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Question about KVarh

  1. Apr 6, 2009 #1
    Hi there,

    I'm not an expert on electricity by a long shot but I do understand most of the basics (i.e. AC voltage follows a sine wave, etc. etc.). Anyway, I'm helping my company analyse the data we're gathering from our various utility meters and our electricity meters are giving us a reading in KVarH in addition to KWH, which we are recording but so far have have absolutely no idea exactly what it signifies. I did a little research on it today and so far what I've come up with is that the KVARH is basically the energy that is wasted starting an inductive load like a motor, and that the ratio between KVarH and KWh is what determines power factor.

    What I'm not getting is that it doesn't seem to relate very well to the actual numbers we're getting. I'm just trying to understand how the two measurements relate to each other. For instance, on a given day we'll have 12,000 KVARH and 19,200 KWH. I guess I'd just like to understand a little better what that actually means. Is that good, bad, indifferent, etc.? Do I add the two numbers together in order to get my actual usage? Do I do something different?
     
  2. jcsd
  3. Apr 6, 2009 #2

    dlgoff

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    Welcome to PF brstilson.

    A http://en.wikipedia.org/wiki/Volt-amperes_reactive" [Broken] (the ratio of the real power flowing to the load to the apparent power) is less than one. So what your are measuring over time (in hours) is the real energy Kw-hr and the reactive energy Kvar-hr.

    From the numbers you are seeing, it is clear that your power-factor is not good. In order to improve the pf and decrease the Kvar-hr measurment you might need to add capacatance if you have a lot of motors runing (inductive load).
     
    Last edited by a moderator: May 4, 2017
  4. Apr 7, 2009 #3
    Depending on your region, this kind of power factor can get you penalized. The pF for the numbers that you gave us is about 0.848. If you are being charged a penalty, I would recommend a synchronous condenser if you have any extra synchronous machines. This is a very easy way to improve your power factor, and you don't have to worry about switching capacitors on and off. Most switches "flip" when the current is equal to zero to prevent arcing. Also the voltage of the capacitor will be at its greatest when the current is equal to zero. These two facts alone can show you that any switch you use on the capacitor will require at least twice the voltage rating of your line.

    Lastly, if the poor power factor is costing your company a lot, hire an expert.
     
  5. Apr 16, 2009 #4
    Thanks for your answer. At my location, the electric company charges a penalty for a power factor under 0.80 and gives a credit on the bill if the power factor is 0.90 or above. I've been studying up on this a lot more and I think I understand how it works.

    A inductive load causes the current wave to lag behind the voltage wave, which is called the phase angle. This delay can be counteracted by a capacitor which acts the opposite way, the voltage wave lags behind the current wave. If the capacitor and inductor are put in series and both phase angles are identical, they effectively cancel each other out.

    So the problem with putting capacitors on an entire building is that while it may compensate for the phase angle of the inductive load, the compensation is static while the inductive load changes. If the inductive load disappears (like at the end of the day when most of the motors are turned off), the capacitors actually hurt the pF because they're pushing the phase angle the opposite way the inductive load was.

    Please correct me if I'm wrong about any of this. I'm still learning. Thanks!
     
  6. Apr 18, 2009 #5
    That's correct. To prevent that, you would have to put some kind of control on either the capacitor bank or the synchronous condenser if you expect to have a large variations in your loads KVAR.
     
  7. Apr 21, 2009 #6
    What is the reactive power conception?
    For a conceptual dicaussion about reactive power, you can refer to General Electrical Riddle No.35 in http://electrical-riddles.com
     
  8. May 20, 2009 #7
    Can you explain how you calculated a power factor of 0.848 from those kVarh and kWh readings. Thanks !
     
  9. May 20, 2009 #8
    We know the apparent power S is calculated from [tex]S^2=P^2+Q^2[/tex]. We also know that [tex]pF=\frac{P}{S}[/tex]. From there, it's just plug and chug.
     
  10. May 22, 2009 #9
    sorry i get 0.78 ......maybe I am doing it wrong. If you wouldnt mind demonstrating it I would appreciate......thanks once again.
     
