E
Ed Lee
Guest
On Wednesday, September 7, 2022 at 3:55:16 PM UTC-7, Ricky wrote:
The motor uses extra energy to go up the hill, then regen back the extra energy. But it\'s the same 52kWh, not an additional 35kWh.
> What are you trying to figure out???
Where that extra 35kWh is coming from.
On Wednesday, September 7, 2022 at 4:08:32 PM UTC-4, Ed Lee wrote:
On Wednesday, September 7, 2022 at 1:04:29 PM UTC-7, lang...@fonz.dk wrote:
onsdag den 7. september 2022 kl. 21.18.41 UTC+2 skrev Ed Lee:
On Wednesday, September 7, 2022 at 12:10:02 PM UTC-7, lang...@fonz.dk wrote:
onsdag den 7. september 2022 kl. 17.13.11 UTC+2 skrev Ed Lee:
On Wednesday, September 7, 2022 at 8:07:51 AM UTC-7, lang...@fonz.dk wrote:
onsdag den 7. september 2022 kl. 15.48.45 UTC+2 skrev Ed Lee:
On Wednesday, September 7, 2022 at 3:27:38 AM UTC-7, Ricky wrote:
On Monday, September 5, 2022 at 7:26:43 PM UTC-4, lang...@fonz.dk wrote:
mandag den 5. september 2022 kl. 23.20.18 UTC+2 skrev upsid...@downunder.com:
On Sun, 04 Sep 2022 07:32:01 -0700, jla...@highlandsniptechnology.com
wrote:
High voltage needs thick insulation on the wire, and skinny wire
windings lose window area to insulation more than fat wires. And
insulation doesn\'t conduct heat as well as copper. Triple whammy.
400 Vdc is sufficient to drive a 120/208 V three phase motor with a
sine wave. 800 Vdc is sufficient for 230/400V motors. The insulation
thickness is not an issue for normal 230/400 V motors in kW size
motors.
The preference for 800 Vdc battery voltage is in the inverter. 1200 V
semiconductors are cheaply available and they should survive in a well
designed 800 Vdc system.
Using a higher voltage will reduce the current. The cost of a
transistor is usually proportional to the chip area which depends on
the current rating. Reducing the current (by using higher voltage)
requires smaller chips and may reduce total cost, if the higher
voltage doesn\'t significantly increase costs. Now that 1200 V devices
are cheaply available, why not go for 800 Vdc ?.
Going for 1500 Vdc would require 2 kV devices, which today is
expensive.
FIA Formula E all use a nominal 750V battery, so ~525V to ~880V
350kW rear wheel drive, 600kW four wheel regen
That would be hard on the battery. I wonder how good the warranty is.
Depends on what the actual current is. My 80 kW motor typically draw around 70A (28kW) on local driving. Anyone got data on other EVs?
Formula E is racing, there\'s a 52kWh net energy allowance for a 45 minute race and afair they claim regen adds about 40%
What\'s the typical power/current draws? 52kWh/0.75 = 70 kW average?
it is racing so pretty much constantly switching from max power to max regen
they claim ~40% of the energy used is regen, so 52kWh battery + ~35kWh regen = 87kWh , so ~116kW average
https://youtu.be/0Kk8PyPSdaM
That\'s pulling regen energy out of nowhere. I think it just mean more distance out of the 52kW battery.
it is not energy out of nowhere, it is reusing the energy rather turning it into heat in the brakes
But total energy is still 52kWh, with average of 70 kW.
The motor sees the energy from the battery, then the energy from regeneration, twice, once in each direction.
The motor uses extra energy to go up the hill, then regen back the extra energy. But it\'s the same 52kWh, not an additional 35kWh.
> What are you trying to figure out???
Where that extra 35kWh is coming from.