How do calculate power consumed by a BLDC motor using two-wattmeter method

Question

From https://www.studyforfe.com/blog/steps-for-solving-the-two-method-wattmeter-problem/ :

$$\begin{aligned} W_1 &= V_L \cdot I_L \cdot \cos(30°+\phi) \\ W_2 &= V_L \cdot I_L \cdot \cos(30°-\phi) \\ P &= \sqrt{3} \cdot V_L \cdot I_L \cdot \cos(\phi) \\ (\phi) &= \tan^{-1} \left[ \sqrt{3} \frac{W_2-W_1}{W_1+W_2} \right] \end{aligned}$$

This notation is a bit confusing because using \$V_L\$ and \$I_L\$ for line voltage and line current it seems like only one watt meter is actually needed? Also, I am assuming that I should be using RMS voltage and current for all of these calculations?

My current labview implementation looks like this:

But powering the ESC with a DC bench-top power supply I know that I am supplying only 88W while this logic measures 120W, about twice what I would expect if my ESC is ~65% efficient.

Any tips appreciated

Answer 1

You're taking RMS values before doing the multiplication to compute power. That's not right.

Power is computed by taking instantaneous voltages and currents, multiplying them, and then computing the RMS value.

By computing RMS prematurely, the relative phase between voltage and current is lost, and the power computation will only work out if there's either no reactive power or no real power. And then you still won't know whether the power you read is real or reactive - only that the numerical value is correct.