Electrical Engineering Asked by Bud on December 16, 2020
I am trying to measure the output power for BLDC controller, in order to calculate the efficiency of the controller. I came to know that two watt-meter method is one of the method used to calculate the three phase AC power.
But instead of using two wattmeters, i have…..
Two clamp meters (AC/DC clamp meter) for measuring I_{r} and I_{Y}.
Two multimeters (AC/DC) for measuring V_{RB} and V_{YB}.
My input power is DC 48 volt. P = V × I will give me the input power.
But to calculate the output power has complexity.
W1 = I_{R} × V_{RB} and W2 = I_{Y} × V_{YB}
W1 + W2 gives the output power.
For voltage measurement V_{RB} and V_{YB}:
My bldc controller output is trapezoidal waveform with variable output frequency, Vrb and Vyb can be calculated by using ac multi meter. Because ac voltage does not depends on the Frequency, whatever the Output frequency the Vrms will be the same i guess. Correct me friends if i am wrong.
For current measurement Ir and Iy :
Since the AC current must not be same for different frequency at all time. How to calculate the AC current at my output frequency say (eg: 100 Hz).
Edited…
Sorry i forgot to say something, I am driving the bldc controller with bldc motor in a load setup.
You can always figure out real power by averaging the instantaneous power of an element (V instantaneous * Iinstantaneous).
If you only have RMS meters then you have to sit down with your theoretical waveform and calculate it out real power analytically, then RMS analytically, and then compare the two to know the appropriate fudge factor to apply to your RMS numbers so that you can adjust the numbers from your meter. But phase and waveform shape cannot change from your theoretical or it will throw things off. That is to say, there is no real way to do it with just RMS meters since they do not retain the phase information.
Answered by DKNguyen on December 16, 2020
To measure power in the situation shown, the instrumentation system must be able to determine the instantaneous values of voltage and current, multiply current X voltage at corresponding instants and integrate the results over the period of the waveform. You can not measure the power using the instrumentation described.
Answered by Charles Cowie on December 16, 2020
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