Capacitance of a Transmission line

[jaus tail]

Homework Statement

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[/B]

Homework Equations

C_AN = 2(3.14)(permittivity)/Log_e [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

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 C_AN = 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 C_AB instead of line to neutral capactiance.

 

[anorlunda]

I guess they've used formula for line capacitance CAB instead of line to neutral capactiance.

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.

 

[rude man]

Homework Statement

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[/B]

Homework Equations

C_AN = 2(3.14)(permittivity)/Log_e [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

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 C_AN = 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

The book answer is off by an order of magnitude.

I guess they've used formula for line capacitance C_AB instead of line to neutral capactiance.

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.

 

[CWatters]

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?

 

[rude man]

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?

I think the same ...

 

[Delta2]

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?

 

[jaus tail]

But why should the factor of 2 not be there?
In the book they've derived the formula as:

So as per the book formula the factor of 2 must be there.

 

[Babadag]

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

 

[rude man]

But why should the factor of 2 not be there?
In the book they've derived the formula as:
View attachment 217463
So as per the book formula the factor of 2 must be there.

Your book clearly derives C_AB 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.

 

[CWatters]

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.

 

[jaus tail]

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.

 

[Babadag]

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.

 

[Babadag]

So, jaus tail, you said: “Book answer is 3.48 micro farad”. It could be 3.48E-01 ?

 

[jaus tail]

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

 

[rude man]

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

Your posted book page clearly states C_AB = πε₀/ln(d/r) and is correct.
Assimilate post 9!

 

[Tom.G]

Where does the divide by 2 part come from?

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?

 

[jaus tail]

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.

C_AB = 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.

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.

 

[jaus tail]

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:

 

[jaus tail]

I found this from google:

Here also they haven't divided by 2 for neutral wire.

 

[Tom.G]

I'm still not getting it

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

 

[jaus tail]

I found the book online also. Here is the link
http://www.brainkart.com/article/Capacitance-of-a-Single-Phase-Two-Wire-Line_12362/
I have hard copy of same book that is there in link.

 

[rude man]

Speaking only of line-to-line capacitance, the answer is what I said in post 15, i.e. C = πε/ln(d/r). This can be derived by invoking two Gaussian surfaces, one for each charged wire, with superposition for the two wires, each charged with charge Q.

I have further confirmed this by computing inductance L, setting Z₀ = √(L/C) which gives Z₀ = 120 ln(D/r) and which agrees with my radio engineers' handbook.

But if one of the wires has zero charge, which is how you can think of a "neutral", the capacitance is doubled since the potential difference between the two wires would now be halved and C=Q/V.

 

[rude man]

There is a problem with the word 'neutral'. A real 'neutral' carries no current, whether it be a third grounded wire or a fictional plane. In my (older) house the wiring is 2 wires, not 3, & one of them (the white one) is called 'neutral' but it carries all the return current & so is in every sense a 2-wire system with no real 'neutral', and the inter-wire capacitance is πε/ln(D/r). It's called 'neutral' because it's connected to Earth ground so its potential is zero.

 

[rude man]

I found the book online also. Here is the link
http://www.brainkart.com/article/Capacitance-of-a-Single-Phase-Two-Wire-Line_12362/
I have hard copy of same book that is there in link.

Their integrals (limits from r to ∞) do not exist! They are sheer nonsense.

 

[jaus tail]

Thanks.