1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    (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


    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:

    • fig.docx
      File size:
      154.6 KB
  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

    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


    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted