Discuss 3 phase power calculations. in the UK Electrical Forum area at ElectriciansForums.net

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Dear all..

May be some body can help me to explain what happened with my generator set module. At my attach picture. According to S2 = P2 + Q2 is not correct. Can some body explain it.
 

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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
 
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.
 
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))
 
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))

So......that's what them pointy arrow things on vector diagrams were all about.....?
Nice one L.
Hope things are going OK, by the way.
 
It's mean triangle power can not use?

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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