Potential Gradient in capacitors easy explanation 106

Potential Gradient in a Cylindrical Capacitor

The dielectric stress at any point in a cable is in fact the potential gradient (or electric intensity) at that point. Under operating conditions, the insulation of a power cable is subjected to electrostatic forces. This is known as dielectric stress.

Mathematical expression

Let us consider a single core cable with core diameter d and internal sheath diameter D. The electric intensity at a point x metres from the centre of the cable is

Ex = Q / 2π ε₀ ε_r x volts/m.

By definition, electric intensity is equal to the potential gradient. Therefore, the potential gradient g at a point x metres from the centre of the cable is g = Ex

g = Q / 2π ε₀ ε_r x volts/m. …..(i)

The potential difference V between the conductor and sheath is

V = (Q / 2 π ε0 εr) log_e D/d volts

Or, Q = 2 π ε0 εr V / log_e D/d …..(ii)

Substituting the value of Q from exp. (ii) in exp. (i), we get,

It is clear from expression (iii) that the potential gradient varies inversely to the distance x. Therefore, the potential gradient will be maximum when x is minimum i.e., when x = d/2 or at the surface of the conductor. On the other hand, the potential gradient will be minimum at x = D/2 or at the sheath surface.

Maximum potential gradient, g_max = 2V / (d log_e D/d) volts/m, Putting x = d/2 in exp. (iii)

Minimum potential gradient, g_min = 2V / (D log_e D/d) volts/m, Putting x = D/2 in exp. (iii)

Therefore, g_max / g_min = [2V / (d log_e D/d)] / [2V / (D log_e D/d)] = D/d

The variation of stress in the dielectric is shown above in the figure. It is clear that dielectric stress is maximum at the conductor surface and its value goes on decreasing as we move away from the conductor. It may be noted that maximum stress is an important consideration in the design of a cable.

For instance, if a cable is to be operated at such a voltage that maximum stress is 5 kV/mm, then the insulation used must have a dielectric strength of at least 5 kV/mm, otherwise breakdown of the cable will become inevitable.

Most Economical Conductor Size in a Cable

As we already know now maximum stress in a cable occurs at the surface of the conductor.

Therefore, for the safe working of the cable, the dielectric strength of the insulation should be more than the maximum stress. Rewriting the expression for maximum stress, we get,

g_max = 2V / (d log_e D/d) volts/m

The values of working voltage V and internal sheath diameter D have to be kept fixed at certain values due to design considerations. This leaves conductor diameter d to be the only variable in expression.

For given values of V and D, the most economical conductor diameter will be one for which g_max

has a minimum value. The value of g_max will be minimum when d log_e D/d is maximum i.e.

Therefore, log_e (D/d) – 1= 0

Or, log_e (D/d) = 1

Or, D/d = 2.718

Hence, the most economical conductor diameter, d = D/2.718,

and the value of g_max under this condition is g_max = 2V/D volts/m …. Putting log_e D/d = 1 in expression (i)

For low and medium-voltage cables, the value of conductor diameter arrived at by this method (i.e., d = 2V/g_max) is often too small from the point of view of current density. Therefore, the conductor diameter of such cables is determined from the consideration of safe current density.

For high voltage cables, designs based on this theory give a very high value of d, much too large from the point of view of current carrying capacity and it is, therefore, advantageous to increase the conductor diameter to this value.