Volt Amps Reactive?

Factual Questions
1

The New York Times had a pretty much incomprehensible article a couple days about these so-called VARs, which it claimed electric companies had to supply along with kilowatt-hours, but got money only for the latter.

So my first question is: Is the load on a long-distance transmission line inductive or capacitive, if you do nothing special. Is it the inductance of a long wire (along with resistance, of course) or the capacitance between wires that dominates? And what is it that the electric company can do to compensate? And what does this compensation consist of? Is it just a matter of trying to keep the voltage and current in phase, or is there something more to it? Apparently the electric company in Ohio where the recent blackout began did not do what they were supposed to do. As I said, the article was so garbled as to confuse more than explain.

2

Not having seen the original article, this response is more on the lines of what happens in the local (not long-distance) part of the network.

Most loads are either resistive (incandescent lamps) or resistive with an inductive component (motors, transformers.) There are few if any (to my knowledge) capacitive loads.

On the ‘what to do’ front, utilities position capcitor banks strategically around the network to offset the inductive component. This means making the total inbound voltage and current being in phase. A load balanced in this manner will draw less total current, meaning less voltage loss in the powerlines, meaning the utility does not have to jack up the voltage at their end (expensive) in order to insure that the voltage delivered to the worst-case end-user does not drop below specs.

BTW In some cases the capacitor banks are switched on-off by timers to reflect changing daily load trends. In others, they can be remotely controlled by radio or other means.

Residential and Industrial customers are treated differently. The watt-hour meter on your house only records resistive load. Industrial customers can be charged for reactive loads as well as resistive. (Or, at least, they could at one time.) I have seen VAR-Hour-Meters and Power-Factor meters that I was told were used to bill industrial customers. I was also led to understand that some industrial customers install and manage their own capactior banks to reduce their bill for Reactive VAR-Hours.

The above is based on experience of quite a few years ago, but much about electricity at least hasn’t changed in that time. So, since the utilities can and do manage VARs in some instances, and can bill some customers for using VARs, I’m not sure why the utilities are saying they have a problem unless it lies exclusively in the long-haul network and for some reason they can’t fix it there.

3

It depends.

The VArs generated by a line is given by

Q[sub]C[/sub] = V[sup]2[/sup]B,

while the VArs absorbed by a line is given by

Q[sub]L[/sub] = I[sup]2[/sup]X

where

V = voltage
I = current
B = shunt (capacitive) susceptance
X = series (inductive) reactance

The V term tends to be pretty constant, while the I term varies with the load being supplied by the line. So whether one dominates the other depends on system conditions at the time.

The voltage and current can’t be kept in phase. It’s more useful to think or it in terms of controlling voltages by appropriately matching the VArs being generated around the system with the VArs being absorbed. Devices commonly used to do this are:

  1. Shunt connected reactors (inductors) and capacitors, which can be switched in our out as required
  2. Thyristor switched versions of (1), which can achieve very precise and automatic voltage control
  3. The excitation systems of synchronous generators that are connected to the grid
  4. Series connected reactors (inductors) and capacitors, which can be bypassed when not required
  5. Series connected thyristor switching systems, which are effectively AC/DC/AC convertors, which can achieve precise control of the flow of both Watts and VArs.
4

The classic way of representing what is happening is to draw what is caled a phasor diagram.

Whenever you make any measurement, you have to have a start point, a referance, and so it is with phasor diagrams.

If you were to take the power in your network that is due to the current and voltage acting together as the referance, then you can add other elements where the current and Voltage do not act together and examine the relationship.

The useful work you get in your transmission line is simply the product of the Current squared* Voltage, and this is our ‘in phase’ referance, or ‘true power’.

Power due to reactance is called ‘out of phase’

We can then add the other elements, capacitance and inductance are opposite ends of the reactance scale and will point in opposite directions on our phasor diagram.

If we take our useful power and use it as our baseline, capacitance works on an axis at right angles to the origin or the true power, inductance works exactly 180 degrees in the other direction as capacitance. Which also means it is at right angles to true power.

If there were, lets imagine, a huge amount of capacitance, such that the reactive power line length were equal to the true power length, we could make a vector sum which would be at 45degrees to both reactive power and true power.

If you had done this to scale, you could measure the length of this new line and you would find it was longer than either true power, or reactive power, and this is the load that the network has to carry.
It is easy using pythgoras to work out the actual lengths.

This longer line is called the ‘apparent power’, and all the network has to be designed to carry it, which could and usually does, mean physically larger components.

Bear in mind that high voltage switchgear can be very large physically, one switch taking up over an acre of land and the scale of engineering means greater cost and maintenece, it is very desirable to reduce the element of reactive power.

