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What is tangent of power factor angle for an alternator?

  1. Jan 7, 2017 #1
    My book had these two solved examples that gave contradictory replies.

    The power output of an alternator is 100 kW. Now if the tangent of pf angle is 0.8 lagging, the KVAR rating must be -80KVAR.

    I drew phasor and got this.
    upload_2017-1-7_13-20-20.png
    I took it that since tangent is negative. Now tangent is opposite divide by adjacent, so negative tangent mean opposite component angle - adjacent component angle is negative, it means that opposite lags adjacent component.


    But then the next section had this example:

    The power output of an alternator is 40 kW and KVAR component is - 25. if theta is power factor angle then the value of tangent of theta, is 0.625 leading.
    upload_2017-1-7_13-22-4.png
    Here if we see then KW is positive in phasor and KVAR is negative in phasor. So tangent of theta is tangent of (-25/50) is -0.625. But it should be lagging since KVAR is lagging KW. So is book wrong in saying tangent of power factor angle is leading 0.625.

    Is first phasor right or second phasor? Is tangent of power angle leading different than power factor leading?
     
  2. jcsd
  3. Jan 7, 2017 #2

    NascentOxygen

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    Should be lagging, I'd say.

    Maybe this isn't a negative sign? Maybe it was printed as KVAR component is:- 25
    and the punctuation was mistakenly read as a negative sign?
     
  4. Jan 7, 2017 #3

    jim hardy

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    One must decide whether he will assign positive to leading or lagging vars. Your author appears to have got confused between those two statements ?

    In my power plant we called lagging megavars positive for a practical reason- more excitation gives you more of them.
    I flipped the varmeter from the way it'd been installed
    so that raising the voltage regulator adjustment made varmeter move same direction as field current meter, UP .
    The meters were adjacent one another so should move together. Raise voltage setting and everything moves up.
    That's important during a system disturbance for operator is under pressure and needs to see consistency in his indications. No time for ambiguous instrument response. I call it "Operator Friendly Design".

    old jim
     
  5. Jan 7, 2017 #4
    I read this article that said:
    if active power and reactive power are both in same direction then it means machine is operating at lagging pf.
    and if active and reactive power are in opposite direction, then it's leading pf.
    upload_2017-1-8_0-3-55.png
    Above is screenshot. It's solved example. The answer to first was D, and second was B. I didn't really get it. Does tangent power factor angle lagging mean power factor is lagging right?
     
  6. Jan 7, 2017 #5

    jim hardy

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    I still think there's a mistake in your textbook.
    An author must be meticulous about defining his terms and sign conventions.

    My Mother-in-Law was an editor for Wiley. She said it is common for authors to assign to graduate students the writing of exercise examples and homework problems. Editors too hire grad students to check the examples and homework problems for correctness.
    So an inconsistency might have slipped past the editors.

    One can confuse himself thoroughly by switching between authors who define their conventions differently.
    I still think it's a mistake.
    Have you inquired of your instructor ?
     
  7. Jan 8, 2017 #6
    This statement of active and reactive power same direction then lagging pf.... and opposite direction then leading pf..... this i read in internet. i tried to find the comment but couldn't.

    It was like, in case of synchronous motor if active and reactive power are flowing into the motor then the motor is at lagging pf, and if active is flowing into the machine but reactive is flowing out of machine, then motor is running at leading pf.

    This is true since:
    Sync motor is underexcited, so it absorbs reactive power and is working at lagging pf. Active and reactive power are going into the motor.
    Sync motor is overexcited, so it gives out reactive power and is working at leading pf. Active power is going into motor while reactive power is coming out of motor.

    Alternator is underexcited, so it absorbs reactive power and is working at leading pf. Active power is going out of alternator, while reactive power is going into alternator. Like source of active power and load for reactive power.
    Alternator is overexcited, so it acts as source of both active and reactive power and is said to be working at lagging pf.

    I found this very helpful.
    We don't have an instructor. I am an Electrical Engineer, 2011 pass out and am now aiming for higher studies. So I have to study on my own, and give an exam. If I get good rank in exam, then I can get admission into college to study masters of electrical engineering.
     
  8. Jan 8, 2017 #7

    NascentOxygen

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    I glanced through a few web sites. One said that the power triangle is the same as the current triangle with the sides are all multiplied by the same constant, viz., V.

    Another site said the power triangle is the same shape as the impedance triangle.

