Illustrative worked example. The inputs are assumptions, not a customer project. The arithmetic is real.
The scenario
A home with rooftop solar wants to cover a 3 kW evening load for 5 hours from stored solar power. Peak demand is 5 kW.
Energy needed
3 kW × 5 h = 15 kWh. If 90% of the battery is usable, the nominal capacity is 15 ÷ 0.9, about 16.7 kWh.
Choosing the configuration
- Pack voltage. A 51.2 V pack (16 LiFePO₄ cells) is in the 48 V class used for home storage.
- Capacity. A 100 Ah pack holds 5.12 kWh. Four in parallel give 20.48 kWh, with about 18.4 kWh usable at 90%.
- Current. 3 kW is about 59 A at 51.2 V, and the 5 kW peak is about 98 A. The BMS, inverter and cabling must carry both.
- Fit. 20 kWh sits inside Eldtek's 5 to 30 kWh home range.
| Item | Value |
|---|---|
| Nominal voltage | 51.2 V |
| Nominal energy | 20.48 kWh (4 × 5.12 kWh) |
| Usable energy at 90% | About 18.4 kWh |
| Peak current at 5 kW | About 98 A |
What Eldtek confirms
- The module sizes and the parallel limit.
- Inverter compatibility and communication with the BMS.
- The solar size and charging profile.
- Mounting, protection rating and installation.
See the battery energy storage systems page for the full range.
Frequently asked questions
How big a battery do I need for a 3 kW load over 5 hours?
15 kWh of energy, which is about 16.7 kWh nominal if 90% is usable. A 20 kWh battery covers it with margin.
What current does a 5 kW load draw at 51.2 V?
About 98 A, since current equals power divided by voltage. The BMS, inverter and cabling must all support it.
Have a similar requirement?Send your figures and an Eldtek engineer will size it for your site.
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