 # Doug S

Members

3

## Community Reputation

0 Neutral

• Rank
Starting Member
1. This is exactly what I was asking. I have seen this method done 2 ways: With, and without, multiplying by the power factor of 0.8 and I have never gotten confirmation on which is correct. I assumed that the power factor was inherent in the generator while running, so that multiplying by .8 would not be necessary. I thought the power factor only came in while figuring kVa to kW?
2. Marc: The purpose of trying to determine existing load with this method is for a month-to-month record of load to show that it is above 30%. If we document a load over 30% every month, we do not have to perform an annual load bank test. I know the method is not great, but for this purpose, what method could you recommend for existing load calculation? I truly appreciate the help.
3. I am trying to determine the most accurate way to calculate the existing load on an emergency generator from the amperage reading, while the generator is running under load. I will use a generator with the following nameplate information as an example : 208 V, 500 kVa, 400 kW, 0.8 Power Factor, 1388 Amps. If my generator reads 600 Amps while running with the load, I know I take 600 amps x 208 v x 1.732 (3 phase power) which gives me 216,153 Watts. I can divide by 1000 to give me approximately 216 kW while under load. My question is do I take that number and divide by 400 kW, or do I divide by 500 kVa to get the load? Could I also simply take that 600 Amp reading and divide by the 1388 Amp from the generator nameplate? These 3 ways would give me 1 of 2 answers, and I do not know which one is correct. Any help is very much appreciated.
×
• Hire Inspector