On the calculation of active and reactive power
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By adjusting the power factor from the natural value of 0.85, a reward of 433.36 yuan was obtained from the power supply company in the form of a reduced electricity bill. Adjusting the power factor from 0.85 also reduced energy losses; how much does the loss decrease per hour? ? ? ? ? November 1: Active power – 1398.26, Reactive power – 434.17. November 30: Active power – 1566.89, Reactive power – 477.54. Can someone help me calculate how much electricity I saved this month? ? ? ? ?
Adjusting the power factor does not save electricity; its purpose is to reduce reactive current, decrease losses in transformers and lines, and improve the quality of the power grid. Therefore, it is not possible to calculate the amount of electricity saved based on the data you provided.
Active power = I*U*cosφ, that is, the rated voltage multiplied by the rated current followed by the power factor; its unit is watts or kilowatts. Reactive power = I*U*sinφ, and its unit is var or kvar. I*U represents capacity, with units of volt-amps or kilovolt-amps. When reactive power decreases or increases, active power remains unchanged. However, when reactive power decreases, the current also decreases, resulting in reduced line losses; conversely, line losses increase
All losses in power systems are inductive losses, and reactive power compensation can reduce these losses, but I’m unable to calculate it based on what you’ve provided