Electrical Engineering - co axial cylinder geometry

Author

co axial cylinder geometry

pfaisalbe@gmail.com

2007-08-14, 1:25 pm

Hi

I am using a co axial cylinder geometry. I wish to calculate the inner
diameter of the conductor in such a way that it will not cause air
breakdown.outer diameter is 3.75 cm.Which formula i have to use

Thanks

Paul Hovnanian P.E.

2007-08-16, 3:25 am

pfaisalbe@gmail.com wrote:
>
> Hi
>
> I am using a co axial cylinder geometry. I wish to calculate the inner
> diameter of the conductor in such a way that it will not cause air
> breakdown.outer diameter is 3.75 cm.Which formula i have to use
>
> Thanks

Hoping that this isn't homework, here goes:

Given a coaxial capacitor (I'm assuming that's what you are referring
to) with length much greater then diameter, outside diameter of inner
cylinder = a, inside diameter of outer cylinder = b and voltage across
the plates (inner and outer cylinder) = V

Ea = V/( a * ln ( b/a ) )

which is the voltage field strength at the surface of the inner
conductor. Assuming you know the system voltage V and the maximum field
strength (humidity and other factors will affect his for a practical
problem), I'll leave it up to you to solve for a.

--
Paul Hovnanian mailto:Paul@Hovnanian.com
------------------------------------------------------------------
Opinions stated herein are the sole property of the author. Standard
disclaimers apply. All rights reserved. No user serviceable components
inside. Contents under pressure; do not incinerate. Always wear adequate
eye protection. Do not mold, findle or sputilate.

2007-08-16, 9:25 am

The "rule" is this class of homework problems was to understand what was
"conserved" (i.e.: constant).

In this case we have two "constants." The first, is the voltage drop
between the two electrodes and that is just the integral of the E-filed
between the two equi-potential surfaces. The second is the "flux" which is
the surface integral of E-field. The "surface" would be an imaginary
cylinder between the two electrodes.

When I "studied" this 45 years ago, I actually enjoyed solving these
problems.

"Paul Hovnanian P.E." <paul@hovnanian.com> wrote in message
news:46C3CD4B.FD6D7C41@hovnanian.com...
> pfaisalbe@gmail.com wrote:
>
> Hoping that this isn't homework, here goes:
>
> Given a coaxial capacitor (I'm assuming that's what you are referring
> to) with length much greater then diameter, outside diameter of inner
> cylinder = a, inside diameter of outer cylinder = b and voltage across
> the plates (inner and outer cylinder) = V
>
> Ea = V/( a * ln ( b/a ) )
>
> which is the voltage field strength at the surface of the inner
> conductor. Assuming you know the system voltage V and the maximum field
> strength (humidity and other factors will affect his for a practical
> problem), I'll leave it up to you to solve for a.
>
>
> --
> Paul Hovnanian mailto:Paul@Hovnanian.com
> ------------------------------------------------------------------
> Opinions stated herein are the sole property of the author. Standard
> disclaimers apply. All rights reserved. No user serviceable components
> inside. Contents under pressure; do not incinerate. Always wear adequate
> eye protection. Do not mold, findle or sputilate.

Paul Hovnanian P.E.

2007-08-16, 5:25 pm

nni/gilmer@nni.com wrote:
>
> The "rule" is this class of homework problems was to understand what was
> "conserved" (i.e.: constant).
>
> In this case we have two "constants." The first, is the voltage drop
> between the two electrodes and that is just the integral of the E-filed
> between the two equi-potential surfaces. The second is the "flux" which is
> the surface integral of E-field. The "surface" would be an imaginary
> cylinder between the two electrodes.
>
> When I "studied" this 45 years ago, I actually enjoyed solving these
> problems.

I don't think the flux is constant for varying cylinder geometry. It
would depend on the capacitance between them and the resulting charge
stored for a given voltage. The capacitance varies with the geometry.

--
Paul Hovnanian mailto:Paul@Hovnanian.com
------------------------------------------------------------------
Why are so many towns named after water towers?

