Question

A 60-Hz single-phase, two-wire overhead line has solid cylindrical copper conductors with 1.5 cm diameter. The conductors are arranged in a horizontal configuration with 0.5 m spacing. Calculate in mH/km (a) the inductance of each conductor due to internal flux linkages only, (b) the inductance of each conductor due to both internal and external flux linkages, and (c) the total inductance of the line.

Answer 1

Answer:

(a) the inductance of each conductor due to internal flux linkages only is 0.05 mH/km per conductor

(b) the inductance of each conductor due to both internal and external flux linkage is 0.8894 mH/km

(c) the total inductance of the line is 1.78 mH/km per circuit

Explanation:

(a) the inductance of each conductor due to internal flux linkages only $L_{int}$  

$L_{int} = \frac{1}{2} * 10^{-7} H/m = \frac{1}{2} * 10^{-7} H/M (\frac{1000mH}{1H})(\frac{1000m}{1km} ) = 0.05 mH/km$ per conductor.

(b) the inductance of each conductor due to both internal and external flux linkage is given by

$L_{x} =L_{y} = 2*10^{-7}*ln(\frac{D}{r^'} } ) H/m$

but $r^{'} = e^{\frac{-1}{4 }}*\frac{d}{2} }$

where D is the spacing between conductors = 0.5 m

d = diameter=1.5 cm= 0.015 m

$r^{'} = e^{\frac{-1}{4 }}*\frac{0.015}{2} }=0.00584$

$L_{x} =L_{y} = 2*10^{-7}*ln(\frac{0.5}{0.00584} } ) =(8.8994*10^{-7} ) H/m= 8.8994*10^{-7} * 1000mH*1000km=0.8894 mH/km$

$L_{x} =L_{y} = 0.8894 mH/km$ per conductor

(c) the total inductance of the line L

$L=L_{x} + L_{y} =0.8894+0.8894=1.7788mH/km$

L≅ 1.780 mH/km per circuit