PW-B2 · Power and power factor

Power-factor correction: erasing reactive power with a capacitor

In B1 reactive power did no work yet enlarged the apparent power and took up equipment. Cancelling the lagging reactive power of an inductive load with a capacitor, which supplies leading reactive power, lowers the height of the triangle; grow the capacitor yourself and learn how the same work is delivered with less current.

Grow the capacitor to shrink the triangle

The base, active P, is fixed because it stands for the same work. Dragging the handle down grows the capacitor's reactive power Q_C, so what remains after subtracting Q_C from the load's lagging reactive power Q_L shrinks, and the hypotenuse, apparent S, and the line current shrink together.

Drag the handle up and down to change the amount of capacitance. The first view is P = 1 and Q_L = 0.75, power factor 0.8, and bringing the remaining reactive power down to 0 cuts the line current to 80%. Going below 0 is overcompensation.

Power factor cosφ and line current (vs before)

cosφ ≈ 0.800

I ≈ 100%   S ≈ 1.25   Q_net ≈ 0.75

Why reactive power is a loss

Inductive loads such as motors and transformers draw lagging reactive power Q_L to set up their magnetic fields. This reactive power does no net work, yet it flows as current in the lines and enlarges the apparent S, so wires and transformers must carry that S, and the line loss I²R grows with it. This is why utilities charge more to customers with a low power factor.

A capacitor supplies leading reactive

A capacitor, the opposite of a coil, has current leading voltage, so it supplies leading reactive power Q_C. It points exactly opposite to the load's lagging reactive power, so the two cancel on the spot. Q_net = Q_L − Q_C. Fit the capacitor right next to the load and the reactive power travels only between the load and the capacitor, not along the line; rather than hauling reactive power from far away, it is borrowed on site.

Just enough, not too much

With less reactive Q, the apparent S = √(P² + Q²) shrinks, and so does the line current I = S/(√3 V_L) that delivers the same active P, freeing room in wires and transformers and lowering losses and charges. Reaching a target power factor cosφ2 takes Q_C = P(tanφ1 − tanφ2), and at pf=1, Q_C = Q_L wipes out the reactive power completely. Add more than that and the power factor becomes leading and worsens again, and at light load at night the leading reactive power can push the voltage up. So in practice the target is not 1 but about 0.9 to 0.95, and only as much capacitance as needed is fitted.

ObserveQ_net = Q_L − Q_C

Net reactive = load reactive minus capacitor reactive.

ChooseQ_L = P · ?

The load's reactive power is the active power times tanφ1.

Fill inQ_C = P (tanφ1 − ?)

The target leaves only tanφ2.

On your ownpf = 1 → Q_C = ?

Full cancellation (pf=1) means capacitor equals load reactive.

Back to the first screen

Dragging the handle down to grow the capacitor shrank the height of the triangle, Q_net, and the hypotenuse S and the line current became smaller together. The base, active P, stayed the same from start to finish, so the same work was delivered with less current. If reactive power enlarged S in B1, B2 cancels that reactive power on site with a capacitor and pulls the hypotenuse back close to P; when the handle comes down to the height of the base P, the reactive power is 0 and the power factor is 1.

Power-factor correction: the lagging reactive power Q_L of an inductive load is cancelled by the leading reactive power Q_C of a capacitor. As the remaining reactive power Q_net = Q_L − Q_C shrinks, the apparent S and the line current shrink, delivering the same active P with less current, and the capacitance needed to reach a target power factor is Q_C = P(tanφ1 − tanφ2). At pf=1, Q_C = Q_L (complete cancellation); beyond that, the power factor is leading.

The next step

So far we have dealt with the power of a single load, but in a real system countless loads switch on and off at different times. The next unit (PW-B3) works out loads and demand factor: why the actual maximum demand is smaller than the sum of all installed capacity (demand factor), and the margin that appears when the peaks of several loads occur at different times and are combined (diversity factor).