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DigitalShoestringSolutions / PowerMonitoring

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Power calculation maths #24

Closed tobyaharris closed 3 months ago

tobyaharris commented 4 months ago

I'm unsure if the analysis module is doing total machine power calculations correctly.

A real example machine is a 3 phase motor at a customer factory. Its plate states 33.7A, 18.5kW. Power factor is suspected to be approx 0.9. The supply voltage has been measured as 421V phase-phase. It is understood the motor draws 33A from each supply phase.

The Historic (all) dashboard shows the machine drawing 100A, which makes sense if it is adding up the current from each phase. The power graph on the same dashboard reports 23kW. Hence I suspect the codebase is assuming 230V supply (seperate issue - it's not obvious how to change this).

This power value is still well over the 18.5kW. If it were multiplying by 421 it would report 42.1kW.

Does the 33.7A 18.5kW plate just not add up, or is the method of summing the phase currents then multiplying by the voltage at fault?

tobyaharris commented 4 months ago

At 400V, a 88% efficient 18.5kW 3 phase motor at 0.9 cos phi should draw 33.715A. I don't think the motor plate is wrong.

tobyaharris commented 4 months ago

To put my overthinking this in writing:

Consider a 8.58 ohm ideal resistor. Under DC 230V it would draw I = V/R = 230/8.58 = 26.8 A. The power dissipated would be P = V^2/R = 52900 / 8.58 = 6166 W. Under AC 230 VRMS the calculated results would have the same values in RMS units.

Consider 3 such resistors in parallel on the same supply. Together they would dissipate 3 6166 = 18498 W. The total current drawn is 3 26.8 = 80.4 A. Consider instead this trio star wired (one end of each resistor connected to N) into a 3 phase supply, with the same 230V phase-neutral - no change in RMS values. The phase-phase voltage is 400V, but using the phase-neutral 230V the correct power value is obtained with 80.4 * 230 = 18492 W (rounding error only).

A previously discussed, as we are only monitoring the supply, not the motor coils, if the 3 phase system uses delta wiring + higher resistances to achieve the same supply draw this has no impact on power.

tobyaharris commented 4 months ago

The calculation process is fine, we assume this motor has an electrical imput of 23.25 kVA and discrepancies are down to power factor and inefficiencies. However, what we can take from this is we need: