1% more energy losses as the Hx drops.
That is easy city driving. Now let's look at DC fast charging at 120 Amps
The new pack heat generation will be 0.5*120*120*0.05 = 16*22.5 = 360 watts,
And a pack with 50% greater internal battery resistance will be 540 watts
Addendum: small edits for clarity