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Leaf E+ driven. If you have Google Translate installed it will automatically translate the page from Japanese for you.

動力性能の向上にダマされるな! 日産リーフe+登場の本当の意味〈新型NISSAN LEAF e+ 試乗記〉|MotorFan[モーターファン]

The text below is Google translated from the E+ Battery link at the bottom of that page. It might sound like gobbledegook as you read it, but the end of it definitely indicates that heat when charging is halved compared to the 40kW battery, so no rapidgate issue even without liquid cooling. Genius.

"I asked if the difficulty in cooling does not occur by stuffing it in the crown. Then, the benefits of 3 parallelization are great, the engineer explains. "If 2 parallels are made in 3 parallels, when we try to flow the same current, each 1/2 becomes 1/3 each, so the flowing current becomes 2/3, which is disadvantageous because it is in close contact, Since it becomes 2/3, it becomes proportional to the square of the current in I ^ 2R, so it is about half, which means that the amount of heat is halved, which is a huge difference."
 

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The reporter asked and the engineer explained.

Is that any different to Nissan saying that they have fixed it, or have I interpreted it wrong?

I think we should wait until a Bjorn or a Vicky test it.
 

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Isn't this bit clear enough?

"which means that the amount of heat is halved, which is a huge difference"

Rapidgate is caused by too much heat, halve the heat and you have no rapidgate.
 

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It's very clear.

A Nissan engineer telling a reporter that the way the cells are arranged halves the amount of heat, right?

My point was that we have heard stuff from Nissan before (but usually a deafening silence) but it is normally better to wait for independent testers to try it in the real world.
 

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What the engineer describes is of course factually correct - adding a third set of parallel cells reduces the current and reduces the power dissipation due to resistive losses.

Per Cell.

What they fail to mention is that all else being equal (and it's not - we'll come to that), the power dissipation per pack is the same. In other words, if the 60kWh pack is squashed up into the same space as the 40kWh pack, the amount of power dissipated in that space is the same. Assuming the 60kWh pack to be heavier and with a similar heat density, then it will heat up a few degress less - all well and good.

However, all else is not equal. We already know that the e+ will offer higher charging rates - in other words they've put the current back up and the power dissipation goes back up to where it was before on a per-cell basis (and will generate more heat in the same space).

Of course, since Chademo chargers over 50kW won't be desperately common (assuming that significant numbers of new charging stations come online with only CCS), this may not be seen in the real world.

So in summary - new pack capable of generating more heat, but under existing charging infrastructure it should generate less heat per cell. If charged at higher rates then it will behave similarly to before.
 

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Leaf E+ driven. If you have Google Translate installed it will automatically translate the page from Japanese for you.

動力性能の向上にダマされるな! 日産リーフe+登場の本当の意味〈新型NISSAN LEAF e+ 試乗記〉|MotorFan[モーターファン]

The text below is Google translated from the E+ Battery link at the bottom of that page. It might sound like gobbledegook as you read it, but the end of it definitely indicates that heat when charging is halved compared to the 40kW battery, so no rapidgate issue even without liquid cooling. Genius.

"I asked if the difficulty in cooling does not occur by stuffing it in the crown. Then, the benefits of 3 parallelization are great, the engineer explains. "If 2 parallels are made in 3 parallels, when we try to flow the same current, each 1/2 becomes 1/3 each, so the flowing current becomes 2/3, which is disadvantageous because it is in close contact, Since it becomes 2/3, it becomes proportional to the square of the current in I ^ 2R, so it is about half, which means that the amount of heat is halved, which is a huge difference."
OTOH if you charge at 100Kw, then that multiplies the heat generation by four.
 

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This would also imply that it would only need 50+ 50×1/3 ie sub 70kW get the Dave level of currents per cell and hence the heat.

Sent from my SM-G960F using Tapatalk
 

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What the engineer describes is of course factually correct - adding a third set of parallel cells reduces the current and reduces the power dissipation due to resistive losses.

Per Cell.

What they fail to mention is that all else being equal (and it's not - we'll come to that), the power dissipation per pack is the same. In other words, if the 60kWh pack is squashed up into the same space as the 40kWh pack, the amount of power dissipated in that space is the same. Assuming the 60kWh pack to be heavier and with a similar heat density, then it will heat up a few degress less - all well and good.
If you have three sets of parallel cells then the overall internal resistance of the pack will be lower, and hence, for the same current the resistive heating will also be lower. Or, to put it another way, the current is divided into three instead of 2.

Let's say the internal resistance of one row of cells is R, then the resistance of two in parallel = 1 / (1/R + 1/R) = 1/2R

For three it would be 1/(1/R + 1/R + 1/R) = 1/3R. The bad news is that the current appears as a squared term.
 

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If you have three sets of parallel cells then the overall internal resistance of the pack will be lower, and hence, for the same current the resistive heating will also be lower. Or, to put it another way, the current is divided into three instead of 2.

Let's say the internal resistance of one row of cells is R, then the resistance of two in parallel = 1 / (1/R + 1/R) = 1/2R

For three it would be 1/(1/R + 1/R + 1/R) = 1/3R. The bad news is that the current appears as a squared term.
How is interval resistance reduced? A cell's resistance or a pack of parallel cells would have the same resistance. Adding 3rd cell shouldn't reduce it unless I'm losing it here and have no idea how electrical circuits work.

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How is interval resistance reduced? A cell's resistance or a pack of parallel cells would have the same resistance. Adding 3rd cell shouldn't reduce it unless I'm losing it here and have no idea how electrical circuits work.

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Each row of cells has a particular resistance. Because they are connected in parallel the row's conductivity, the reciprocal of resistance, is summed to make the overall conductivity. Or, to put it another way, the current in each row is 1/3 of the total current instead of 1/2.
 

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Ah, but the voltage involved is not constant, because the voltage due to internal resistance is added to the EMF. You charge batteries with a constant current, not a constant voltage. The voltage on the charger terminals is the EMF + (the desired current x the internal resistance).
 

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What the engineer describes is of course factually correct - adding a third set of parallel cells reduces the current and reduces the power dissipation due to resistive losses.

Per Cell.

What they fail to mention is that all else being equal (and it's not - we'll come to that), the power dissipation per pack is the same. In other words, if the 60kWh pack is squashed up into the same space as the 40kWh pack, the amount of power dissipated in that space is the same. Assuming the 60kWh pack to be heavier and with a similar heat density, then it will heat up a few degress less - all well and good.
The power dissipated as heat is not the same, as power increases in proportion to total current, but heat with the square of the string currents.

Or to look at it in the same way as one of the later posters, the overall pack resistance has reduced through the addition of a third string.

For constant charging voltage (a pessimistic assumption), we have 1.5x as many strings (with more surface area) each producing roughly half the heat as before.

So for the same charging speed (and equally importantly, discharge rate when driving), the total heat dissipated is reduced. Which presumably also brings a small improvement in drivetrain efficiency...
 
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