# Battery Sizing for AC system

Jeza43

## Main Question or Discussion Point

Hi,

Interested in sizing some DC lithium-ion batteries (through a controller) to power the following system: 90KVA, 200/115 Volts AC, 400 Hz, 3-phase.

The current approaches I've taken have been drastically oversized so I'm clearly missing something.

Cheers,
J.

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russ_watters
Mentor
How long does it need to produce 90 kva?

My understanding is that Tesla does 88kvah by paralleling thousands of small standard cells...

CWatters
Homework Helper
Gold Member
Hi,

Interested in sizing some DC lithium-ion batteries (through a controller) to power the following system: 90KVA, 200/115 Volts AC, 400 Hz, 3-phase.

The current approaches I've taken have been drastically oversized so I'm clearly missing something.

Cheers,
J.
Perhaps show us your calculations. They might be correct.

Jeza43
How long does it need to produce 90 kva?

My understanding is that Tesla does 88kvah by paralleling thousands of small standard cells...
It needs to produce for around two hours, yes I'm currently looking at utilizing Nissan Leaf battery packs which have a capacity of approximately 24 kWh.

Jeza43
To elaborate, CWatters, I'll step you through my process back to front. To obtain the effective power derived from the AC system I calculated the rms current and voltage, which then using a power factor of 0.9 (assumption as cannot find specific to the system) and a added factor of sqrt(3) for the three-phase aspect provided the following:

V_rms=V_max / sqrt(2) = 200 V/sqrt(2)=141.4 V
I_rms=90 KVA / V_rms = 636.4 A
P_eff = V_rms * I_rms * PF * sqrt(3) = 140 kW

Which would indicate for two hours of operation I would need a capacity of at least 280 kWh.

Any feedback would be appreciated, come from a mechanical background so am not confident with my methodology.
Cheers,
J.

russ_watters
Mentor
To elaborate, CWatters, I'll step you through my process back to front. To obtain the effective power derived from the AC system I calculated the rms current and voltage, which then using a power factor of 0.9 (assumption as cannot find specific to the system) and a added factor of sqrt(3) for the three-phase aspect provided the following:

V_rms=V_max / sqrt(2) = 200 V/sqrt(2)=141.4 V
I_rms=90 KVA / V_rms = 636.4 A
P_eff = V_rms * I_rms * PF * sqrt(3) = 140 kW

Which would indicate for two hours of operation I would need a capacity of at least 280 kWh.

Any feedback would be appreciated, come from a mechanical background so am not confident with my methodology.
Cheers,
J.
I would think you are given voltage in RMS (it almost always is), but I'm not familiar with a 200V system...

In either case, since you already have kVa, you don't need to do anything other than correct it for power factor: 90*.9=81 kW....and for 2 hours, is 162 kWh.

• CWatters
Jeza43
I would think you are given voltage in RMS (it almost always is), but I'm not familiar with a 200V system...

In either case, since you already have kVa, you don't need to do anything other than correct it for power factor: 90*.9=81 kW....and for 2 hours, is 162 kWh.
Thanks for the feedback russ_waters, still larger than I was expecting but it's good to know the maths is sound.
Cheers,
J.

Not enough info:

What is the "system" - 90 KVA may be the rating but at what rate does it use real power, not just Watts vs LVA, but I doubt it runs at FULL power for 2 hours. That spec looks like a UPS? -- The UPS, if it is controlling the battery, needs to be connected to the same battery technology. So if it was built for Lead Acid - you can not just change it to LiPO. The issue is the charging of the LiPo - discharge not as much of a concern.

The load - may use 90 KVA and use very little power - so the system can be running at maximum rating, but the battery supplying little power.

In an AC system - KVA can be "no" real power, but a confusing point is that in a DC system ( battery) - VA is effectively the real power. Batteries are rated in Amp-Hours, but you have to look at the datasheet to see where this rating was developed, the RATE of discharge affects the total A-H delivered, and the rating is to what Voltage.