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50 Hz High-Voltage Transmission Line

  1. Jan 10, 2015 #1

    M P

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    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:

    mimetex.cgi?%20V_S=V_R(A_1+jA_2)+I_R(B_1+jB_2).gif

    mimetex.cgi?%20I_S=V_R(C_1+jC_2)+I_R(D_1+jD_2).gif

    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 mimetex.cgi?%20V_S.gif and mimetex.cgi?%20I_S.gif and hence the power mimetex.cgi?%20P_{SO}.gif 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
    formulae.png

    3. The attempt at a solution
     

    Attached Files:

    Last edited by a moderator: Apr 16, 2017
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  3. Jan 10, 2015 #2

    M P

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    Regarding question b) Do I need to go down the route of Z1=Z3=1/2(R+j omega L) x length
    and Y2 = (G + 1 omega C) x length ??? any help appreciated...
     
  4. Jan 10, 2015 #3

    M P

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    And then form matrix [ A B ]
    [ C D ] ???
     
  5. Jan 12, 2015 #4

    rude man

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    If the xmsn line is open-circuited at R, what are VR and IR?

    And to get PSO, what is the expression for the input impedance of an unterminated ABCD network? Combine the answer for that with the answer to the previous question to get the answer to this question.
     
  6. Feb 24, 2015 #5
    This question has got me baffled, so if anyone could point me in the right direction, it would be much appreciated.

    With question (a)

    If the receiving end is open circuit Ir=0, therefore the equation becomes:

    Vs=Vr(A1+jA2)

    Vs=88900(0.8698+j0.03542)

    Vs= 77325.22+j3148.84


    Is=Vr(C1+jC2)

    Is=88900(0+j0.001349)

    Is=j119.93


    Now the power should be calculated by

    P=Real{VsIs*}

    P=Real{(77325.22+j3148.84)(j119.93)}

    P=377565w


    I'm hoping I'm on the right track with this?


    Not sure where to start with (b)
     
  7. Feb 24, 2015 #6

    rude man

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    This problem may be unsolvable. The receiving voltage VR's magnitude is given but not its phase. Was it intended that the phase be zero?
     
  8. Feb 25, 2015 #7
    Hi RM

    I'm not sure to be honest, but the question does only say estimate. I've been reading through various text books, but cannot find any similar questions to this. I'm trying to form an ABCD matrix from the lumped components.

    Initially I was concerned that my Vs was incorrect due to it being lower than Vr, but after researching open ended transmission lines, this appeared correct. How do my figures look? That's quite a bit of power consumed @ 37.8 kW, but it is however 50 km in length.

    Any pointers?
     
  9. Feb 25, 2015 #8

    rude man

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    You already have the ABCD parameters, both numerically (your table) and in relation to the T network Z1, Z3 and Y1.

    For part (b), as I said you already have the relations between the ABCD parameters and T-network components, so just solve for Z1, Z3 and Y3 from those.
    I will compute power myself a bit later and tell you qualitatively how close your answer is to mine.
     
    Last edited: Feb 25, 2015
  10. Feb 25, 2015 #9

    rude man

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    Up to here all looks 100% unless you made a math error.
    Real power is (1/2)(VsIs* + Vs*Is). Vs is complex since we've defined Vr as having zero phase. P is a real number.
    Repeat your computation of P using this formula.
    see my penultimate post for part (b)
     
  11. Feb 25, 2015 #10
    Hi RM

    I recalculated the power absorbed to get an answer of 377590W. My text books only showed the formula I used above, where * indicates the complex conjugate. I've been trying to find reference to the formula you provided which extended this, but could not find any. Could you possibly point me in the direction of a reference as I'm keen to learn. So I was slightly out with my initial figure.

    Thanks
     
  12. Feb 25, 2015 #11

    rude man

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    You did very well! I apologize for not paying closer attention to your work earlier.
    And guess what - our two formulas are identical! Which I found out after some simple math.
    So you should have gotten the same answer: P = VsrIsr + VsiIsi.
    where r means real part and i means imaginary part.
    EDIT: note I changed the expression for P (sign change from last post).

    PS - looked at part (b) yet?
     
