220/240 to 200v dc power supply
To get 10 kW out of an inverter with 85% efficiency your input current at 24 V would be \$ I = \frac {P}{V \times e} = \frac {10000}{24 \times 0.85} = 490 \ \text A\$.

If, instead, you arrange your batteries in series connection of eight you get 192 V nominal and the current from the batteries would be \$ I = \frac {P}{V} = \frac {10000}{192} = 52 \ \text A \$ with no inverter losses.

Most of us, given the option, would opt for the 192 V setup to avoid the very high currents and inverter losses. Either option requires a high level of understanding of the risks to personnel and equipment involved.