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2014 Energy Storage Symposium - JB Straubel's Keynote

JB Straubel speaking at the Keynote 2014 Energy Storage Symposium.

https://www.youtube.com/watch?v=zWSox7mLbyE

Got a summary for those of us who do not have time to watch the video?

No. It's too info-dense. You have time, believe me.

I think this technology: ARES (Advanced Rail Energy Storage),would complement Tesla's storage solution for renewable resources applications very well. Particularly wind turbines.

https://vimeo.com/48546631

http://www.aresnorthamerica.com/

@frmercado

Thanks for that!

Interesting.
Maybe they will be able to make a practical ARES system that is cost-competitive.

I'm thinking that there may be better ways of using gravity in similar systems.

Ron :)

2014 Energy Storage Symposium - JB Straubel's Keynote (after 18 minutes).

JB Straubel regarding the Gigafactory:

"The overall target goal here is cell production in the order of about 35 GWh per year, and pack production of about 50 GWh per year.

This is just to meet the needs of what we see in the Fremont plant for Tesla only.

This is not sort of a merchant plant trying to satisfy the whole world car demand.

It's satisfying the demand of one reasonably large car plant.

So, we have intentionally sized the pack output here a little bit more than the cell output, to try and focus on other markets, and stationary (storage) is the biggest one of those, so we are basically allocating about 15 GWh per year of capacity in this plant to build stationary battery packs.

So, you know when come only back to sort of the California mandate for energy storage by 2020, it seems like a huge amount and it seems really remarkable, and drives the industry faster than it could ever be driven before, but we are actually quite a lot more bullish than that.

I think that that mandate will get met and far exceeded well before the timeframe on it expires.

This is something like ten times more allocated capacity for packs than just the California mandate by the same timeframe it's supposed to end.

So, you know, I guess again we should all be thinking bigger. The grid oppertunity here is huge and the sort of economics are going to drive this forward and happening much faster than many people expect."

Etc.

--------------------------------------------

I think that the most important sentence is:

"So, we have intentionally sized the pack output here a little bit more than the cell output, to try and focus on other markets, and stationary (storage) is the biggest one of those, so we are basically allocating about 15 GWh per year of capacity in this plant to build stationary battery packs."

But what if the demand for the Tesla EV's in the coming years increases much sooner and much faster than that they are currently expecting it to increase? How about: 700,000 (S+X+3)in 2018?

Should they then decide to make battery packs for their cars or should they still be making battery packs for stationary storage (according to the current plan)?

@Benz | AUGUST 4, 2014

JB Straubel regarding the Gigafactory:

"The overall target goal here is cell production in the order of about 35 GWh per year, and pack production of about 50 GWh per year."

"So, we have intentionally sized the pack output here a little bit more than the cell output, to try and focus on other markets, and stationary (storage) is the biggest one of those, so we are basically allocating about 15 GWh per year of capacity in this plant to build stationary battery packs."

There is a glaring math error in these statements. They say the GF will make 35 GWh of cells and 50 GWh of battery packs. That would not leave 15 GWh of cells for making stationary batteries. It would require them to get 15 GWh of cells from other sources to be able to make 50 GWh of battery packs.

Given this math error or misstatement, I don't know how to figure out what JBS meant.
(I was an honors math major in college.)
Can someone please try to explain this? Or offer a guess about what JBS intended to say?

I can think of one intriguing possibility: Maybe Tesla hopes to improve the battery chemistry enough to transform a physical plant designed for 35 GWh capacity with current battery chemistry to be able to store 50 GWh with the new chemistry. That would require nearly a 50% improvement in the energy density of the cells. I'd sure like to ask JBS is that's what they expect to do.

Any other thoughts?

Go Tesla!
Ron :)

@ Grinnin

Panasonic will continue to be shipping battery cells from Japan to Tesla Motors, both before and after the Gigafactory comes online. So, Panasonic will not stop producing battery cells for Tesla Motors once the Gigafactory comes online. And those battery cells should be about 15 GWh in total. If Panasonic for whatever reason will not be able to produce that many, than Tesla Motors will look for other partners (Samsung SDI, LG Chem, etc.).

Have I answered your question correctly?

I think the original press releases, etc. about the gigafactory plans had the same 35 kWh/50 kWh numbers, and Tesla indicated that the 15 kWh gap would be shipped in from elsewhere.

