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How is Tesla charging the Panasonic 18650PF at over a 1C rate?

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It can't be less. The entire battery pack contains 85 kWh of energy. So an 85 kW Supercharger would fill an empty pack in one hour. Which is 1C.

135 kW of power would charge it at 135/85= 1.6C

EDIT : The info Posted Below this line looks to be totally wrong...sorry for the misinformation.....move on past this post.... :)

I am probably mistaken and have no clue what I'm talking about, but I thought that the "C" indicated the maximum discharge/charge rate for the pack. So say, on my S85 when I floor it, the inverter demands about 270Kwh from the pack. Let's just say that is the actual max discharge rate for the pack (it's probably MUCH more). That means that the 135kw SC is only charging at about .5C.

I swear I read this on the forum at some point.....then again.....I might have misunderstood/imagined it....
 
I am probably mistaken and have no clue what I'm talking about, but I thought that the "C" indicated the maximum discharge/charge rate for the pack. So say, on my S85 when I floor it, the inverter demands about 270Kwh from the pack. Let's just say that is the actual max discharge rate for the pack (it's probably MUCH more). That means that the 135kw SC is only charging at about .5C.

I swear I read this on the forum at some point.....then again.....I might have misunderstood/imagined it....

Yes, you misunderstood. 1C = the rate/power at which the pack is discharged/charged in 1 hour.

Battery charger - Wikipedia, the free encyclopedia
 
I am probably mistaken and have no clue what I'm talking about, but I thought that the "C" indicated the maximum discharge/charge rate for the pack. So say, on my S85 when I floor it, the inverter demands about 270Kwh from the pack. Let's just say that is the actual max discharge rate for the pack (it's probably MUCH more). That means that the 135kw SC is only charging at about .5C.

I swear I read this on the forum at some point.....then again.....I might have misunderstood/imagined it....

You are confusing kWh and kW. When you floor your Model S, it uses 270 kW of instantaneous power. If you could continue to draw that amount of power, the pack, which has 85 kWh of energy, would be depleted in about 85/270*60= 19 minutes. Which is a 60/19= approx. 3C discharge rate. Which, by the way, the Model S battery can only sustain for brief periods of time before the car's computer will limit power output due to battery over heating issues.

C doesn't indicate max. discharge/charge rate. Spec sheets tell you that. C indicates the rate at which the battery is charging or discharging relative to its maximum energy storage.
 
If you could continue to draw that amount of power, the pack, which has 85 kWh of energy, would be depleted in about 85/270*60= 19 minutes. Which is a 60/19= approx. 3C discharge rate. Which, by the way, the Model S battery can only sustain for brief periods of time before the car's computer will limit power output due to battery over heating issues.
My understanding is that this is a popular theory not substained fact. Perhaps you have some newer information you could link to? Thanks!


Running with that theory a bit, the acceleration limiter on my P85 puts a hard line at 160 kWh.
(Acceleration limiter, some charts from Pacific Raceways)

Limited @ 160 kW
60/(85/160*60)=1.88C ?

Full @ ~348 kW (local maximum in my data)
60/(85/348*60)=4.09C ?
 
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My understanding is that this is a popular theory not substained fact. Perhaps you have some newer information you could link to? Thanks!

It is basic knowledge about battery management. If you discharge a battery at such a high rate you are going to run into trouble pretty quick. The losses in the battery itself and the entire electrical drive train is substantial. At normal driving the losses are 10%. Double the power and you quadruple the ohmic losses. Drawing 200-300 kW out of the battery will probably have total losses around 50 kW or more. The Model S is clearly not designed to handle that much cooling and will limit the power. Also the voltage in the cells will drop below a healthy level and cause permanent damage. It's fine do do it for short periods of time like it is common in normal driving. But honestly I can' think of a situation, other than a large circular race track that allows full throttle 100% of the time, where you could possibly run the car at full load for a extended time.

Yes some electric powered remote cars and helicopters discharge at that rate but they don't come with an 8 year warranty and those batteries die within a few months when used that way on a daily basis.
 
It is basic knowledge about battery management. If you discharge a battery at such a high rate you are going to run into trouble pretty quick. The losses in the battery itself and the entire electrical drive train is substantial. At normal driving the losses are 10%. Double the power and you quadruple the ohmic losses. Drawing 200-300 kW out of the battery will probably have total losses around 50 kW or more. The Model S is clearly not designed to handle that much cooling and will limit the power. Also the voltage in the cells will drop below a healthy level and cause permanent damage. It's fine do do it for short periods of time like it is common in normal driving. But honestly I can' think of a situation, other than a large circular race track that allows full throttle 100% of the time, where you could possibly run the car at full load for a extended time.

Yes some electric powered remote cars and helicopters discharge at that rate but they don't come with an 8 year warranty and those batteries die within a few months when used that way on a daily basis.

Our German fried in this thread Discharging at High Speed (100 mph) drove 180 km at an average speed of 165 km/h which means he must have used slightly over an hour to discharge the battery (just under 1 C on average). He calculates Ohmic losses elegantly, please let me quote him:

Very interesting was the amount of energy I was able to get out of the battery. 0 Typical Range was reached 100m (90yrds) in front of the garage. So the Zero Mile protection as well as the Block Protection were untouched. 71,2 kWh came out of the battery. With these 9 kWh protection the battery just gave 80,2 kWh (instead of 85 kWh).

This is normal, because Peukert's Law says, that higher currents result in more losses. I compared this high speed discharging with a typical discharging of mine at an average of 204 Wh/km (328 Wh/mi). I got more (74,3 kWh) out of the battery during slow discharging. I also calculated the Kapa of the Peukert Function. It resulted in a value between 1.03 and 1.04, which is significantly better than a standard Li-Ion battery (1.05). This shows the high end chemistry of the cells in our cars.

I'm not sure if or how often he got power limited, he doesn't say in his post.

David99 I know you know this (I see your reply in horst's thread) but I thought it would be of interested in this thread to those who haven't seen it.
 
I would assume that the inverter is the first thing to reach temperature limits since the motor can operate at much higher temperatures than the IGBT's in the inverter. Plus I'd think the highest efficiency losses in the system would be from the power conversion going on in the inverter. Inverter output current should be higher than battery current until the motor reaches base speed, I think.
 
Actually C is the charge and discharge rating, it does stand for capacity. But if you want to know what for instance 2C charge rate on your cell is you multiply it by your cells capacity (stated in mAh). In your cells datasheet you will find charge rate and discharge rate stated separately in numbers of C.