In the previous, lets imagine that the reactive power was due to capacitance, how would you counter it?

Add in an element whose action works at 180degrees to it is the answer, so you would switch in an inductive element into a capacitive network(also called a reactor) and in a network that was inductive, you would switch in banks of capacitors.

Unfortunately network conditions do not remain constant, and are continually changing as the load changes. This means that the reactive power correction also has to be able to change.

Another way to change the power to compensate for reactive elements is to change the way the generator operates, it can be made to produce power that can counter for the circuit conditions.

5

Thank you all for very informative answers. I didn’t know about the banks of inductances or capacitors switched in as needed and that is what the article omitted to mention. I imagine then that the Ohio power company was simply not doing its job of countering reactance, or at least that was what was alleged.

It is amazing how badly they explain technical things in the popular press. I mean why bother? People who already understand it don’t need the explanation and people who don’t aren’t going to learn anything from the garbled versions they do print.

6

Hari Seldon, the problem is this: a customer has a 100 W lightbulb and uses 0.1 Kwh/h of active power for which he pays. Another customer has fluorescent tubes which use the same active power for which he pays but, in addition, use reactive power, for which he does not pay because it is not “real” power. But, in spite of it not being real power, it forces the electric company to have larger wires, larger transformers etc. In other words: it costs money to deliver and they are not charging for it. Charging fo it is not practical for small residential customers but huge consumers like factories often have separate meters for active and reactive power and then they have an interest in diminishing the reactive component which is easily done with capacitors (as most reactive power is inductive due to motors, transformers, ballasts etc).

7

Reactive power is real.

However, the reactive part of electric power reverses its direction of flow. It reverses twice per cycle. Normal power (non-reactive or “real” power) does not reverse: it flows smoothly from the generator to the load.

In other words, when the power is 100% reactive, they send you some energy, but you send it right back again 120th second later. Reactive power is “unreal” because it averages out to zero in the long run. That’s why electric meters don’t detect it and why the utility companies don’t charge you for it.

On the other hand, reactive power is still real because it involves voltage and current just like any normal watt. Therefore the generators, transmission lines, transformers, etc., still get hot.

If a utility company sends out a KWH of reactive energy, and a few percent is lost because of inefficiency… and then the customer’s load sends all the received energy back again 120th of a second later (and again a few percent is lost,) then the customer isn’t charged, since the customer didn’t consume any energy at all. Yet the utility companies have to shovel fuel into the boilers, since some electrical energy was consumed on the fly by the miles of power lines. Reactive power represents a financial loss to the utility companies.

8

bbeaty’s poost brings up another issue here.

The amount that the apparent power lags or leads the true power is called the power factor and is usually quoted as the Cosine of differance angle between them (remember the phasor diagram).

Steps can be taken at differant physical locations within a network to correct this, hence power correction.

However, with all the reactive energy bouncing around the behaviour is not easily predictable, and if you carry out all your correction near the source of supply rather than the load which could be many many miles away, you could need larger equipment just to carry both the out of phase element and the correction elements.

In practice it is best to try to correct the power factor nearer to the load that is causing the problem, which is why you have all these banks of capacitors, inductors etc.

The problems of running networks, and the protection systems are huge, there will be sensors that show overload, as you would expect, leakage to earth, reverse current, current surges, overvoltage and others.

If one part of a network trips out, the load is redistributed, and this would also have to include the way the power factor was being dealt with.
If that part that tripped out was also responsible for a large part of the power factor correction then the load might be transferred to the rest of the network, without the corresponding corrction.
The VAR’s could increase suddenly and dramatically causing various safety devices to trip.

Modern networks should be able to cope with this, the monitoring is extensive but can only be as effective as the information available to the computor programmers who write the software for it.

9

More than one, I’d say. One in particular I find noteworthy, i.e., the utter confusion regarding terminology:

Which is followed up by:

To pick another quote, this time from the OP:

Which is precisely the problem with bbeaty’s post. It’s only comprehensible to people who know more about the subject than bbeaty himself.

It’d be better to shut up altogether rather than juxtapose two entirely different senses of the word “real” in trying to explain what’s real and what isn’t.

10

Maybe the wording was confusing due to the use of the word “real” in more than one sense. Strictly speaking, both active and reactive power are just as “real”.

11

Are you for real, sailor? I thought I just said exactly the same thing.

They’re obviously both real in the sense that they exist, but I would ask you to justify your use of the words “strictly speaking”.

12

One thing I can say, although my posts on electrical/electronic issues are never as precise and accurate as Desmostylus, and also knowing this poster will often appear in such threads, it certainly has made me very careful about what I state.

