Electrical question no.20 - Battery rating calculation for UPS
To be more specific, I was looking for a 2 KVA UPS to cater 6 PCs for a back-up tme period of 1 hr. I found each & every manufacturer recommending different rating of Battery AH & quantities. Please help me with the calculation to make right judgment.

#1
Mon, September 1, 2008 - 17:33
Two methods are used to size stationary batteries, constant current and constant power.
Constant current is used for applications such as generating stations, substations, and
industrial control.
Constant-power sizing is used almost exclusively for UPS system applications. Unlike constant-current loads, the current of a constant-power load (e.g., an inverter or dc motor) increases as the battery terminal voltage decreases during a discharge, so that  P =E  x I  remains constant. Load profiles are not normally drawn for constant-power sizing since the load, once applied, remains the same throughout the battery discharge. A more detailed discussion of constant-power sizing is included
in IEEE Standard 1184.
Electric utility and industrial control-voltage windows generally use the standard cell end voltage but are sometimes limited by the minimum system-voltage limit. UPS applications can offer the largest voltage window, since the inverter input can be designed to accept a wide range of voltage without affecting the inverter’s output voltage. The standard discharge rate for lead-acid stationary batteries is 8 h in North America and 10 h in other parts of the world. There are some exceptions to this. UPS batteries are often rated at the 15-min rate.

For example determination the battery required to supply a UPS system rated 600 kVA, 0.8 pf, for 15 min to an end voltage of 1.67 Vpc. The battery has 200 lead-acid cells in 50, 4-cell. Inverter efficiency is 0.92. Assume temperature correction and design margin are equal to 1.00 and that an aging factor of 1.25 will be used. We can write:

1- Cell Size (watts) = (UPS dc input/number of cells) x design margin x temp
corr x aging factor.

2. Compute the dc Input Watts Required by the Inverter

W = (VA x pf)/efficiency = 600 000 x 0.8/0.92 = 521,739 W

3. Compute the Cell Size

Cell size = (522 kW/200) x 1.00 x 1.00 x 1.25 = 3.26 kW/cell.

4. Select a Standard Cell
Using a battery manufacturer’s discharge tables or curve, in watts, select a cell that has a minimum discharge capability of 3.26 kW/cell at the 15-min rate to 1.67 Vpc.

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