Roy Lewallen wrote:
: As it happens, you've caught me right in the midst of trying to program
: just that calculation. But it turns out to be a toughie. It involves a
: couple of Bessel functions -- which aren't a problem in themselves, but
: the problem is that when the tube diameter is the skin depth, the
: formulas I've come across so far require subtracting two huge numbers of
: limited accuracy (even if the wall itself isn't particularly thick in
: terms of skin depth). So the results I've gotten so far, even with
: double precision math, are garbage. I've been working at reducing and
: rearranging the equations, and using asymptotic approximations, but so
: far no joy.

: Anyway, I believe the phenomenon you mention. I've come across it a
: couple of places. A similar thing happens with the plating of wires,
: where a relatively poorly conductive plating of the right thickness
: actually improves the RF conductivity of the wire. The effect,
: unfortunately, is quite small, as for the case of wall thickness.

: Roy Lewallen, W7EL

: Reg Edwards wrote:
: Take an ordinary round solid conductor. For well known reasons at
: alternating currents it will have a higher resistance per unit length that
: at DC.
:
: Now, if a round hole is drilled down the centre of the conductor, ie., its
: highly-conductive center is removed, the AC resistance will be REDUCED.
: There are, of course, better ways of producing tubes.
:
: The effect is at a maximum when the wall thickness of the tube is about 1.6
: times skin depth in the material, ie., when an appreciable fraction of the
: current flows on the internal surface. The reduction in resistance is only a
: few percent and for a conductor of given dimensions it affects only the
: lower frequencies.
:
: It has something to do with internal inductance and the relative phases of
: the inside and outside surface currents.
:
: It is likely the effect is considered only when transmission line efficiency
: is of overriding importance, perhaps at power frequencies, the 'hole' in the
: conductor's center being occupied by a steel tensile strength member.
:
: My one and only reference book is Robert Chipman's "Theory and Problems of
: Transmission Lines", 1968, McGraw Hill, where the effect is described in a
: little more graphical detail than above. But the likelihood of obtaining a
: copy of this book is remote
: ----
: Reg, G4FGQ

The calculation of the skin effect current in a pipe is straightforward
but has a few numerical problems as Roy noticed. Please note that I have
not checked my results carefully. To get a handle on the problem it
is simplest to analyze something like a coaxial cable with a perfectly
conducting outer conductor and a hollow inner conductor made of a good
conductor like copper. You can then show that the outer conductor does
not change the current distribution measurably. Further since you know
the fields in coaxial cable pretty well it is easy to identify good
approximations.

Perfect coax is a waveguide operating in the transverse electromagnetic (TEM)
mode. With imperfect conductors, there must be an electric field along
the conductors to drive the current from Ohm's law J = sigma E. Therefore
this can no longer be a transverse electromagnetic mode since E is no
longer transverse. The current however is still just in the direction of
the axis, so the magnetic field will still be transverse and the mode
with imperfect conductors is transverse magnetic. The usual waveguide
simplifications can be made. The fields for a wave propagating in the
z direction have an e^{i k z} dependence (I assume e^{- i w t} time
dependence; substitute i=-j everywhere if you prefer e^{j w t} time
dependence). Everything can be calculated in terms of E_z the z component
of the electric field. In particular in SI units it satisfies a Helmholtz
equation,
1/rho d/drho rho d E_z/drho + gamma^2 E_z = 0,
where gamma^2 = mu epsilon w^2 -k^2 and the magnetic field is
H_phi = i epsilon/gamma^2 dE_z/drho.
E_z and H_phi must be continous at the inner conductor boundaries, while
E_z = 0 on the perfect conductor and is well behaved at the origin.
As Roy says the solutions of the Helmholtz equation are combinations of
two Bessel functions in each region. Matching the boundary conditions
then finds the coefficients and the appropriate value of k, and the
solution is complete.

Now let's make some approximations. Since we know the mode is close
to the TEM mode we know that in the vacuum regions, gamma^2 must be
very nearly zero since it would be exactly zero for the TEM mode. In
the conductor epsilon has an imaginary part proportional to sigma/w,
which gives a term enormously bigger than the k^2 term or the real
part. Therefore, we find that the k^2 terms can be dropped from the
calculation. Once that is done, the outer conductor no longer plays a
part in the solution, and we no longer have to consider it.

Look at solution in the interior of the inner pipe. Normally it would be
proportional to the Bessel function J_0(gamma rho), but we saw that gamma
in vacuum for a near TEM mode is very close to zero. J_0(x) goes like
1-x^2/2 for small x, so the electric field inside is nearly constant,
and the magnetic field is nearly zero. Matching boundary conditions at
the inner surface of the inner conductor therefore requires to a good
approximation a nonzero E_z with a zero derivative. The current density
is simply J = sigma E_z, so the form of the current inside the pipe is
J(rho) = A J_0[(1-i) rho/delta] + B N_0[(1-i) rho/delta]
where J_0 and N_0 are the Bessel and Neuman functions (N_0 is also
written Y_0 sometimes) and delta is the skin depth. The coefficients A
and B must be chosen so that the rho derivative is zero on the inside
of the pipe, and the integral of J gives the total current desired. The
power loss is then the integral of |J|^2/2*sigma.

As Roy noticed, for a reasonable size inner conductor, rho is
always an enormous number of skin depths so the arguments of the Bessel
functions are large, and the Bessel functions have large exponentially
growing parts so that it can be numerically difficult to calculate the
coefficients A and B. It is therefore best to use the large argument
asymptotic expressions:
J_0(x) = sqrt(2/pi x) cos(x-pi/4)
N_0(x) = sqrt(2/pi x) sin(x-pi/4)
and isolate the divergent part. However, in the usual case, we can ignore
the curvature of the pipe to a good approximation. That is, if the pipe
is many skin depths thick, the result will be the same as a solid pipe,
which is B=0. If on the other hand it is only a few skin depths thick,
then rho will change very little across the pipe, and we can replace
the rho in the square root with the value at the outer radius. In that
case, it is easy to satisfy the boundary conditions and the result is
approximately:
J(rho) = I [(1-i)/(2 pi b delta) ]
[cosh((rho-a)(1-i)/delta)/sinh((b-a)*(1-i)/delta)]
where b and a are the outer and inner radii of the conductor.
The total current is 2*pi*b int_a^b drho J(rho) = I by construction.
Calculating the effective resistance from the power loss gives:

R_effective = R_infinity *(sinh(2*t)+sin(2*t))/(cosh(2*t)-cos(2*t))

where R_infinity is the resistance for a solid pipe, and t is the
pipe thickness in skin depths = (b-a)/delta,
R_infinity = 1/(2*pi*sigma*delta*b). The minimum is at t=pi/2 which
is approximately 1.6 skin depths as Reg mentioned. It takes the value
R_infinity*tanh(pi/2) or about 0.92*R_infinity. For t3, the difference
between the pipe and the solid wire is less than a half a percent.

I would be pleased to hear from Reg if the plot of the equation above
agrees with that in his reference.

73 Kevin