# Conductor Material Required in Single-Phase Overhead AC Transmission System

## Single-Phase AC Transmission System

The electric power transmission system in which two conductors viz. phase conductor and neutral wire are used to transmit the electric power is known as single phase AC transmission system.

The single phase AC transmission system can be classified into following three types viz. −

• Single-phase two-wire system with one conductor earthed

• Single-phase two-wire system with mid-point earthed

• Single-phase three-wire system

## Conductor Material Required in 1-Phase 2-Wire System with One Conductor Earthed

The single phase two wire AC system with one conductor earthed is shown in Figure-1.

Let,

• $\mathrm{Power\: transmitted\mathrm{\, =\, }\mathit{P}}$

• $\mathrm{Maximum \: voltage \: between\: conductors\mathrm{\, =\, }\mathit{V_{m}}}$

• $\mathrm{RMS\: value\: of \: voltage\mathrm{\, =\, }\frac{\mathit{V{_m}}}{\sqrt{2}}}$

• $\mathrm{Power\: factor\: of\: load\mathrm{\, =\, }cos\, \phi }$

Then, the load current is given by,

$$\mathrm{\mathit{I_{\mathrm{1}}\mathrm{\, =\, }\frac{P}{\frac{V_{m}}{\sqrt{\mathrm{2}}}\mathrm{cos}\, \phi }\mathrm{\, =\, }\frac{\sqrt{\mathrm{2}P}}{V_{m}\, \mathrm{cos}\, \phi} } }$$

If a1 is the area of cross section of each conductor, then the resistance per conductor is given by,

$$\mathrm{\mathit{R_{\mathrm{1}}\mathrm{\, =\, }\frac{\rho \, l}{a_{\mathrm{1}}}}}$$

Hence, the line losses are given by,

$$\mathrm{\mathit{W\mathrm{\, =\, }\mathrm{2}I\mathrm{_{1}^{2}}\, R_{\mathrm{1}}\mathrm{\, =\, }\mathrm{2}\times \left ( \frac{\mathrm{\sqrt{2}}P}{V_{m}\mathrm{cos}\, \phi } \right )^{\mathrm{2}}\times \left ( \frac{\rho \, l}{a_{\mathrm{1}}} \right )}}$$

$$\mathrm{\mathit{\Rightarrow W \mathrm{\, =\, }\frac{\mathrm{4}P^{\mathrm{2}\, }\rho l}{V_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi \,a_{\mathrm{1}}} }}}$$

$$\mathrm{\therefore Cross\: sectional \: area,\mathit{ a_{\mathrm{1}}\mathrm{\, =\, }\frac{\mathrm{4}P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}}$$

Now, the volume (say K) of conductor material required in the single phase two wire overhead AC transmission system with one conductor earthed is given by,

$$\mathrm{\mathit{K\mathrm{\, =\, }\mathrm{2}a_{\mathrm{1}}l\mathrm{\, =\, }\mathrm{2}\times \left (\frac{\mathrm{4}P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }\right )\times l}}$$

$$\mathrm{\mathit{\therefore K\mathrm{\, =\, }\frac{\mathrm{8}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}\, \, \, \cdot \cdot \cdot \left ( 1 \right )}$$

## Conductor Material Required in 1-Phase 2-Wire System with Mid-Point Earthed

A single phase 2-wire AC transmission system with mid-point earthed is shown in Figure-2.

Here, the two wire have equal and opposite maximum voltages (Vm) with respect to earth. Therefore, the maximum voltage between the two wires is 2Vm.

$$\mathrm{RMS\: value\: of \: the \: voltage\mathit{\mathrm{\, =\, }\frac{\mathrm{2}V_{m}}{\mathrm{\sqrt{2}}}\mathrm{\, =\, }\sqrt{\mathrm{2}}V_{m}}}$$

$$\mathrm{Load \: current,\mathit{I_{\mathrm{2}}\mathrm{\, =\, }\frac{P}{\sqrt{\mathrm{2}}V_{m}\, \mathrm{cos\, \phi}}} }$$

If 𝑎2 is the area of cross section of each conductor, then the total power loss in the transmission line is

$$\mathrm{Line\: losses,\mathit{W\mathrm{\, =\, }\mathrm{2}I\mathrm{_{2}^{2}}\, R_{\mathrm{2}}\mathrm{\, =\, }\mathrm{2}\times \left ( \frac{P}{\mathrm{\sqrt{2}}V_{m}\mathrm{cos}\, \phi } \right )^{\mathrm{2}}\times \left ( \frac{\rho \, l}{a_{\mathrm{2}}} \right )}}$$

$$\mathrm{\mathit{\Rightarrow W \mathrm{\, =\, }\frac{P^{\mathrm{2}\, }\rho l}{V_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi \,a_{\mathrm{2}}} }}}$$

$$\mathrm{\therefore Area\: of \: cross \: section,\mathit{ a_{\mathrm{2}}\mathrm{\, =\, }\frac{P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}}$$

