[Robotbeat]

Yet again, the useless milliamp-hours per gram for one component of a cell instead of actual energy per unit mass for a whole cell. What is the Watt-hours per kilogram????

[rsfern]

The uni press release sort of buried the lede, but if you read the methods section of the paper they are doing these measurements in a coin cell with a sodium anode.

This is early stage cathode optimization research, everyone reports current density and charge-discharge curves because the power density requires both high current density and stable operating voltage. Also you want minimal degradation with charging cycles. You can integrate those discharge curves to get power density if you want, but I’m not sure what it’s supposed to really add at this point in the research pipeline. You need to scale up from the lab form factor anyway to get something reasonable to work with

https://www.nature.com/articles/s41586-021-03757-z

[FabHK]

Need to multiply by the (average) voltage, and incorporate a fudge factor for the other components of the cell, I think.

(P = UI, where P is power in Watt, U voltage in Joule per Coulomb, I current in Coulomb per second. And what we seek is specific energy, ie energy per mass, in J/kg or Wh/kg.)

[BenjiWiebe]

Watts = Volts * Amps

Any reason you've defined it as: (Amps * Seconds * Volts) / (Amps * Seconds) * (Amps * Seconds * Seconds) ?

Fairly sure but not positive that I've reduced the units to base SI units correctly...

[FabHK]

ha, you're right, the Ampere is the SI base unit for purposes of definition, but the Coulomb is arguably the conceptually simpler one... (though Volt is definitely derived). So, in terms of "intuition" (left) and SI base units (right) you have:

  Watt   = J s^-1 = N m s^-1 = kg m^2 s^-3
  Volt   = J C^-1 = N m A^-1 s^-1 = kg m^2 s^-3 A^-1
  Ampere = C s^-1 = A

[LinAGKar]

And J/cm³. That's usually the limiting factor for phones, I think.