Stationary battery packs would logically be optimized with a different type of battery cell than what is best for a car.

Also, Tesla does currently make some amount of stationary battery packs, and they have indicated in the past that these use different cells (than what the Model S uses), so making those stationary battery packs isn't further decreasing the supply of batteries used to build cars.

All the battery cells produced by Panasonic at the Gigafactory will be the next generation battery cells (different geometry and better energy density) and will all be used to make battery packs for Tesla EV's.

@ EQC:

"I think the original press releases, etc. about the gigafactory plans had the same 35 kWh/50 kWh numbers, and Tesla indicated that the 15 kWh gap would be shipped in from elsewhere."

"Stationary battery packs would logically be optimized with a different type of battery cell than what is best for a car."

Makes sense. Possibly I misunderstood what they said in the CC.

Ron :)

@Benz | AUGUST 4, 2014

"All the battery cells produced by Panasonic at the Gigafactory will be the next generation battery cells (different geometry and better energy density) and will all be used to make battery packs for Tesla EV's."

I see a need for a new cell, optimized based on Tesla's research. They have said it will be about 10% taller and 10% wider. Those new cells couldn't fit in the shells of the current battery packs. Hence, we can expect a new generation of battery packs, sized to contain suitable numbers of the new cells.

I'm not so sure about "old battery cells and packs made in the existing assembly factory" and "new battery cells and packs made in the GF". If that works out for the timing, numbers of packs needed and capacities of the GF and the existing factory, that would be preferable.

I'd guess that they might need some of the new battery packs before the GF is ready to build them. They might make some of them at the existing factory.

Ron :)

Benz;
Yep, "oppertunity" knacks! Anser de door! ;p

I think it's been "known" for some time that there're markets for all the proven, low-cost static storage and "load-leveling" anyone can produce, but no one has dared to grasp the nettle before, and actually put their cash into actually filling it.

Makes much sense. Possibly I misunderstood what they said in the CC.

How I read the new dimensions of the battery cell was that the increase in height(10%) compensates for the decrease in relative surface area of the pack with the 10% increase in diameter. So you end up with the same battery pack surface area for heat transfer requirements. The packing factor of the cells remains the same regardless of diameter (same volume of battery). So the true benefit of the small increase in diameter (10%), is it would reduce the number of cells in a pack by 10% e.g 700 cells for a model S. There would be a cell manufacture cost for each of those cells which of course is saved. So if you take the cell chemistry improvement of 15% and a saved manufacturing cost of 10%. There is 25% just there before you do anything else. I believe the cost reductions planned once all the saving are added together are going to be way more than has been suggested. At a guess, 10 good ideas added together, 200% reduction in cost.

Whoops, what I meant to say is a 67% reduction in cost.

@ alanemay

"How I read the new dimensions of the battery cell was that the increase in height(10%) compensates for the decrease in relative surface area of the pack with the 10% increase in diameter. So you end up with the same battery pack surface area for heat transfer requirements. The packing factor of the cells remains the same regardless of diameter (same volume of battery)."

I don't understand that. Maybe a bit more detailed explenation will make me understand it better. Thanks

@alanemay | AUGUST 5, 2014

"So the true benefit of the small increase in diameter (10%), is it would reduce the number of cells in a pack by 10%"

I think that when the dimensions (height & width are increased by 10%:

* The cell's volume increases by a factor of 1.331 while the surface area increases by a factor of 1.21.
* The energy content of the battery is proportional to its volume.

Consequently, the 10% increase in size of the cells would result in:

* A reduction of about 30% in the number of cells needed for a given capacity battery pack.
* Approximately no change in the size of the battery pack.
* A reduction of about 10% in the cell walls and their weight.

I presume that the larger battery cells could be assembled for approximately the same cost as the current battery cells.
To make battery packs weighing substantially less per kWh, the battery density (chemistry) must be improved.

If I'm missing something, or misunderstanding something, please explain my error(s). Thanks.

Ron :)

Maybe I'm doing something wrong? When I calculate a 10% increase in diameter, plus a 10% increase in height for a cylinder, I keep getting a factor of 1.58:1 as an increase in volume... Should I delete the volume if the anode & cathode? Or is everyone only increasing the radius by 5%, instead of 10%...?

Red, it's a cubic increase, so 58% increase in volume sounds in line with a 10% increase in one dimension

okay okay, I'll do a calculation:
(pi x radius x radius) x height.

