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Complex ABCD equations

  1. Nov 5, 2013 #1
    FIGURE 3(a) represents a 50 Hz, three-phase, high-voltage, transmission
    line. For one phase, the relationships between the sending end voltage
    and current and the receiving end voltage are given by the complex
    ABCD equations:

    Vsp = Vrp(A1 + jA2) + Irp(B1 + jB2)

    Isp = Vrp(C1 + jC2) + Irp(D1 + jD2)


    where VSP is the sending-end phase voltage, ISP the sending-end phase
    current and VRP is the magnitude of the open-circuit receiving end phase
    voltage.


    (a) Given the parameter values in TABLE A and if the magnitude of the
    receiving-end line voltage VRL is measured as 154 kV when feeding
    a balanced load of 40 MVA at a power factor of 0.9, calculate the
    value of the sending-end phase voltage VSP and sending-end phase
    current ISP.

    [N.B. VSL = √3 × VSP and the total power in a three-phase load is
    given by P = √3VI cos θ.]

    (b) Hence or otherwise calculate the sending-end power and thus the
    power lost in the cable.

    (c) If the line is modelled by the Π-circuit of FIGURE 4(b), see if you
    can estimate the primary line coefficients R, L, G and C. The line is
    50 km long.


    A1= 0.8698
    A2= 0.03542
    B1= 47.94 Ω
    B2= 180.8 Ω
    C1= 0 S
    C2= 0.001349 S
    D1= 0.8698
    D2= 0.03542

    so

    a) P=sqrt(3)VI
    I=40000000/(sqrt(3)*154000) = 149.961 A
    Vrp=Vrl/sqrt(3) = 88912 V

    Using formulas given for ABCD.
    Vsp =84524.79 + j30262.16 = 89778.8∠19.70
    Isp = 130.44 +j125.252 = 180.83∠43.84

    b) p= sqrt(3)VI cos θ = sqrt(3)*154000*149.961*0.9 =36 MW

    Vsl = sqrt(3)*89778.8 = 155501.4*149.961*0.9 = 36.3MW

    350960.2 W lost in the cable.

    c) I don't have a clue?
     

    Attached Files:

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  2. jcsd
  3. Apr 14, 2014 #2
    Hi mate, Did you ever find the solution to part (C)? I cannot find for the life of me find any info on how to solve it. I have very similar answers to you for the rest of 4 but (c) has become a road block. Any help would be much appreciated, it almost feels like something is missing from the question!!
     
  4. Apr 18, 2014 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    I think the OP has come & gone. I'd like to look at this but can't open the .docx with my office 2003. Could you make a pdf file out of it? Or describe fig. 4(b) in words?


    I do know how to produce the ABCD parameters for a transmission line given R,L,G and C per-unit-length quantities.
     
  5. Apr 18, 2014 #4
    Abcd

    Hi mate, I am out of the country at the moment. But basiacally i have a table of values from A1,A2 through to D1, D2 but I cant find the formulas i need to find R,L,G,C.. The values in the table do have the relevant units ie ohms, S, etc with their respective imaginary part. The tranmission line is contructed as a pi circuit with Y being the 2 parallel resistors and z being the top one. It is a 50 hz 3 phase line with a 50 km length..

    This probably isn't much use but i didn't want you to think i was ignoring you, I appreciate the offer to help.
     
  6. Apr 19, 2014 #5

    NascentOxygen

    User Avatar

    Staff: Mentor

    Not especially convenient, but there are some online document viewers, e.g., http://www.docspal.com
     

    Attached Files:

  7. Apr 19, 2014 #6
    Thanks mate,

    I do want to stress however that i am not looking for the answer here. I just need a nudge in the right direction or better still a good resource to look at. Im not finding my books very useful...
     
  8. Aug 4, 2014 #7
    Hello everyone,
    I'm struggling with the same here, I don't know how to find R, L, G, C from given data. Has anyone got an idea?
     
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