Inductance Calculation of Transmission Lines

Demonstrative Video

Inductance calculation
- Inductance due to internal flux
- Inductance due to external flux
- Inductance of a single phase two wire line
- Flux linkage of one conductor in a group
- Inductance of composite conductor line
- Inductance of $3-\phi$ line with equilateral spacing
- Inductance of $3-\phi$ line with unsymmetrical spacing

Inductance of the conductor

Inductance is defined as \[L=\dfrac{\lambda}{I}\]
Some lines of flux exist inside the conductor
Filaments on the surface of the conductor are not linked by internal flux
Flux linking the filament near the surface of the conductor is less than that of interior

ac flux induce higher voltage at the interior filament than that of surface
by Lenz’s law, induce voltage oppose change of current producing it
higher induced voltage at the inner filament cause higher current density at near the surface
causes higher effective resistance
skin effect is significant in large conductors

Inductance due to Internal Flux

Assuming return path of the current is very far, it does not effect the magnetic field, hence lines of flux are concentric with the conductor
by Ampere’s law \[\mbox{mmf}=\oint H\centerdot ds=I~ At\]
$H_x =$ field intensity at a distance $x$ m from the center of the conductor, is constant and tangent to the circular path

\[\begin{aligned} \oint H_{x}\centerdot ds & =I_{x}\\ H_{x}2\pi x & =I_{x} \end{aligned}\] where $I_x$ is the current enclosed.

Assuming uniform current density \[I_{x}=\dfrac{\pi x^{2}}{\pi r^{2}}I~ At/m\]
The flux density at $x$ m from the center of the conductor \[B_{x}=\mu H_{x}=\dfrac{\mu xI}{2\pi r^{2}}~Wb/m^2\]
The flux per m of length \[d\phi=\dfrac{\mu xI}{2\pi r^{2}}dx ~ Wb/m\]
The flux linkage $d\lambda$ per m of length \[d\lambda=\dfrac{\pi x^{2}}{\pi r^{2}}d\phi=\dfrac{\mu x^{3}I}{2\pi r^{4}}dx ~ Wbt/m\]

The total flux linkage inside the conductor \[\lambda_{int}=\intop_{0}^{r}\dfrac{\mu x^{3}I}{2\pi r^{4}}dx=\dfrac{\mu I}{8\pi} ~ Wbt/m\] For $\mu_{r} = 1$ and $\mu_{0} = 4\pi \times 10^{-7}$ H/m

$$\boxed{L_{i n t}=\frac{1}{2} \times 10^{-7} \mathrm{H} / \mathrm{m}}$$

Inductance due to External Flux

At a distance $x$ from the center of the conductor, the mmf around the element is \[2\pi xH_{x} = I\]
The flux density in the element is given by \[B_{x}=\dfrac{\mu I}{2\pi x} ~ Wb/m^{2}\]
The flux $d\phi$ in the tabular element $dx$ \[d\phi=\dfrac{\mu I}{2\pi x}dxWb/m\]

The flux linkage between $P_1$ and $P_2$ \[\lambda_{12}=\intop_{D_{1}}^{D_{2}}\dfrac{\mu I}{2\pi x}dx=\dfrac{\mu I}{2\pi}ln\dfrac{D_{2}}{D_{1}}~Wbt/m\]

$$L_{12} = 2\times 10^{-7}\ln\dfrac{D_2}{D_1}~\mathrm{H/m}$$

Inductance of a single-phase two-wire line

Line with two-conductors, one conductor being the return of the other
Flux set up due to the current in conductor 1 at a distance equal to or greater than $\left(D+r_{2}\right)$ does not link the circuit
For $\left(D+r_{2}\right)$ the fraction of the total current linked by the flux is 1.0.

Therefore, when $D$ is much greater than $r_1$ or $r_2$, it is logical to use $D$ instead of $\left(D - r_{1}\right)$ or $\left(D - r_{2}\right)$

Inductance (internal + external) of the circuit due to current in conductor 1 \[L_{1}=\left(\dfrac{1}{2}+2ln\dfrac{D}{r_{1}}\right)\times10^{-7}~H/m\]
In compact form \[L_{1}=2\times10^{-7}\left(ln~e^{1/4}+ln\dfrac{D}{r_{1}}\right)~ H/m\] or \[L_{1}=2\times10^{-7}\left(ln\dfrac{D}{r_{1}e^{-1/4}}\right)=2\times10^{-7}\left(ln\dfrac{D}{r_{1}^{'}}\right) ~ H/m\] where $r_{1}^{'} = 0.7788r_{1}$ (only for solid round conductor)

The fictitious conductor, $r_{1}^{'}$ assumed to have no internal flux but with the same inductance as the actual conductor

