Question

A 400 mm2 single core copper conductor cable operating at 400 kV, 50 Hz has the following parameters:

εr = 2.0 and tanδ = 0.0003, ri = 30.3mm, C = 0.250 [μF / km]
(a) calculate the external radius of the insulation (mm)
(b) if the dielectric loss of 1 km is 1800 [w/km], using the same given data, calculate the new capacitance ( μF / km) and the external radius of the insulation (mm)

Answer 1

The external radius of the insulation is 10.3 mm. The new capacitance is approximately 0.25025 μF/km.

Step-by-step explanation:

(a) To calculate the external radius of the insulation, we can use the formula:
ro = ri + x
where ro is the external radius of the insulation, ri is the internal radius of the conductor, and x is the thickness of the insulation. We can rearrange the formula to solve for x:
x = ro - ri
Plugging in the given values, we have:
x = 30.3 mm - 20 mm = 10.3 mm
Therefore, the external radius of the insulation is 10.3 mm.

(b) To calculate the new capacitance, we can use the formula:
C_new = C_initial * (1 + tanδ * loss_factor)
where C_initial is the initial capacitance, tanδ is the dielectric loss angle, and loss_factor is the loss in watts per kilometer divided by the voltage squared. We can plug in the given values to find the new capacitance:
C_new = 0.250 μF/km * (1 + 0.0003 * (1800 W/km / (400 kV)^2))
C_new ≈ 0.250 μF/km * 1.001
C_new ≈ 0.25025 μF/km
Therefore, the new capacitance is approximately 0.25025 μF/km.

To calculate the external radius of the insulation, we can use the formula:
ro = ri + x
where ro is the external radius of the insulation, ri is the internal radius of the conductor, and x is the thickness of the insulation. We can rearrange the formula to solve for x:
x = ro - ri
Plugging in the given values, we have:
x = 30.3 mm - 20 mm = 10.3 mm
Therefore, the external radius of the insulation is 10.3 mm.