# 220/240 to 200v dc power supply

I have a 12kva (10kw) ups and I want to power it from a 24v battery bank. The dc input is 192dc input from the battery side. I was thinking off getting a 24 to 220 Inverter and rectifing it to output 200v dc if possible. I know a little about electronics so not to much off a noob. Thank you for your time

This is the ups that I have 8 to 18kw https://www.jsfleming.com/wp-content/uploads/2016/07/sl-24870.pdf

• How many batteries do you have in your 24v battery bank and what capacity are they? – brhans Dec 27 '18 at 20:27
• 220 AC rectified is 311V... So not a good plan. Given the voltages and powers, this seems like a dangerous project given your experience level. – MadHatter Dec 28 '18 at 0:19
• So to recap, you need to provide information on the UPS and the batteries you intend to use. The UPS likely includes a charger. All cabling you use should be properly rated for the maximum current of the UPS, regardless of what you intend to plug in. The UPS won't limit it's output until 10Kw(Which really isn't that much as domecile power consumption goes) so you must be sure the input is rated to sustain that full 10kw. If rated current is 490A, you must use 490A or better wire and batteries that can sustain that draw. Use a suitable battery bank jfor the UPS and contain it properly. – K H Dec 28 '18 at 0:31
• I don't have any batteries but I can get them, my plan was to build a power supply with a low voltage between 24 to 48 volts and output the required dc voltage to run the ups system. – Computer Tech Dec 29 '18 at 2:06
• About the 220 to 200 dc power supply, yes it's 311v but I was going to use a step down transformer to 120 or so then rectify it, caps and filter etc. and it should be in the ball park off the voltage that I'm looking for.(yes, no)???? – Computer Tech Dec 29 '18 at 2:20

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.