# "reactive power"

• posted

Suppose we have 120 volts across a motor Current = 5 amps at phase = 33.6 degrees lagging Apparent power S = EI = 600 watts Real power P = EI cos 33.6 = 499.75 watts Reactive VA Q = EI sin 33.6 = 332.03 watts Power factor = P/EI = 0.833

So the motor is dissipating 499.75 watts, and creating what is sometimes referred to as "reactive power" of 332.03 vars. It is said the utility company must supply the "reactive power" as well as the real power used by the motor, or perhaps better put - supply the current to circulate the "reactive power" in the lines between the source and the motor.

The question then: Is this "reactive power" dissipated in the resistance of the lines between the source and the motor? If so, then it is not real power as seen by the source?

• posted

Generally, a motor doesn't 'create' reactive power, we in the utility industry say that it 'consumes' reactive power. And a generator somewhere must 'generate' that reactive power. (reactive power can also be 'generated' with other devices, but let's keep this simple)

At a particular moment in the AC cycle, electrical energy from the generator is used to establish a magnetic field around a given set of windings in the motor. But later in the cycle, when that magnetic field collapses, that energy is re-induced into the windings as a short burst of electrical energy flowing *back* to the generator. On the next half-cycle, another bit of electrical energy from the generator is used to establish a different magnetic field around another set of windings, (or perhaps the same winding but with opposite polarity). Again, a short time later in the cycle, that energy is re-induced into the windings as another short burst of electrical energy flowing *back* to the generator again.

So the 'reactive power' is a measure of the amount of energy that is flowing

*back* and *forth* between the generator and a reactive load.

To see this graphically, plot the voltage and current on the same X-Y axis. Pick several points through the cycle of the voltage, and multiply the instantaneous value by the instantaneous current value at the same moment in time. Most of the time, the results will be a positive number (either because both voltage and current are positive, or both voltage and current are negative). But for a small period of time in each half cycle, the product will be negative (because the voltage and current don't cross the x-axis at the same instant). This small negative portion is the reactive power 'flowing' back to the source. If you calculate the average of all the time the V*A waveform is above the axis, you will find it's average is

*larger* than the 499.75 watts of 'real power' that is consumed in the motor. The excess is reactive power 'flowing' from the source to the load.

For this energy to flow back and forth, you have to have an electric current. This current is 90 degrees out of phase with the applied voltage. When you combine this current, with the current associated with the 'real power' flowing from the generator to the motor, you get a total current. Because they are out of phase, you have to use some trig (or Pythagorion's theorem) to combine them. That's why you have (apparent power)^2 = (real power)^2 + (reactive power)^2.

Now, if all the rest of the circuitry were 'ideal', then that would be the end of things. The generator puts a small amount of energy into the magnetic field of the motor, and the motor returns that energy to the generator a short time later (1/2 a cycle).

But the conductors, wires and cables carrying all this current have a resistance. This dissipates power from the system in the form of I^2*R losses. To minimize losses, you want to minimize I and R. You can only do so much to minimize R for a given investment. You can minimize I by reducing the reactive load as much as possible.

So, back to your example, the motor you describe is expending 499.75 watts of power in overcoming friction, I^2R losses in the resistance of the windings (not the connecting wires, just the internal windings), *and* in turning the shaft output connected to some load. The 332 'vars' is saying that 332 watts of power is flowing to and fro between generator and motor

*every* 1/2 cycle. The combination of the two results in a total current of 5A.

To figure out the power dissipated in the lines between generator and motor, we have to know what the resistance of those wires is. Let us assume 1000 ft of 12 AWG copper wire (500 ft each way). That has a resistance of about

1.64 ohms. So the power dissipated in the wire is (5A)^2 * 1.64 ohms = 41 watts.

If we were able to correct the power factor *at the load*, then the total current could be dropped to just 4.16A (499.75w/120V). Then the total power dissipated in the same 500 ft feeder (1000 ft total wire length) would be just (4.16A)^2 * 1.64 ohms = 28.4 watts.

On this scale, it isn't hardly worth it. But on a massive scale like a large industrial customer, the savings can be significant.

Hope this helps...

daestrom

• posted

The motor is consuming 500W but some of it is converted into mechanical power output. The difference between the input power and the mechanical output power is what is dissipated.