    So it seems there may not be a consistent convention to follow, as Jim suggests.
     
  9. Jan 8, 2017 #8

    jim hardy

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    Best bet is to go back to basics and draw a phasor diagram of minimal system:
    just two machines connected to a bus both at zero real power. Raise excitation on one , adjust excitation of other to restore system voltage.

    I made this drawing last Spring for another thread. It represents one machine delivering power to an infinite bus .
    I started to redraw it for above thought experiment, got flustered . Will probably try again later

    Numbers are arbitrary, for simplicity i'd change the 1.2 ohms to 1 ohm and let it represent per-unit Zsynchronous. syncgen1.jpg
    Ef phasor needs to be redrawn in phase with Vnetwork.

    Reactive current phasor will be perpendicular to Vnetwork.,

    Then add a second machine on the right side in different color maybe purple call it Ef prime, Ef ', .
    Phasors for ( Ireactive X j1) will subtract from horizontal Ef and add to horizontal Ef '
    because by KVL one is +j2 Iline and one is -j2 Iline.

    I have to go back to basics, define everything in my own terms (in other words be my own author) to get my thinking straight. (That's what Lavoisier said we have to do. )

    Probably you are better with graphics than i and can daw it quickly. I fought Paint for two hours and had to declare a ceasefire.
    Here's what it looked like after i erased my mistakes...
    syncgen3.jpg
    old jim
     
    Last edited: Jan 8, 2017
  10. Jan 8, 2017 #9

    jim hardy

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    wow you can see i'm no painter.

    syncgen4.jpg

    As we know, reactive current directly aids or opposes field current

    and that's reflected in the phasor diagram


    solve both machines for Vnetwork, which is machine terminal voltage
    Vnetwork = 1.1 - I , lagging reactive current opposes internal voltage
    Vnetwork = Ef ' + I, lagging reactive current aids the internal voltage

    Now you should repeat the thought experiment to see what happens if neither machine has a voltage regulator,

    ie when you change excitation of one machine, other's excitation remains constant and network voltage is allowed to come to a new value. .

    i don't feel i've made a good explanation here. Needs to be smoothed.

    But i hope it helps. Point is , define your conventions and stick with them. Become fluent enough to work with somebody else's conventions, translating them into yours if need be.
     
  11. Jan 8, 2017 #10

    jim hardy

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    Seems self contradicting, doesn't it ?
    EE is that way - one must be aware and rigorous with his conventions and assumptions.

    Observe i assigned directions and polarities in my Paint sketch, despite its primitive appearance.

    If one wants to make a general memory aid,
    i'd add direction to those statements
    lagging reactive current EXITING a machine opposes its excitation
    lagging reactive current ENTERING a machine aids its excitation

    When one studies armature reaction it becomes intuitive.
     
    Last edited: Jan 8, 2017
  12. Jan 8, 2017 #11

    anorlunda

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    Shouldn't one of those be "leading" @jim hardy ?

    It probably won't make the OP feel better because he has to satisfy his professor. But I managed a 45 year career in power engineering analysis without ever learning the lead/lag sign convention.

    If you work with P and Q, (MW and MVAR both signed), then there is no need for the concept of power factor or lead/lag. For example, the generator capability curve below is the way that generator designers look at things. You see only MW and MVAR, not power factor.

    th_coord%2003.gif

    Motor people might want to define the sign convention for current the other way, which could flip the lead/lag convention also. If you analyze things as @jim hardy showed you, you can correctly calculate P and Q and arbitrarily define plus to be flow to the left or right, and still get the correct answers, no matter which is the generator and which is the motor.
    • The best thing you can learn as a student is that these sign conventions (all the way back to minus charge on the elecron) are arbitrary. You can flip the signs, and as long as you are consistent, nothing changes but the plus/minus label on the answer.
    • The best thing you can learn as an engineer is to never assume that the other guy uses the same sign convention as you. Double check polarities every time. In three-phase systems, the same applies to A-B-C phase labeling, never assume, always double check.
     
  13. Jan 8, 2017 #12

    jim hardy

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    Thank You , @anorlunda

    Ahhh it sure seems so.
    As you pointed out it's all in the assignment of direction and signs.
    I think i said it right for the primitive sketch that i used

    to make that sketch i did something real sneaky - i said
    zero real power - so they're neither motors nor generators , thereby avoiding the decision you pointed out when you said
    and i defined lead/lag with respect to terminal volts.
    If i define Positive current entering a node 90 degrees behind that node's voltage as lagging, I∠-90,
    then current flowing other way would be accurately described as either -I∠-90(lagging vars going opposite way) or +I∠+90(Leading vars going same way.)
    which is why it's so important to draw your current direction arrows on your sketch.

