# Power factor correction calculation for 3-phase inductive load

#### fireman313

Joined Jan 2, 2017
7
Hi all,

A newbie here so pardon my understanding of electronics.
I have a question about power factor correction for three phase delta, inductive load. Let me first walk through my math. For a balance load, Phase A, B, C has resistance of 2 ohm and inductance of 5 mH, and let frequency be 100 Hz.

1. I find the inductive reactance, Xl = 2*pi*f*L

2. The impedance in each phase is then, Z = 2 Ohm + j 3.14 Ohm

3. Convert #2 into phasor form to find the angle, Thetha = tan-1(3.14 /2 Ohm) = 57.5 degrees

4. If the phase voltage, Vab = 60 V, 0 degrees, then PF = cos (0 – 57.5) = 0.53

My question is how do I calculate the value of capacitor required to do a power factor correction to have PF of 0.8? From my understanding, because current leads in capacitive loads, this helps compensate the fact that current lags in inductive load. On another web site, I’ve seens a PF correction equation that is C = 1/(2*pi*f*X), Is this formula correct? And if so, is X the inductive reactance?

#### WBahn

Joined Mar 31, 2012
26,398
That formula just gives the value of capacitance, C, that corresponds to X where X is the capacitive reactance of that capacitor.

It's basically the same as X = 2*pi*f*L you used above in which the X is the inductive reactance of an inductor having L inductance.

#### fireman313

Joined Jan 2, 2017
7
You are right, the capacitive reactance formula is Xc = 1/(2pi*f*C). so If I used the inductive reactance, Xl, in the formula does this means Im calculating for a capacitance value that helps to correct the power factor?

Basically if Z = R + jL - jC, and if JL = JC then I would get a unity power factor. is this correct if one is to do this experimentally?

#### WBahn

Joined Mar 31, 2012
26,398
Since you are working with complex impedance, it is far, far better for you to deal with reactance as a signed quantity. A positive reactance is inductive and a negative reactance is capacitive. This falls out naturally since

$$Z \; = \; R \; + \; jX$$

Where Z is the impedance, R is the resistance, and X is the reactance.

$$Z_L \; = \; j \omega L Z_C \; = \; \frac{1}{j \omega C}$$

Therefore

$$X_L \; = \; \omega L \; = \; 2 \pi f L X_C \; = \; \frac{-1}{\omega C} \; = \; \frac{-1}{2 \pi f C}$$

Trying to have both be positive and then having to keep track of whether it is inductive or capacitive is a recipe for disaster.

Your thinking if fine if the goal is to get unity power factor. But here your goal is to get a power factor of 80%. So your starting point is the definition of power factor, which is....?

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#### fireman313

Joined Jan 2, 2017
7
The difference in phase angle between phase voltage and phase current. That's my basic understanding. So then my goal is to reduce the angle from 57.5 degrees to 35 degrees. which means reducing X (X = XL+XC).

#### MrAl

Joined Jun 17, 2014
8,386
Hi,

If you start out with some inductive and resistive load and then add capacitance to help the power factor, do you know how to calculate the angle of lead or lag from that information?
You need to calculate the lead or lag, then the power factor. You will then be in a position to calculate the required capacitance for a given power factor (like the 0.8 for example).

#### WBahn

Joined Mar 31, 2012
26,398
The difference in phase angle between phase voltage and phase current. That's my basic understanding. So then my goal is to reduce the angle from 57.5 degrees to 35 degrees. which means reducing X (X = XL+XC).
So power factor is measured in degrees? Or in radians?

It is a fraction. It is defined as the ratio of real power to apparent power.

#### MrAl

Joined Jun 17, 2014
8,386
So power factor is measured in degrees? Or in radians?

It is a fraction. It is defined as the ratio of real power to apparent power.
Hi,

It sounds like he wants to use the angle between current and voltage which can then be used to calculate the power factor too.

#### WBahn

Joined Mar 31, 2012
26,398
Hi,

It sounds like he wants to use the angle between current and voltage which can then be used to calculate the power factor too.
He needs to understand what the concepts are that he is working with.

#### fireman313

Joined Jan 2, 2017
7
WBahn, Sorry if I'm a bit lost. Im using degrees to calculate power factor. I understand that PF = P/S (P = V*i*PF and S=V*I).

Mr. AI, I have learned to draw phasor diagram to help visualize lag and lead of voltage and current. But I would first calculate total impedance Z with all the loads. Z = R + jX. ZL = j2pifL. ZC = 1/j2piC = -j2piC. then phase angle, thetha, is tan-1 (X/R). Cos(Phase voltage - thetha) = PF

Capacitor prevents change in voltage (noise) while allowing current to flow through hence the current leads, while a current flowing through an inductor creates a magnetic flux which prevents change in current (AC, but still allows DC current to flow through) hence current lags. Am I getting close to this?

I appreciate the help from both of you.

#### MrAl

Joined Jun 17, 2014
8,386
Hi,

You have the right idea, now you just have to apply the right math for the job.