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Capacitance of a Transmission line

  1. Dec 26, 2017 #1
    1. The problem statement, all variables and given/known data
    A single phase transmission line has 2 parallel conductors 1.5 metres apart. The diameter of each conductor being 0.5 cm. Calculate the line to neutral capacitance for a line 80 km long



    2. Relevant equations
    CAN = 2(3.14)(permittivity)/Loge [D/r] This gives capacitance per metre
    Where D is the distance between conductors and r is the radius.
    Multiplying above answer into length of line gives capacitance

    3. The attempt at a solution
    D = 1.5 metres = 150 cm
    r = radius = 0.5 / 2 = 0.25 cm
    Substituting in above equation I get CAN = 8.69 * 10-12 F / m
    Multiplying this with 80 km or 80,000 m I get 0.69 micro farad.
    Book answer is 3.48 micro farad

    I guess they've used formula for line capacitance CAB instead of line to neutral capactiance.
     
  2. jcsd
  3. Dec 26, 2017 #2

    anorlunda

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    What is your definition of line capacitance CAB and your definition of line to neutral capacitance?

    Note that the problem did not give you the height of the line above the ground, so it can't be the line to ground capacitance.
     
  4. Dec 26, 2017 #3

    rude man

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    The book answer is off by an order of magnitude.
    That is correct. I think they assumed the "neutral" is the second wire. They did specify a single-phase system which will have 2 wires, not 3.
     
  5. Dec 26, 2017 #4

    CWatters

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    I'm a bit rusty but..

    If the question is about a phase and neutral pair then I don't think there should be a factor of 2 in the equation in the OP?
     
  6. Dec 26, 2017 #5

    rude man

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    I think the same ...
     
  7. Dec 26, 2017 #6

    Delta²

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    Still, without the factor of 2 the book answer is ahead by a factor of 10 (if I did the calculations correctly). MAYBE the line length is 800km instead of 80km???
     
  8. Dec 26, 2017 #7
    But why should the factor of 2 not be there?
    In the book they've derived the formula as:
    upload_2017-12-27_8-37-2.png
    So as per the book formula the factor of 2 must be there.
     
  9. Dec 27, 2017 #8
    upload_2017-12-27_9-12-5.png
    I’ll try not to boring you with this theory and I’ll endeavor to be as brief as I can.

    Let’s define:

    E(x)=electrostatic field at radius x from the first conductor center

    D(x)=the electric flux density[displacement] at the same radius x

    Q=electric charge

    V=voltage between the two wires

    Then :

    Q=∫D.ds=D(x).2.π.x.length

    D(x)=ε.E(x)

    E(x)=Q/ ε /2/π/x/length

    V=∫E(x).dx|x=r to x=Dist|

    Since the electric field is more complex in the inner part of the conductor a correction it is required. So instead of r we use r’=e^(-1/4)r'=0.778801.r

    V=Q/length/ ε /2/π.ln(Dist/r') or

    C/length=Q/V= 2.π. ε/ln(Dist/r') [F/unit length]

    For more information see-for instance:

    http://alphard.ethz.ch/Hafner/Vorles/PhysicalMod/chapter1.pdf
     
  10. Dec 27, 2017 #9

    rude man

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    Your book clearly derives CAB without the factor of 2.
    The confusion arives because your problem uses "neutral" as the second, not the third, wire. There is no third wire.
     
    Last edited: Jan 2, 2018
  11. Dec 27, 2017 #10

    CWatters

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    The example on that website calculates the capacitance between two wires that have opposite polarity eg when one is positive the other is negative. It then modifies that equation to give one for the capacitance between either wire and a virtual neutral mid way between them.

    Your situation is different. In your case one of the wires is neutral not the opposite polarity. So the neutral point in your case is physically twice as far away from the line as their example.
     
  12. Dec 27, 2017 #11
    Can I assume that the book answer is wrong?
    I thought 0.7788 factor was for inductance and not for capacitance. That's what the book says.
     
  13. Dec 27, 2017 #12
    You are right, jaus tail. My mistake. Thank you. Since the electric field it is E=ro.J and the current is zero in the conductor core due to skin effect- in a.c.- the electric field is also zero inside the wire.
    On the other hand this reasoning is valid only for a single charged wire. If there are two opposite charged wires the electric field is different. See Figure 1.4. Graphical representation of the electrostatic field from the above mentioned article.
     
  14. Dec 31, 2017 #13
    So, jaus tail, you said: “Book answer is 3.48 micro farad”. It could be 3.48E-01 ?
     
  15. Dec 31, 2017 #14
    Where does the divide by 2 part come from? I'm getting answer as 0.69 micro farad. I've pasted the book page in post#7
     
  16. Jan 2, 2018 #15

    rude man

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    Your posted book page clearly states CAB = πε0/ln(d/r) and is correct.
    Assimilate post 9!
     
  17. Jan 2, 2018 #16

    Tom.G

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    The book gives the distance between "A" and "B" conductors and shows an "N" conductor half way between them. It then asks for Capacitance between a main conductor (A or B) and "N".

    What happens to the capacitance between conductors when the distance between them changes?
     
  18. Jan 2, 2018 #17
    The capacitance increases when the distance between plates/wires reduces.

    I'm still not getting it. This is what the book page says:
    Consider a single phase overhead transmission line consisting of 2 parallel conductors A and B placed d metres apart. Let their respective charge be +Q and -Q C/metre length.
    upload_2018-1-3_9-0-45.png
    CAB = 3.14(permittivity)/log(d/r) F/m

    Now for Capacitance to neutral:
    Book says:
    Above equation gives Capacitance between the conductors of a two-wire line. Often it is desired to know the capacitance between one of the conductor and a neutral point between them. Since the potential of the mid-point between conductors is zero, the potential difference between each conductor and ground or neutral is half the potential difference between the conductors. Thus the capacitance to ground or capacitance to neutral for the two-wire line is twice the line-to-line capacitance as shown in the figure below.
    upload_2018-1-3_9-7-0.png
    There is no separate neutral wire. The Neutral is a point at zero potential and Can = 2Cab. In the main question posted in post#1 they are asking to calculate line to neutral capacitance. And they are saying a single phase line has 2 charged conductors. So one would be +Q and the other -Q. Just like in diagram above. I'm really confused. There is one solved example but that requires to calculate line to line capacitance and they've substituted formula Cab.
     

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  19. Jan 2, 2018 #18
    Oh I guess you'll are saying that in the question in post#1 one conductor is charged whereas other is neutral. But how? There is one solved example like this:
    upload_2018-1-3_9-22-43.png
     
  20. Jan 2, 2018 #19
    I found this from google:
    upload_2018-1-3_9-27-3.png
    Here also they haven't divided by 2 for neutral wire.
     
  21. Jan 3, 2018 #20

    Tom.G

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    Hmmm. Me neither. I hope someone here can come up with a good explanation. Sorry. :frown:
     
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