A single-core lead-sheathed cable, with a conductor diameter of 2 cm i

1.	A single-core lead-sheathed cable, with a conductor diameter of 2 cm is designed to withstand 66 kV. The dielectric consists of two layers A and B having relative permittivities of 3.5 and 3 respectively. The corresponding maximum permissible electrostatic stresses are 72 and 60 kV/cm. Find the thicknesses of the two layers
Answer» g_max1/g_max2 = (ε_r2 x r₂)/(ε_r1 x r₁) or 72/60= (3 x r₂)/(3.5 x 1) or r₂ = 1.4cm Now, g_max = (V₁ x √2)/(2.3r₁log₁₀(r₂/r₁)) where V₁is the r.m.s. values of the voltage across the first dielectric. ∴ 72 = (V₁ x √2)/(2.3 x 1 x log₁₀1.4) or V₁ = 17.1kv Obviously, V₂= 60 − 17.1= 48.9 kV Now, g_max2 = (V₂x √2)/(2.3r₂log₁₀(r₃/r₂)) ∴ 60 = 48.9/(2.3 x 1.4log₁₀(r₃/r₂)) ∴ log₁₀ (r₃/r₂) = 0.2531 = log₁₀ (1.79) ∴ r₃/r₂ = 1.79 or r₃= 2.5cm Thickness of first dielectric layer = 1.4 − 1.0 = 0.4 cm. Thickness of second layer = 2.5 − 1.4 = 1.1 cm.

A single-core lead-sheathed cable, with a conductor diameter of 2 cm is designed to withstand 66 kV. The dielectric consists of two layers A and B having relative permittivities of 3.5 and 3 respectively. The corresponding maximum permissible electrostatic stresses are 72 and 60 kV/cm. Find the thicknesses of the two layers

Answer»

g_max1/g_max2 = (ε_r2 x r₂)/(ε_r1 x r₁) or 72/60= (3 x r₂)/(3.5 x 1) or r₂ = 1.4cm

Now, g_max = (V₁ x √2)/(2.3r₁log₁₀(r₂/r₁))

where V₁is the r.m.s. values of the voltage across the first dielectric.

∴ 72 = (V₁ x √2)/(2.3 x 1 x log₁₀1.4) or V₁ = 17.1kv

Obviously, V₂= 60 − 17.1= 48.9 kV

Now, g_max2 = (V₂x √2)/(2.3r₂log₁₀(r₃/r₂))

∴ 60 = 48.9/(2.3 x 1.4log₁₀(r₃/r₂))

∴ log₁₀ (r₃/r₂) = 0.2531 = log₁₀ (1.79)

∴ r₃/r₂ = 1.79 or r₃= 2.5cm

Thickness of first dielectric layer = 1.4 − 1.0 = 0.4 cm.

Thickness of second layer = 2.5 − 1.4 = 1.1 cm.