The reactive power would not be a problem if it were not for the conductive losses the extra reactive current produces in the transmission conductors and the windings of the alternator. That I^2*R loss produces real heat that limits how much the utility's equipment can deliver without overheating.

Bill

• posted

----------- Daestrom has it right.

The fact that we have inductance (as in a motor) or capacitance means that there is energy being stored and then returned to the source at a different part of the cycle. [Draw a voltage wave and a current wave with the current out of phase with the voltage. Plot the product at each instant of time. This gives the instantaneous power. The result will be a constant term plus a sinusoidal term. The constant term is the real (average)power and the sinusoidal component has a zero average. The "reactive power" is a measure of the magnitude of the 0 average component which shuttles from source to load and back] This shuttling of energy has nothing to do with the net real power (which is generally considered as the average over a cycle) delivered as its average is 0 over a cycle.

Reactive "power" is NOT dissipated in resistance and is NOT real (average as that is what we are dealing with in general power seen by the source.

However, there is an associated component of current (e.g. at 0.8 pf, the current will be 1.25 times the current at 1.0pf for the same power) which increases circuit loss and (if inductive) voltage drops in the circuit. In that way, the reactive component does increase real power dissipation in line resistance but it is not correct to say that the reactive "power" is dissipated in the line resistance. The utility generator sees the effects of reactive as well as real power but the turbine driving the generator sees only the real component and in a motor, only the real component (less real losses) is converted to mechanical energy.

• posted

Your answer is an example of what a tool the internet can be, as well as the goodwill that should be more common. Thanks for your time and sharing of knowledge.

So the bottom line regarding power dissipated due to reactive power is that it causes I^2 R power loss ln the lines above what that power loss would be with a resistive load, i.e power factor = 1. Would this be an accurate way to put it?

Seems to me that since reactive power causes additional current flow above that required to produce real power, that it creates real power in the lines going one way or the other at any given instant. However since this real power received by the load during a half cycle is returned to the source the next half cycle, then over a whole cycle it sums to zero real power. Therefore it would require no extra energy from the source above that driving the real power associated with the load.

JC

• posted

Thanks Don. You're right in there with the good technical stuff as usual.

JC

• posted

Yes, I was lax in the wording.

Thanks.

JC

• posted

Add - other than that due to added I^2R loss in the lines.

JC

• posted

Yes. And to supply this extra current safely, requires slightly larger wires, breakers, and other equipment. There is an obvious tradeoff between correcting a very poor power factor (say, 0.3) to reduce the current-carrying requirements of the equipment, and correcting a 'nearly perfect' power factor (0.85). Sort of a 'guns & butter' economic tradeoff.

And it doesn't. The only energy that has to be input to the generator shaft is the 'real power' needed to overcome I^2R losses, windage/friction in the generator *and* motor, and the motor output energy.

Well, there *is* a little bit more energy needed in the generator's excitation system because a poor power factor (lagging) requires more excitation to maintain the generator output voltage. But nothing near the amount of power that is flowing back/forth every 1/2 cycle as vars.

But a utility will get after a customer with a very poor power factor. Not because they require more *energy* (other than the increased I^2R losses), but because a large load with low power factor requires larger equipment, with larger current capacity for the same amount of kwh the utility sells.

Given two customers, both using 1MW-hour a month, normally the utility would garnish the same revenue from each customer in the form of billable MW-hour's consumed. But if one customer has a 0.3 power factor, the utility has to maintain equipment that is three times the current capacity to supply that customer versus the customer with a 0.9 power factor. Larger transformers, transmission lines, extra voltage regulating equipment, fuse/breakers with higher capacity, all sorts of things (if it truly is a

*large* customer).

So the public service commission (or whatever regulatory body is responsible) allows the utility to levy an extra charge against the customer that has a very poor power factor. They recognize that such a customer represents a higher cost to the utility, and allow the utility to recoup those higher costs from that customer.

And to avoid this extra charge, a customer may purchase their own 'power factor correction' equipment. With such equipment, the utility doesn't 'see' the poor power factor, and doesn't levy the extra fees.

daestrom

• posted

You are welcome and from what you said to Daestrom, you have it right.

• posted

To Daestrom

Again, thanks a lot for the assistance.

JC

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