High voltage transmission line with ABCD complex equations

Discussion in 'Homework Help' started by bizuputyi, Aug 6, 2014.

  1. bizuputyi

    Thread Starter New Member

    Aug 3, 2014
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    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:

     V_S=V_R(A_1+jA_2)+I_R(B_1+jB_2)

     I_S=V_R(C_1+jC_2)+I_R(D_1+jD_2)

    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  V_S and  I_S and hence the power  P_{SO} 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.

    What I'm thinking is  I_R=0 as this is an open-circuit and given  V_R ;  V_S and  I_S can be calculated with matrix.

     V_S=77325+j3149 KV
     I_S=j119.9A

    But I don't think that is correct because  V_S should not be lower than  V_R , 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  Zo 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.

    Any comments are appreciated.
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Regarding (a).

    77325+3149j kV is greater than 88.9 kV.
    Or did you mean 77325+3149j V? That would be 77.325+3.149j kV, and yes, this is less than 88.9 kV.
     
  3. bizuputyi

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    Aug 3, 2014
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    I'm sorry, that's a typo. It's 77325+j3149 V. That's where it gets suspicious, it shouldn't be less than 88.9kV.
    I got this result by using matrix equations shown in attachment with  I_R=0 .
     
  4. The Electrician

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    Oct 9, 2007
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  5. bizuputyi

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    Aug 3, 2014
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    Wow, that's a new aspect. From what I read about Ferranti effect I reckon I've got the voltages and currents all right. And the power becomes  P=Re[VI^*]=377565.1W
    What I'm thinking if the impedance in the stem is  \frac{V_R}{I_S} then I get j741.45Ω from which

     X_c=\frac{1}{\omega lC} (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 4.29μF.

    Resistance can be neglected, so can conductance, from http://www.electrical4u.com/ferranti-effect-in-power-system/ source.

    I don't feel I'm on the right track. I only found methods to calculate R,L,G,C with ∏-model not T like this one.

    I greatly appreciate your help.
     
  6. The Electrician

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    Consider the T network in Fig. 3(b).

    Label the two series resistors R/2, the two series inductors L/2, the shunt conductance G and the shunt capacitance C.

    Now form the ABCD matrix for that T network. Have a look here:

    http://en.wikipedia.org/wiki/Two-port_network

    about halfway down under the heading "ABCD-parameters". There you will find info (under "Table of transmission parameters") about how to form an ABCD matrix from lumped components.

    Having formed the symbolic ABCD matrix from R, L, C, G you can then equate each symbolic element in your ABCD matrix for the T network to the corresponding numeric values you have for the line. That will give you 4 equations in 4 unknowns. Solve them and you will get values for the line coefficients.
     
  7. bizuputyi

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    Aug 3, 2014
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    Please see attachments, I think that's the same approach. This is fantastic, I'm getting so close. However, I need time to figure those 4 equations.
     
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  8. The Electrician

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    Oct 9, 2007
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    You're on the right track.

    However, because the T network is symmetric, the A and D elements are the same, so you will only have 3 independent equations. That means that one of R,L,G,C will have to be chosen arbitrarily. A good starting point would be to set G=0.

    If you try a solution with 4 equations, you'll probably get an error. You might be able to do a least squares solution with 4 equations.

    I notice that the numeric values for A,B,C,D form a matrix whose determinant should be exactly 1, but it isn't, so you haven't been given an artificial problem with an exact mathematical solution; the numbers perhaps come from a real transmission line.
     
  9. bizuputyi

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    Aug 3, 2014
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     A=D=0.8698+j0.03542=1+\frac{1}{2}(R+j\omega L)(G+j\omega C)l^2
     B=47.94+j180.8=(R+j\omega L)+\frac{1}{4}(R+j\omega L)^2(G+j\omega C)l^3
     C=j0.001349=(G+j\omega C)l

    I'm sorry for my blindness but I don't see where to go from here and also where the least squares solution comes in. I'm so close just need a bit more advise.
     
  10. The Electrician

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    After giving a try to solving those equations, I think an easier approach would be this:

    Consider a general ABCD matrix:

    [ A B ]
    [ C D ]

    Now if you have a T network that has that matrix, the elements of the T network will be:

    Left series element = (A-1)/C

    Shunt element = 1/C

    Right series element = (D-1)/C

    You can work this out by considering how the ABCD matrix is formed from a T network.

