co axial cylinder geometry

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
Reply to
pfaisalbe
Loading thread data ...
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.
Reply to
Paul Hovnanian P.E.
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.
Reply to
<nni/gilmer
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.
Reply to
Paul Hovnanian P.E.
----------------------------
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).
Reply to
Don Kelly
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.
Reply to
Paul Hovnanian P.E.
----------------------------
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
Reply to
Don Kelly

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.