Inductance of a Transmission Line – Complete Derivation

We mark the conductors as A, B, and C. Additionally, we mark the position of the conductors in the transmission line as 1, 2, and 3. The distance between conductors A and B is $D_{12}$​. Similarly, we can say, the distance between conductors B and A is $D_{21}$.

Therefore,$$D_{12}=D_{21}$$Similarly, the distance of conductor B to C and C to B are $D_{23}$​ and $D_{32}$,​ respectively. Hence,
$$D_{23}=D_{32}$$Lastly, the distance of conductor C to A and A to C are $D_{13}$​ and $D_{31}$, respectively. Therefore,
$$D_{13}=D_{31}$$

Fictitious Radius of a Conductor

Also consider the radius of the conductors is r. Therefore, the fictitious radius will be,
$$r’ = r \times 0.7788$$

Balanced Currents

Also, we assume that conductors A, B, and C are carrying current $I_1$​, $I_2$​ and $I_3$​. As it is a three-phase balanced system, we know that$$I_1+I_2+I_3=0$$

Flux Linkage of Conductors

Now, the flux linkage of conductor A for its own current and the currents on the other two conductors is$$\lambda_A = 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} + I_2\ln\frac{1}{D_{12}} + I_3\ln\frac{1}{D_{13}} \right]$$Now, along the length of this transmission line, the conductors are transposed at two intermediate points.

For section – 1, the flux linkage will be represented as$$\lambda_{A1} = 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} + I_2\ln\frac{1}{D_{12}} + I_3\ln\frac{1}{D_{13}} \right]$$

Obviously, for section – 2, the flux linkage of conductor A for its own current and the current of other conductors will be,$$\lambda_{A2} = 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} + I_2\ln\frac{1}{D_{23}} + I_3\ln\frac{1}{D_{21}} \right]$$

Similarly, for section – 3, the flux linkage of conductor A for its own current and the current of other conductors will be,$$\lambda_{A3} = 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} + I_2\ln\frac{1}{D_{31}} + I_3\ln\frac{1}{D_{32}} \right]$$

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Average Flux Linkage

Therefore, the average flux linkage of conductor A will be$$\lambda_A = \frac{\lambda_{A1}+\lambda_{A2}+\lambda_{A3}}{3}$$$$= \frac{2\times10^{-7}}{3} \Bigg[ 3I_1\ln\frac{1}{r’} + I_2\left( \ln\frac{1}{D_{12}} +\ln\frac{1}{D_{23}} +\ln\frac{1}{D_{31}} \right)$$$$+ I_3\left( \ln\frac{1}{D_{13}} +\ln\frac{1}{D_{21}} +\ln\frac{1}{D_{32}} \right) \Bigg]$$ $$= 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} + I_2\ln\frac{1}{\sqrt[3]{D_{12}D_{23}D_{31}}} + I_3\ln\frac{1}{\sqrt[3]{D_{13}D_{21}D_{32}}} \right]$$

Again,$$\sqrt[3]{D_{12}D_{23}D_{31}} = \sqrt[3]{D_{13}D_{21}D_{32}} = D_{eq}$$Therefore,$$\lambda_A = 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} + I_2\ln\frac{1}{D_{eq}} + I_3\ln\frac{1}{D_{eq}} \right]$$$$= 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} + (I_2+I_3)\ln\frac{1}{D_{eq}} \right]$$Since,$$I_1+I_2+I_3=0$$$$I_2+I_3=-I_1$$Hence,$$\lambda_A = 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} – I_1\ln\frac{1}{D_{eq}} \right]$$$$= 2\times10^{-7}I_1 \ln\left(\frac{D_{eq}}{r’}\right)$$Therefore, the inductance of the conductor A is,$$L_A=\frac{\lambda_A}{I_1} = 2\times10^{-7} \ln\left(\frac{D_{eq}}{r’}\right) \quad \text{H/m}$$

Special Case: Vertical Conductor Configuration

If the spacing between adjacent conductors is h, $$D_{12}=D_{21}=D_{23}=D_{32}=h$$Also, the distance between the top and bottom conductors is$$D_{13}=D_{31}=2h$$Therefore,$$D_{eq} = \sqrt[3]{D_{12}D_{23}D_{13}} = \sqrt[3]{2h^3} = \sqrt[3]{2}\,h$$Substituting into the inductance expression,$$L_A = 2\times10^{-7} \ln\left(\frac{\sqrt[3]{2}\,h}{0.7788\,r}\right)$$​Since,$$\frac{\sqrt[3]{2}}{0.7788}=1.6177$$The final expression becomes$$\boxed{ L_A = 2\times10^{-7} \ln\left(1.6177\frac{h}{r}\right) \ \text{H/m} }$$

Numerical Example of Inductance of a Transmission Line Conductor

Suppose, the clearance between adjacent conductors is $$h=3\,m$$The conductor is ACSR Panther hence, the conductor diameter$$d=21\,mm=0.021\,m$$Hence,$$r=\frac{0.021}{2}=0.0105\,m$$Substituting,$$L = 2\times10^{-7} \ln \left( 1.6177\times\frac{3}{0.0105} \right)$$$$= 2\times10^{-7} \ln(462.2)$$$$= 2\times10^{-7}\times6.136$$$$\boxed{ L = 1.227\times10^{-6}\;H/m }$$or$$\boxed{ L = 1.227\;mH/km }$$​

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