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Transformers and Active/Reactive power question from HW

  1. Jan 28, 2017 #1
    1. The problem statement, all variables and given/known data
    A 7200-240 V, 60 Hz transformer is connected for step up operation, and a 144∠46° Ω Load is connected to the secondary. Assume the transformer is ideal and the input voltage is 220 V at 60 Hz.
    Determine
    a)secondary voltage
    b)secondary current
    c)primary current
    d)input impedance
    e)1)active
    2)reactive
    3)apparent power input to the transformer

    2. Relevant equations
    since its operating as "Step Up" transformer...the turns ratio aka "a' = 30

    240-7200
    1:30 step up

    3. The attempt at a solution
    a)secondary voltage =220*30= 6600 V∠0°
    b)secondary current = 6000V/144Ω ∠46°= 45.8333 A∠-46°
    c)primary current= a x I sec = 45.83333 * 30 = 1375 A∠-46°
    d)Z input= 220 V / 1375A = 0.16∠46°Ω

    Active / reactive and apparent Power
    are all listed as answers in the back of the book
    I am having trouble keeping these numbers straight in my head! Please help me understand my errors on this problem...

    210.1 kW =active power is my true power
    217.6 kVAR = Is my reactive power
    302.5 kVA 1375*220 = 302500 /1000(for the K) = 302.5 kVA

    Can you kindly show me what I am forgetting about active and reactive power guys?

    Thanks Joe
     
  2. jcsd
  3. Jan 28, 2017 #2

    goodphy

    User Avatar
    Gold Member

    One-cycle-averaged power is conveniently calculated by using complex representations of voltage and current such as [tex]\tilde S = \frac{{\tilde V{{\tilde I}^*}}}{2} = P + iQ[/tex], where P and Q are an active and reactive power respectively, and absolute value of S is an apparent power.

    Here, I choose that V and I are for the secondary side of the transformer so, [tex]\tilde V = 6600{e^{i0}} = 6600,{\rm{ }}\tilde I = 45.8333{e^{i\left( { - {{46}^ \circ }} \right)}}[/tex]

    Calculation shows that P ≅ 105 kW, Q ≅ 109 kW, and the apparent power is 151 kW.

    Your values for powers are twice than what I calculated. I guess you missed a factor of 1/2 somehow.
     
  4. Jan 28, 2017 #3
    210.1 kW
    217.6 kVAR

    These are the answers from the book
     
  5. Jan 29, 2017 #4
    Since you are dealing with effective or rms voltages, the power equation is:

    S = V * I' = P + j Q

    I' should be I-conjugate

    There is no factor of 2 involved.
     
  6. Jan 30, 2017 #5
    GUYS, I figured it out...My problem was that once I solved for "S"..and I had the angle @ 46°...then I can solve for active and reactive power
    Power triangle means
    s=302.5 theta=46°
    Active=210.13 =302.5cos 46°
    Reactive=217.34 =210.13tan 46

    Answer
     
  7. Jan 30, 2017 #6

    cnh1995

    User Avatar
    Homework Helper

     
  8. Jan 30, 2017 #7
    210.13 tan 46 gets you 217.59 which is what I was looking for..

    Answers from the book are
    210.1 kW Active Power
    217.6 kVAR Reactive Power

    This is per the power triangle
     
  9. Jan 30, 2017 #8

    cnh1995

    User Avatar
    Homework Helper

    Oh..Right!
    You took Q=PtanΦ and I misread it as Q=StanΦ.
    Q=S*sinΦ=P*tanΦ..
     
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