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Technical Battery Discussion

So I am an engineer and really want to understand the battery. It is my understanding that there are 6831 NCR18650 Panasonic batteries in the "Battery pack". 11 moduals in series each with 9 "bricks" in series and each brick with 69 18650's in parallel. So 69 * 9 * 11 = 6,831. Are you with me so far?

Now the hard part. On Panasonic's website i get specs for the NCR18650 showing a nominal voltage and capacity of 3.6 VDC and 2.9 AH, respectively. Lets do the voltage first. 9 bricks X 3.6 volts X 11 moduals = 356.4 volts. But the Tesla specs say the battery is 375 volts. Backing into the nominal voltage needed to get 375 volts each battery has to have a nominal of 3.78 volts. If someone knows the answer to this puzzle I would greatly appreciate an explanation. Thanks in advance.

Now lets do kWH. 69 cells X 2.9 AH = 200.1 AH/brick X 9 bricks X 11 moduals X 3.78 volts /1000 = 74.9 kWH. So how do I match this up to the 56 kWH claimed by Tesla? At 75% the number is 56kWH. Does that mean that only 75% of the batteries nominal capacity is "usable"? Is this because there is about a 25% loss between input energy (from the HPC) and actual usable stored and and delivered energy from the battery to the inverter/motor?

Finally, the new 18650 battery is supposed to be 4 AH instead of 2.9 AH in 2012. The 2.9 AH battery weighs 44 g while the new 4 AH battery weighs 54 g. 6831 * 44 g = 300.564 kg or 663 lbf. Since the entire battery weighs 990 lbf, the battery enclosure, coolant and electronics must weigh 990 - 663 = 327 lbf. Therefore if the new batteries weigh 6831 * 54 g = 368.874 kg or 814 lbf, then a new battery pack for my Roadster with the new batteries should weigh 814 + 327 = 1,141 lbf. For the extra 151 lbf of weight, one should get an increased nominal range of 245 * 4.0/2.9 = 338 miles minus a little for hauling around the extra weight. Do I have it right on all counts?

As for weight, you have to consider that there is more in the Tesla battery than just the Panasonic NCR18659 in a pile. At the very least, there is a system for pumping coolant through the assembly, and the assembly itself probably weighs a substantial amount. In other words, a new battery pack made with new batteries will probably weigh more, but they could easily change more than just the Panasonic part when revising the pack.

You have wrong battery. The one used in Roadster is 2.1Ah battery, not 2.9. The one used in Model S is probably 3.1Ah battery.

Also that 4Ah battery weights only 46g not 54g. That is unless you have some better and newer information than I do. Those that go to Tesla are also modified somehow, which probably does something to the weight, not sure what though.

Thanks Timo.

The 2.1 AH clears up the kWH. I found the 54g on a website somewhere. 48g is a lot better news. Looks like all of them are the same physical dimensions (2.1, 2.9, 3.1 and 4.0). Do they all have the same discharge ability? Do you know what the xC rate is? Wouldn't the 3.1's give 47% more range (3.1/2.1 = 1.47)? In any event, it is exciting to think that much more range and or lower weight for the same range is just over the horizon. Can't wait.

Any clarification on the nominal voltage? Is the 2.1 AH battery nominal 3.78 volts? My model airplane LiPo's are all 3.7 v.

rsdio,

If you look at the calculations again the batteries themselves made up 663# of the 990# weight so I assumed the balance of the weight was due to the enclosure, cooling system and electronics (about 327# worth).With the 48 gram weight for the 4 AH I get 1050# total pack weight (663*48/44 = 723 + 327 = 1050#

The roadster batteries are 2200mah,3.7V and 44g.

Model S 300 mile pack batteries are supposedly the 3100mah, 3.6v and weigh 44.5g.

Thx to all. Calculations all work now (accept for voltage). I will be watching the battery technology and feel certain that someday we will all be able to get new batteries that either weigh a lot less for the same range or have considerably more range for the wieght when the time comes for replacement. 6 weeks left for #1364.

we will all be able to get new batteries that either weigh a lot less for the same range or have considerably more range for the wieght when the time comes for replacement.
I'd say it's a safe bet that the batteries will be imporved with respec to power/weight ratios, but far more important will be improved costs.Frankly, right now, with 300 miles of range, and a 45 minute recharge, they are good enough to be fully competitive with gas powered jobs.

The Roadster Innovations / Battery page has a typographical error. It says that a brick contains sixty-nine cells, a sheet contains ninety-nine bricks, and a pack contains 11 sheets. That would be 75,141 cells in all, but we know it's only 6,831. Based upon the opening message of this thread, I'm assuming that the "Ninety-nine bricks" should read "Nine bricks" - but maybe they mean that the 99 bricks in series make 11 sheets. It's basically a little confusing the way it's worded.

