Since unit of specific capacitance of a supercapacitor is (F/g), and the energy density is given as E=0.5CV^2, it would be helpful for graduate student to shw them how can be the unit of the energy denisty given as (Wh/kg)?
To derive the energy density unit (Wh/kg) for a supercapacitor starting from the energy formula E=0.5CV2, where the specific capacitance is given in F/g, follow these step-by-step unit conversions. This process is essential for graduate students to understand unit transformation in energy storage systems.
Core Formula & Units
1.Energy Formula: E=0.5CV2
- C: Capacitance (Unit: F)
- V: Voltage (Unit: V)
- E: Initial unit is joules (J) (since 1F⋅V2=1J).
2.Specific Capacitance: Cs (Unit: F/g)
- Definition: Capacitance per unit mass (gram).
3.Step Derivation to Wh/kg
3.1 Energy → Mass Normalization
- Specific Energy=E/m=0.5CsV2(Unit: J/g)
- Substitute C=Cs⋅m into the formula, canceling mass m.
3.2 Gram → Kilogram (1 kg = 1000 g)
- Energy Density=0.5CsV2×1000 (Unit: J/kg)
3.3. J → Wh (1 Wh = 3600 J)
- Energy Density=0.5CsV2×1000/3600=7.2CsV2 (Unit: Wh/kg)
- Badges
- Science topic