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I have a three phase power line that uses flat configuration for the phase conductors, $a$, $b$ and $c$. The situation is shown in fig 1.

My goal is to calculate the charge per unit length, $\lambda$, for each phase conductor: $\lambda_a$, $\lambda_b$, $\lambda_c$, given in coulomb/m. From here the goal is to calculate the electric field as shown in this article: (waset.org/publications/4137/electrical-field-around-the-overhead-transmission-lines).

The article uses a formula, shown in table 1, that uses AC resistivity of the conductors to calculate the charge per unit length. Is the formula correct, and why/why not?

I need to find the charges to calculate the electric field around the power line... Thank you! (:

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Elaboration / ANSWER TO FLORIS:

I was looking at the exact article as the one you found. I should have linked it here. The root to my question is that I’m trying to calculate the electric field, $E_{tot}(x,y)$, some distance away from the power line, using the same analytical procedure as in the article.

To try and answer your questions:

1) I know the dimension of my conductor, so that should help me find my own $Y$, with some value close to the one in TABLE 1.

2) I guess I don’t care about the surface field of the conductors (if that was what you were asking?), due to the fact that I’m trying to find the field at some distance away from the power line.

3) The formula in the article does not include voltage, only charge, so I’m trying to find the same charge per length, $\lambda$

4) I'm (as they’re doing in the article) using the method of image charges for the «ground plane»

5) Hm… The dimension analysis is a good point, strange…

My main question is really:

Do you think the approach for calculating the E-field in the article is correct? If so – will $\lambda$ be a constant depending on the conductor dimension only? And if so – will $\lambda$ be possible to calculate, or do I need to find this «AC resistivity» from some conductor data table?

Furthermore, why is this technique only applicable for calculating the electric field close to ground? An lastly, it seems that they calculate the E-field from only one of the three conductors, $L_i$. Does that mean you find the type of equation as in $(5)$, but for all three conductors: $E_1$, $E_2$ and $E_3$, followed by summing them using superposition: $E_{tot}(x,y) = E_1 + E_2 + E_3$, to find the final total E-field?

Equation (1) and (4) also seems to be slightly wrong, where $r_1$ should be $ \sqrt{x^2+\left( \mathbf{h-y} \right)^2}$ and $E_y$ should be $ -d \phi / d \mathbf{y} $, respectively.

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closed as off-topic by peterh, Kyle Kanos, John Rennie, Jon Custer, SchrodingersCat Jun 7 '17 at 13:24

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  • $\begingroup$ Do you have the dimension of the conductor? Do you care about the surface field, or the field at some distance? The charge leads to the potential - but if you already know the voltage, why do you need the charge... dimension of the conductor is key. $\endgroup$ – Floris Jun 5 '17 at 16:18
  • $\begingroup$ Incidentally - if you do a dimensional analysis of your expression you find that the dimensions of $\lambda$ are not $C/m$. That should be a red flag. $\endgroup$ – Floris Jun 5 '17 at 16:53
  • $\begingroup$ Your table appears in this article. The configuration there is a little different - specifically they are looking for the electric field at 1 m above ground, and they assume that ground is a "ground plane" meaning that they include method of images to compute the electric field. What are you really asking? You can't "just" apply the equations from one study in another without clearly understanding how the assumptions do / do not apply. $\endgroup$ – Floris Jun 5 '17 at 18:31
  • $\begingroup$ Thank you very much, I appreciate you taking the time! Sorry for being far from precise. My answer is a little too long to post here, so I posted it at the bottom of my initial question, under "ANSWER TO FLORIS". $\endgroup$ – Jon L Jun 5 '17 at 21:09
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Now that you have elaborated your question, we can try to help you answer it.

First - the field MUST be a function of the voltage - an expression that doesn't contain the voltage cannot be calculating the field (in fact, I think that they are computing the capacitance per unit length, not the charge - and that you would need to multiply by the voltage to get the charge).

However - if you instead consider that you have a line at a certain potential, then computing the electric field is not very hard at all. In the case you are considering you have three lines with their potential out of phase by 120° - you need to take this into account when you do your vector sum.

The electric field due to a line of charge is

$$E = \frac{\lambda}{2\pi\epsilon_0 r}$$

And the potential is

$$V = \frac{\lambda}{2\pi\epsilon_0}\log{\frac{r_0}{r}}$$

If you know the diameter $r_0$ of the conductor, and the (peak) potential (that is, the voltage running on the power lines) then you can calculate the charge per unit length from this expression.

With the earth as a ground plane, you have a total of 6 "wires" to consider, with the charges out of phase.

Do you think you can now do the calculation?

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  • $\begingroup$ Thank you Floris – the phase shift is a good point that I need to consider. I think my fundamental understanding of the problem is not there yet, not your fault. If I were to try and do this, I would have used the E-field equations (3) and (4), and calculated the $\lambda$ from your formula, $$ \lambda = \frac{V \times 2 \pi \epsilon_0}{ln \left( \frac{r_0}{r} \right)}, $$ where $r_0$ is the conductor diameter, and $r$ is the distance to some other conductor/image conductor (?). $\endgroup$ – Jon L Jun 6 '17 at 12:17
  • $\begingroup$ I could then find $$\hat{E}_{tot}sin(2\pi ft + \Phi) = \hat{E}_{a}sin(2\pi f t + \Phi_a) + \hat{E}_{b} sin(2\pi f t + \Phi_b) + \hat{E}_{c} sin(2\pi f t + \Phi_c),$$ where $a$, $b$ and $c$ are the conductor subscripts. The phase shift in $E_{a}$ would be introduced through the total line charge on conductor a, $\lambda_a$. The real problem would then have been to calculate $\lambda_a$, who include all real and image conductors. $\endgroup$ – Jon L Jun 6 '17 at 12:17
  • $\begingroup$ It could look something like $$ \lambda_a (t) = \lambda_{aa}(t) + \lambda_{ab}(t) + \lambda_{ac}(t) + \lambda_{aa’}(t) + \lambda_{ab’}(t) + \lambda_{ac’}(t), $$ where $’$ is the subscript for an image conductor. The same would have had to be done for conductor $b$ and $c$. Is any of my thinking correct? $\endgroup$ – Jon L Jun 6 '17 at 12:18
  • $\begingroup$ Your sum for the E field needs to be over all 6 conductors. Each conductor has the same $\lambda$ but with a different time dependence (phase). I wouldn't try to sum the charges - instead, sum the electric fields (in vector form). $\endgroup$ – Floris Jun 6 '17 at 13:03
  • $\begingroup$ Okey, I will try that. To find $\lambda$: $r_0$ is the diameter of the conductor - is $r$ just the height above ground? $\endgroup$ – Jon L Jun 6 '17 at 13:31

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