Pumps & Systems , July 2007
Last month we took a close look at the flow of voltage and current in purely resistive and inductive circuits. We showed how inductive reactance can inhibit the flow of primary current in a circuit and how this lagging effect determines the value of power factor (PF).
Although the purely resistive and inductive examples we used helped us understand what was happening, they do not represent a typical electrical circuit. Almost all circuits are a combination of resistive and inductive loads and, in some cases, those that are capacitive as well. Motors, arc welders, transformers, heaters, incandescent lighting, ballast lighting, solid state power supplies, and many other electrical devices contribute to the overall load on the circuit.
Figure 1 shows a more typical circuit that is feeding both resistive and inductive devices. The impedance (Z) of the circuit (in Ohms) is √R2 + XL2, where R is the resistance and XL is inductive reactance.
Figure 1
This example assumes that these values are relatively equal and results in current lagging voltage by about 45-deg. The power factor of this circuit is about 0.7 (cosine 45 = .707), which means that about 70 percent of the actual power in the circuit is available to do real work. The remainder or "reactive" power is returned to the circuit as the inductive fields decay.
Now, you will have to admit that this circuit looks a whole lot better than the purely reactive one we saw last month. But almost 30 percent of the power in the circuit is still unusable, and that can result in some undesirable consequences.
Consequences
So what are the consequences of low PF? Basically, there are two.
The first one involves power generation. We have all seen what happens when there is a voltage drop in a circuit feeding an electric motor. The voltage drop creates a net reduction in watts needed by the motor, but the motor couldn't care less. It simply increases its current demand in amps and gets watts back to their original level.
A similar thing occurs when inductive reactance causes current to lag voltage. The circuit uses additional current to make up for those areas of the curve where power is unavailable. And where does that additional current come from? The utility, of course, and if that additional current is not already in the system, they have to generate it. Now the last thing a utility wants to do is add generating capacity, so they will typically charge consumers a surcharge if their total PF is below some required minimum.
The second consequence has to do with power distribution and it can affect both the utility and the consumer. In an electrical circuit, the energy expended (as heat) in maintaining current flow increases as the square of current intensity. Double the current and losses increase fourfold. This is the reason why long distance power transmission lines employ very high voltages (and lower currents) and then reduce that voltage at the point of use.
As PF declines, current increases and so do losses due to heat. At some point, the wire size must be increased in order to prevent further losses. For example, suppose an electrical system can use 0 AWG cable to supply 100-KW at 460-V at a PF of 100 percent. If PF is reduced to 70 percent, the cable size will increase to 000 AWG! And it's not just wire size. Every other component in the system (supply transformer, switch gear, etc) must also be oversized if it is to accommodate the additional current required by a low PF installation.
At a PF of 100 percent, that same electrical system could utilize a transformer rated at 100-kVA. Reduce PF to 70 percent and 143-kVA would be required.
Motor Power Factor
No AC induction motor can have a PF rating of 100 percent. Why? Because a
