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Homework Help: Single phase transformer

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data

    A single phase transformer has the following rating: 120 kVA, 2000 V/100 V, 60 Hz with 1000 primary turns.

    Determine:

    a) the secondary turns
    b) the rated primary and secondary currents
    c) the maximum flux
    d) given a maximum flux density of 0.25 T, the cross sectional area of the core.

    2. Relevant equations

    V2/V1 = N2/N1
     
  2. jcsd
  3. Apr 16, 2012 #2
    trying to work out the answer for part a

    N2 = N1*V2 / V1


    = (1000 * 100) / 2000


    N2= 50 turns


    Is this correct?
     
  4. Apr 16, 2012 #3

    NascentOxygen

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    1000 turns : 50 turns ≡ 2000V : 100V
     
  5. Apr 16, 2012 #4
    Thanks, can you offer any advice on solving part b?

    Im stuck trying to work out the current in either N1 or N2

    once i have one of these values I guess i can use V2/V1 = N2/N1 = I2/I1
     
  6. Apr 16, 2012 #5
    ok here is an attempt at b

    I1=120000/2000 = 60A

    I2 = 60/0.05 = 1200A



    how does this look?
     
    Last edited: Apr 16, 2012
  7. Apr 16, 2012 #6

    NascentOxygen

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    What does it mean when it says a single phase transformer has the following rating: 120 kVA?
     
  8. Apr 16, 2012 #7
    120kVA is the maximum power the transformer can output. its a measure of volt-amps.

    so... 120kVA / 100v = 1200A on the secondary?
     
  9. Apr 16, 2012 #8

    NascentOxygen

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    Close enough. :smile:
    Yes. And the same idea applies to the primary, too.
     
  10. Apr 17, 2012 #9
    I'm a bit stuck with part c. Any advice on this one?
     
  11. Apr 17, 2012 #10

    NascentOxygen

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    What equations do you know that involve magnetic flux?
     
  12. Apr 18, 2012 #11
    Would this be appropriate to use?

    Magnetic Flux = B.A.cosθ
     
  13. Apr 18, 2012 #12

    NascentOxygen

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    Φ=B.A.cosθ is a good equation to memorize, and we'll need that for part (d) which relates flux to flux density and area.

    For part (c) we need an equation relating flux, voltage and number of turns for a transformer.
     
  14. Apr 18, 2012 #13
    ok I see. Would this be more appropriate?

    Φ = V × T / N

    Φ = Flux (weber)
    V = Voltage
    T = Time
    N = No. of turns
     
    Last edited: Apr 18, 2012
  15. Apr 18, 2012 #14

    NascentOxygen

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    In an operating transformer the flux is constantly changing, and in a sinusoidal manner. It's the changing flux linking the secondary that induces the secondary voltage. :smile:
    You're getting close. It's based on that, but others have done all the necessary fiddle-factoring and you've almost certainly been presented with the final equation in lectures or textbook reading.

    Hint: there's a 4.44 factor in it.
     
    Last edited: Apr 18, 2012
  16. Apr 18, 2012 #15
    right then I think I may have got it...

    Vp = 4.44 * f * Np * Φmax


    2000 = 4.44 * 60 * 1000 * Φmax

    2000 = 266400 * Φmax


    2000 / 266400 = Φmax


    Φmax = 7.51 mWb
     
  17. Apr 18, 2012 #16

    NascentOxygen

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    :smile:
     
  18. Apr 18, 2012 #17
    Great thanks for you help and advice. Its making more sense now. Last part to go.....

    Φ = B * A * cosθ

    would θ = 0 ?



    7.51x10-3 = 0.25 * A * cos(0)

    7.51x10-3 / 0.25 = A

    A = 0.03003mm2
     
  19. Apr 18, 2012 #18

    NascentOxygen

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  20. Apr 18, 2012 #19

    NascentOxygen

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    EDIT: Oops. Misread the Greek. :tongue:
    Yes, windings are perpendicular to the core.
    Check those units.
     
  21. Apr 18, 2012 #20
    would it be m2? (standard SI base unit)
     
  22. Apr 18, 2012 #21

    NascentOxygen

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    It should be.

    0.03 m² sounds better. 120kVA would be a heavy industrial transformer. If the core were to have a circular cross-section it would be 20cm. diameter, without the windings, so I guess that sounds in the ballpark.

    So, you have the transformer equations all connected up now? i0om5.gif
     
  23. Apr 18, 2012 #22
    so that would mean if the core was circular, the diameter would be roughly 19.5cm? That is huge.

    yes its all a lot clearer now. thank you for your help and time.
     
  24. Feb 26, 2013 #23
    Hi could somebody expand on 4.44 factor if you dont mind.
     
  25. Mar 17, 2013 #24
    hmmm

    I'm guessing that the equation used to find the max flux is the universal EMF equation :-

    E = 4.44 x f x N x a x B

    where
    E = rms voltage of winding
    f = frequency
    N= number of turns
    a = cross=-sectional area of core
    B = peak flux density

    Just wondering how we can actually use this to find 'B' as we don't know what 'a' is either? It looks like it was just left-out. Surely this would give a wrong answer then??

    I've reached this question too, and my notes have no mention of this universal equation so I'm struggling a bit to work it out..

    Could anyone please explain why it's been left out and yet the answer is still right..
     
  26. Mar 17, 2013 #25

    NascentOxygen

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    If you are asked to find maximum flux, then you are to find flux, Φ, not flux density, B. https://www.physicsforums.com/showpost.php?p=3871394
     
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