Current in conductor 2 flows in opposite direction ($180^{0}$ phase shift), flux linkage produced by conductor 2 alone is in the same direction as conductor 1
Resulting flux is sum of the mmfs of both conductors
For constant permeability, flux linkage (or inductance) can be added

$$L=L_1+L_2 = 4\times 10^{-7}\left(\ln \dfrac{D}{\sqrt{r_1^{\prime}\cdot r_2^{\prime}}} \right)$$

If $r_{1}^{'} = r_{2}^{'} = r^{'}$, then

$$L= 4\times 10^{-7}\left(\ln\dfrac{D}{r^{\prime}} \right)$$

Flux linkages of one conductor in a group

Conductors $1,2,3..,n$ carry currents $I_{1}, I_{2}, I_{3},...I_{n}$
Flux linkage of conductor 1 due to $I_1$ \[\lambda_{1P1}=2\times10^{-7}I_{1}ln\dfrac{D_{1P}}{r_{1}^{'}}~Wbt/m\]
Flux linkage of conductor 1 due to $I_2$ \[\lambda_{1P2}=2\times10^{-7}I_{2}ln\dfrac{D_{2P}}{D_{12}}~Wbt/m\]

Flux linkage of conductor 1 due to all conductors \[\lambda_{1P}=2\times10^{-7}\left(I_{1}ln\dfrac{D_{1P}}{r_{1}^{'}}+I_{2}ln\dfrac{D_{2P}}{D_{12}}+I_{3}ln\dfrac{D_{3P}}{D_{13}}+....I_{n}ln\dfrac{D_{nP}}{D_{1n}}\right)\]
On expanding and regrouping the logarithmic terms \[\begin{aligned} \lambda_{1P} & =2\times10^{-7}\Bigg(I_{1}ln\dfrac{1}{r_{1}^{'}}+I_{2}ln\dfrac{1}{D_{12}}+I_{3}ln\dfrac{1}{D_{13}}+....I_{n}ln\dfrac{1}{D_{1n}}\\ & +I_{1}ln~D_{1P}+I_{2}ln~D_{2P}+I_{3}ln~D_{3P}+....+I_{n}ln~D_{nP}\Bigg) \end{aligned}\]

Since, $I_{1}+I_{2}+ I_{3}+...+I_{n} = 0$

Therefore, $I_{n} = -(I_{1}+I_{2}+ I_{3}+...+I_{n-1})$

Substituting $I_n$ in the second term and regrouping \[\begin{aligned} \lambda_{1P} & =2\times10^{-7}\Bigg(I_{1}ln\dfrac{1}{r_{1}^{'}}+I_{2}ln\dfrac{1}{D_{12}}+I_{3}ln\dfrac{1}{D_{13}}+....I_{n}ln\dfrac{1}{D_{1n}}\\ & +I_{1}ln\dfrac{D_{1P}}{D_{nP}}+I_{2}ln\dfrac{D_{2P}}{D_{nP}}+I_{3}ln\dfrac{D_{3P}}{D_{nP}}+....+I_{\left(n-1\right)}ln\dfrac{D_{\left(n-1\right)P}}{D_{nP}}\Bigg) \end{aligned}\]
$P$ is very far, the set of terms containing logarithms of ratios of distances from $P$ becomes infinitesimal, since the ratio becomes 1

$$ \boxed{\lambda_{1 P}=2 \times 10^{-7}\left(I_1 \ln \frac{1}{r_1^{\prime}}+I_2 \ln \frac{1}{D_{12}}+I_3 \ln \frac{1}{D_{13}}+\ldots . I_n \ln \frac{1}{D_{1 n}}\right)~\mathrm{Wb}} $$

Inductance of composite conductor line

Stranded conductors comes under the classification of composite conductors
Assumption stands are identical and carry equal current
Conductor $X$ $\Longrightarrow$ $n$ identical parallel filaments, each carrying $I/n$
Conductor $Y$ $\Longrightarrow$ return circuit for $X$ carry $-I/m$

Flux linkage of filament $a$ \[\begin{aligned} \lambda_{a} & =2\times10^{-7}\dfrac{I}{n}\left(ln\dfrac{1}{r_{a}^{'}}+ln\dfrac{1}{D_{ab}}+ln\dfrac{1}{D_{ac}}+...+ln\dfrac{1}{D_{an}}\right)\\ & -2\times10^{-7}\dfrac{I}{m}\left(ln\dfrac{1}{D_{aa^{,}}}+ln\dfrac{1}{D_{ab^{'}}}+ln\dfrac{1}{D_{ac'}}+...+ln\dfrac{1}{D_{am}}\right) \end{aligned}\] Rearranging \[\begin{aligned} \lambda_{a} & =2\times10^{-7}I\cdot ln\dfrac{\sqrt[m]{D_{aa^{'}}D_{ab^{'}}D_{ac^{'}}...D_{am}}}{\sqrt[n]{r_{a^{'}}D_{ab}D_{ac}...D_{an}}} \end{aligned}\]
Inductance of filament $a$ \[\begin{aligned} L_{a}=\dfrac{\lambda_{a}}{I/n} & =2n\times10^{-7}ln\dfrac{\sqrt[m]{D_{aa^{'}}D_{ab^{'}}D_{ac^{'}}...D_{am}}}{\sqrt[n]{r_{a^{'}}D_{ab}D_{ac}...D_{an}}}~ H/m \end{aligned}\]