  11. May 12, 2010 #10
    As mentioned by ravioli above S2 = P2 + Q2 i.e. s = rt((19200)2 + (12000)2)
    s = 22641.55 kVA

    pF = p/s i.e. 19200/22641.55
    pF = 0.85 (0.847)
     
  12. May 11, 2011 #11
    Hello
    Please, i want to ask the following question and give me the answer as much as possible
    What is the difference between the watt hour meter and Var hour meter in terms of internal connection and work theory .
    thanks
     
  13. May 11, 2011 #12

    dlgoff

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    Volt Amp Reactive (VAR) are measured by determining the phase angel between the current and voltage waveforms. This measured angel is used to calculate the Power Factor which determines the VAR value.

    Transducers are used to meter such parameters in the power industry and can be bought with as "2 element", "2 1/2 elements", or "3 elements".

    You might be interested in looking into these devices. They preform the same calculations of modern Watt/Var meters.

    Here's an earlier post on the subject:

    https://www.physicsforums.com/showpost.php?p=2619637&postcount=4"

    Regards
     
    Last edited by a moderator: Apr 25, 2017
  14. May 12, 2011 #13
    Thank you dear
    I understand the triangle of power components.
    But my question for the analog meter with a dynamic disk. Do the same meter measures kw and kvar or there is a difference in internal structure and theory of work between each meter.
    ie how can i know the meter if it is watt hour meter or varh meter with out specification lable fixied on it.
    Thank you
     
  15. May 12, 2011 #14

    dlgoff

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    I took this quote from the paper http://www.google.com/search?q="var...=org.mozilla:en-US:official&client=firefox-a".

    I have emphasized the answer to the relevant part of your question in Bold Type.
     
    Last edited by a moderator: Apr 25, 2017
  16. May 15, 2011 #15
    Thank you my friend
    Do you mean that interconnect such as Star or Delta connections are Determined the work of the meter
    If it internalconnected as a star for voltage coils it records kWh and if internalconnected is delta it records kilo varh ????????
     
    Last edited by a moderator: Apr 25, 2017
  17. May 15, 2011 #16

    dlgoff

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    No and No.

    The electromechanical meter (for 3 phase power) feeds the power of each line (phase) through the meter and the disk spins for either Watt-Hr or Var-Hr measuring. The only difference is for Var-Hr meters, there's "equipment for the displacement by 90° of the voltage; this can be an autotransformer or a RC circuit".
     
  18. May 15, 2011 #17
    you mean now that the kvarh meter has adesplacment angle ( -90 ) for current coil lagging the voltage in voltage coil .
    and the kwh meter has adesplacment angle ( o) ie in phase ?
    And nothing to do with the internal link
    And the settings are but by special equipment at the factory to adjust the above ?????????
    please correct.
    thank you
    very much
     
    Last edited: May 15, 2011
  19. Sep 8, 2011 #18
    I'm also interested in this question, and it would be nice if someone posted the corrected result, please!!!
     
  20. Feb 2, 2012 #19
    Hi there. I did thru same calculations with my electric bills.
    I compare my numbers for october or last year where i have
    79920 kWH and 36400 kVARH
    S=√79920^2+36400^2
    S=87818
    ph=79920/87818=0.91, which is not bad, right?
    Then we replaced Cooling Tower and my numbers droped to
    62960 kWH and 11920kVARH... uh, huge drop for the second number!
    But then when i do the same calculation i come up with pF=0.97...
    So, how is such a big drop of kVARH give me only 6% of pF?
    Do that number even look right?
     
  21. Feb 2, 2012 #20

    jim hardy

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    draw your right triangles and check the trig.
    at small angles the hypotenuse isn't much affected by opposite side.
    SRT scale on sliderule demonstrates that, as does cosine graph it's pretty flat at top.

    i use a slightly different power factor algorithm that lets me do a sanity check at each step, and might be less keystrokes...

    ARCTAN of (VARS/WATTS) gives angle between volts and current ,
    COS of that angle gives power factor
    after division it's only two keystrokes.

    And if a get a large angle i've mis-typed something.

    try it and see if if is comfortable for you.

    36400kVARhrs/79920kWH = 0.455455 etc repeating decimal
    arctangent of that = 24.487 degrees, reasonable enough
    cosine of that = 0.91005

    awareness of the angle might help you - i dont know...
     
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