Battery backup design with inverter at 48 V

Assumptions: inverter efficiency 90%, battery depth of discharge 80% (given), all loads may run at the same time, DC bus 48 V, back-up for 1 day as listed.

Step 1: Load and daily energy

Step 2: Inverter rating

Connected load        = 9400 W
With 25 % margin      = 11 750 W
At pf 0.8 → kVA       = 11 750/0.8 = 14.7 kVA

Choose a 15 kVA, 48 V DC input, pure sine-wave inverter (pumps need surge capability of about 3 × running current; a soft starter/VFD for pumps is advised).

Step 3: Battery energy and Ah

Energy from battery = 52 980 / (η_inv × DoD)
                    = 52 980 / (0.9 × 0.8)
                    = 73 583 Wh
Ah at 48 V          = 73 583 / 48 = 1533 Ah

Step 4: Battery bank options

Option 1: 12 V, 200 Ah

Series  = 48 / 12     = 4
Parallel = 1533 / 200 = 7.67 → 8 strings
Total   = 4 × 8       = 32 batteries
Bank    = 48 V, 1600 Ah
Usable  = 1600 × 48 × 0.8 × 0.9 = 55 296 Wh ≥ 52 980 ✓

Option 2: 2 V, 1500 Ah

Series   = 48 / 2      = 24 cells
Parallel = 1533 / 1500 = 1.02 → 2 strings
Total    = 48 cells, 48 V, 3000 Ah

One string (24 cells, 1500 Ah) gives 1500 × 48 × 0.8 × 0.9 = 51 840 Wh, about 2% short (enough only if inverter losses are ignored: 52 980/(0.8 × 48) = 1380 Ah).

Selection

Option 1 (32 × 12 V, 200 Ah, 4 series × 8 parallel) matches the need closely (1600 Ah vs 1533 Ah). Option 2 with one string is cheaper in wiring and 2 V tubular cells last longer, but slightly undersized; with two strings it is oversized (3000 Ah). Option 2 (single string) can be chosen only if the load hours are reduced a little.

Step 5: Charger

To recharge 1600 Ah in about 10 h: charging current ≈ 0.1 C = 160 A at 48 V (≈ 8–9 kW charger, or the inverter's built-in charger / solar charge controller).

AC ─►[Charger 48V,160A]─┬─►[15 kVA Inverter]─► Loads
                        │
              [48 V battery bank]

Answer: Daily energy 52.98 kWh; 15 kVA inverter; battery 1533 Ah at 48 V → 32 nos. 12 V 200 Ah (4S × 8P) recommended.