It should be noted, however, that the self-induction of either circuit is not materially altered by these arrangements, being dependent upon the average distance between the two wires of each circuit, and not upon the presence of the other circuit. "Skin Effect" is the name given to the phenomenon according to which alternating currents tend to have a greater density near the surface than they have along the axis of a conductor. If we imagine a wire to be made up of elementary filaments parallel to its length, it is evident that the central or axial filament will be surrounded by a greater number of magnetic lines than an element at the surface, since each filament tends to set up lines around. itself. This fact produces no effect upon a steady current after it has been established, there being no variation in the number or position of the lines. Hence a steady current has a perfectly uniform distribution throughout the entire cross-section of a conductor having a uniform specific resistance. In the case of an alternating current, the additional lines of force that inclose the filaments near the axis are reversed twice during each period, the effect being to generate a greater back E.M.F. of self-induction than for the outer filaments of the wire. Consequently the current density is less near the axis than it is near the surface. With high frequency and large conductors this action may be so great that there is actually a back flow of current at or near the axis. But with ordinary sizes of wire and frequencies, the effect is small. This "skin effect" is generally treated as an increased apparent resistance of a conductor, being called its virtual resistance; and since it involves a larger drop in voltage and a greater loss of energy, it is practically the same as true resistance. In Fig. 107, which shows graphically the values of virtual resistance, R is the apparent or virtual resistance for a given alternating current, R is the true ohmic resistance of a copper conductor at 20° C. (68° F.), A is the area of cross-section of the latter in circular mils, and f is the frequency. d A conductor one inch in diameter has a cross-section of one million circular mils, so that at a frequency of 100, the product of A and f, is 100,000,000. Referring to Fig. 107, we find that Ra Ra 1.21; that is, the virtual resistance is 21 per cent greater than the true resistance, consequently this is too large a = conductor to use at that frequency. On the other hand, No. 0 0 10 20 PRODUCT OF A & F IN MILLIONS 80 90 100 30 the conductor may be made hollow or in the form of a flat bar, the With iron conductors the virtual resistance is much greater than with copper or other non-magnetic metal; but iron is not often. used to carry alternating currents, and the exact value of the permeability is not easily determined,* so that formulas will not be given. Capacity of Overhead, Underground, and Submarine Conductors. *Merritt, Physical Review, November, 1899. trical conductors, and almost all cases will come under one of the following heads: Case 1. Insulated conductor with metallic protection; for example, an iron-armored submarine cable, or a lead-covered underground conductor, having the metallic sheathing connected with the earth, which is the usual condition. Case 2. Single aërial conductor with earth return. Case 3. Metallic circuit consisting of two parallel aërial conductors. In the following expressions, K is the capacity in farads, k is the dielectric constant, D is the internal diameter of the metallic covering, d is the diameter of the conductor, h is the height above the ground of an aerial wire, and A is the interaxial distance between two parallel wires. In cases 2 and 3, the medium being air, k = 1, and does not appear in the equations. This assumes that the conductors are bare; but if they are covered with insulation of ordinary thickness it would only slightly increase the capacity, k being greater than 1 for insulating materials. The proximity of other conductors may increase the capacity considerably, but their effect is difficult to calculate. Examples. What is the capacity of one mile of No. 0 (A. W. G.) leadcovered cable, with rubber insulation .15 inch thick? Substituting in (61) for d, the diameter of No. 0 wire = .325 inch, and for D the external diameter of the insulation = .325 + (2 × .15) = .625 inch, and for k the dielectric constant of pure rubber = 2.5, we have: K per mile = log 38.8310-9 342. × 10-9 farad = .342 microfarads. What is the capacity of one mile of single overhead bare No. 0 wire, 10 feet above the ground, with earth return? Substituting in (64) the values of h and d, both in inches, we have: K per mile = log 38.83 × 10-9 = 12.2 × 10-9 farad = .0122 microfarad. What is the capacity of two parallel overhead bare No. 0 wires, 12 inches apart, and each one mile long? Substituting in (67), we have: K= 2 x 19.42 × 10-9 log 2 × 12 20.8 × 10-9 farad = .0208 microfarad. Means of Reducing Capacity. It is evident from equation (59) that the capacity of a given length of insulated conductor with metallic covering is decreased by diminishing k, the dielectric constant of the insulation, by increasing D, the internal diameter of the metallic covering, or by reducing d, the diameter of the conductor. Since the capacity varies in direct proportion to k, the insulating material should have the minimum dielectric constant. Unfortunately the best insulators usually have high values for k, notably india rubber, gutta-percha, paraffin, and mica. The dielectric constant of paper is comparatively low, and largely for that reason it is used for insulating the wires in a telephone cable. Paper is also used for the insulation of electric light and power cables, and would have special advantages when it is desired to make the capacity as low as possible. This question will be considered further under the head of insulated and underground conductors. The reduction of capacity by diminishing d, the diameter of the conductor, is limited in practice by the necessity for using a certain size in order to give sufficient current capacity, and not have too much resistance. It is also a fact that it is not practicable to materially reduce electrostatic capacity by augmenting D, or in other words, by increasing the thickness of the insulation. Fig. 108 represents a lead-covered cable, in which d is the diameter of Figs. 108 and 109. Reducing Capacity by Increasing Thickness of Insulation. That is, the capacity is reduced only 36 per cent by doubling the thickness of insulation. The volume of insulation in the two cases would be in the proportion (D- d2): (D' — d'2) = 3: 8, which is an increase of 267 per cent, or almost three times as much. Since the amount of insulating material affects directly the cost and size of the cable, it would seldom pay to nearly treble this material in order to diminish the capacity to the extent of only 36 per cent. Hence in almost all cases the thickness of insulation is determined by its insulating qualities, and strength to withstand breakdown by electrical and mechanical pressures. |