Internal resistance calculations. You add resistances when they are in series and when the resistances have identical values and are in parallel you divide the resistance by the number of parallel circuits.

So the individual string of 20 cells in series has a combined resistance of 20 x 5 milliohms = 100 milliohms.

You have two strings in parallel and so the resultant resistance = 100/2 = 50 milliohms

if you had three strings in parallel the resultant resistance = 100/3 = 33.3 milliohms

And 4 strings gives 25 milliohms and so on.

The internal resistance is important because the current flowing through the strings of cells in parallel will cause a voltage drop across the end of the strings. To get the amount of power into the UPS assumimg it is 100 % efficient and providing 140 kW you have to supply 140000/120= 1167A from the two strings in parallelI. The voltage drop inside the battery strings will be 1167 x 50/1000 = 58 V and if you measure the “240Vdc” across the string ends it would be 240-58 = 182V.

You can extend these ideas to the case where you have 3 strings of 20 batteries and 4 strings of 20 batteries. For these cases you get 1167 x 33.3/1000 = 39V and 1167x25=29V respectively giving you 201V and 211V respectively at the terminals.

The UPS has to work with the loaded voltage of the batteries as calculated in the above two paragraphs and the maker will have it designed to work with a minimum voltage which will be in their specification. I don’t know what that value is.

A few words about power factor and kW. With alternating voltages and current they both have a sinusoidal wave shape if seen on a display but they may not be exactly in synch and there could be a phase difference between them of a number of degrees. The Cosine of this angle is the power factor in your case 0.8. If the waves shapes were exactly in phase the Cosine of 0 is 1 and if they were 90 deg out of phase then the Cosine is 0. The power taken by an AC system is volts x amps x power factor. In your case the 140KW has already taken the power factor into consideration. You can’t have it twice.

I know nothing about the nature of the 140KW load on the UPS. It would be useful to commission some measurements as this sounds like a guesstimate and its value is what all the calculations are based on.

With regard to efficiencies. I don’t know what they mean by battery efficiency. Do they mean you get 95% energy out compared to what you supply? And how is it measured? I guess that the quoted UPS efficiency of 98% means that you get 98 % of the energy out compared to what you put in but that will only be the right number for specific working conditions. In any event the quoted efficiencies are high enough that for practical purposes you can assume 100% and just apply a reasonable safety factor to make sure all will be well in practice.

Hi,

We have 6 Dell PowerEdge T430 servers with 2 750W PSU on each server (one connected to the wall socket and one to a UPS system)

What is the best way to calculate my UPS needs?

I’ve checked in a few vendors automated calculators and each one gave me a different result.

I do know the basics about Amps, Volts, and Watts but can not figure out how to do this calculation.

Thanks in advance. I appreciate your help : - )

@Eaton @APC_by_Schneider_Electric