When doing the load flow study in the Simulink environment, we need to put the transmission line parameters in the Simulink block. In the question, typically, the parameters are given as per unit values. But when we put the values into the Simulink block, we need to put the actual resistance, inductance, and capacitance value or value/km as parameters. To reduce the hassle of those long calculations, I have made a calculator for the transmission line parameters. In that calculator, we need to put the value of voltage level and the apparent power and per unit values and the frequency. The result section will give back the actual resistance, inductance, capacitance value, and per km values.

Transmission line length formula:

 the velocity of light in km/s

Now suppose  a transmission line parameter data is as follows

R= 0.010 pu, X= 0.085 pu, B=0.176 pu.

The base value of power and voltage is given asVb=230 kV , Sb= 100 MVA

Then the length is

Length = 97.33 km

Resistance Value Calculation:

R_{actual} = R_{base}\  × \ R_{pu} 

R_{base} = X_{base} = \frac{V_{b}^2}{S_{b}} = \frac {(230 \times 10^{3})^2}{100 \times 10^6} = 529 \\ . \\R_{actual} = 529 \times 0.01 = 5.29 \ Ω \\.\\ R_{Ω/km} = R_ {actual} / Line\ Length = 5.29 / 97.33 = 0.0543 Ω/km \\.\\ Zero \ Sequence \ Resistance\ , R_{0} = 0.0543 \times 3 \ = \ 0.0163 Ω/ km

Inductance Value Calculation:

X= 0.085 pu

X_{base} = R_{base} = 529

X_{actual}= 529\times 0.085 = 44.965

L_{actual} = X_{actual}/2πf \\ .\\ = 44.965/2πf  \\.\\ = 0.1192

So L actual is 0.1192 Henry

Want to get the value in per kilometer, so we will divide this value by the total length of the transmission  line(97.33 km)

L_{H/km} = 0.1192/97.33 \\.\\ = 1.225 \ × \ 10 ^ {-3} {H/km} 

Zero sequence Inductance will be

L_{0} =1.225 \ × \ 10 ^ {-3} \times 3 \\.\\ = 3.676 \ × \  10 ^ {-3}{H/km} 

Capacitance Value Calculation:

In order to find the value of capacitance, first, we have to get the per unit value of susceptance from the datasheet, which is 0.176.

B_{base} = \frac {100 \ × \ 10^6} {(230 \  × \  10^3)^2} \\.\\ = 1.890 \  × \   10^{-3}

B_{actual} = B_{base} \times  \ B_{pu} \\.\\  =  1.890 \times 10^{-3} \times 0.176 \\.\\ = 3.3264 \times \ 10 ^{-4} 

C_{actual} \ = \ 3.3264 \times \ 10^{-4} / 2 π f \\.\\ = \ 8.8235 \times \ 10^ {-7}  

C_{F/km} \ = \  8.8235 \times \ 10^ {-7} / \ 97.33 \\.\\ = \ 9.065 \times \ 10^{-9} F/km

C_{0} \ = \ C_{F/km} \times \ 3 \\ .\\= \ 2.719 \times 10 ^ {-8} F/km