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from the power systems department...
Reactive power
Engineering and workplace issues. topic
Posted by Anonymous on 13 August, 2007 - 12:26 pm
Please, can someone explain Reactive power in layman's language?


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Posted by Bruce Durdle on 14 August, 2007 - 12:45 am
Oops - that can of worms has been re-opened! Here's my version:

In an electrical circuit containing inductance, voltage across the inductive elements is proportional to the rate of change of current. So if the current is a sine function of time, the voltage across the inductive element is a cosine function. The instantaneous power applied to the inductive element is the product of instantaneous current and instantaneous voltage. Over one complete cycle of the current waveform, the power completed 2 complete cycles and has an average value of zero. This represents energy being stored in and removed from the inductance. Although the energy involved is not used to heat or drive a load, it is essential to allow the load to operate. I've tried to come up with a useful mechanical analogy for this and it's not easy. However, if you are into rowing at a gym, you may be able to relate to this: A rowing machine has a relatively large flywheel and some elastic bits. When you pull on the handle, some of the effort does to driving the fan or other load - this relates to active power. Some of the effort goes into accelerating the flywheel or stretching the elastic - this energy is then recovered during the return stroke and is the reactive component.

Hope this helps,

Bruce


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Posted by CSA on 14 August, 2007 - 12:59 am
The first place you should look is http://www.wikipedia.org; search for 'reactive power'.

There are contributors to this site who will insist that reactive power doesn't "flow", or that there's no such thing as reactive "power". But for the purposes of this contribution (and per commonly acepted convention--see http://www.wikipedia.org) we will consider that reactive power does indeed exist and that it flows. And remember: the originator of this thread asked for an explanation in layman's terms, which for me, anyway, does not initially include vector diagrams and trigonometry and mathematical formulae. All of these things are necessary for a complete and proper understanding of the phenomenon, but not for a rudimentary understanding on which to base all of the formulae and vector diagrams, etc., which just serve as mathematical proofs of the principles.

In an AC power system, there are resistive loads and reactive loads. Reactive loads include inductive and capacitive loads--and the majority of loads on any grid are inductive (from induction motors used for everything from vibrators to elevators to refrigeration compressors to fans to all manner of pump motors (potable water, sewage, chemical, and so on)). Transformers are also an inductive load on the system.

So, since they're the predominant load, let's focus on inductive loads. In an induction motor, a magnetic field has to be induced on the rotor of the motor in order for it try to follow the apparently rotating magnetic field of the stator that is connected to the AC power source in order to produce torque. The act of inducing the magnetic field on the rotor of the induction motor causes a shift to occur between the AC current sine wave and the AC voltage sine wave of the AC power system. Some of the total amount of energy that's being consumed from the AC power system by the induction motor is used in the induction of the magnetic field on the rotor of the motor which allows the motor to produce torque. In other words, the total amount of energy being consumed by the motor does not end up as torque (real power).

The magnitude of the shift between the AC voltage- and current sine waves is measured or expressed in many ways: power factor, VArs, phase angle, etc.

The key thing to remember is that if nothing is done to try to contain or reduce the phase shift between the AC voltage- and current sine waves they can eventually shift so much that the ability of the AC system to supply real power will be reduced (such as when brownouts occur).

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 primarly 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 reactve 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 everyhwere, 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 idiosyncracies.

Real power, watts, can be converted in to torque and work which is something we can understand and visualize. Reactive power, VArs, have no such tangible quantity that we can see or envision. A lot of people like to say that, "VArs are like foam on beer; it's just there but it doesn't do anything." Foam on a beer tells the drinker the beer is not flat. There's very little worse than a cold, flat beer especially on a hot day.

Okay, a hot, flat beer is worse. And if AC power systems aren't properly managed, the beer could be flat and hot.


Posted by Robert on 16 August, 2007 - 12:23 am
Since you gave such a good explanation of Reactive Power and how inductance works to consume some of the availble power how about an explanation on how capacitors counter act or improve a low power factor?


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Posted by CSA on 16 August, 2007 - 11:11 pm
Relatively easy. Capacitance shifts the angle between the voltage and current sine waves of an AC power system in the opposite direction than inductance does, in effect "countering" the effects of inductance.


Posted by j.white on 19 September, 2007 - 12:56 pm
You gave a great explanation for what VARS does but as an operator I have always had one point of confusion. Everything I have read so far on how "real" power is supplied says it is done by adjusting excitation. Also, in seperate texts "reactive" power is adjusted by adjusting excitation. Yet when I'm operating our units I can adjust reactive power and real power independently. So I am confused how excitation can be increased or decreased to control MW or VARS independently from one another.


