Reactive power

C
That wouldn't be how I would use that data, unless we had a mandate to maintain a given PF for financial reasons. Some power suppliers impose a penalty and there are other cost implications. The ratio will typically bounce around as different loads are switched on and off so detecting a problem that way would require that you know a lot about what's going on. So you might use it to justify PF correction gear. But since it is almost entirely dependent on the loads connected, it is generally something you live with, or correct to the extent that the variation allows.

Regards
cww
 
cww is exactly correct: We don't know what kind of "problems" you are trying to diagnose. Generator problems? Exciter (AVR) problems? Problems maintaining VAr setpoint or power factor setpoint? Generator heating problems? Generator vibration problems? Real power stability issues? Reactive power stability issues?

Without understanding what problems you are trying to diagnose or solve, no one can say if the information will be practically useful or not.
 
Can you please explain more, on reactive power, and its effect..........., how does it affect the winding temperature of a generators..........and in order to minimize the heating effect on a windings, what should i do.........decrease or increase the VAR components...........for a consistent load on a generator.
 
When power is being produced at a lagging power factor (lagging VAr indication), the excitation provided to the generator field (the rotor) is usually in excess of what's required. That excess current flow results in heating (current flowing in a conductor produces heat) and the ability of the generator cooling system to remove that heat from the rotor so that the insulation and windings aren't damaged is the limiting factor for the lagging VAr "production".

When power is being produced at a leading power factor (leading VAr indication), the magnetic fields in the generator become slightly distorted (from their normal configuration) and the stator windings and end-turns can become overheated damaging insulation and windings and stator punchings, etc.

The best way to avoid excessive heat generation is to operate the generator in accordance with the reactive capability curve provided by the generator's manufacturer, which depicts the limits of operation for lagging and leading and zero VAr conditions based on generator temperatures.
 
Dear All,

I try to explain reactive power compare with our body. if we have to do some work we need energy. for example, if we had 20 calories, 15 calories will be used for do useful work (moving the table from one place another, lifting an equipment etc...) 5 calories will be used for move the hand and foot. This 5 calories like reactive power though it's not useful but without that we can do effective work.

Like above if a generator produce power or during transmission of power some of the energy will be wasted as heat that is reactive power.

Regards
G.Rajesh
 
A

Anthony Giorgianni

This thread is interesting, so I thought I'd post to tell an interesting story about vars or VArs or however it's spelled.

I'm a journalist who years ago covered Northeast Utilities in the Northeast. As a writer for the Hartford Courant at the time, I had a tip that New England - or at least Connecticut - might undergo its first brownouts in years. So I called Bernie Fox, an engineer who also was then president of NU.

He invited me into his office in Berlin, Conn. and gave me a big talk about vars, which I had never heard of. I remember his telling me how vars are needed to operate motors and how they they didn't travel over great distances, if travel is the right word.

I remember his saying that burying the transmission cables (or maybe distribution cables) enabled vars to travel further or increased them or something. It had to do with the cable and ground and magnetism, or something. Maybe induction between the cable and ground? Specifically I remember his drawing Boston on his blackboard, where I guess they intentionally buried the cables to increase this effect.

Anyway, I wrote the page 1 story, breaking the news. And sure enough New England experienced its first brownout soon after, maybe that summer.

I never forgot about vars, not that I really understand them or electricity in general (As I kid, I used to take the "How and Wonder Book of Electricity" out of the school library again and again. There was a story about a blackout that I like to read over and over, oddly.)

I never forgot Bernie either. He died a few years back. He was a good guy. He'd always explain the engineering or utility financing (placing Millstone 3 into the rate base) behind stuff before he'd let me ask questions about anything. And he never complained about my writing stories critical of NU as long as I had my technical stuff right.

Anyway, thought anyone still monitoring this thread all might find that interesting.

Regards,
Anthony Giorgianni
 
Anthony... not bad for a lay person. Furthermore I'll give you a "plus" since your name ends in a vowel!

If are looking for further enlightenment you might want to peruse Control.Com Thread # 1026242714

Regards, Phil Corso
 
What are the causes of reactive power? I have records of 15 min interval of kwh and kvarh and observed a difference of the trend of kvarh after the changed of the substation from 69/34.5kv to 69/13.2kv. Actually, we had a double transformation, 69/34.5kv, then 34.5/13.2kv. The record shows that during off-peak,the kvarh is zero, at peak is below 100. After we switch to 69/13.2kv, it registered 100 below at off-peak and 100-300 at peak. Please advice.

-lmgs
 
LMGS... I suggest you post a new thread outlining your problem since it is not typical of what this thread provides regarding education of Control.Com members.

