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13% out does look a bit high.
You would need the meter manual, it is likely the kWh and kVarh are instantaneous, whilst the kVAh may be averaged over some time period, or all likely different periods
You would need the meter manual, it is likely the kWh and kVarh are instantaneous, whilst the kVAh may be averaged over some time period, or all likely different periods
The instantaneous values of S, P and Q (kVA, kW, kVAr) add in quadrature so S² = P² + Q². But the time integrals (kVAh, kWh, kVArh) do not add in this way. The generator kVAh figure is the integral of the vector sum, not the vector sum of the integrals.
can you translate that into english? ?
Sure.
If you walk one mile east and then one mile north, you will be one mile east and one mile north of your starting point, having walked a total of two miles.
If instead you walk 1.4 miles north-east you will still arrive at the same point one mile east and one mile north of your starting point, having only walked 1.4 miles.
You cannot calculate the journey distance (kVAh) from the distance moved east (kWh) and the distance moved north (kVArh) without knowing on which bearings you travelled along the way (cos (φ))
If you walk one mile east and then one mile north, you will be one mile east and one mile north of your starting point, having walked a total of two miles.
If instead you walk 1.4 miles north-east you will still arrive at the same point one mile east and one mile north of your starting point, having only walked 1.4 miles.
You cannot calculate the journey distance (kVAh) from the distance moved east (kWh) and the distance moved north (kVArh) without knowing on which bearings you travelled along the way (cos (φ))
Not on the cumulative totals, only on the instantaneous value, unless the power factor is exactly constant. If the power factor changes during the counting of the totals, the hypotenuse of the 'triangle' stops being a straight line.
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