Reactive power loss

[LarryFine]

. . . the energy is stored in the wire for 1/2 of the cycle, . . .

Stored in the wire?

 

[rcwilson]

That makes sense. Thanks!

Will a series circuit with just a resistor and inductor have both a real and reactive current portion?

Yes. As Besoeker said, splitting the current into real and reactive parts is a mathematical modeling trick that simplifies calculations.

A 1 ohm resistor and 1 ohm inductor in series will have a total impedance = 1 + j1 ohms = 1.414 ohms @ 45 degrees. Put 1 amp through that circuit and it will consume 1 watt and 1 var which = 1.414 VA. The resistor will have 1 volt = (1 A x 1 Ohm) across it as does the reactor. The total circuit will have 1.414 volts across it since the reactive voltage is 90 degrees out of phase with the resistive voltage. (1V@ 0 degress + 1V @ 90 degrees = 1.414V @45)

The same current flows in the complete series circuit. Mathematically we can divide it into its reactive and real component. Note that just looking at a current, we can't tell how much is real or reactive unless there is a source voltage to get a phase angle reference.

But we can caclulate the wire's I^2R losses if we know the wire resistance. The wire only carries amps and doesn't try to split them into real and reactive.

 

[rcwilson]

Stored in the wire?

Stored in the electo-magnetic field around the wire.

 

[LarryFine]

Stored in the electo-magnetic field around the wire.

I've been under the impression that the wire carries the reactive current back to the source (or any capacitance closer), which is why the POCO is concerned with larger customers' PF's.

 

[K8MHZ]

I've been under the impression that the wire carries the reactive current back to the source (or any capacitance closer), which is why the POCO is concerned with larger customers' PF's.

Kinda....

Customers are only charged for true power. The wattmeter only records true power. Any power generation used for reactive power does not provide for revenue. One way to put this into check is to charge customers with large reactive loads a premium, which indeed is done and fairly commonplace.

 

[philly]

Yes. As Besoeker said, splitting the current into real and reactive parts is a mathematical modeling trick that simplifies calculations.

A 1 ohm resistor and 1 ohm inductor in series will have a total impedance = 1 + j1 ohms = 1.414 ohms @ 45 degrees. Put 1 amp through that circuit and it will consume 1 watt and 1 var which = 1.414 VA. The resistor will have 1 volt = (1 A x 1 Ohm) across it as does the reactor. The total circuit will have 1.414 volts across it since the reactive voltage is 90 degrees out of phase with the resistive voltage. (1V@ 0 degress + 1V @ 90 degrees = 1.414V @45)

The same current flows in the complete series circuit. Mathematically we can divide it into its reactive and real component. Note that just looking at a current, we can't tell how much is real or reactive unless there is a source voltage to get a phase angle reference.

But we can caclulate the wire's I^2R losses if we know the wire resistance. The wire only carries amps and doesn't try to split them into real and reactive.

Great example. Thank You!

Just to add to my understanding of this example, how would the 1A current that you used break into its real and reactive components of current?

 

[Besoeker]

That makes sense. Thanks!

Will a series circuit with just a resistor and inductor have both a real and reactive current portion?

It will have current. With a bit of trigonometry, you can resolve it into in-phase and quadrature components if you know the phase relationship between the current and the voltage. But the circuit just carries current. It doesn't really take two kinds and add them together.

 

[Besoeker]

Anyhow, I'm pretty sure (but not 100% positive) that the core losses would appear as a resistive load.

Agreed??

I believe you are correct.
Here is the transformer equivalent circuit I mentioned previously:

Ic is defined as the core loss current. Note that it flows through a resistive element.

 

[LarryFine]

Kinda....

Customers are only charged for true power. The wattmeter only records true power. Any power generation used for reactive power does not provide for revenue. One way to put this into check is to charge customers with large reactive loads a premium, which indeed is done and fairly commonplace.

All that, I knew. I'm just questioning that the wires provide energy storage. In fact, without PF corrective devices, the entire reactive current flows all the way back to the generators (less any incidental system capacitance.)

That's what PF correction devices are for: They shorten the pathway that must carry reactive current, sort of "shunting" it from the rest of the supply system, leaving the rest of the supply to only carry productive current.

 

[Hameedulla-Ekhlas]

when the field is created/collapsed in the coil.

by the way, the transformer by itself generates reactive power due to it's inductive nature If for the active power (correctly the energy) exists a net flow from one point of the network to another, for the reactive power there is a continuously flow back and forth (to and fro), but the net flow is zero for a complete cycle, as the amount of energy flowing in one direction for half a cycle is equal to the amount of energy flowing in the opposite direction in the next half of the cycle. The reactive power is exchanged by different parts of the network ? capacitors and reactors ? permanently, but is never
consumed or produced. In reality, we can say that this reactive power is produced once, when the network is energized (after a collapse) and the same reactive power is consumed once, when the network collapses again. In between these two major events, the reactive energy stays constant. Of course the situation is changing when equipment is switched on or off.

 

[K8MHZ]

All that, I knew. I'm just questioning that the wires provide energy storage. In fact, without PF corrective devices, the entire reactive current flows all the way back to the generators (less any incidental system capacitance.)

That's what PF correction devices are for: They shorten the pathway that must carry reactive current, sort of "shunting" it from the rest of the supply system, leaving the rest of the supply to only carry productive current.

Well, to be precise, there is some energy storage in the magnetic (not electromagnetic) field around a conductor, but at 60 Hz, it's practically negligible except on long transmission lines. As such, they aren't responsible for much in the way of reactive losses, especially when compared to resistive losses along the same set of lines.

