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!

Wave propagation velocity?

  1. Jan 10, 2015 #1
    Hello, I'm stuck on how to find the propagation velocity of the signal in the Line as stated in Question C on the attached image below.

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

    Table one (in the attached image) shows typical values of Z and Y for an overhead line and underground cable. Please note that this is an engineering problem, thus all complex numbers are expressed in a+jb rather than a+ib.

    2. Relevant equations
    Using simplifying assumptions calculate [tex]Z_0, \gamma[/tex] and the wave propagation velocity for each case.


    3. The attempt at a solution

    So far I've only attempted the solution for the overhead line, I've found [tex]Z_0[/tex] and [tex]\gamma[/tex], however, I'm not entirely sure how to find the wave propagation velocity. I've attempted it, but to no avail.

    [tex]Z_0 = \sqrt{\frac{Z}{Y}} = \sqrt{\frac{0.03 + j0.38}{0 + j5{x10^{-6}}}} = \sqrt{76{x10^3} \angle-4.5^o}[/tex]

    [tex]= 275.7\angle-2.25^o \Omega = 275.5 - j10.82 \Omega[/tex]

    [tex] \gamma = \sqrt{ZY} = \sqrt{(0.03 + j0.38)(0 + j5x10^{-6})} = \sqrt{(0.4 \angle85.5^o)(5x10^{-6} \angle90^o)}[/tex]

    [tex]= 1.41x10^{-3} \angle{87.75^o} = 5.5x10^{-5} + j1.41x10^{-3}[/tex]

    At this point I'm not sure how to get the wave propagation velocity. I've tried making the assumption that propagation velocity [tex] U = \frac{1}{\sqrt{LC}}[/tex] the only problem here is that this is for a lossless line and I'm not entirely sure if this assumption is allowed considering this isn't really a lossless line. In any case I get:

    [tex] U = \frac{1}{\sqrt{\frac{(0.38)(5x10^{-6})}{1000}}} = 725.5x10^3 ms^{-1}[/tex] Note that I divided by 1000 because the stated values in the table are given per km.

    Am I on the right track here or am I way off?
     

    Attached Files:

    Last edited: Jan 10, 2015
  2. jcsd
  3. Jan 10, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Aren't the imaginary parts of Z and Y equal to ωL and ωC, respectively? So, for example, 0.38 Ω/km is the value of ωL rather than the value of L. Or am I mistaken?

    If you include the units given for Z and Y in the table when you calculate U, do you get m/s?
     
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: Wave propagation velocity?
Loading...