The Most Economical Conductor Diameter for Cables | Electricity

For a given voltage and overall diameter of a single core cable there is a certain value of the conductor radius (or core) diameter that gives a minimum potential gradient at the surface. A small conductor radius allows a greater insulation thickness, but on the other hand, the smaller radius of curvature tends to increase the stress; while the effect of too large a diameter of core is to cause an increase of the stress through reduction of the total thickness of the insulation.

For fixed values of the voltage V and overall diameter D, the g_max will be minimum when d log_e D/d is maximum and d log_e D/d will be maximum if:

For the low and medium voltage cables, the diameter of the conductor determined from the above consideration, i.e. d = 2V/g_max is somewhat less than that determined from the consideration of the safe current density, therefore, the main criterion for determining conductor diameter for such cables is the current carrying capacity.

On the other hand for high voltage cables the value of core diameter d determined from the above expression would, in general, give a conductor of cross-section much larger than required from the consideration of current carrying capacity.

The desired overall diameter of the conductor without increas­ing the cross-sectional area can be had in two ways, by making the conductor hollow or by building it with a lead sheath. The latter method has the advantage of eliminating at the same time the increase of stress due to stranding which may be as high as 20%.

Example:

Show that for a given voltage V and maximum stress E_max in the dielectric of a single core cable, the sheath diameter D is a minimum when D: d = e where d is conductor diameter and e is the base of natural logarithms. Determine D and d for V = 10 kV and E_max = 23 kV/cm.

Solution:

Maximum stress, g_max = E_max = 23 kV/cm = 23 × 10⁵ V/m (assume rms)

Applied voltage, V = 10,000 V

∴ Diameter of conductor, d = 2V/g_max = (2 × 10,000) / (23 × 10⁵) = 0.0087 m or 8.7 mm Ans.

Internal diameter of sheath, D = e × d = 2.71828 × 8.7 = 23.64 mm Ans.