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High voltage transmission line with ABCD complex equations

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Figure shows a 50 Hz, high-voltage, transmission line. The relationships between the sending and receiving end voltages and currents are given by the complex ABCD equations:

    [itex] V_S = V_R (A_1+jA_2)+I_R (B_1+jB_2) [/itex]

    [itex] I_S = V_R(C_1+jC_2)+I_R(D_1+jD_2) [/itex]

    where 'S' stands for sending-end and 'R' stands for receiving-end

    (a) Given the parameter values in TABLE C and an open-circuit received voltage measured as 88.9 kV, calculate the values of [itex] V_S [/itex] and [itex] I_S [/itex] and hence the power [itex] P_{SO} [/itex] absorbed from the supply by the transmission line on open circuit.

    (b) If the line is modelled by the T-circuit of FIGURE 3(b), see if you can estimate the primary line coefficients R, L, G and C. The line is 50 km long.

    2. Relevant equations

    [itex]
    \begin{bmatrix}
    V_S\\
    I_S
    \end{bmatrix}
    =
    \begin{bmatrix}
    A_1+jA_2 & B_1+jB_2\\
    C_1+jC_2 & D_1+jD_2
    \end{bmatrix}
    \begin{bmatrix}
    V_R\\
    I_R
    \end{bmatrix}
    [/itex]

    3. The attempt at a solution

    What I'm thinking is [itex] I_R=0 [/itex] as this is an open-circuit and given [itex] V_R [/itex]; [itex] V_S [/itex] and [itex] I_S [/itex] can be calculated.
    Now,
    [itex] V_S=77325+j3149 KV [/itex]
    [itex] I_S=j119.9 A [/itex]
    But I don't think that is correct because [itex] V_S [/itex] should not be lower than [itex] V_R [/itex], also 50Hz frequency is given for a reason, I'm sure it has to be used somewhere.

    And as for question (b) I have only got ideas. I can find coefficient of propagation γ and [itex] Z_o [/itex] from R,L,G,C but I don't know how to produce them vice versa (from ABCD), I don't see where I could go from there either.

    Or another idea:
    [itex] \frac{V_R}{V_S}=e^{-αl} [/itex]
    where
    [itex] α=\sqrt{RG} [/itex]

    Any comments are appreciated.
     

    Attached Files:

  2. jcsd
  3. Aug 5, 2014 #2
    I wouldn't worry about Vr > Vs.

    How much current is flowing through the resistor and inductor on the right leg of the circuit?
    Do the values of those two components matter?
    What is the voltage at the top center node of the circuit?

    You can compute the total loss of power through the other two resistors. They're the only components that will generate real heat.
     
  4. Aug 6, 2014 #3
    A transmission line is usually modelled in my learning material by the figure shown in attachment, I reckon this should be the case in this exercise.

    Right, the current in the right leg is zero which means that resistor and inductor are out of consideration, hence, the voltage in the top centre node is [itex] V_R [/itex].

    Now, [itex] P = Re[VI^*] = 377565.1 W [/itex]
    Is that correct so far?

    R can be calculated via [itex] P = \frac{V_S^2}{R} [/itex]

    Where do frequency and line length come in?
     

    Attached Files:

  5. Aug 6, 2014 #4
    Yes, it is!
    My skills at this are seldom practiced. It looks good to me.
     
  6. Aug 7, 2014 #5
    I've been googling and figured that this is Ferranti effect:

    http://en.wikipedia.org/wiki/Ferranti_effect
    http://www.electrical4u.com/ferranti-effect-in-power-system/

    If the impedance in the stem is [itex] \frac{V_R}{I_S} [/itex] I get j741.45Ω, from which

    [itex] X_c=\frac{1}{\omega lC} [/itex] (where l is length in metres)

    C=85.86pF that is far too low for such a long line.
    Without putting 'length in metres' in that equations capacitance is 4.29[itex]\mu[/itex]F.

    Resistance can be neglected, so can conductance, from that source.

    I don't feel I'm on the right track. I've got the voltages and currents all right I think but I'm lost from there.
     
  7. Aug 7, 2014 #6
    I've got new ideas, see the pictures attached. The T-circuit can be modelled by ZYZ circuit shown in figure.

    We know ABCD, frequency and length of line, the question is how to find R,L,G,C.

    I'm so close, just need the last nudge.
     

    Attached Files:

    • ZYZ.jpg
      ZYZ.jpg
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      13.9 KB
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      229
    • line.jpg
      line.jpg
      File size:
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  8. Aug 7, 2014 #7
    This may be the blind leading the blind, but this is what I did for the base leg:

    The voltage over the base leg is 88900V and the current is 119.9jA. so I calculated the resistance:
    R = 88900V/119.9jA = -741.5jOhms.

    I went to this site:
    http://chemandy.com/calculators/ac-networks/capacitor-and-resistor-in-parallel-calculator.htm
    and entered 0.05KHz, 4.293uF, and 99,999,999MOhms, to get -741.5jOhms.

    So, I conclude (with less than full confidence) that the resistor in the base leg is open (infinite resistance) and the capacitor is 4.293uF.

    The voltage across the left arm is (77325+3149j)V-88900V and the current is 119.9jA.
    R = (-11575+3149j)V/119.9jA = (26.26 + 96.54j)Ohms.

    I went to this site:
    http://chemandy.com/calculators/ac-networks/inductor-and-resistor-in-series-calculator.htm
    and entered 0.05KHz, 307.3mH, and 26.26Ohms, to get (26.26+96.54j)Ohms.

    So I concluded that the left arm inductor is 307.3mH and the resistor is 26.26ohm. Since this is suppose to be a symmetric T circuit, I would use those same values for the right arm.

    Do my numbers work better for you?

    --- edit
    I just noticed your last post. I hope we don't have to go in that direction.
    Also, if you need a complex calculator, there is one here:
    http://www.mathsisfun.com/numbers/complex-number-calculator.html
     
    Last edited: Aug 7, 2014
  9. Aug 7, 2014 #8
    Yes, that's great, thank you. Let's go down the other line I just figured and see if the values match. I think I'll need 4 equations with 4 unknowns. I need some time to figure them out.
     
  10. Aug 8, 2014 #9
    I think I've got it. Referring to my latest attachments the left or right series element [itex] Z = \frac{A-1}{C} [/itex], shunt element [itex] Y = \frac{1}{C} [/itex]
    Real part of Z = R/2, imaginary part of Z = L/2, divide by ω for L value.
    1/C=jωCs (shunt capacitor)
    Divide all values by line length.
     
  11. Aug 8, 2014 #10
    I get the same L and R values as you did. Thank you very much indeed!
     
  12. Aug 8, 2014 #11
    Very good. Does that give you an estimate of the line length?
     
  13. Aug 9, 2014 #12
    Line length is given in the exercise.
     
  14. Mar 2, 2015 #13
    Why does Z=A-1/C and Y=1/C. How are these left/right and shunt elements obtained?
     
  15. Dec 8, 2016 #14
    Hi Guys,

    I am struggling with this question as well. Could you explain, how we get from this ABCD matrix formed from a T - network to the Z=A-1/C and Y=1/C ??
    How the above equations are obtained?



    line-jpg.71986.jpg

    Please help
     
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