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Supercapacitors

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Based on E = 1/2 CV^2 I find the capacitance necessary for an energy of 100kWh at 400V to be 4500 F.

When the voltage drops to 350V the remaining energy is 76.56 kWh.

A Tesla model S gets about 3 miles per kWh and its range is about 300 miles. Driving at 60 miles per hour takes 5 hours to cover 300 miles.

Assuming a constant power output from the supercapacitor the 300 miles and the time 5 hours will be the same as for the battery (energy is energy). Note the voltage of the supercapacitor drops as the square root of the time.

Right, so if the inverter input voltage range is 350 to 400V, you only get 100-76.56=23.44kWh usable from the capacitor bank. If you have a boost converter to keep the voltage at 350, you can utilize more of the stored energy (albeit with rising current draw from the capacitor). Alternatively, switching subsections from parallel to series would boost voltage also. 4500/23.44×100 gets to the 20k Farad number I provided.
 
Right, so if the inverter input voltage range is 350 to 400V, you only get 100-76.56=23.44kWh usable from the capacitor bank. If you have a boost converter to keep the voltage at 350, you can utilize more of the stored energy (albeit with rising current draw from the capacitor). Alternatively, switching subsections from parallel to series would boost voltage also. 4500/23.44×100 gets to the 20k Farad number I provided.
Sorry but I don't know anything about "boost converters". Do you mean DC to AC conversion? I thought modern converters could handle a wide range of voltages. I assumed a *constant* DC current to yield a 300 mile range from 100kWh supercapacitor.
 
Sorry but I don't know anything about "boost converters". Do you mean DC to AC conversion? I thought modern converters could handle a wide range of voltages. I assumed a *constant* DC current to yield a 300 mile range from 100kWh supercapacitor.

A boost converter increases a DC voltage. Power out = power in × efficency, so less current out than in. It would be needed to get a lower capacitor bank voltage up to the level needed for the motor/inverter.
Regarding draw from the cap bank, 4500 F at 400V is 100kWh of energy, but the current draw will not be constant. Power draw would be constant at the 20kW rate, but as the cap bank voltage drops, the current increases. Bank voltage drop rate is proportionatel to current draw, so the voltage drops slowly at the beginning as shown by the first 50V (12.5%) representing 23kWh (23%) of the energy (and the squared term in the energy formula) and faster at the end (last 23% is a drop of 200V)
 
A boost converter increases a DC voltage. Power out = power in × efficency, so less current out than in. It would be needed to get a lower capacitor bank voltage up to the level needed for the motor/inverter.
Regarding draw from the cap bank, 4500 F at 400V is 100kWh of energy, but the current draw will not be constant. Power draw would be constant at the 20kW rate, but as the cap bank voltage drops, the current increases. Bank voltage drop rate is proportionatel to current draw, so the voltage drops slowly at the beginning as shown by the first 50V (12.5%) representing 23kWh (23%) of the energy (and the squared term in the energy formula) and faster at the end (last 23% is a drop of 200V)
Thank you.
I just read up on Wikipedia about "boost converters".
I should have said constant power (20kW).
 
My worry is the retention time which is not mentioned in any of the reports I have read. High charge and discharge rates mean nothing if the energy leaks away in a few seconds or a day.

How about a hybrid system where fast charging elements continue to charge the slower charging parts even after you are unplugged and back on the road. Best of both worlds?
 
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It would also depend on the energy density of the capacitor. And the cost. Say both were equal. (They aren't) Would you trade half your battery for immediate charging? You'd only get half as far. If you were on a road trip you might as well just use the capacitor.
 
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Scientists from Imperial College London and University College London (UCL) have used lignin to create a new generation of low-cost, high-energy supercapacitors to power electric vehicles.

In the study, published in Advanced Science, teams at Imperial College London and used lignin – a bio-based by-product of the paper industry – to create free-standing electrodes with enhanced energy storage capacity.

The researchers believe this could be a game-changer for existing supercapacitor technology and emphasise the importance of reducing the production cost of carbon-based electrodes and the reliance on critical materials if free-standing supercapacitors are to play a major role in decarbonising transport alongside batteries and fuel cells.