# Converting Instantaneous kW to kwH

I'm breaking my head over this. I've got 1200 samples of true RMS current using a BK Precision 5492B over a period of 10 minutes (this means a sample every 0.5 seconds). I'm using a power factor of .8 for this calculation.

P (apparent) = Irms x Vrms

P (real) = P (apparent) x Power Factor.

If I sum all of my 1200 samples of P (real) I get 0.8523W

Now, since I took samples every 0.5 seconds, that means in 1 hour there are 7200 samples, so if I divide that number by 7200 I get: 0.000118385kWh

Am I correct in my calculation?

I'm trying to figure out the cost of running this device for 1 hour.

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 I arrive at the same value, does this mean that 0.000118385kWh are consumed in 10 minutes. Should I multiply this by 6 to get the khW in 1 hour? I'm trying to figure out the cost of running this device for 1 hour. – Gustavo Corona Dec 12 '12 at 1:57 Correct, but you're in the Wh range or lower, so you're looking at 0.710Wh. How confident are you in the accuracy of your power measurements (i.e. was the BK Precision 5492B set to appropriate measurement ranges for current and voltage)? Does less than a watt align with your test expectations? – HikeOnPast Dec 12 '12 at 2:15 Yes, sir. The device was in stand-by mode, it's basically a very small PC, so I expect it to consume very little power in that mode. Alright, so that means I consume 0.000710311kWh in one hour. At $0.10 per kWh it would cost me$0.000071 USD to run it. Correct? – Gustavo Corona Dec 12 '12 at 2:18 Subject to measurement accuracy of the metering device, yes, it will theoretically cost \$0.000071 USD/hour to operate in standby per your data. – HikeOnPast Dec 12 '12 at 2:24 I'm going to add another comment, because you made me think a bit about my results, and yeah they were a little too low, I remeasured using the lower range and now got a reading of 0.019A rms, I'm going to take another 1200 samples, this looks more reasonable. – Gustavo Corona Dec 12 '12 at 2:27