That said, here is my suggestion:

This kind of a problem is best solved by developing an OMT or Operations Matrix Table, similar but more extensive than what you provided!

It should contain, Date, Time, Electrical parameters, i.e., MVA, MW, MVAr, kV, A, PF, Excitation values, Grid exchange parameters, etc, and pertinent mechanical parameters, i.e., fuel flow, temps, etc.

Regards, Phil Corso
 
D

david garnett

Hello there,

With reference to the text below and my understanding of VAr's can someone explain how by increasing machine (rotor) excitation current to increase terminal voltage above the "system" voltage i.e. over-excitation a lagging power factor is achieved. My understanding is that by attempting to increase generator terminal voltage it is the magnitude of the voltage sine wave that is increased? this is different to pf / var production as there is no phase shift between current and voltage sine waves- just an increased magnitude of the voltage waveform, what actually generates the shift - the same question is true for the leading region (under-excited)?

"If we think of the induction of the magnetic field on the rotor of the induction motor as needing "power" and we call that reactive power, then we can think of reactive power as being consumed, in the same was as real power is consumed. We can measure the amount of power required to induce the magnetic field on the induction motor rotor in VArs. And, to counter the effect of the phase shift on the system of lots and lots of induction motors and transformers (which also induce magnetic fields on secondary windings) we can produce reactive power to reduce the phase shift and maintain the AC power system's ability to provide power.

We can produce reactive power to "compensate" for the inductive load on the system primary by increasing excitation current to the rotors of synchronous generators over and above the amount required to maintain the generator terminal voltage equal to the voltage of the grid to which the generator is connected. This causes reactive power, called VArs to flow out of the generator terminals on to the AC power system, reducing the phase shift and supporting the AC power system's ability to transmit real power.

So, VArs are "required" or "consumed" to induce the magnetic field on the induction motor rotors commonly used everywhere, and to increase or decrease voltage through transformers used to distribute AC power. The effect of VARs on the system is to shift the AC voltage- and current sine waves out of phase with each other. If they get too far out of phase, the ability of the AC power system to supply power is reduced. VARs can be produced by synchronous generators--but since that requires DC power, and that DC power usually comes from the AC power system, that power is not available for sale to consumers. Power factor correction capacitors can be used to help with controlling the phase angle between the AC voltage- and current sine waves, but they are usually expensive and have their own idiosyncrasy"
 
R
Concerning the phase angle between voltage and current produced by inductive loads, I'm laboring under the impression that this lagging effect is caused by the slip the rotor rotating asynchronously with the rotating magnetic field produced by the stators. If it is torque-load on the motor's shaft that ultimately creates this phase angle, can it not be corrected by using an engine to spool the motor slightly above its panel-rated rpm? This would create negative slip and thus correct (depending on the load or percent slip) the phase angle, would it not? Also, since watt-hour meters measure power factor, would it not slow the meter down, stop it, or even reverse it?
 
And what kind of "engine" would you use to increase the motor's speed? What would be the energy source for the "engine"? How would you arrange for the energy to be delivered to the "engine"? How would you control the energy being delivered to the "engine"?
 
Ryan,

The lagging is due to the properties of the inductor. The relationship between the current and voltage of an inductor is: the voltage across the inductor is a product of the inductance (L) and the rate of change of current (i) with respect to time, or V=L*(di/dt).

If you have a sinusoidal current (60Hz (or 377rad/sec) in the US), your current can be expressed by i=I*sin(wt) where w is the frequency (377rad/sec in the US)

So, plugging that into the formula for voltage:
V=L*(d(sinwt)/dt)

If we focus on evaluating the derivative:
d(sinwt)/dt = w*cos(wt)

and a cos wave is the same as a sin wave if we shift it by 90 degrees:
w*cos(wt) = w*sin(wt-90)

Which means that:
V = L*w*sin(wt-90)

See the -90 term above? This is where the lagging comes into play. The current lags the voltage by 90 degrees

Because the voltage is related to the rate of change (derivative) of current, and the rate of change of a sinusoid causes a phase shift of 90 degrees, the voltage across an inductor leads the current by 90 degrees (which is the same as the current lagging the voltage by 90 degrees).

> If it is torque-load on the motor's shaft that ultimately creates this phase
> angle, can it not be corrected by using an engine to spool the motor slightly
> above its panel-rated rpm?

You are confusing the current voltage phase angle relationship with the phase angle relationship between an internally generated voltage and the voltage seen at the terminals of a generator. Changing the torque input to a generator will shift the phase angle between the internally generated voltage and the terminal voltage, but has little affect of the relationship between the current and voltage phase angle.

