Power Factor Correction
concept
Having a low power factor (one near \(0\)) is bad for cost and efficiency reasons. In this topic we'll cover why that is and how you can improve the power factor of your load to lower your costs and increase efficiency.
If you have some load connected to the outlet in your home you have a fixed AC voltage source and some fixed real power output you need to provide.
The lower the power factor is the more out of phase your voltage and currents are because of a high reactance compared to the resistance.
This means that if you have a fixed voltage and a fixed real power you will need a very high current to achieve these.
fact
Poor power factors cause unnecessarily high currents to be drawn in order to provide the needed real power to a load. High currents require large, costly conductors.
Utilities will charge you an added fee if your power factor is poor because of the extra costs involved in providing overhead lines capable of handling the large currents that poor power factors cause.
We typically only correct lagging power factors since the vast, vast, vast majority of industrial loads are inductive in nature.
fact
To correct a lagging power factor we need to add a capacitor in parallel to our load.
fact
The method for calculating the capacitor needed to correct the power factor is:
- Draw power triangle for the current load
- Draw desired power triangle for wanted power factor
- Find the reactive power the capacitor needs to provide to reach the desired level \(Q_c = Q_{new} - Q_{old}\)
- Find the reactance the capacitor needs to deliver that power at the operating voltage \(X_c = \frac{V^2}{Q_c}\)
- Calculate the capacitance needed to provide that reactance at the circuit frequency \(C = \frac{-1}{\omega X_c} = \frac{-1}{2\pi f X_c}\)
example
A load has an apparent power of \(S = 100\)kVA and a real power of \(P = 80\)kW. The supply voltage is 120V rms at a frequency of 60Hz Find the capacitance necessary to make the power factor 0.95
Our first step is to draw out the power triangle for this load. We can see from the problem that pf\( = \frac{80}{100} = 0.8\) And \(Q_{old} = \sqrt{S^2 - P^2} = 60\)kVAR
Now we redraw the power triangle as we want it to be, with the same real power but the power factor we want.
pf\( = 0.95\)
\(S = \frac{P}{0.95} = 84.2\)kVA
\(Q_{new} = \sqrt{S^2 - P^2} = 26.3\)kVAR
Now our capacitor has to supply \(Q_c = Q_{new} - Q_{old} = -33.7\)kVAR
The reactance necessary is given by the formula:
\(Q_c = \frac{V^2}{X_c} \implies X_c = \frac{Q_c}{V^2}\)
\(X_c = -280\)
And now, finally, we can use the capacitor's formula for reactance to find the required capacitance:
\(X_c = \frac{-1}{2\pi f C}\)
\(-280 = -\frac{1}{120\pi C} \implies C = \frac{1}{120\pi \cdot 280} = 9.5\mu\)F
So we need a capacitor of \(C = 9.5\mu\)F in parallel with our load to bring our power factor up to 0.95
There's a lot of steps to power factor correction but none of them are tricky, just write all the steps out on a cheat sheet and make sure to practice plenty of problems and you'll have no troubles at all.
practice problems