    Note that the numerical value your problem gave you for C is 0 + .001349 j

    This means that G is zero, which is what I suggested you assume when solving those equations above.

    But, now if the shunt element of the T network is 1/C , we just need to solve (because G=0):

    1/C = 1/(.001349 j) = 1/(j ω Cs), where Cs (I designate it Cs instead of just C to avoid confusion with the C of the matrix) is the shunt capacitance of the T network.

    R/2 will be the real part of (.8698 + .03542 j - 1)/(.001349 j)

    and L/2 will be the imaginary part of (.8698 + .03542 j - 1)/(.001349 j) divided by ω.

    These values of R, L and Cs are for the whole line; divide by the line length for the final result.
     
    Last edited: Aug 8, 2014
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  11. bizuputyi

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    Aug 3, 2014
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    Is it really that simple? That's great, thank you very much indeed.

    Do these appear to be correct?

    R=0.0005252Ω
    L=6.14\mu F

    1/C = 1/(.001349 j) = 1/(j ω Cs)

    Are you sure about this? Isn't 1/C = jωCs?
     
  12. The Electrician

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    Here's what I get for the T network components before dividing by the line length. The R and L values are for each series arm of the T network. They need to be doubled for the actual R and L for the line:

    [​IMG]
     
  13. bizuputyi

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    Aug 3, 2014
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    Of course. What's wrong with me?! I wanted this so bad I made silly mistakes.
    Thank you for your effort to help!
     
  14. CNC682

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    Jan 23, 2015
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    Left series element = (A-1)/C

    Shunt element = 1/C

    Right series element = (D-1)/C

    How do you derive these equations?
     
  15. The Electrician

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    I said in post #10:

    "You can work this out by considering how the ABCD matrix is formed from a T network."

    Look halfway down the page here: http://en.wikipedia.org/wiki/Two-port_network

    Under the heading "Table of transmission parameters" you can see how to form the [a] matrix for a series impedance. Follow this with the [a] matrix for a shunt impedance, and finally the [a] matrix for another series impedance. Multiply the three matrices and get the overall ABCD matrix for a T network.

    Then, you can work backwards to see what the T network elements would be if you're starting from an ABCD matrix.
     
  16. viky1104

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    Aug 9, 2015
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    Why we transmit power at high voltage ? If there is reduction in power loss i.e. I^2*R losses , then power is also depends on voltage as power P=(V^2)/R . Hence with high voltage power loss should be more with respect to voltage and power loss should be less with respect to current. Which one is correct ? and why ?
     
  17. koupitan

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    Dec 6, 2016
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    Hi All,

    I am still struggling to understand how the left series element = (A-1)/C, Shunt element = 1/C , Right series element = (D-1)/C are obtained.
    I found in my notes that ABCD matrix formed from the T-network looks like this:
    [​IMG]

    What should be done next? Should i Plug for A in the matrix 0.8698+j0.03542 and so on and then equate A with?
    Please help, I am completely stuck!
     
  18. The Electrician

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    What is your goal?
     
  19. Papabravo

    Expert

    Feb 24, 2006
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    Ohms law says they are the same. The reason we use high voltage transmission is that transformers allow us to easily increase voltage while reducing current and vice versa.

    https://en.wikipedia.org/wiki/Electric_power_transmission

    Scroll down to the section on Losses
     
  20. koupitan

    New Member

    Dec 6, 2016
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    Hi The Electrician,
    Sorry for not giving enough details.
    I am referring to part b) of the original question which is:
    " 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". There is also a table with values :

    What I understand so far is to I have to obtain an ABCD matrix for the T network that looks like the one i posted in my firs post.
    Reading your earlier replies, I understand you recommended doing so using [a] matrices for series impedance and shunt impedance from wikipedia Two port network article. Assuming we don't have to consider right hand impedance as this is an open circut , is it done by multiplying only these two matrices?
    upload_2016-12-9_7-32-47.png

    Is using the ABCD matrix obtained from the above matrices equivalent to the one in posted in my previous post?

    upload_2016-12-9_7-23-5.png

    I home the explanation of my problem is clear enough:)

    Cheers
     
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