Maximum Apparent Power

Your question is some what confusing. You can calculate your apparent power by using the formula App Power = voltage x amps x 1.73 for 3 phase installations. You would take a measurement of the amperage using a clip on meter while taking a reading of the phase to phase voltage at the service location. You may need to take several measurement to get a close approximation for the maximum. If your utility meter is reading the the actual power, KW, then use the inductive reactive reading associated with the maximum KW reading.

The Max Demand KVA = Sqrt( KW² + KVAR²). You can compare this the you actual measurements.

Thanks for your response

I noted your comment on the question and I will try to reformulate it in another form:

Can I say that in order to calculate MaxP = Maximum Demand Apparent Power, having

MaxA = Maximum Demand Active Power and

Q = Reactive Power at the time of Maximum Demand of Active Power,

it is correct to say that

MaxP = √(MaxA)²+Q²)

?

Remember that we are speaking about the maximum value in a period. In a nutshell, this would mean that the maximum of the apparent power is always concurrent with the maximum value of the Active Power.

The Max Demand KVA = Sqrt( KW² + KVAR²) and MaxP = √(MaxA)²+Q²) appear to be the same.

Does your meter also record the time at which the readings were taken ?

The maximum of apparent, active and reactive power may not appear at the same moment. If you have the measurements for active (P) and reactive (Q) power at the SAME moments, you can calculate the corresponding apparent power (S) at that moment.

If you want to determine the maximum demand, you have to do it for each of S,P and Q separately. It may be different for each of them, both as value and time recorded/calculated.

S^2 = P^2 + Q^2.