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Posted by CSA on 19 September, 2007 - 11:09 pm
When you say you can adjust reactive power and real power independently, you should be able to do so. Reactive "power" and real power are different, VArs being controlled by excitation and watts being controlled by torque input to the generator.

At small real power outputs, adjusting reactive power *can* have an effect on real power.

But it's not clear exactly what your confusion is. Real power is a function of torque; reactive "power" (VArs and/or power factor) is a function of excitation. Other than when lightly loaded (real power output) there should not be too much effect on real power when adjusting excitation when connected to a large (infinite) grid.


Posted by Intern on 24 May, 2010 - 3:11 pm
So let's get down to practical usage.

I have a list of kwh and kvarh, polled at 15min intervals; can I compare the two to identify problems?

If so, what types of problems and how do I go about doing so?


Posted by curt wuollet on 25 May, 2010 - 4:00 am
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


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Posted by CSA on 26 May, 2010 - 11:17 am
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.


Posted by kinley on 11 July, 2010 - 2:08 am
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.


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Posted by CSA on 12 July, 2010 - 1:54 pm
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.


Posted by LMGS on 9 December, 2010 - 2:38 am
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


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Posted by Phil Corso on 9 December, 2010 - 10:05 am
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


Posted by Anonymous on 14 August, 2007 - 6:12 pm
If power is a glass of beer, reactive power is the foam. It takes up space, but you can't drink it.


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Posted by CSA on 14 August, 2007 - 10:14 pm
Okay; if that's true, why do we care about reactive power? Why not just let the generator and the grid do whatever it bloody well pleases?

For that matter, why not just break the top off the beer bottle and dump the beer in a cold mug? Because most of the beer will turn to foam! Why not shake the can of beer before you open it? It usually takes a new bartender wannabe a few pulls to get the hang of pulling a draft without having to keep pouring the foam off. Why do people take such care to keep the foam on beer to a minimum? Because it's not to be ignored, or, the glass will not contain as much alcohol.


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Posted by Phil Corso, PE on 15 August, 2007 - 10:59 pm
Responding to CSA's not so "white" paper on Reactive Power:

Although a small step for man, you appear to be leaning in the right direction, i.e., substituting the proper term "VAr" for the oxymoron expression "Reactive Power!"

Thanks, Phil Corso (cepsicon@aol.com)


Posted by Anonymous on 15 August, 2007 - 11:51 pm
Drink straight from the bottle and there will be no foam. Perhaps we should discuss the effects of reactive power.

For instance, it is ineffective to attempt to "flow" VArs over long distance transmission lines. Drink straight from the bottle. Large inductive loads need their capacitive counterpart near the inductive source to be effective. A generator may be able to produce VArs and have a manufacturer published reactive curve, but voltage limitations on the generator prevent producing maximum VAr output in relation to terminal voltage limitations.

Generator and transformer limitations can cause watt output to be clamped while producing a large VAr output. When phase angle is shifted by an inductive load; the generator should produce more power, both real and/or imaginary (VAr). The additional voltage and/or current has a fuel cost associated with it.

Transmission system operators have a dilemma. The voltage is low on a portion of the grid(where the inductive load resides) and they have to attempt to raise the voltage over a significant distance of tranmission line and transformers. Sometimes, these operators find that the VAr they need cannot be produced by a generator due to terminal voltage limitations of the generator.

Come on, folks! Most people do not want to know the math, sine of phase angle. They want to know why we make a VAr and what we do with it.

Start typing Now! No beer drinking until complete.


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Posted by CSA on 16 August, 2007 - 12:32 am
I found this to be an excellent site:

http://www.the-power-factor-site.com

An excellent discussion of the fundamentals, with graphs and graphics, and maths, all in one nice little package.

If I drink from the bottle too fast, the beer will turn to foam. I only do that on a really hot day. Otherwise, I like to savor my hops and barley and yeast--and keep my bottle close.


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Posted by CSA on 17 August, 2007 - 10:34 am
VArs are required for the magnetizing currents required for induction motors and transformers.

The effect of too many induction motors and transformers on the system is to cause the AC voltage and current sine waves to shift away from each other, reducing the ability of the AC power system to deliver power (real power--watts).

By producing VArs with synchronous generators or synchronous condensers or synchronous motors, or introducing capacitance using power factor correction capacitors (which serve to shift the current- and voltage sine waves in the opposite direction as the inductive load, countering the effects of the inductances), the relationship between the current- and voltage sine waves will be maintained at a level which allows the AC power system to deliver optimal (not ideal) real power--watts.

It's always been my understanding that utilities with large hydroturbine plants use these plants for VAr production during summer months when air conditioning loads add to the inductive loading of the system.

