Question

Question: For a completely transposed three-phase line identical conductors, each with GMR denoted with conductor distance D₁₂,D₂₃ , and D₃₁ give expressions for GMD between phases and the average per-phase inductance.

Short answer

Answer

Therefore, the expression for GMD between the phase $\sqrt[3]{D_{12}{D}_{23}{D}_{31}}$is and the average per-phase inductance is $L=2\times {10}^{-7}\ln \left(\frac{\sqrt[3]{D_{12}{D}_{23}{D}_{31}}}{D_s}\right)$.

Step-by-step solution

Determine the GMD between the phase.

Let assume and are the current through the conductors and .

Assume the balance condition,

$I_a+I_b+I_c=0$ ……(1)

Total flux linkage of the phase a conductor in position 1,

${\lambda }_{a1}=2\times {10}^{-7}\left[I_a\ln \left(\frac{1}{D_s}\right)+I_b\ln \left(\frac{1}{D_{12}}\right)+I_c\ln \left(\frac{1}{D_{31}}\right)\right]$ ……(2)

Total flux linkage of the phase a conductor in position 2,

${\lambda }_{a2}=2\times {10}^{-7}\left[I_a\ln \left(\frac{1}{D_s}\right)+I_b\ln \left(\frac{1}{D_{23}}\right)+I_c\ln \left(\frac{1}{D_{12}}\right)\right]$ ……(3)

Total flux linkage of the phase a conductor in position 3,

${\lambda }_{a3}=2\times {10}^{-7}\left[I_a\ln \left(\frac{1}{D_s}\right)+I_b\ln \left(\frac{1}{D_{31}}\right)+I_c\ln \left(\frac{1}{D_{23}}\right)\right]$ ……(4)

The average flux linkage is given by

${\lambda }_a=\frac{{\lambda }_{a1}+{\lambda }_{a2}+{\lambda }_{a3}}{3}$ ……(6)

Substitute the expression from the equations (2), (3), and (4) in equation (5),

$$\begin{gathered} \frac{\left[2\times {10}^{-7}\left[I_a\ln \left(\frac{1}{D_s}\right)+I_b\ln \left(\frac{1}{D_{12}}\right)+I_c\ln \left(\frac{1}{D_{31}}\right)\right] \\ +2\times {10}^{-7}\left[I_a\ln \left(\frac{1}{D_s}\right)+I_b\ln \left(\frac{1}{D_{23}}\right)+I_c\ln \left(\frac{1}{D_{12}}\right)\right] \\ +2\times {10}^{-7}\left[I_a\ln \left(\frac{1}{D_s}\right)+I_b\ln \left(\frac{1}{D_{31}}\right)+I_c\ln \left(\frac{1}{D_{23}}\right)\right] \\ \right]}{3} \\ {\lambda }_a=\frac{2\times {10}^{-7}}{3}\left[3I_a\ln \left(\frac{1}{D_s}\right)+I_b\ln \left(\frac{1}{D_{12}{D}_{23}{D}_{31}}\right)+I_c\ln \left(\frac{1}{D_{12}{D}_{23}{D}_{31}}\right)\right] \\ {\lambda }_a=\frac{2\times {10}^{-7}}{3}\left[3I_a\ln \left(\frac{1}{D_s}\right)+\left(I_b+I_c\right)\ln \left(\frac{1}{D_{12}{D}_{23}{D}_{31}}\right)\right] \end{gathered}$$

From equation (1),

$I_b+I_c=-I_a$

Substitute the value and solve as

$$\begin{gathered} {\lambda }_a=\frac{2\times {10}^{-7}}{3}\left[3I_a\ln \left(\frac{1}{D_s}\right)-I_a\ln \left(\frac{1}{D_{12}{D}_{23}{D}_{31}}\right)\right] \\ {\lambda }_a=\frac{2\times {10}^{-7}{I}_a}{3}\left[3\ln \left(1\right)-3\ln \left(D_s\right)-\ln \left(1\right)+\ln \left(D_{12}{D}_{23}{D}_{31}\right)\right] \\ {\lambda }_a=2\times {10}^{-7}{I}_a\left[\frac{1}{3}\ln \left(D_{12}{D}_{23}{D}_{31}\right)-\ln \left(D_s\right)\right] \\ {\lambda }_a=2\times {10}^{-7}{I}_a\left[\ln {\left(D_{12}{D}_{23}{D}_{31}\right)}^{\frac{1}{3}}-\ln \left(D_s\right)\right] \end{gathered}$$

Solve the expression further as

……(6)

From the equation (6), the GMD between the phase is .

Determine the average per-phase inductance.

The average per-phase inductance can be calculated as

$L_a=\frac{{\lambda }_a}{I_a}$

Substitute the expression from equation (6) and solve as

$$\begin{gathered} L_a=\frac{2\times {10}^{-7}{I}_a\left[\ln \left(\frac{\sqrt[3]{D_{12}{D}_{23}{D}_{31}}}{D_s}\right)\right]}{I_a} \\ {L}_a=2\times {10}^{-7}\left[\ln \left(\frac{\sqrt[3]{D_{12}{D}_{23}{D}_{31}}}{D_s}\right)\right] \end{gathered}$$

Hence the average per-phase inductance is.