    Last edited: Feb 26, 2015
  13. Jun 7, 2015 #12
    In the table we had A1, B1, C1, D1 and A2, B2, C2, D2 - would i be correct in thinking that:

    A1......B1......=.....0.8698.....47.94
    C1......D1............0.............0.8698

    represents the series impedance - Z1 in the T circuit in the first post and that:

    A2......B2......=......0.03542............180.8
    C2......D2..............0.001349......0.03542

    represents the shunt admittance - Y2 in the T circuit?

    So to get ABCD matrix for the ZYZ circuit i need to multiply the 3 together to get a single matrix?

    If that's the case i have the single matrix to be:

    0.083...................142.84
    0.00102..................0.083

    but then i cant get that to tally with

    1+Z1Y1.............Z1+Z2+(Z1Y2Z3)
    Y2................................1+Y2Z3

    Am i going about this in the correct way or am i totally off-course? Also apologies for the formatting, i dont know all of the tags to space correctly.
     
    Last edited: Jun 7, 2015
  14. Jun 7, 2015 #13
    Or if my assumption above is correct, looking at my notes, Z1=47.94 & Y2=0.001349
     
    Last edited: Jun 7, 2015
  15. Jun 7, 2015 #14

    rude man

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    @Gremlin, you can solve for Z1, Y2 and Z3 in terms of the ABCD parameters.
    There are only 4 ABCD parameters:
    A1 + jA2,
    B1 + jB2,
    C1 + jC2,
    D1 + jD2.
    You are not to separate real and imaginary parts of the four parameters the way you tried to do. You should solve for the three T-equivalent parameters, each of which will (probably) be complex. From that T-network you can then compute the power furnished by the source.
    Look some more at posts ## 9 and 11.
    EDIT: you can use the T-equivalent network as given in fig. 3(b) of the OP's post 1. The series resistors are R, the parallel resistor is 1/G, and the L and C are obvious.
     
    Last edited: Jun 7, 2015
  16. Jun 7, 2015 #15
    Thanks rude man,

    Just so we dont get our wires crossed here i've already completed part a) and my posts #12 & #13 relate to part b) - don't your posts #9 & #11 relate to part a)?
     
  17. Jun 7, 2015 #16
    Ok i've got Z1, Y2 & Z3.
     
  18. Jun 7, 2015 #17
    Once you have the impedance how do you go about splitting them into L & R? And the same with splitting Y2 into G & C? You say it's obvious, which slightly unnerves me as it isn't obvious to me at the moment.

    Z1 = 0.5 (R +jωL) x length of line.
    Z1 = 26.26 + j96.52
    f = 50
    l = 50km

    26.26 + j96.52 = 0.5 (R + j 2*pi*50 L) * 50

    Not sure from that a) how you can isolate R or L and b) whether it should be 50km or 50000m

    Similarly
    Y2 = (G + jωC) x length of line.
     
  19. Jun 7, 2015 #18
    After a bit of thought and following this equation through:

    (26.26 + j 96.52) x 2 = (R + j 2*pi*50*L) * 50

    (52.5128 + j 193.0318) / 50 = R + j 2*pi*50*L

    1.05 + j 3.86 = R + j 2*pi*50*L

    R = 1.05Ω per km
    L = j 3.86 / j 2*pi*50 = 0.0122 H per km

    Does that look like the right way of doing it?
     
  20. Jun 7, 2015 #19

    rude man

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    Z1 = R + jwL = Z3
    Y2 = G + jwC
    Each impedance/admittance has a real and an imaginary part. The relationship with R,L,C and G should be obvious once you've computed Z1, Z3 and Y2 in terms of A,B,C and D.
    BTW don't mix up C of the ABCD's with the line capacitance C.
    The length of the line is implied in the ABCD parameters. You should have Z1, Z3 and Y2 strictly in terms of the complex ABCD parameters. You have that relationship given to you, albeit indirectly (cf. your post 1, section 2)..
    Remember f = 50 Hz.
    Hint: since the line is unterminated, does Z3 matter at all? You can make life much simpler by deleting Z3. Your VR voltage is then measured at the mid-point of the original T-equivalent network.
    One thing: they ask for R,L.G and C. Usually, these parameters are per unit length of line. Since they don't give you the length of the line you have to assume R,L,C and G are for the WHOLE line. That is unusual and misleading IMO.
    Yes. Post 14 and this one are the only ones addressing part (b).
     
  21. Jun 7, 2015 #20

    rude man

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    The answers are not "per km" unless the line is given as 1 km long. Are you using SI units faithfully? I can't check all your math, sorry.
     
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