TOPIC-04 · Engineering guide
Battery Runtime from Watt-Hours
Deliverable energy is voltage × ampere-hours × state-of-charge window × health × discharge efficiency. The locked fixture delivers 864 Wh and runs 8.64 hours at 100 W.
Formula, variables and conditions
E = V × Ah × (socStart − socEnd) × soh × efficiency runtime hours = E / loadW
Variables and units
- V: nominal volts, greater than 0.
- Ah: capacity, greater than 0.
- socStart and socEnd: 0 to 1, and the end must not exceed the start.
- soh and efficiency: greater than 0 and at most 1.
- loadW: constant watts, greater than 0.
Conditions
- Voltage and load stay constant for the whole interval.
- This is not the UPS runtime model. UPS energy omits efficiency, then divides the AC watts by efficiency to get battery-side power.
- SRC-010 URL on the calculation record is null. No manufacturer discharge curve is implied.
Worked example
Locked TOOL-016 fixture.
- Battery
- 12 V, 100 Ah
- State of charge
- 1 to 0.2
- Health / efficiency
- 1 / 0.9
- Load
- 100 W
- Deliverable energy
- 864 Wh
- Runtime
- 8.64 hours
E = 12 × 100 × 0.8 × 1 × 0.9 = 864 Wh. Runtime = 864 / 100 = 8.64 hours.
Boundaries and common mistakes
Boundaries
- Peukert effect, temperature, chemistry, BMS limits and voltage cutoff are not modeled.
- The result is not an installation approval.
- A 48 V UPS teaching fixture belongs to the UPS guide. Do not relabel 8.64 hours as that 3.84 hour result.
Common mistakes
- Omitting efficiency here, or applying it a second time to the load.
- Using a state-of-charge end above the start. The model rejects that order.
- Converting ampere-hours to hours by dividing by current when the known load is watts and the efficiency term is required.