This has to be a good thing.

I saw those terms about real power in bbeaty’s post and thought to comment, but then knew I had better leave it alone for fear of screwing it up.

It can be hard to explain some issues in ways that those not familiar with the subject can easily understand, you often know what you want to say, but to put it in easily digestable form is a challenge.
It’s all too easy to try explain something technical in other technical terms that may be easy for the expert to understand but don’t really provide much illumination to to others.

Technical authoring takes a great deal of consideration.

13

I apologise for agreeing with what you said and I am sorry for any distress this may have caused you.

Strictly speaking they are both real in the sense that they exist and are not just an imaginary convention. Active power is power which the user receives and uses. Reactive power is power which the user receives and returns.

14

The term ‘real’ in this context is unfortunate since those who have the technical knowledge will also be familiar with complex notation.

In such terms we have ‘real’ and ‘imaginary’ parts and these can be used to assist in calculation.

The in phase element is assigned the real quantity, and the out of phase or reactive element is given the ‘imaginary’ quantity.

It would be better to use the terms,

True power
Apparent power
Reactive power.

…and reserve use of the word imaginary for use in complex calculation.

15

You know I understood it at least as well after the first few posts as after reading them all. I know all about out of phase power and how you pay for the integral of I*E, which is 0 if they are 90 deg out of phase. What I still am not sure about is that NYTimes claimed all these problems were exacerbated by the fact that so much power is sent long distance. Leaving aside resistive losses (but of course, that’s what the’re trying to minimize) which are a linear function of distance, I was curious whether the extra phase change arises from the inductance of the wires or the capacitance between wires. Or could you design transmission lines so that these cancel?

It may be of interest for the experts to know that Hydro-Quebec ships a significant amount of DC power to aluminum smelters. I do not know the voltage, but there would be no point if it weren’t the same as the voltage it is used at. DC power would have entirely different properties, I assume. The reactance would be irrelevant, for one thing.

16

I may be getting in to this a little late but I hope not.

When you build a system there are a lot of things that have to be’excited’ before actual load current can flow.

Energizing the equipment with all of it’s inductances and capacitance values is what we used to call “wattless power” and it has to be there

Such energies can work aaiinst you and are then known as reactives

It can be measured as "charging "current ----this then is a measurable value that has to be deducted from the metered watt sum in order to determine what the actual load figure is.

Vars are the sum of the of all of the charging currents converted into watts.

No fancy formulas----------no text book jargon-----it’s what you have to provide for the system before those little punched plates in the watt meters start rolling.

And even the metering coils are set at an angle to adjust for that part of the load is Vars

Have a goodun!

17

This gets a little complicated but here goes. Transmission lines have a property known as their characteristic impedance which depends upon the geometry of the lines. For example, most ordinary two wire cables like an extension cord have a characteristic impedance, at high frequency, in the vicinity of 90 Ohms. This property is only important for long lines and the length, measured in wavelengths of the frequency involved, are always short for 60 cycle power lines. So no, you can’t really design a power transmission line to get rid of the reactance because the reactance that counts is that of the load and not of the line.

To illustrate. The 1-Hp motor in my air conditioner takes 12 amps at 120 v. At full load the power into the motor for which the utility company charges is 746 watts and at 120 v. this is an in-phase current of 6.2 amp. In order to compute the quadrature current I can use the Pythagorean formula and take the square root of 12.6[sup]2[/sup] - 6.2[sup]2[/sup] which equals 10.3 amp. This means that at full load the motor has a voltage-current phase lag of 58[sup]o[/sup] or a power factor of the cosine of that or just about 0.5.

So no wonder the utility wants large consumers to have as high a power factor as possible. For my motor they have to send 12 amps to get paid for 6.

18

Aww, I can’t believe I missed this question.

Incase it hasn’t been answered already, wires have inductance, capacitance, and resistance. Normally, they have inductance and resistance in series with the wire, but if the frequency of oscillation is sufficiently high, there will be a capacitance in parallel with the load.

I believe the question of power factors and complex power has already been answered, so I’ll leave it at that.

19

There are indeed high voltage DC transmission lines. Once the voltage gets very high AC has more losses due to the corona effect and DC becomes more attractive: http://www.hydro.mb.ca/our_facilities/ts_nelson.shtml
DC is not used at the transmission voltage (in the range of 1 megavolt) but is transformed down. It is a bit more complex than just using a transformer but it is done with solid state switches.

20

A minor nitpick: there will always* capacitance present, both between lines and from line-to-Earth. At low frequencies, the capacitive reactance is high and therefore has little effect on the circuit.