# Volume of Conductor Material Required in Underground Single-Phase AC System

Power SystemsUtilization of Electrical EnergyTransmission of Electric PowerDistribution of Electric Power

## Underground Single-Phase AC System

The electric transmission system which transmits the single-phase AC electric power through the underground cables is termed as the underground single-phase AC system of transmission.

The underground single-phase AC system of electric power transmission is classified into three types viz. −

• Single-Phase Two-Wire AC System

• Single-Phase Two-Wire AC System with Mid-Point Earthed

• Single-Phase Three-Wire AC System

The various systems of the electric transmission require different volume of conductor material. The volume of conductor material required in the above 1-phase underground AC system is described below.

## Conductor Material Required in 1-Phase 2-Wire Underground AC System

The typical single-phase 2-wire AC underground AC system is shown in Figure-1.

In this system, the voltage between the conductors is Vm.

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

Let the power factor of the load is cos𝜙. Then, the load current is given by,

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

If a1 is the cross sectional area of each conductor and R1 is the resistance per conductor. Then, the Lines losses are given by,

$$\mathrm{\mathit{W\mathrm{\, \mathrm{\, =\, }\, }\mathrm{2}I\mathrm{_{1}^{2}}\, R_{\mathrm{1}}}}$$

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

$$\mathrm{\therefore Area\: of\: cross\: section \: ,\mathit{ a_{\mathrm{1}}\mathrm{\, \mathrm{\, =\, }\, }\frac{\mathrm{4}P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}}$$

Hence, the volume of conductor material required in the single-phase two-wire underground AC system, say K, is given by,

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

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

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

Figure-2 shows the underground single-phase two-wire AC system with mid-point earthed.

Here,

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

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

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

Thus, the load current is given by,

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

If 𝑎2 is the cross-section area per conductor and R2 is the resistance per conductor. Then, the total power loss in the line conductors is

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

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

$$\mathrm{\therefore Area\: of\: cross\: section \: ,\mathit{ a_{\mathrm{2}}\mathrm{\, \mathrm{\, =\, }\, }\frac{\mathrm{4}P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}}$$

Hence, the volume of conductor material required in the underground single-phase two-wire AC system with mid-point earthed, say K1, is given by,

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

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

Hence, from equations (1) and (2), it is clear that the volume of conductor material required in the 1-phase 2-wire underground AC system with mid-point earthed is same as that required for the 1-phase 2-wire underground AC system.

## Conductor Material Required in the 1-Phase 3-Wire Underground AC System

The underground single-phase three-wire AC system is shown in Figure-3. In this system, an additional neutral wire is provided. The cross-sectional area of the neutral wire is to be half of the either of the outer wire.

Also, if the load connected to the system is balanced. Then, no current flows in the neutral wire. Therefore, the line losses are given by,

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

$$\mathrm{\mathit{\Rightarrow W\mathrm{\, \mathrm{\, =\, }\, }\frac{\mathrm{4}P^{\mathrm{2}\, }\rho l}{V_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi \,a_{\mathrm{3}}} }}}$$

$$\mathrm{\therefore Area\: of\: cross\: section \: ,\mathit{ a_{\mathrm{3}}\mathrm{\, \mathrm{\, =\, }\, }\frac{\mathrm{4}P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}}$$

Hence, the volume of conductor material required in the single-phase three-wire underground AC System, let say K2, is given by,

$$\mathrm{\mathit{K_{\mathrm{2}}\mathrm{\, \mathrm{\, =\, }\, }\mathrm{2.5}a_{\mathrm{3}}l\mathrm{\, \mathrm{\, =\, }\, }\mathrm{2.5}\times \left (\frac{\mathrm{4}P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }\right )\times l}}$$

$$\mathrm{\mathit{\therefore K_{\mathrm{2}}\mathrm{\, \mathrm{\, =\, }\, }\frac{\mathrm{10}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}\, \, \, \cdot \cdot \cdot \left ( 3 \right )}$$

Updated on 25-Feb-2022 10:26:59