Nanoparticles are interesting, as capacitance is directly proportional to surface area and inversely proportional to the distance between the plates.

This may be why so many people are investigating this.

I wonder what the operating range is. Most capacitors can go up to 105 C, but seem to have no practical lower limit.

Lithium/NiMH go down to -30C NiCd =-40, but they can’t deliver a lot of current at that temp.

Canadians are hoping the automotive industry goes with super capacitors, we tire of having to plug in our cars.

Nanoparticles are interesting, as capacitance is directly proportional to surface area and inversely proportional to the distance between the plates.

This may be why so many people are investigating this.

I wonder what the operating range is. Most capacitors can go up to 105 C, but seem to have no practical lower limit.

Lithium/NiMH go down to -30C NiCd =-40, but they can’t deliver a lot of current at that temp.

Canadians are hoping the automotive industry goes with super capacitors, we tire of having to plug in our cars.

Many people on the New Scientist website thought that 2500J seemed very low and someone did the same maths I did.

The difference is that he assumed cm to the power of -2 meant cm squared. He also used 10 Volts and came up with 2500J. Volts are multiplied by twice in the conversion from Farads to Joules so multiplying up to 100 Volts changes the final value by an order of 10,000 which gives a value somewhere in between the two values I arrived at. He gets that for 1kg and I get it for 100kg but that can be explained by the cm squared difference.

Another way of looking at it is that a AA battery holds about 10,000J and is a lot less than a kilogram.

Many people on the New Scientist website thought that 2500J seemed very low and someone did the same maths I did.

The difference is that he assumed cm to the power of -2 meant cm squared. He also used 10 Volts and came up with 2500J. Volts are multiplied by twice in the conversion from Farads to Joules so multiplying up to 100 Volts changes the final value by an order of 10,000 which gives a value somewhere in between the two values I arrived at. He gets that for 1kg and I get it for 100kg but that can be explained by the cm squared difference.

Another way of looking at it is that a AA battery holds about 10,000J and is a lot less than a kilogram.

I can’t help but think that the figure of 2500 joules per kilogram must be a typo. Considering the following sentence from the New Scientist article, mistakes and typos seem probable.

“Now a prototype capacitor has been made that manages to store power as densely as a supercapacitor, but deliver it at speeds comparable with electrostatic capacitors.”

For those who didn’t study physics at high school, power is not stored. Power is delivered. If it is stored, it is energy, not power. One would expect a journalist working for New Scientist to know this and to get it right.

The original paper (or at least, the abstract) measures the capacitance in Farads which is normal for capacitance but not easily comparable to the figure of 2500J. It gives a value of 10 micro Farads per cm but rather than cm squared they have used cm to the power of -2. This notation seems confusing to me but I am going to assume that cm ^ -2 is the same as mm ^ -1 or m ^ -4 so if we multiply by 1000 throughout we get 10cm squared. Multiplying the rest of the values to get towards my guess at 1kg we get 10 Farads for a 10cm squared slice of 10 millimetere thick aluminium oxide. This is 100mL of the capacitor which, being aluminium oxide weighs about 250g. 1kg will be 4 times this and therefore will have 40 Farads which is quite a bit better than traditional capacitors which would struggle to get 1 Farad into this volume.

Converting Farads into Joules requires knowing the Voltage that the system will run at… which I don’t. The voltage in electric cars varies between around 100 and 300 Volts. Using these two values, I get either 400kJ/kg or 3,600kJ/kg. Either way, it’s a long way from 2,500J/kg. 100kg of the 300 Volt sort would give us 360MJ or 100kWh. That’s 10 hours of driving at 10kW power usage. Average driving will use more than that but however you look at it, these are impressive figures.

Please feel free to do the maths again to make sure I didn’t get it wrong myself.

Dagnabbit ! I just spotted an error myself. I switched cm squared and squared cm right near the start. I think a quick division be 10 will correct the error but I can’t be sure without going through the whole lot again. This would leave us with 10kWh for 100kg of batteries which is still quite impressive.

I can’t help but think that the figure of 2500 joules per kilogram must be a typo. Considering the following sentence from the New Scientist article, mistakes and typos seem probable.

“Now a prototype capacitor has been made that manages to store power as densely as a supercapacitor, but deliver it at speeds comparable with electrostatic capacitors.”

For those who didn’t study physics at high school, power is not stored. Power is delivered. If it is stored, it is energy, not power. One would expect a journalist working for New Scientist to know this and to get it right.

The original paper (or at least, the abstract) measures the capacitance in Farads which is normal for capacitance but not easily comparable to the figure of 2500J. It gives a value of 10 micro Farads per cm but rather than cm squared they have used cm to the power of -2. This notation seems confusing to me but I am going to assume that cm ^ -2 is the same as mm ^ -1 or m ^ -4 so if we multiply by 1000 throughout we get 10cm squared. Multiplying the rest of the values to get towards my guess at 1kg we get 10 Farads for a 10cm squared slice of 10 millimetere thick aluminium oxide. This is 100mL of the capacitor which, being aluminium oxide weighs about 250g. 1kg will be 4 times this and therefore will have 40 Farads which is quite a bit better than traditional capacitors which would struggle to get 1 Farad into this volume.

Converting Farads into Joules requires knowing the Voltage that the system will run at… which I don’t. The voltage in electric cars varies between around 100 and 300 Volts. Using these two values, I get either 400kJ/kg or 3,600kJ/kg. Either way, it’s a long way from 2,500J/kg. 100kg of the 300 Volt sort would give us 360MJ or 100kWh. That’s 10 hours of driving at 10kW power usage. Average driving will use more than that but however you look at it, these are impressive figures.

Please feel free to do the maths again to make sure I didn’t get it wrong myself.

Dagnabbit ! I just spotted an error myself. I switched cm squared and squared cm right near the start. I think a quick division be 10 will correct the error but I can’t be sure without going through the whole lot again. This would leave us with 10kWh for 100kg of batteries which is still quite impressive.