Inductance of a Transmission Line – Complete Derivation

We mark the conductors as A, B and C. Additionally, 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 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,$$\lambda_{A3} = 2\times10^{-7} \left[ I_1\ln\frac{1}{r’} + I_2\ln\frac{1}{D_{31}} + I_3\ln\frac{1}{D_{32}} \right]$$

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$$\boxed{ L = 1.227\;mH/km }$$​