Question

A single-phase transmission line has two conductors each of 10 mm radius. These are fixed at a center-to-center distance of 1m in a horizontal plane. This is now converted to a three-phase transmission line by introducing a third conductor of the same radius. This conductor is fixed at an equal distance D from the two single-phase conductors. The three-phase line is fully transposed. The positive sequence inductance per phase of the three-phase system is to be 5% more than that of the inductance per conductor of the single-phase system. The distance D, in meters, is__.

• A. 1.427

Answer 1

The correct option is A 1.427
$L_1=2\times {10}^{-7}\ln \frac{D}{r^{\prime }}=2\times {10}^{-7}ln\left(\frac{100}{0.7788}\right)$
$L_1=0.97\mu H/m$
$L_2=1.05\times 0.97=1.0185\mu H/m$
$L_2=2\times {10}^{-7}\ln \left(\frac{\sqrt[3]{3D^2\times 100}}{0.7788}\right)$
Solving, we get
$\ln \left(\frac{\sqrt[3]{3D^2\times 100}}{0.7788}\right)=\frac{1.0185\times {10}^{-6}}{2\times {10}^{-7}}$

$\ln \left(\frac{\sqrt[3]{3D^2\times 100}}{0.7788}\right)=5.0925$

$e^{5.0925}=\frac{\sqrt[3]{3100D^2}}{0.7788}$

$D=142.7cm=1.427m$