Answer / kuzhali

ups back up time=(AH rating*volt*power factor)/total load

Answer / velmurugan, so/d, barcf, kalpa

Consider the 2 KVA online UPS.
specifications are as follows: output voltage: 230V, Pf:
0.8, Load is 1600W. DC bus voltage is 96V.

Now Total output is = 2000 VA
PF of the load is = 0.8
There fore output in watts = 2000x0.8= 1600W.
Efficiency of the inverter is 80% therefore power input to
the inverter will be = 1600/0.8= 2000W
DC current of the inverter section= Watts / Voltage
Consider the fully charged voltage = 13.5X8=108V
The DC bus voltage
up to which the battery can deliver= 10.5x8=84 Volts.

therefore maximum current drawn by the inverter when battery
is in line = 2000/84=23.8A

Now consider the backup required for the load, i.e 1 hour
back up is required to safe shutdown loads when power supply
is not available.

Consider a class 10 type of battery where it has a
efficiency of 90%, 300Ah battery.

the effective AH available for the load is 300x0.9=270AH.
For 1 hour duration the discharge current will be 27.0A
which is higher than the required value of 23.5A.

There fore for 2 KVA online UPS having the load of 1600W
with 0.8 pf load, it requires 300AH of 8 nos of battery for
1 hour backup.

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