Question

A three-phase overhead transmission line is designed to deliver 190.5 MVA at 220 KV over a distance of 63 KM, such that the total transmission line loss is not to exceed 2.5% of the rated line MVA.

Given the resistivity of the conductor material to be 2.84*10$$\begin{gathered} -8{ \\ }^{} \end{gathered}$$$\Omega -m$, determine the required conductor diameter and the conductor size in circular mils. Neglect power losses due to insulator leakage currents and corona.

Short answer

Therefore, the conductor diameter is 556,000 cmil..

Step-by-step solution

Given data

The overhead line delivers S=190.5 MVA at $V_L$=220KV.

The distance of line is I= 63 KM.

The power loss is 2.5% of delivered power.

The resistivity of conductor is p = 2.84*${10}^{-8}$$\Omega -m$${10}^{-8}$.

Determine the formula of power loss, current, three-phase power and resistance of conductor

Write the formula of power loss in the line.
$P_L$= 2.5%S……. (1)

Here, S is the MVA power.

Write the formula of line current.

……. (2)

$I_L$= $\frac{S}{\sqrt{3VL}}$

Here, S is the MVA power and$V_L$ is line voltage.

Write the formula of power loss in line.

$P_L=3{I_L}^{2}R$ ……. (3)

Here, R is resistance in the line.

Write the formula of resistance of line.

$P=\frac{A}{I}$ ……. (4)

Here, $\rho$is resistivity, $I$is length of line and $A$is cross section area of conductor.

Write the formula of area of conductor.

${\rm A}={\left(\frac{d}{2}\right)}^2$ ……. (5)

Here, $d$is diameter of conductor.

Determine the power loss, current and resistance.

Substitute 190.5 MPA for $S$in equation (1).

$$\begin{gathered} P_L=2.5\%\left(190.5MVA\right) \\ =4.7625MW \end{gathered}$$

Substitute $190.5MVA$for $S$and $220Kv$for $V_L$in equation (2).

$$\begin{gathered} I_L=\frac{190.5MVA}{\sqrt{3}*220Kv} \\ =500A \end{gathered}$$

Substitute $500A$for $I_L$and 4.7625 MW for $P_L$in equation (3).

$$\begin{gathered} 4.7625MW=3{\left(500A\right)}^{2}R \\ R=\frac{4.7625MW}{3{\left(500A\right)}^2} \\ =\Omega 6.35 \end{gathered}$$

Determine the conductor cross-section area and diameter.

Substitute $\Omega 6.35$for $R$, 63KM for $I$and $2.84\times {10}^{-8}\Omega -m$for $p$in equation (4).

$$\begin{gathered} \Omega 6.35\left(2.84\times {10}^{-8}\right)\left(\frac{63KM}{A}\right) \\ A=\frac{\left(2.84\times {10}^{-8}\Omega -m\right)\left(63KM\right)}{\Omega 6.35} \\ =2.817\times {10}^{-4}{m}^2 \end{gathered}$$

Substitute $2.817\times {10}^{-4}{m}^2$for$A$in equation (5).

$$\begin{gathered} 2.817\times {10}^{-4}{m}^2=\pi {\left(\frac{d}{2}\right)}^2 \\ d=2\sqrt{\frac{2.817\times {10}^{-4}{m}^2}{\pi }} \\ d=1.894cm \end{gathered}$$

In inches, the diameter is,

$$\begin{gathered} d=1.894\times 2.54 \\ =0.7456in \end{gathered}$$

In circular mils, the diameter is,

$$\begin{gathered} d=0.7456\times 745708 \\ =556,000cmil \end{gathered}$$

Therefore, the conductor diameter is$556,000cmil$.