Material rho rho_rel K1 ----------------------------------- aluminum 2.69 1.51 1.23 copper 1.77 1.000 1.000 lead 22. 12.4 3.52 silver 1.59 0.90 0.94 tin 11.5 6.49 2.55 Tungsten 5.5 3.11 1.76 Brass ~7.0 3.95 1.99 (depends on alloy and temp.) Phos-bronze 7.8 4.4 2.1 Bronze 17. 9.6 3.1 Units of rho are *1e-8 ohm*meter For any material, rho_rel = rho / 1.77e-8 ohm*meter rho_rel is unitless. Values extracted from "Industrial Electronics Reference Book", 1948.
------------------------------ Date: Sun, 29 Mar 1998 14:20:06 -0700 (MST) From: Tesla List
To: firstname.lastname@example.org Subject: Re: Primary and copper (fwd) Message-ID: Content-Type: TEXT/PLAIN; charset=US-ASCII ---------- Forwarded message ---------- Date: Sun, 29 Mar 1998 09:16:29 -0800 From: Jim Lux To: Tesla List Subject: Re: Primary and copper (fwd) > Or: > skin depth=sqrt(resistivity/(pi*frequency)/u0) meters > where u0=pi*4e-7. > For copper (resistivity=1.724e-8 Ohm-m): > skin depth=0.06608/sqrt(frequency) meters, or > skin depth=2602/sqrt(frequency) mils. > > This value is valid for flat sheets. For round wires the usual approach > is to consider a sheet wrapped around the wire, what results in: > resistance=(length/diameter)*8.31e-8*sqrt(frequency) (any length unit). > This works if the skin depth is much smaller than diameter/2, but fails > totally for thinner wire or lower frequency. > Do someone know the expression for wire resistance as function of frequency > when the wire radius is not much greater than the skin depth? > > Antonio Carlos M. de Queiroz > http://www.coe.ufrj.br/~acmq > This would be a pretty tricky problem, because the current in a particular point is affected not only by the layer on the side of the circle closest to it, but also across the diameter. The usual caveat: "valid for conductor radius >> skin depth" is a manifestation of this. The whole skin effect phenomena is due to the propagation of the EM radiation in the conductor (i.e. it takes time to propagate in), so some sort of solution of Maxwell's equations (but of course...) with the right boundary conditions should give it. I suspect that there is a fairly simple empirical equation around. At some frequency, the wire starts to act more like a transmission line than a conductor with a skin. The question would be at what frequency. FWIW here are some things from "Reference data for engineers" Rac equation is valid for conductors spaced at least 10 diameters from adjacent conductors. when spacing is 4 diameters, Rac is increased about 3%, were depth of penetration is small. it then gives an equation: factor = D * sqrt(f) * sqrt(Permeability * conductivity of Copper / conductivty) if factor is >40 then Rac/Rdc = .0960 * factor + 0.26 For an isolated wire or tubular conductor for T < D/8 or T1 < D/8 then Rac = FactorA * sqrt(f) / D * (permeability * conductivity of Copper/ conductivity) * 1E-6 A value of T (thickness of tubular conductor in inches) that just makes A = 1 indicates penetration of current below the surface. Therefore, T1 (depth of penetration) = 3.5 / sqrt(f) * sqrt( perm * condCu / cond) where T1 in inches. There is a table in the book of factor A, ranging from 1 to 2 for solid conductors. It would appear that for D*sqrt(f) * sqrt(perm * condCu/cond) < 3.0, the ac resistance is approximately equal to the dc resistance. permeability is relative to free space.