Don Kelly

2007-08-19, 8:25 pm

----------------------------
"Paul Hovnanian P.E." <paul@hovnanian.com> wrote in message
news:46C4A42C.380B58B3@hovnanian.com...
> nni/gilmer@nni.com wrote:
>
> I don't think the flux is constant for varying cylinder geometry. It
> would depend on the capacitance between them and the resulting charge
> stored for a given voltage. The capacitance varies with the geometry.
>
> --
> Paul Hovnanian mailto:Paul@Hovnanian.com
> ------------------------------------------------------------------
> Why are so many towns named after water towers?

The "flux" would be constant as E depends on rho/r where rho is the charge
per unit length and r is the radius and the surface area per unit length
would be proportional to the radius. It all works out (Shades of potential
coefficients for capacitance of power lines parallelling the inductance
"flux based" model).
--

Don Kelly dhky@shawcross.ca
remove the X to answer

As for the towers -this brings up the question "Why did Shakespeare named
many of his characters after sidings on the (defunct) Kettle Valley
Railway?". Can we assume that he read Nostrademus as fervently as the
tabloids?

Paul Hovnanian P.E.

2007-08-19, 9:25 pm

Don Kelly wrote:
>
> ----------------------------
> "Paul Hovnanian P.E." <paul@hovnanian.com> wrote in message
> news:46C4A42C.380B58B3@hovnanian.com...
>
> The "flux" would be constant as E depends on rho/r where rho is the charge
> per unit length and r is the radius and the surface area per unit length
> would be proportional to the radius. It all works out (Shades of potential
> coefficients for capacitance of power lines parallelling the inductance
> "flux based" model).

The capacitance of a coaxial line per unit length is

C = 2PIe / ln ( B/A )

where B is the outer cylinder radius and A is the inner cylinder radius.

For a fixed V across the plates and a fixed B, as A increases so does C.
As C increases, the charge per unit length increases (for the same V).
Since we assumed that B, and therefore the outer cylinder surface area
is constant, the charge per unit area and therefore flux increases.

--
Paul Hovnanian mailto:Paul@Hovnanian.com
------------------------------------------------------------------
"There's something vewy scwewy going on awound here." -- Elmer Fudd

Don Kelly

2007-08-21, 3:25 am

----------------------------
"Paul Hovnanian P.E." <paul@hovnanian.com> wrote in message
news:46C8F9DD.D1592459@hovnanian.com...
> Don Kelly wrote:
>
> The capacitance of a coaxial line per unit length is
>
> C = 2PIe / ln ( B/A )
>
> where B is the outer cylinder radius and A is the inner cylinder radius.
>
> For a fixed V across the plates and a fixed B, as A increases so does C.
> As C increases, the charge per unit length increases (for the same V).
> Since we assumed that B, and therefore the outer cylinder surface area
> is constant, the charge per unit area and therefore flux increases.
----------------
Look at "gilmers" statement. If you have coaxial cylinders with a voltage
between them of V, then you can postulate a "flux" which happens to be the
charge per unit length on the inner conductor[Flux=2pirE =2pir(rho/2pir)
=rho] . At any equipotential surface where the voltage is less than V, the
field between this surface and the outer surface is the same as if the
charge rho is (somewhat more thinly spread) on this intermediate surface. No
more than that. Certainly, if the dimensions A or B change, for a fixed V,
the capacitance changes and the flux does change .
I personally would not use this approach. I would rather simply find the
charge per unit length knowing the dimensions and the voltage and once this
is known, find the field. This is what was done to get the field expression
that you gave. However, to find capacitance, the latter step is not needed
as you are looking at charge/voltage (which is independent of "flux").
[ rho=2pieV/ln B/A so rho/V=2pie/ln B/A =capacitance per unit length].
Gilmer's flux reduces to the charge per unit length of the inner conductor.

If there was an "a priori" assumption that B was fixed and A variable,
(other than A<B) I certainly missed it. Certainly there is a change in
capacitance and "flux" as the dimensions change. I don't think that gilmer
was considering changes in "flux" or capacitance with dimensional changes.
Possibly the "flux" approach was what was used in some places rather than
simply charge per unit length. Teaching approaches vary depending on when
and where.

Don Kelly dhky@shawcross.ca
remove the X to answer

>
> --
> Paul Hovnanian mailto:Paul@Hovnanian.com
> ------------------------------------------------------------------
> "There's something vewy scwewy going on awound here." -- Elmer Fudd

LinkBot