A question please.
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
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?
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
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
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...
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.
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.
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
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.
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
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.