Volume of Conductor Material Required in Underground Three-Phase AC System

Underground Three Phase AC System

When the three phase AC electric power is transmitted through the underground cables, then the system of electric power transmission is called the underground three phase AC system.

Depending upon the number of conductors used, the underground three-phase AC system of electric power transmission is classified into two types −

• Three-Phase Three-Wire System
• Three-Phase Four-Wire System

Conductor Material Required in Underground Three-Phase Three-Wire System

The circuit diagram of the underground 3-phase 3-wire AC system is shown in Figure-1. In this system, the three line conductors are taken from the three phase windings of the alternator.

Let the maximum value of voltage between line conductors is Vm. Then, per phase maximum voltage is 𝑉𝑚/√3. Therefore, the RMS value of the per phase voltage is given by,

$$\mathrm{RMS\: voltage\: per\: phase\mathrm{\, =\, }\mathit{\frac{V_{m}/\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}}\mathrm{\, =\, }\frac{V_{m}}{\sqrt{\mathrm{6}}}}}$$

Also,

$$\mathrm{Power\: transmitted\: per \: phase\mathrm{\, =\, }\mathit{\frac{P}{\mathrm{3}}}}$$

Therefore, the load current per phase is given by

$$\mathrm{\mathit{I_{\mathrm{1}}\mathrm{\, =\, }\frac{P/\mathrm{3}}{\frac{V_{m}}{\sqrt{\mathrm{6}}}\mathrm{cos\, \phi }}\mathrm{\, =\, }\frac{\sqrt{\mathrm{6}}P}{\mathrm{3}V_{m}\,\mathrm{cos\, \phi }} }}$$

If 𝑎1 is the area of cross section per conductor and R1 is the resistance of each line conductor, then the total power losses in the lines are,

$$\mathrm{\mathit{W\mathrm{\, \mathrm{\, =\, }\, }\mathrm{3}I\mathrm{_{1}^{2}}\, R_{\mathrm{1}}\mathrm{\, =\, }\mathrm{3}\times \left ( \frac{\mathrm{\sqrt{6}}P}{\mathrm{3}V_{m}\mathrm{cos}\, \phi } \right )^{\mathrm{2}}\times \left ( \frac{\rho \, l}{a_{\mathrm{1}}} \right )}}$$

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

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

Hence, the volume of conductor material required in the underground 3-phase 3-wire AC system, say K, is given by,

$$\mathrm{\mathit{K\mathrm{\, \mathrm{\, =\, }\, }\mathrm{3}a_{\mathrm{1}}l\mathrm{\, \mathrm{\, =\, }\, }\mathrm{3}\times \frac{\mathrm{2}P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }\times l}}$$

$$\mathrm{\mathit{\therefore K\mathrm{\, \mathrm{\, =\, }\, }\frac{\mathrm{6}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}\, \, \, \cdot \cdot \cdot \left ( 1 \right )}$$

Conductor Material Required in Underground Three-Phase Four-Wire System

The underground 3-phase 4-wire AC system for electric power transmission is shown in Figure-2.

There are three line conductors taken from the outer terminals of the windings of alternator and the fourth one is the neutral wire taken from the common terminal of three windings. If the load connected to the system is balanced, then no current flows through the neutral wire. As a result, the four system reduces to 3-wire system.

Let,

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

• $\mathit{\frac{V_{m}}{\sqrt{\mathrm{2}}}\mathrm{\mathrm{\, =\, }Maximum\: voltage\: per\: phase}}$

• $\mathit{\frac{P}{\mathrm{3}}\mathrm{\mathrm{\, =\, }Power\: transmitted\: per\: phase}}$

Thus, the RMS value of the per phase voltage is given by,

$$\mathrm{RMS\: voltage\: per\: phase\mathrm{\, =\, }\mathit{\frac{V_{m}/\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}}\mathrm{\, =\, }\frac{V_{m}}{\sqrt{\mathrm{6}}}}}$$

Hence, the load current per phase is,

$$\mathrm{\mathit{I_{\mathrm{2}}\mathrm{\, =\, }\frac{P/\mathrm{3}}{\frac{V_{m}}{\sqrt{\mathrm{6}}}\mathrm{cos\, \phi }}\mathrm{\, =\, }\frac{\sqrt{\mathrm{6}}P}{\mathrm{3}V_{m}\,\mathrm{cos\, \phi }}}}$$

$$\mathrm{\therefore Line\: losses, \mathit{W\mathrm{\, \mathrm{\, =\, }\, }\mathrm{3}I\mathrm{_{2}^{2}}\, R_{\mathrm{2}}\mathrm{\, =\, }\mathrm{3}\times \left ( \frac{\mathrm{\sqrt{6}}P}{\mathrm{3}V_{m}\mathrm{cos}\, \phi } \right )^{\mathrm{2}}\times \left ( \frac{\rho \, l}{a_{\mathrm{2}}} \right )}}$$

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

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

Assuming the area of cross section of neutral wire to be half that of line conductor.

Therefore, the volume of conductor material required in the underground 3-phase 4-wire AC system, say K1, is given by,

$$\mathrm{\mathit{K_{\mathrm{1}}\mathrm{\, \mathrm{\, =\, }\, }\mathrm{3.5}\, \, a_{\mathrm{2}}l\mathrm{\, \mathrm{\, =\, }\, }\mathrm{3.5}\times \frac{\mathrm{2}P^{\mathrm{2}\, }\rho l}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }\times l}}$$

$$\mathrm{\mathit{\therefore K_{\mathrm{1}}\mathrm{\, \mathrm{\, =\, }\, }\frac{\mathrm{7}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, }}\, \, \, \cdot \cdot \cdot \left ( 2 \right )}$$

Now, comparing equations (1) & (2), we get,

$$\mathrm{\mathit{\frac{K_{\mathrm{1}}}{K}\mathrm{\, \mathrm{\, =\, }\, }\frac{\left ( \frac{\mathrm{7}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, } \right )}{\left ( \frac{\mathrm{6}P^{\mathrm{2}\, }\rho l^{\mathrm{2}}}{WV_{m}^{\mathrm{2}}\,\mathrm{cos^{2}\phi }\, } \right )}}\mathrm{\, =\, }\frac{7}{6}}$$

$$\mathrm{\mathit{\therefore K_{\mathrm{1}}}\mathrm{\, =\, }\frac{7}{6}\times \mathit{K}\: \cdot \cdot \cdot \left ( 3 \right )}$$

Hence, the conductor material required in the underground 3-phase 4-wire AC system is (7/6) times of that required in the underground 3-phase 3-wire AC system.