When I say "little affect" there still is some relationship, however that becomes a much more complicated concept to explain, and very minor compared to the voltage to voltage phase angle relationship. If you're interested in knowing, let me know, I can try and take a crack at writing something comprehensible.

Hope that helps,
Nic
 
B
Nope.

The phase angle between voltage and current in an inductive (or capacitive) load is nothing to do with the source of the energy. In an inductance, the voltage is proportional to the rate of change of current, so for a sine wave of voltage applied the current is a cosine wave.

With an induction motor, the stator produces a rotating magnetic field that induces a voltage in the rotor because the rotor bars are moving (relatively slowly) relative to the stator field - the speed difference is the slip. The induced voltage sets up a current that develops its own magnetic field - it's the interaction of the two magnetic fields that develops the torque. No slip - no induced rotor voltage - no torque.

Driving the rotor to eliminate the slip would mean that the "engine" used to drive the rotor will provide all the power, plus the small amount needed to overcome the losses in the induction motor - so it actually becomes a drag on the system.
 
M

Mikethesignguy

Reactive power flows as current right along with the real power current. It does not register on a watt meter so the power company does not get paid for it. The current loads up the generators, the transformers and the transmission lines. That is where the extra heating comes from. If there is unity power factor all of the amps on the system are measured and billed.

Power factor correction cuts down on the current flow in the system.
 
This is a somewhat funny analogy, but here goes: apparent power is like a total volume of beer just poured in a glass. The reactive power is the foam that forms at top. The real power is the liquid under the foam.

Hope it helps the layman's picture.
 
How does this analogy explain reactive power?

If you were to be invited to my home on a fine summer afternoon and I served you your favorite beer in a cold glass with no head on it, what would be your first impression? "WOW! CSA is excellent at pouring beer--there's no head on it at all! He must have a LOT of experience pouring beer to have perfected his technique to this degree!"

You then take a large swallow of beer from the frosty cold mug and immediately realize the beer is entirely devoid of any carbonation whatsoever. What would you think than? "How fast can I find a place to spit out this swill? And what kind of host serves a person his favorite beer with no carbonation on a summer day in a cold glass? Does CSA know nothing about beer, not to mention his lack of manners?"

ih, my friend, foam on has a very significant purpose and while too much of it is not a good thing, none of it is also not a good thing. If I accidentally did pour a flat beer into a cold glass with no head I would immediately empty the glass down the drain and proceed to open another of your favorite beers in the hope it would not be flat. (I would NEVER invite someone over without having at least six of their favorite beers (if it weren't also my favorite beer of which I rarely have less than five in the house at any time).) I would also have other beer to offer in the unlikely event all of your favorite beer was flat. "Be prepared!" Is one of my favorite mottos. And I am a good host if nothing else.

And the moral of this story is: Foam on beer has an extremely important purpose and anyone who says otherwise is itching for a fight. And while it may seem to be useless, it is absolutely not. Without reactive power that beer would never have gotten cold and so reactive power is also extremely important--bit too much of it can also be a bad thing.

Beer foam and reactive power are not analogous--not in the least--and while it sounds cool to describe it thus it does absolutely nothing to answer the question of what is reactive power. Nothing at all.

I'm sure you'll agree, even if you don't drink beer--which just means there's more for the rest of us to drink and enjoy and solve the world's problems while enjoying the taste of a carbonated beer, as evidenced by the slight head which a good host knows how to pour!

Here's hoping you'll think thrice before using that alleged analogy again. ¡Salud!
 
I
To explain better understanding of apparent power, you may want to consider this analogy. someone who pull the cart with rope as depicted on below image:

http://i430.photobucket.com/albums/qq21/aanjpg/VAR analogy_zpsyamvcphv.png

On this analogy, this man pull the cart to the left by implementing force that have angle to the horizontal line. Force that implemented to the cart (APPARENT POWER) can be divided between two parts, vertical part and horizontal side, that perpendicular each other. Because of the cart only be moving to the left side, the
horizontal force is the only useful force (REAL POWER) that made the cart moving to the left, while the vertical force is not useful force (REACTIVE POWER) .

Hope this help
 
I agree, CSA, that it's not a real (technical, physical) analogy. It's just a picture I saw years back, and I thought this is something anyone could understand. You have something in your glass that is no liquid, so you have "something else", and if you have to much of it, you probably wouldn't end up drinking a lot at all. Sorry if I misslead anyone. If I ever travel to where ever you live, I would like to argue and learn more while enjoying a beer or two with you :)

Cheers!
 
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