3-phase power calculation when phase to phase voltages are different

Question

I have a 3-phase power supply rated at 230VAC per phase. However, the phase to phase voltages are not equal. Suppose \$V_{RY}\$, \$V_{YB}\$ and \$V_{BR}\$ be the three phase-to-phase voltages between the R-Y, Y-B and B-R phases respectively. I also have the measurements of the currents for each phase: \$I_R\$, \$I_Y\$ and \$I_B\$. These are all RMS values. I am applying resistive load on each phase, so the power factor can be considered to be 1.

How do I calculate the total power consumption of this setup? I do not know the load resistance values (and they may be variable). The three voltages can be considered to be phase separated by 120 degrees from each other. However, the three phase to phase voltages are not the same between each pair of phases.

^{simulate this circuit – Schematic created using CircuitLab}

The typical formula that I see either assumes that the load is balanced, or that the phase to phase voltages is the same. For me, neither is true. If an exact formula for this is difficult, I can manage with a good guideline/rule-of-thumb formula for calculating the power.

PS: The power supply is the mains power supply, but due to poor distribution/load-balancing etc. the voltages vary substantially between each phase.

Answer 1

If you have resistors from red to blue, from blue to yellow and from yellow to red (i.e. a delta formation load) then the power is the sum of the individual line voltages (squared) divided by the individual resistance across each line:

$$P = \frac{V_{RB}^2}{R_{RB}} + \frac{V_{BY}^2}{R_{BY}} + \frac{V_{YR}^2}{R_{YR}}$$

If you have resistors in a star formation this is more difficult unless you have a neutral wire commoning the three resistors. If you do then measure the individual phase voltages and do individual power calculations then add the three powers to give total load power.

If you don't have a neutral then you will have to calculate the star-point voltage relative to red, blue and yellow respectively. Then you'll have three voltages and three resistors and power is the sum of the individual powers.

Answer 2

Since you know the currents in each phase, you also need to know the voltages across each resistor as they are different (i.e. Phase voltages R-N, Y-N, B-N). You don't need to know the values of resistive loads which may vary. They will certainly vary from cold (lower resistance) when off-line to hot (higher resistance) when energized. Compute power as:

$$P_{total} = V_r I_r + V_yI_y + V_bI_b$$ where each voltage is with respect to N.

The above \$V^2/R\$ formula is fine for theory but not practical in your case because you don't know the values of load resistance. If you can measure each phase-to-neutral voltage and phase current, then you can computer total power using the \$P_{total}\$ formula. Note that this formula works even if the phases are not exactly \$120^{\circ}\$ apart.