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If a load is connected to an AC power source, and the current wave on it is shifted from the voltage wave of the power source, for some periods of time the instantaneous real power (Vinst x Iinst) will give a negative instantaneous power value, due the voltage value be positive and the current value be negative at the same instant, or vice versa. How this "negative power" affect the Wh real power consumption measurement? How it accounts in cumulative energy measurement?

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How this "negative power" affect the Wh real power consumption measurement? How it accounts in cumulative energy measurement?

Here are a few waveforms to consider: -

enter image description here

The power waveform is in red and, it is the average value of this waveform that indicates true power consumption. If the instantaneous power goes negative then it just reduces the average power. At 90 ° phase shift between voltage and current, the average power is zero because positive parts exactly cancel negative parts.

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  • \$\begingroup\$ Thanks. My doubt: I know when the average power is positive it does increase the value of Wh consumption accumulator... but when the average power is negative? It does decrease or also increase the value of Wh consumption accumulator? \$\endgroup\$ – abomin3v3l Dec 31 '20 at 12:08
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    \$\begingroup\$ When average power is negative then you are returning power from your load to your supply such as in a solar farm where it is putting power onto the grid. Or, as in a motor that turns into a generator and returns energy back to the source. For most normal loads, it can never happen @abomin3v3l \$\endgroup\$ – Andy aka Dec 31 '20 at 12:11
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    \$\begingroup\$ A very bad power factor occurs when the load is purely capacitive or inductive and the average power is then 0 because PF = 0. Low PF can mean a penalty. If Wh is negative AND your energy meter is capable of running in reverse AND you have the right sort of agreement with your energy provider means your bill can reduce. \$\endgroup\$ – Andy aka Dec 31 '20 at 12:33
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    \$\begingroup\$ @abomin3v3l this should answer your question. And also this one. And also this one. \$\endgroup\$ – Andy aka Dec 31 '20 at 13:13
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    \$\begingroup\$ Sum each 1 second average and divide by 3600 gives watt hours. \$\endgroup\$ – Andy aka Dec 31 '20 at 15:04
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How this "negative power" affect the Wh real power consumption measurement? How it accounts in cumulative energy measurement?

The true energy delivered is defined as the time integral of the instantaneous product of the voltage and the current.

How well any particular measuring device measures that, rather than the VA, or something else, depends on how well it has been implemented.

In the old-fashioned spinning disc meter, the disc is driven by a special motor that forms the VI product to generate the torque. The disc is dynamically braked, so speed is proportional to torque. The total angle it turns then forms the time integral. This type of meter records the real power very well, to within its calibration accuracy and friction.

In a full-price modern digital meter, ADCs read instantaneous voltage and current, and the product is summed. This also reads very accurately, subject to all the significant waveform harmonics falling inside the digitisation bandwidth of the ADCs, or to put it another way, the ADCs being fast enough to capture all the significant waveform dynamics.

There are lower cost ways that digital meters may attempt to read true power. One type reads the current waveform, but only measures the phase of the voltage waveform, assuming its amplitude to be nominal, to compute a power factor due to the phase shift between the two waveforms. Other types sacrifice bandwidth for slower ADCs, so miss some high frequency dynamics.

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