2
\$\begingroup\$

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.

schematic

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.

\$\endgroup\$
  • \$\begingroup\$ If you do not know the voltage AND you don't know the resistance, knowing the current will not let you calculate either. You must know at least two of U=I*R to calculate the third. \$\endgroup\$ – medivh Aug 7 '13 at 13:43
1
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ @Szymon - thanks for the edit it looks nicer now. I going to have to learn how to do that properly \$\endgroup\$ – Andy aka Apr 7 '13 at 17:22
  • \$\begingroup\$ Thanks for the answer. I do have a neutral common to the three loads, but I do not have the value of the loads. I only have the current measurements of each phase. I also have voltages between phases (not with neutral). I'll try to add a schematic to make it clear. \$\endgroup\$ – mayank Apr 9 '13 at 11:18
  • \$\begingroup\$ If you don't know the voltage between neutral and at least one of the phases, you can't calculate the power dissipated in any of the loads. You're going to need some kind of reference... \$\endgroup\$ – Mels Aug 7 '13 at 13:22
  • \$\begingroup\$ If you have control over the placement of the existing voltage meters, remove them from the delta formation they are in and connect them across neutral and each phase. Multiply the voltage for each phase by the corresponding current and add the three together. (Note that you don't "lose" the phase-phase voltage measurements this way; since all three meters share the neutral side, you can look at the difference between measurements to get the differential voltage.) \$\endgroup\$ – Mels Aug 7 '13 at 13:23
1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.