5 cm tall by 2 cm radius cylinder:
(3.14 x 2 x 2) x 5 = 62.8 cm

5.5 cm tall by 2.2 cm radius cylinder:
(3.14 x 2.2 x 2.2) x 5.5 = 83.59 cm

83.59/62.8 = 1.331

So, it's an increase in volume of about 33% for a 10% increase each dimension.

This is a demonstration of calculations, not based on any actual measurement of any real-life cylinders.

@holidayday: Correct. (I was a math major in college.)

But the dimensions of the new battery packs (containing the new battery cells) will be slightly different than the dimensions of the current Tesla Model S battery packs, right? Slightly higher, because the new battery cells are also slightly higher?

Benz: "dimensions of the new battery packs (containing the new battery cells) will be slightly different than the dimensions of the current Tesla Model S battery packs, right?"

Depends on if you call the "pack" the "entire item" that creates the skateboard or the "several objects" that have insulation between them to create a single item.

I would expect the skateboard size to remain consistent (for the same model) no matter the configuration of the individual cells or combinations of cells with insulation in between.

Indeed, it would be better if the size, volume, and dimensions of the new battery packs would remain exactly the same as the current Tesla Model S battery packs. So that an old battery pack can be swapped with a new battery pack, because of the higher capacity of the new battery pack, resulting in a longer range. And that will be good for the current owners of a Tesla Model S.

But will that be possible?

I just thought that if the new battery cells will have a higher hight, then that will automatically result in a higher hight of the new battery pack, because the battery cells will be stacked vertically. Or is that a too simplistic assumption?

To just clarify the increase in the diameter of the cell by 10%. The surface area is proportional to the radius or diameter for a fixed height e.g A= 2 x pi x r, A = r. So if you increase the diameter or the radius by 10% then you increase the surface by 10%. Now if you think of a grid of 10 cells by 10 cells (100 cells an increase the diameter of the cells by 10%, the cells are now 1.1 wide. For the fixed area you can only fit 9 cells one way by 9 cell the other way e.g 9 x 9 = 81 cells. But the surface area has increased by 10% so 81 cells x 1.1 increase in area = 89.1 approximately 90% of the original area of 100 cells. To compensate for that you make the cells 10% longer to get back to the original area of the pack. Now if you look at the volume which is proportional to the radius x radius. So a 10% increase in radius results in a 1.1 x 1.1 = 1.21 or 21% increase in volume for a fixed height. But you only have 81 cells now in your fixed volume. 81 cells x 1.21 is approximately 100 units of volume. Increase the height by 10% and you get 110 units of volume for the same fixed area. That does not save you any money to have more volume, as you have to pay for those materials. But if you have a machine that is producing 100 cells per hour and you can increase the size of the cells by 10% while maintaining the production rate and then need less cells per car. You have just spread your fixed costs over 10% more production.

Interesting math
just to clarify:
independent of the form, a volume can be described in 3 dimensions x,y,z and in some form of x*y*z. Therefore, linear increasing in each dimension leads to: linear improvement in each dimension, quadratic increasing in surface and cubic increasing in volume.

Examples:
Assuming a "building block": volume = x*y*z. Increasing each dimension by 10% leads to 1,1x (length), 1,1x*1,1y = 1,21 xy in surface and 1,1x * 1,1y * 1,1z = 1,331 xyz in volume - done!

Cylinder:
Circumference (not: area!) is 2*pi*r (forget 2pi, it's just a constant)
increasing r by 10% leads to 2pi*1,1r = 2,2pi*r (linear increasing)
Area: pi*r^^2 (forget pi, it's just a constant). Increasing r by 10% leads to pi*(1,1r)^^2 = pi * 1,1r * 1,1r = 1,21pi * r^^2. quadratic increase
Volume (height=h): pi * r^^2 * h. Increasing r by 1,1 and h by 1,1 leads to pi*1,1r*1,1r*1,1h=*1,331*pi*r^^2*h. Done!

Hope this is sorted out - BTW, I have not been at college (coming from Germany :-))

@German_Tesla_Fan | AUGUST 6, 2014

"Interesting math"

"Therefore, linear increasing in each dimension leads to: linear improvement in each dimension, quadratic increasing in surface and cubic increasing in volume.

"Hope this is sorted out - BTW, I have not been at college (coming from Germany :-))

Right. Your math education was OK.

Ron :)


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