Question

. A 132 kV, 3-phase, 50 Hz transmission line 200 km long consists of three conductors of effective diameter 20 mm arranged in a vertical plane with 4 m spacing and regularly transposed. Find the inductance and kVA rating of the arc suppression coil in the system​

Answer 1

Given:

V = 132 kV

f =  50 Hz

l= 200 Km

d = 20 mm

r = 4 m

To find:

1) L =?

2) KVA rating =?

Explanation:

1)

Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it.

∴ L = $\frac{\mu N^2A}{l}$

where

L = inductance in Henery

μ = permeability = 4π x $10^{-7$ N/$A^2$

N = no of turns of the coil

A = area encircled by coil

i.e. A = $\pi r^2$

l = length of coil

Put given values in the above equation

∴ L =  $\displaystyle \frac{4\times3.14\times10^{-7}\times3^2\times3.14 (10\times10^{-3})^2}{200\times10^3}$

     =  1.7747 $\times10^{-10$

     = 0.18 $\times 10^{-9$

L = 0.18 nm

2)

According to ohm's law for inductor

V = $L \frac{dI}{dt}$

132 × $10^3$ = 0.18 $\times 10^{-9$ × $\frac{dI}{1}$

I = 733.33 × $10^{12$ A

Now,

KVA = $\frac{3IV}{1000}$

       = $\displaystyle \frac{3\times132 \times 10^3\times7.33\times10^{12}}{1000}$

        =   2902.6 $\times 10^{12$

KVA = 2.9 $\mathbf{\times 10^{15}}$