45 minute recharge? Even with the 90A charger it takes about 3 hours.

The 45 min recharge is referring to some kind of level 3 charging (such as DC fast charging), not the level 2 charging the Roadster can use. At the 70A level 2 charging level of the Roadster, the 300 mile pack would probably take around 5 hours to fully recharge.

the way you calculate the energy available gives the correct number but its not representing the correct electrochemical event of discharge.

even though you have 6831 cells, only the capacity (Q) of 69 of them (those arranged in parallel) is used for discharge at the nominal voltage of 375V (given by the arrangement in series of the remainder 99). E = Q x V so E = (69 x 2.2)A x 375V = 56.9 kWh. the capacity coming from the cells arranged in series is only there for "balancing". If they were "empty" then the capacity of the 69 arranged in parallel would have to be averaged over the entire number of cells (6831).

as you can see an increase in capacity is most wanted as it allows for both an increase in energy (driving range) and power (acceleration, etc) EVEN if the individual cell operates at a much lower voltage. For example, if you had double the capacity in a single cell you would only need 35 of them in parallel for the same energy (driving range) and 195 in series which would now give a much higher nominal voltage. V = 195 x 3.5 = 683V for a 3.5V cell or 585 for a 3.0V cell, etc.

so lets hope there is interest in post lithium ion batteries such that the chemistries of such batteries are ironed out in due time :).

Does anyone know how much litium the battery contains?

So I have calculated that tesla will be able to manufacture 367 million cars with the 11.000.000 tonnes of lithium in the world. Do you concur?

I think we are safe on the lithium supply until we we are making a billion plus cars. See the very insightful article from Nick Butcher about battery constraints. http://seekingalpha.com/article/654441-ev-myths-and-realities-part-1-the...

Yes, I miscalculated: 1,6 billion!

i see lots of talk about the specs but nothing on cycles, does anybody know how well these things will do in the long run?

the amount of lithium in the battery pack can be estimated by math. Lithium only exists in the LiCoO2 cathode and the LIPF6 electrolyte. The electrolyte is typically 1 mol/L so there is 1 mol of Li in 1L of electrolyte.

Based on the total capacity of the unit cell (18650) of ~3000 mAh and the reversible specific capacity of LiCoO2 of 140 mAh/g, we can estimate 21.5g of reversible LiCoO2 in one cell. however, the total capacity of LiCoO2 is ~280 mAhg, but only ~50% of is reversible, so we have to multiply 21.5 x 2 = 43g LiCoO2 to obtain total amount in the unit cell.

now the molecular weight of LiCoO2 is 97.87 g/mol and that of Li is 6.941 so Li weighs 7.1% of total active material in cathode. Then 43g x 0.071 = 3.05g total Li in the cathode of one 18650 cell with ~3 Ah capacity.

I dont know how much electrolyte one 18650 cell contains but it must be as much as the free volume (~50%) of its celgard separator which must be the same geometric area as the cathode. The geometric area of the cathode in an 18650 cell is 265.65 cm2. So if i assume ~280 cm2 geometric area for the celgard separator with 0.1 mm thickness and ~50% porosity, i get ~ 280 cm2 x 0.01 cm = 2.8 cm3 / 2 = 1.4 cm3 or 1.4 ml of electrolyte.

at a a typical concentration of 1M LiPF6, a volume of 1.4 ml corresponds to 0.269 g LiPF6 (mw of LiPF6 is 191.905 g/mol). there is 3.6% Li in LiPF6 by weight so 0.269 g x 0.036 = 0.0097 g Li.

to add everything up we have 0.0097g + 3.05g = 3.06g Li in one 18650 cell. Model S has 7000 cells so a total of 21.4 kg Li per battery pack or just about 47 lbs of lithium.

i hope i made only a few mistakes in my math.
if anyone can pitch in how much that costs...

Tesla is a leading brand in electric and litium-ion batteries but the mathmatical calculation can not find the capacity and capability of battery, The battery volt is depends its ampere and how load it resolve, Okaya XL6000T 150ah recommended rating 650VA to 1000VA on 2 Fans and 2 Tube lights 4 to 6 hours otherwise we give more load like 4 Fans and 4 Tube lights recommended rating 650VA to 1000VA only 2 to 3 hours. To find brief details go to http://www.batterybhai.com/inverter-batteries/Okaya/5.

I will be watching the battery technology and feel certain that someday we will all be able to get new batteries that either weigh a lot less for the same range or have considerably more range for the gclub when the time comes for replacement.


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