Now, the volume (say K1) of conductor material required in the single-phase two-wire overhead AC transmission system with mid-point earthed is,

$$\mathrm{\mathit{K_{\mathrm{1}}\mathrm{\, =\, }\mathrm{2}a_{\mathrm{2}}l\mathrm{\, =\, }\mathrm{2}\times \left (\frac{P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }\right )\times l}}$$

$$\mathrm{\mathit{\therefore K_{\mathrm{1}}\mathrm{\, =\, }\frac{\mathrm{2}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}\, \, \, \cdot \cdot \cdot \left ( 2 \right )}$$

Comparing equations (1) and (2), we have,

$$\mathrm{\mathit{ \frac{K_{\mathrm{1}}}{K}\mathrm{\, =\, }\frac{\left ( \frac{\mathrm{2}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, } \right )}{\mathrm{\mathit{\left ( \frac{\mathrm{8}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, } \right )}}}}\mathrm{\, =\, }\frac{1}{4}}$$

$$\mathrm{\mathit{\therefore K_{\mathrm{1}}\mathrm{\, =\, }}0.25\times \mathit{K}\: \: \cdot \cdot \cdot \left ( 3 \right )}$$

Hence, the volume of conductor material required in 1-phase 2-wire AC system with mid-point earthed is 0.25 times of that required in 1-phase 2-wire system with one conductor earthed.

## Conductor Material Required in 1-Phase 3-Wire System

The single-phase three-wire AC system is shown in Figure-3. This system consists of two outer wires and one neutral wire taken from the mid-point of the phase winding. When a balanced load is connected to the system, the current through the neutral wire is zero.

Refer the circuit,

• $\mathrm{Maximum\: voltage\: between\: conductors\mathrm{\, =\, }\mathit{\mathrm{2}V_{m}}}$

• $\mathrm{RMS\: value\: of\: voltage\: between\: conductors\mathrm{\, =\, }\mathit{\frac{\mathrm{2}V_{m}}{\mathrm{\sqrt{2}}}\mathrm{\, =\, }\mathrm{\sqrt{2}}V_{m}}}$

• $\mathrm{Power\: factor\: of\: load\mathrm{\, =\, }cos\, \phi }$

Then, the load current of the system is given by,

$$\mathrm{\mathit{I_{\mathrm{3}}\mathrm{\, =\, }\frac{P}{\sqrt{\mathrm{2}}V_{m}\, \mathrm{cos\, \phi}}} }$$

If a3 is the area of cross-section of each outer conductor, then

$$\mathrm{Line\: losses,\mathit{W\mathrm{\, =\, }\mathrm{2}I\mathrm{_{3}^{2}}\, R_{\mathrm{3}}\mathrm{\, =\, }\mathrm{2}\times \left ( \frac{P}{\mathrm{\sqrt{2}}V_{m}\mathrm{cos}\, \phi } \right )^{\mathrm{2}}\times \left ( \frac{\rho \, l}{a_{\mathrm{3}}} \right )}}$$

$$\mathrm{\mathit{\Rightarrow W \mathrm{\, =\, }\frac{P^{\mathrm{2}\, }\rho l}{V_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi \,a_{\mathrm{3}}} }}}$$

$$\mathrm{\therefore Area\: of \: cross \: section,\mathit{ a_{\mathrm{3}}\mathrm{\, =\, }\frac{P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}}$$

Let the area of cross-section of the neutral wire is to be half that of the outer wire. Then, the volume (say K2) of conductor material required in the single phase three wire overhead AC transmission system is given by,

$$\mathrm{\mathit{K_{\mathrm{2}}\mathrm{\, =\, }\mathrm{2.5}a_{\mathrm{3}}l\mathrm{\, =\, }\mathrm{2.5}\times \left (\frac{P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }\right )\times l}}$$

$$\mathrm{\mathit{\therefore K_{\mathrm{2}}\mathrm{\, =\, }\frac{\mathrm{2.5}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}\, \, \, \cdot \cdot \cdot \left ( 4 \right )}$$

Now, comparing equations (1) & (4), we get,

$$\mathrm{\mathit{ \frac{K_{\mathrm{2}}}{K}\mathrm{\, =\, }\frac{\left ( \frac{\mathrm{2.5}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, } \right )}{\mathrm{\mathit{\left ( \frac{\mathrm{8}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, } \right )}}}}\mathrm{\, =\, }\frac{5}{16}}$$

$$\mathrm{\mathit{\therefore K_{\mathrm{2}}\mathrm{\, =\, }}\left ( \frac{5}{16} \right )\times \mathit{K}\: \: \cdot \cdot \cdot \left ( 5 \right )}$$

Hence, the volume of conductor material required in single phase three wire overhead AC transmission system is (5/16) times of that required in single phase two wire system with one conductor earthed.