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There is a common question in my coursework that we are asked to :
Let's say we have an inductive 3 - phase load that consumes $$S=15 + j10$$MVA
"If , we connect parallel to this load, 3 capacitors that form a star, compute the capacity of each in order to achieve, unit power factor. " I know, there is a formula that says $$Q_c = \omega*C*V^2$$ but I don't know the proof( we were just given the formula right away) and I am not able to identify if I am using it correctly.
If V, is the polar voltage ("line voltage" I think it is in English), then does this $$Q_c$$ represent the non - active power that the whole star produces?
So, I can say (??): $$C=\frac{Q_c^{3ph}}{\omega V^2}$$
What, would change if those capacitors were connected such as to form a triangle?

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If you have phase-to-phase (Vpp) and phase-to-neutral/ground (Vpn) voltages, you may calculate reactive power for three phases in the following way

$$ Q_{3ph}=3Q_c=3V_{pn}^2\omega C=3\left(\frac{V_{pp}}{\sqrt3}\right)^2\omega C=V_{pp}^2\omega C $$

By the way, this is the magic of calculating balanced three phase circuits. You draw scheme for just one phase, then pretend you have phase-to-phase voltage instead of phase-to-ground and get three phase powers instead of single phase powers. Very handy. Just remember that currents are square root of 3 times bigger than in reality.

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  • \$\begingroup\$ so , if for example $$ V_{pp} = 380V$$ then , $$V_{pn} \approx 220V$$ and $$Q_{3ph}=220^2 * \omega C$$ (for a star connection) am I right ? \$\endgroup\$ Feb 7 at 11:28
  • \$\begingroup\$ but if it was a triangle connection: then we would have $$Q_{3ph}=380^2*\omega*C$$? \$\endgroup\$ Feb 7 at 11:29
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    \$\begingroup\$ @brucebanner My formulae are for star connection. So \$Q_{3phy}=380^2\omega C\$ or \$Q_{3phy}=3*220^2\omega C\$ if you like. For delta connection sum three capacitors at Vpp voltage, thus \$Q_{3ph\Delta}=3*380^2\omega C\$ \$\endgroup\$
    – AlexVB
    Feb 7 at 11:52
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I don't know the proof( we were just given the formula right away) and I am not able to identify if I am using it correctly.

\$Q_c=ω∗C∗V^2\$ is the reactive power (Q) of a capacitor (C) connected across an AC voltage source (V). It's related to real power in a resistor (\$P = \frac{V^2}{R}\$) but, because we know the reactance of a capacitor is \$\frac{1}{ωC}\$, the reactive power formula becomes as you show it.

does Qc represent the non - active power that the whole star produces?

No, it's the single phase reactive power of a capacitor across a voltage source. If you have three capacitors in star formation and you only have the line voltage then, the reactive power for each of those capacitors is found using \$\frac{V}{\sqrt3}\$.

In fact you might as well convert your first formula to a single phase apparent power by dividing it by 3 then solving as a single phase problem.

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