I have never thought about correction in the manner of shortening a pathway. I have always thought about the process as removing it so there is nothing (hopefully) left to travel down the pathway. I do see your rationale, but prefer to think in terms of resonance (I have some background in music and radio, that helps un-complicate things for me) as removal with the pathway (the conductors) remaining the same length.

I had the opportunity to hear sound being canceled out by 180 degree phase shift transmission and it's pretty cool. Two speakers (transducers, actually) generating sound and using up energy and nothing is heard. The acoustic version of a purely reactive load.

I think our friend in Afghanistan did the best in offering a simple, easy to understand answer to the OP's question.

 

[LarryFine]

In reality, we can say that this reactive power is produced once, when the network is energized (after a collapse) and the same reactive power is consumed once, when the network collapses again. In between these two major events, the reactive energy stays constant.

Doesn't that energy constantly have to be replenished as it's lost to voltage drop, various system capacitances, and other losses?

In theory, I agree with you.

 

[LarryFine]

I do see your rationale, but prefer to think in terms of resonance (I have some background in music and radio, that helps un-complicate things for me) as removal with the pathway (the conductors) remaining the same length.

I have similar backgrounds, too, being both a piano player and an electronics nut since age 5 or so.

 

[K8MHZ]

I have similar backgrounds, too, being both a piano player and an electronics nut since age 5 or so.

Wow! I knew I had heard your name somewhere before!

 

[LarryFine]

Wow! I knew I had heard your name somewhere before!

If only. :roll:

 

[mivey]

Boy! A detailed answer is too much and a simple answer is too little. One day I'll find the middle of the road. Anyway:

The current running back and forth and heating up the wire would be I^2*R as you correctly stated, the R being from the conductor resistance. That would thus be resistive losses.

That the current is higher as a result of the circuit being lower than unity power factor doesn't make those I^2*R being anything other than resistive.

Of course they are resistive. The point, as I'm sure you know, was that the increase in current required by the reactive load is what caused those losses and so the reactive load gets to take the blame for those losses.

So in other words the kW loss is Itotal^R and not just Ireal^R? Is this correct

Correct. But I am blaming a portion of those losses on the reactive load and by doing so, am calling this the reactive loss, or loss due to reactive load.

If I have only a resistor and inductor in a series circuit will this circuit have both a real and reactive current portion to it? If so will it be the combination of the real and reactive currents of the total current that goes across the resistor to calculate I^2R losses across the resistor, and not just the real portion of the current?

If you "look" at that wire with a "current microscope", you will only see one net current. By assigning a portion of that net current to the reactive load, we are assigning responsibility, kind of like splitting a power bill between tenants. In our case, the apartment is the equivalent two-terminal network with and without the reactive load (the extra tenant) and the difference in current is what we say is the impact of the extra tenant (the reactive load).

Anyhow, I'm pretty sure (but not 100% positive) that the core losses would appear as a resistive load.

Agreed. I thought that is what I said, but evidently it did not come across well. I was just trying to present a simple explanation in layman's terms as that is what it appeared the OP wanted.

 

[mivey]

Will a series circuit with just a resistor and inductor have both a real and reactive current portion?

If you want to talk about the loss impact due to the reactive load, look at the current due to the two-terminal network with & without the reactive load. A change in the net current will result in a change in net I^2 * R losses.

Stored in the wire?

Simply put: yes. It is just another reactive load, although spread along the length. Same is true for the cable capacitance.

I've been under the impression that the wire carries the reactive current back to the source (or any capacitance closer), which is why the POCO is concerned with larger customers' PF's.

That is true, but the wire is not a perfect conductor and is a small load itself.

Just to add to my understanding of this example, how would the 1A current that you used break into its real and reactive components of current?

Kind of depends on what you consider to be the incremental reactive load. When applying power factor correction, we are only considering that reactive load that we are going to impact. Usually, those are parallel reactive loads.

From a power supply standpoint, we generally consider our sources to be much stiffer, or stronger, than the loads so we would take this as a constant voltage supply. Given a supply of 1.414 volts, we know the circuit would have 1.414 amps without the series inductor. In the series case, the reactive load actually reduces the current (thus is assigned a negative current contribution to the two-terminal network). Negating this inductance would actually raise our current level. A parallel inductance would have been a better example circuit as those are the ones we are usually correcting for.

 

[mivey]

Ic is defined as the core loss current. Note that it flows through a resistive element.

Not just through "a" resistive element but through resistive elements (plural), including the transformer R1, feeder R, and source R. If I can reduce the core loss, I have also reduced the losses in the supplying circuit and the supply sees a reduction that is more than just the reduction in the actual core loss.

All that, I knew. I'm just questioning that the wires provide energy storage. In fact, without PF corrective devices, the entire reactive current flows all the way back to the generators (less any incidental system capacitance.)

That's what PF correction devices are for: They shorten the pathway that must carry reactive current, sort of "shunting" it from the rest of the supply system, leaving the rest of the supply to only carry productive current.

Unfortunately, we do not have perfect conductors and they become a small part of the total load the supply must service.

 

[mivey]

but the net flow is zero for a complete cycle, as the amount of energy flowing in one direction for half a cycle is equal to the amount of energy flowing in the opposite direction in the next half of the cycle.

Only if you neglect the traveling losses, which I am not.

Doesn't that energy constantly have to be replenished as it's lost to voltage drop, various system capacitances, and other losses?

And so we come full circle back to where I started. You are correct in saying that we must compensate for the delivery losses. If these delivery losses are caused by reactive load, I count that as part of the reactive loss for the system as a whole. In complete disclosure: I usually only isolate that which I can do something about.

 

[Besoeker]

Of course they are resistive.

And thus not reactive losses as you stated.
QED.