    Which drives home your points -


    Honestly, until i set real power to zero to eliminate the motor-or-generator matter it was even more confusing for me.

    @@jaus tail - work lots of phasors . They're real handy tool. You'll be expected to be fluent in them when you get into machinery courses and learn how they derived anorlunda's capability curve.
    Notice on that curve UP is more excitation, lagging vars into system(out of machine), and is labelled Positive.
    Thats why i turned our varmeter upside down, so that raising excitation to drive more lagging vars into system drove the needle UP, same direction as exciter amp meter and generator voltmeters. I figured the instruments on the board ought to move in the direction that's intuitive to operators from the published curves in their procedures.

    Thanks anorlunda.

    old jim
     
    Last edited: Jan 8, 2017
  14. Jan 12, 2017 #13
    The diagrams are very scary. Now I'm like not even sure if I want to take the subject of machines.
     
  15. Jan 12, 2017 #14

    jim hardy

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    That's only because i'm no artist.

    The way to learn is to draw them one step at a time.

    Have you learned phasor notation ? I presumed you had but really don't know. My high school electronics teacher drilled us boys until they became intuitive, and that was so long ago it's a natural mistake to assume everybody uses them with ease.
    Phasors are basic to EE , not just machinery. We did not study machines in high school, we used Phasors for basic AC circuit analysis. When you're in tenth grade doing polar - rectangular conversions by slide rule it sticks because those are one's formative years.


    Come to believe that a sinewave with any phase angle can be represented by two of them 90 degrees out of phase. Pythagoras leads you to real and imaginary components.



    old jim
     
  16. Jan 12, 2017 #15

    jim hardy

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    syncgen4-jpg.111290.jpg

    Black Circles: radius of black circles is amplitude of terminal voltage. They're the same because i copied and pasted.
    Angle of terminal voltage i chose as zero, my reference. Black circles would have horizontal line for their phasor, but it gets overlaid with blue and purple and brown a couple steps further along in the drawing.

    Blue and purple circles have radii equal to internal voltage of the alternators.
    Blue and purple phasors also have angle zero because ( remember my sneaky trick?) real power is zero hence so is power angle, meaning they are in phase with terminal volts so lay atop the horizontal black radii covering them up.

    The brown vertical phasors represent armature current
    which get rotated 90 degrees as they flow through machine's Zsynchronous and become horizontal phasors too
    one of them adds to internal voltage the other subtracts from it (by KVL)

    I could have made a separate sketch for each step but didn't

    you can draw it easily, just go one step at a time.

    old jim
     
  17. Jan 18, 2017 #16
    We had phasor rotation. As in we were taught like:
    suppose you're standing on stator of sync motor and now the stator is moving anti clockwise. So then in case of alternator the field flux will pass first and then the resultant flux will pass you. So we say that flied flux is leading resultant flux and draw phasor such that if flied flux is like at 3 o clock, pointing right, then resultant flux is at 6 o clock, pointing south and we draw anti clockwise arrow.
     
  18. Jan 18, 2017 #17

    jim hardy

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    This thread has my very primitive sketches ..

    Maybe they'll help you along.

    https://www.physicsforums.com/threads/synchronous-generator-excitation.829603/#post-5214090

    https://www.physicsforums.com/threads/armature-reaction-drop.826513/#post-5191404

    You've got to draw these out as you talk yourself through, visualizing in your mind the flux.

    What's confusing is the derivative relation between flux and voltage. Sine is a mathematical oddity, it doesn't change shape when you differentiate it but shifts phase a quarter turn.. That's why voltage peak is 90 degrees to flux peak.
     
  19. Jan 22, 2017 #18

    cnh1995

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    I'd felt exactly like that when I took the subject of machinery.
    But then came @jim hardy with his awesome intuitive explanations and sketches, and machinery is now my favorite subject.

    I believe the threads he has linked in #17 are mine (there are many others). Those sketches are really helpful. Phasor diagrams are indeed a handy tool to analyse machinery , avoiding any complicated alegebraic calculations.
    Good luck with your studies!
     
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