These hydro plants are located hundreds of miles, in some cases, from the urban loads connected to the grid. Since it takes real power (watts) to produce the DC required to produce VArs, they use the hydroturbines to provide the torque to the shaft-driven exciters to reduce fuel costs. Water from a penstock is usually cheaper than fuel in a fossil plant.


Posted by THe original brewer on 19 August, 2007 - 11:40 am
When a capacitor or inductor is connected to an AC source, current flows in the circuit. THis current is out of phase with the voltage. The energy in the system is stored as either a magnetic field, (in the case of an inductor) or in an electric field (in the case of a capacitor). The time rate change of this energy is referred to as 'reactive power'. If the circuit includes pure resistances, as well as reactances, then the total (apparent power) is the vector sum of these two entities.

Just as the foam in a glass of beer doesn't contribute to your ability to drive a car, reactive power can't contribute to the utility's bottom line (for the home owner).

Yes, it usually does for the business owner.


Posted by Oil Analyst on 30 December, 2008 - 1:57 pm
Since there are a lot of water and pipes analogies in electricity... I wonder if you could compare reactive power to line-fill in a pipeline system?


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Posted by CSA on 5 January, 2009 - 10:24 am
Can you elaborate on how you would liken reactive power to line-fill in a pipeline system?

Are you referring to some kind of "accumulator" effect?


Posted by electrics on 30 May, 2009 - 9:13 am
pls tell me when we have a resistive load do we have a reactive current lagging because of the windings of the generator? if so pls tell me how a reactive power will go forth and back through the source?

i mean the inductance of the windings will loaded and unloaded as this energy goes to and out of source ok? so how can generator get back a reactive power? is it possible to say that the energy goes back to the exciter?

With whay kind of a circuit i think is this. There is a 0 degree source and it has a winding impedance at the generator side and a resistive load across its output, so this impedance will have a reactive power but this energy needs to return to the source true? so i can get how can this be? I really dont know how a source can get its energy back? does it store this energy in its windings ? if so my circuit model is wrong? as u see a lot of question


Posted by Dipak Barot on 21 April, 2010 - 11:42 am
> Please, can someone explain Reactive power in layman's language? <

Reactive power can be explained simply with the observation of Induction motor. Remember, Rotor of IM is always shorted. When power is given to the stator, it creates rotating magnatic field which link with rotor.so induced emf produced in rotor, and this 100 % emf circulate the current inside the rotor, which also produces magnetic field opposed to stator magnetic field.This two field always push each other to rotate the rotor.

As the stator is steady and rotor field can be shift with rotor , due to this pushing activity stator field contracts and shift the electron a bit back. so current could not be remain in phase of voltage.

So reactive power is just a opposition of supply power which shift the electron a bit back. so reactive power is just loss of active power nothing else. Syncronous generator/motor uses same principle to improve the PF i.e. shift the field back or forward. that's it.


Posted by curt wuollet on 21 April, 2010 - 2:50 pm
I understood reactive power better before that explanation :^) But, sometimes it's good to get several approaches to find one that "clicks" with the questioner. So here goes.

Reactive power is the product of voltage and the current flowing to charge capacitors or magnetize inductors. These reactances store and release energy but (ideally) do not dissipate it. Lets use a capacitor as an example. With DC this is straightforward, you charge a capacitor which takes current, and the charge is maintained. But with AC, the voltage is constantly changing and even reversing, so the charge must be removed and the capacitor is charged in the opposite direction. These also take current. So you have an applied voltage and you have these alternating currents, but no power is being dissipated.

That is reactive power, but it's not "real" power as no work is being done or power being dissipated. So, it's treated separately. Since it's not "real" power, it is also often referred to as Apparent power. Because it's not a perfect world, almost any AC circuit has both real and apparent power being transferred at the same time.

This was the explanation that helped the most folks when I was teaching electronics. I saved vectors and phase angles for a different day. If all you remember is the first sentence, you will be doing better than 30% of the students. This is really where things get complicated.

Regards
cww (who kinda enjoyed teaching)


Posted by G.Rajesh on 14 July, 2010 - 12:31 am
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


Posted by Anthony Giorgianni on 20 August, 2010 - 6:33 pm
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


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Posted by Phil Corso on 21 August, 2010 - 1:00 pm
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


Posted by Anthony Giorgianni on 23 August, 2010 - 1:12 pm
Thanks Phil! I'll check that out.


Posted by david garnett on 15 December, 2010 - 9:32 am
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"


Posted by Ryan Graham on 22 August, 2012 - 8:55 pm
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?


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Posted by CSA on 23 August, 2012 - 11:26 am
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"?


Posted by nic on 23 August, 2012 - 1:06 pm
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


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Posted by Bruce Durdle on 23 August, 2012 - 4:07 pm
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.

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