Doug S Posted February 10, 2020 Report Share Posted February 10, 2020 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. Quote Link to comment Share on other sites More sharing options...
Marc Posted February 11, 2020 Report Share Posted February 11, 2020 (edited) 6 hours ago, Doug S said: 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. If you've inductive loads, or loads that have a duty cycle, that method would not be advisable. Quote My question is do I take that number and divide by 400 kW, or do I divide by 500 kVa to get the load? The generator size is selected using VA demands. The engine HP is selected by using wattage demands. Edited February 11, 2020 by Marc Quote Link to comment Share on other sites More sharing options...
mtwitty Posted February 11, 2020 Report Share Posted February 11, 2020 Hi Doug. to find the KW under the running load you would calculate: 208 x 600 x .8 x 1.732 =172,923/1000 =173 KW. if your generator is rated for 400 KW, you would be running at about 43% of full capacity. Is that what you are asking? Quote Link to comment Share on other sites More sharing options...
mtwitty Posted February 11, 2020 Report Share Posted February 11, 2020 I defer to Marc. This falls in his wheelhouse and he is much more qualified to give you a quality answer.🙂 Quote Link to comment Share on other sites More sharing options...
Doug S Posted February 12, 2020 Author Report Share Posted February 12, 2020 On 2/10/2020 at 10:11 PM, Marc said: If you've inductive loads, or loads that have a duty cycle, that method would not be advisable. 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. Quote Link to comment Share on other sites More sharing options...
Doug S Posted February 12, 2020 Author Report Share Posted February 12, 2020 On 2/10/2020 at 10:23 PM, mtwitty said: Hi Doug. to find the KW under the running load you would calculate: 208 x 600 x .8 x 1.732 =172,923/1000 =173 KW. if your generator is rated for 400 KW, you would be running at about 43% of full capacity. Is that what you are asking? 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? Quote Link to comment Share on other sites More sharing options...
mtwitty Posted February 12, 2020 Report Share Posted February 12, 2020 KVA=apparent power=power before used. KW is power under load. (power consumed) So, to find actual KW usage, the power factor must be in the calculation. At least that is how I interpret it. Quote Link to comment Share on other sites More sharing options...
Marc Posted February 13, 2020 Report Share Posted February 13, 2020 (edited) KW is a unit of power. KVA is a calculation, equal to amperage multiplied by voltage. The two are the same when no reactive currents are present. There are two types of reactive currents: Inductive and capacitive. Motors and transformers are two of the most common sources of inductive currents. Capacitors create capacitive currents. The only energy transfer inherent in reactive currents are the I square R losses in the conductors. Utilities hate reactive power in their power lines because the I square R losses cost them money. They sometimes penalize users with large motor loads because motors create so much reactive currents. Long ago, utilities would create capacitive currents using 'over excited (field windings were deliberately over driven)' synchronous motors to create capacitive currents, which reduced the inductive currents in the power lines. Reactive power is actually power that is echoed back and forth between source and load with no net transfer of power actually taking place. Doug S, I'd suggest investing in a 3 phase power meter. Takes only a few seconds to connect it and it will show you the actual load. I have a handheld model laying around in my shop somewhere. Must be older than I am. It's about the size of a large clamp-on meter but has 3 leads coming out of it to connect to the lines. Edited February 13, 2020 by Marc Quote Link to comment Share on other sites More sharing options...
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