Electronics Forums > Re: "reactive power"

# Re: "reactive power"

daestrom
Guest
Posts: n/a

 08-30-2005, 10:55 PM

"jclause" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
>
>
> 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?
>

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

daestrom
Guest
Posts: n/a

 09-01-2005, 12:25 AM

"jclause" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> In article <gt5Re.54058\$(E-Mail Removed)>,
> daestrom@NO_SPAM_HEREtwcny.rr.com says...
>>
>>
>>
>>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
>>
>>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
>>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

>
>
> 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?
>

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.

> 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.
>

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

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