Timeline for How to calculate kWh from amps
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 30, 2019 at 23:27 | comment | added | Bdloul | thank you very much, it really helped | |
| Dec 30, 2019 at 23:27 | vote | accept | Bdloul | ||
| Dec 30, 2019 at 22:30 | comment | added | SteveSh | Agree with what muyustan said. What you're doing is just an approximation of the integration function of power over time. If your load (current usage) doesn't vary much with time, and it looks like it doesn't in your example, you're close. | |
| Dec 30, 2019 at 22:14 | comment | added | muyustan | the summation approaches to what it is known as integration. | |
| Dec 30, 2019 at 22:14 | comment | added | muyustan | If you accept(you should not) the power "4.49kW" constant all over between 11:00 and 12:00 and same for other 1 hour intervals, then your calculation in your example is correct. See, you have power which is a measure of how much joules per one second you consume, then multipliying that power with the time duration of consumption, you get total consumed energy. However, in real cases, the power consumption will be different all the time, which leads, for a better estimation, you have to narrow the time between measurements, as much as possible. As your sampling period approaches to zero,.... | |
| Dec 30, 2019 at 22:06 | comment | added | Bdloul | @SteveSh I've added an example, can you tell me if it's right ? | |
| Dec 30, 2019 at 22:06 | comment | added | Bdloul | @muyustan I don't understand how to go from watt to watt hour if I'm getting instatenous power, that's my issue ! | |
| Dec 30, 2019 at 21:57 | history | edited | Bdloul | CC BY-SA 4.0 |
added 182 characters in body
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| Dec 30, 2019 at 21:51 | answer | added | Elliot Alderson | timeline score: 2 | |
| Dec 30, 2019 at 21:45 | comment | added | SteveSh | You're almost there. You've got the power. Now just multiply by the time that power is used. In your example, if your device uses 10 watts for 1 hour, the energy usage 0.01 Kw-hr. That's basically integrating the power (10 watts) over the time (1 hour). | |
| Dec 30, 2019 at 21:44 | comment | added | muyustan | I * V gives instantenous power at any time in watts. So if you accept your measurements are constant all over the time, which will not be, (I * V * time elapsed) could be a good approximation. However, for a realistic approach, you monitor the data for a period, and then integrate the I * V vs time graph to get total energy consumption. The conversions from watts or watt-hours to kilowatt or kilowatt-hours range is just a mathematical modification. Note that, watt(joules per sec) is the unit of power and energy(joules) is just power times the amount of time elapsed. | |
| Dec 30, 2019 at 21:44 | history | edited | JYelton | CC BY-SA 4.0 |
Removed unnecessary thanks (show thanks by up-voting and accepting answers).
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| Dec 30, 2019 at 21:40 | review | First posts | |||
| Dec 30, 2019 at 21:44 | |||||
| Dec 30, 2019 at 21:36 | history | asked | Bdloul | CC BY-SA 4.0 |