Question

If the per-phase line loss in a $\mathbf{70}\mathbf{km}$-long transmission line is not to exceed$\mathbf{65}\mathbf{kW}$while it is delivering $\mathbf{100}\mathbf{}\mathbf{A}$per phase, compute the required conductor diameter if the resistivity of the conductor material is$\mathbf{1}\mathbf{.}\mathbf{72}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\Omega }\mathbf{‐}\mathbf{m}$.

Short answer

Therefore, the diameter of the conductor is$15.3\mathrm{mm}$

Step-by-step solution

Write the given data by the question.

The length of the transmission line$/=70\mathrm{km}$

The power loss in the transmission line$P_L=65\mathrm{kW}$.

Resistivity of the conductor material$p=1.72\times {10}^{-8}\mathrm{\Omega }-\mathrm{m}$

Current in the conductor.$/=100\mathrm{A}$

Determine the formulas to calculate the value of the diameter of the conductor.

The expression for the power loss in the transmission line is given by,

${\mathbf{P}}_{\mathbf{L}}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{R}$ …… (1)

Here, is the current in the conductor and is the resistance of the conductor.

The expression for the resistance of the conductor is given by,

$\mathbf{R}\mathbf{=}\frac{\mathbf{pI}}{\mathbf{A}}$ …… (2)

Here is the cross-section area of the conductor.

The expression for the cross-section area of the conductor is given by,

$\mathbf{A}\mathbf{=}\frac{{\mathbf{\tau \tau d}}^{\mathbf{2}}}{\mathbf{4}}$ …… (3)

Here, is the diameter of the conductor.

Calculate the diameter of the conductor.

Calculate the resistance of the conductor.

Substitute $65\mathrm{kW}$for $P_L$and $100\mathrm{A}$for$/$into equation (1).

$$\begin{gathered} 65\times {10}^3={100}^2\times R \\ R=\frac{65\times {10}^3}{100\times 100} \\ R=\frac{65}{10} \\ R=6.5\Omega \end{gathered}$$

Calculate the cross-section area of the conductor.

Substitute$1.72\times {10}^{-8}\Omega -\mathrm{m}$for $p$, $70\mathrm{km}$for $6.5\Omega$and $R$for into equ$$\begin{gathered} 6.5=\frac{1.72\times {10}^{-8}\times 70\times {10}^3}{A} \\ A=\frac{1.72\times {10}^{-8}\times 70\times {10}^3}{6.5} \\ \mathrm{A}=\frac{1.204\times {10}^{-3}}{6.5} \\ \mathrm{A}=1.85\times {10}^{-4}{\mathrm{m}}^2 \end{gathered}$$

Calculate the diameter of the conductor.

Substitute $1.85\times {10}^{-4}{\mathrm{m}}^2$for$A$ into equation (3).

$$\begin{gathered} 1.85\times {10}^{-4}=\frac{\pi \times d^2}{4} \\ {d}^2=\frac{1.85\times {10}^{-4}\times 4}{\pi } \\ {d}^2=2.355\times {10}^{-4} \\ {d}^2=0.0153\mathrm{m} \end{gathered}$$

Hence, the diameter of the conductor is$15.3\mathrm{mm}$ .