Similarly, inductance of filament $b$ \[\begin{aligned} L_{b}=\dfrac{\lambda_{b}}{I/n} & =2n\times10^{-7}ln\dfrac{\sqrt[m]{D_{ba^{'}}D_{bb^{'}}D_{bc^{'}}...D_{bm}}}{\sqrt[n]{D_{ba}r_{b^{'}}D_{bc}...D_{bn}}}~ H/m \end{aligned}\]
The average inductance of the filaments of conductor $X$ \[L_{av} = \dfrac{L_{a}+L_{b}+L_{c}+...+L_{n}}{n}\]
Since, all filaments have different inductance, the inductance of conductor $X$ \[L_{X} = \dfrac{L_{av}}{n} = \dfrac{L_{av}}{n^{2}}\]
Substituting the logarithmic expression

\[L_{x}=2\times10^{-7}\times ln\dfrac{\sqrt[mn]{\left(D_{aa'}D_{ab'}D_{ac'}\cdots D_{am}\right)\left(D_{ba'}D_{bb'}D_{bc'}\cdots D_{bm}\right)\ldots\left(D_{na'}D_{nb'}D_{nc'}\cdots D_{nm}\right)}}{\sqrt[n^{2}]{\left(D_{aa}D_{ab}D_{ac}\cdots D_{an}\right)\left(D_{ba}D_{bb}D_{bc}\cdots D_{bn}\right)\ldots\left(D_{na}D_{nb}D_{nc}\cdots D_{nn}\right)}}\]

Geometric Mean Distance, GMD $(D_m)$ : The $mn^{th}$ root of the product of $mn$ distance between conductor $X$ and $Y$
Geometric Mean Radius, GMR $(D_s)$ : product of the distances from every filament in the conductor to itself and to every other filament.

$$L_x = 2 \times 10^{-7} \ln \dfrac{D_m}{D_s}$$
Total inductance of the line

\[L = L_{x}+ L_{y}\]

Inductance of 3-phase line with equilateral spacing

Assuming balanced three-phase currents $I_{a}+I_{b}+I_{c} = 0$
Because of symmetry $L_{a} = L_{b} = L_{c}$

\[\lambda_{a} = 2 \times 10^{-7} \left(I_{a}~ln \dfrac{1}{D_{s}}+I_{b}~ln\dfrac{1}{D}+I_{c}~ln \dfrac{1}{D}\right)\]

Since $I_{a} = -(I_{b}+I_{c})$, then

\[\lambda_{a} = 2 \times 10^{-7} \left(I_{a}~ln \dfrac{1}{D_{s}}-I_{a}~ln\dfrac{1}{D}\right) = 2 \times 10^{-7} I_{a}~ ln \dfrac{D}{D_s}\]

and \[L_{a} = 2 \times 10^{-7}~ln \dfrac{D}{D_s}\]

Inductance of 3-phase line with unsymmetrical spacing

Flux linkage and inductance of each phase are unequal, finding inductance is difficult
Results in unbalanced circuit
Transposition: Balance of $3-\phi$ can be restored by exchanging the position of the conductors at regular intervals along the line so that each conductor occupies the original position every other conductor over an equal distance.

Transposition results in each conductor having the same average inductance over the whole cycle.
The avg. value of the flux linkage for $a$ occupying different position \[\lambda_a = \dfrac{\lambda_{a1}+\lambda_{a2}+\lambda_{a3}}{3}\]

\[\lambda_{a1}=2\times10^{-7}\left(I_{a}ln\dfrac{1}{D_{s}}+I_{b}ln\dfrac{1}{D_{23}}+I_{c}ln\dfrac{1}{D_{12}}\right)\] \[\lambda_{a2}=2\times10^{-7}\left(I_{a}ln\dfrac{1}{D_{s}}+I_{b}ln\dfrac{1}{D_{12}}+I_{c}ln\dfrac{1}{D_{31}}\right)\] \[\lambda_{a3}=2\times10^{-7}\left(I_{a}ln\dfrac{1}{D_{s}}+I_{b}ln\dfrac{1}{D_{31}}+I_{c}ln\dfrac{1}{D_{23}}\right)\]

Using $I_a = -(I_b+I_c)$, we get \[\lambda_{a}=2\times10^{-7}~I_a \cdot ln \dfrac{\sqrt[3]{D_{12}D_{23}D_{31}}}{D_s}\]

Avg. inductance per phase \[L_{a}=2\times10^{-7}\cdot ln \dfrac{D_{eq}}{D_s}\] where $D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}$