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!

50Hz HV Transmission line question

  1. Feb 3, 2017 #1

    David J

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    This question was raised on the 11th January 2015 but that thread appears to be no longer active now. There are 2 parts to the question (a) and (b). Initially I am only trying to work on (a)

    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.gif

    mimetex.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.gif and mimetex.gif and hence the power mimetex.gif absorbed from the supply by the transmission line on open circuit.

    The Table C values are :-

    ##A_1=0.8698##
    ##A_2=0.03542##
    ##B_1=47.94\Omega##
    ##B_2=180.8\Omega##
    ##C_1=0 S##
    ##C_2=0.001349 S##
    ##D_1=0.8698##
    ##D_2=0.03542##

    2. Relevant equations

    ##V_s=V_R\left(A_1+jA_2\right)+I_R\left(B_1+jB_2\right)##

    ##I_s=V_R\left(C_1+jC_2\right)+I_R\left(D_1+jD_2\right)##

    3. The attempt at a solution

    ##V_s=V_R\left(A_1+jA_2\right)+I_R\left(B_1+jB_2\right)##

    The receiving end is open circuit so there cannot be any current flow so ##I_R=0##

    ##V_s=88900\left(0.8698+j0.03542\right)+0\left(47.94\Omega+j180.8\Omega\right)##

    ##V_s=\left(77325.22+j3148.84\right)+0##

    So ##V_S=77325.22+j3148.84## or ##77389.31\angle2.33^0##

    ##I_s=V_R\left(C_1+jC_2\right)+I_R\left(D_1+jD_2\right)##

    ##I_s=88900\left(0+j0.001349\right)+0\left(0.8698+j0.03542\right)##

    So ##I_s=j119.93##

    To find the absorbed power

    ##P_{SO}=V_SI_S##

    So ##\left(77325.22+j3148.84\right)\left(j119.93\right)##

    So ##P_{SO}=-377640.38+j9273613.63##

    In the post from 2015 this same calculation was made and I have shown it below

    Now the power should be calculated by

    P=Real{VsIs*}

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

    P=377565w


    In the earlier post answer they only included the real value (P=377565w) which I suppose is ok so my calculation, in that case, is pretty close apart from the fact that my value is negative.
    The question is asking for a value of power in watts so do I just assume that even though the calculation appears negative it is actually positive ??

    My question is why is my value different ?? Have I made a mistake somewhere with this calculation ??

    Appreciated as always


     
  2. jcsd
  3. Feb 3, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    Remember that when calculating power using complex values for voltage and current that you should use the complex conjugate of the current :wink:
     
  4. Feb 3, 2017 #3

    David J

    User Avatar
    Gold Member

    ok so the complex value of current is ##0+j119.93## so the complex conjugate of ##0+j119.93## would be ##-j119.93## ????

    In that case

    ##\left(77325.22+j3148.84\right)\left(j119.93\right)##

    becomes

    ##\left(77325.22+j3148.84\right)\left(-j119.93\right)##

    which gives me ##377640.38 + (-j9273613.63)##

    am I heading in the right direction with the above ??? It makes more sense but I am still not getting the same values as others got and also, if this is correct then what do I do with the imaginary part, ##-j9273613.63## ???
     
  5. Feb 3, 2017 #4

    gneill

    User Avatar

    Staff: Mentor

    Yes.
    Your result looks okay to me. Given the original three significant figure of the received voltage you should be good with three figures for the power. I'd attribute any differences to rounding/truncation of intermediate values. If you can supply a link to the original thread I can take a look to see what happened there.

    The imaginary part of the power represents reactive power. While the power supply has to provide this power to its load, the load cannot convert it to useful work (only real power does work). Power companies don't like high reactive power components because it represents wasted "effort" to supply it for no benefit to the customer (or themselves). The associated concept is "power factor" which represents how much a given load skews the phase of the current waveform from the voltage waveform. You don't have to do anything with this component of the power in the context of the current problem.
     
  6. Feb 4, 2017 #5

    David J

    User Avatar
    Gold Member

    Good Morning and thanks for the info. I worked out a final answer of 378000 watts or 378KW

    The initial post I was looking at can be reached via the link below. It was created on the 11th January 2015. I think it may actually still be active but there are no recent updates to any of the questions

    https://www.physicsforums.com/threads/50-hz-high-voltage-transmission-line.791633/

    Thanks again for your help with this
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: 50Hz HV Transmission line question
Loading...