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Rkamu Rkamu
wrote...
Posts: 171
Rep: 1 0
12 years ago
let us consider we are going to design an UPS for a system.client wants the battery back up time is 4 hrs for 500 KW. the next day the same client wants to upgrade the back up time to 7 hrs. In this case we know that our battery discharge should be less when compared to 4 hours back up.here doubt is how to calculate the ampere hour of battery for 4 hrs ,and 7 hrs for the same load.(500 KW).
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wrote...
12 years ago
huh?

you are designing this? You are asking some basic questions.

500 kW is a very large UPS. Very! enough to run 100 homes. The largest UPS that APC makes is 9.6 kW and that costs $10000, and it has a backup time of 30 minutes. You would need 50 of these!

If you want the operating time on battery to go from 4 hr to 7hr, you need a bigger battery. No way around that.

500 kW is increased due to efficiency losses in the inverter, so figure 600 kW. For 7 hours that is 4200 kw-hrs.

A large lead acid battery has a capacity of 100 amp hours, which translates to 1000 watt-hrs or 1 kW-hr.

You will need *** 4200 *** large lead acid batteries to hold for 7 hours, and 2400 batteries for 4 hours.  Enough for a large submarine.



I think some of your numbers are wrong. First rule: get the numbers correct!
wrote...
12 years ago
Double it(ok  multiply it by 7/4ths which is a little less than double).Twice the time at the same load.
wrote...
12 years ago
Volts is missing from your problem so we'll assume 120volts.
p = i times e,  so i equals 500000/120=4166 amps per hour.
So your battery, which is rated in amp hours will be 4 X 4166 amp hours or 7 X 4166 amp hours. of course if your battery is some other voltage, say 12 VDC, then multiply the above  times a factor of 10.
wrote...
12 years ago
4h x 500kW = 2000 kWh
7h x 500kW = 3500 kWh

kiloAmpere-hours = kWh divided by volts.

Assuming 120V,

2000 kWh / 120V = 16.6 kAh
3500 kWh / 120V = 29.2 kAh

I would suggest a diesel generator the size of a mini-van, complete with auto-transfer switch as your UPS.
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