Conductor Material Required in Overhead DC Transmission System

Overhead DC Transmission System

The overhead transmission system is the one in which the conductors are hanged with the help of pole supports. When the transmission lines carry direct current, then the system is called the DC transmission system.

There are three types of overhead DC transmission systems viz. −

• Two wire DC system with one conductor earthed
• Two wire DC system with mid-point earthed
• Three wire DC system

Conductor Material Required in Two-Wire DC System with One Conductor Earthed

Consider a two wire DC system with one conductor earthed as shown in Figure-1. Here, one is the positive wire and the other is the negative wire and the load is connected between the two wires.

Here,

• $\mathrm{\mathit{V_{m}}}$ = Maximum voltage between conductors
• 𝑃 = Power to be transmitted

Therefore, the load current is given by,

$$\mathrm{\mathit{I_{L\mathrm{1}}\mathrm{\, =\, }\frac{P}{V_{m}}}}$$

Let, 𝑅₁ is the resistance of each line conductor, then,

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

Where, 𝑎₁ is the cross-sectional area of the conductor.

Hence, the total power loss in the line is given by,

$$\mathrm{\mathit{W\mathrm{\, =\, }\mathrm{2}\,I_{L\mathrm{1}}^{\mathrm{2}}\,R_{\mathrm{1}}\mathrm{\, =\, }\mathrm{2}\times \left ( \frac{P}{V_{m}} \right )^{\mathrm{2}}\times \frac{\rho l}{a_{\mathrm{1}}}\mathrm{\, =\, }\frac{\mathrm{2}P^{\mathrm{2}}\rho l}{V_{m}a_{\mathrm{1}}}}}$$

⇒ Cross sectional area,

$$\mathrm{\mathit{a_{\mathrm{1}}\mathrm{\, =\, }\frac{\mathrm{2}P^{\mathrm{2}}\rho l}{WV_{m}}}}$$

Therefore, the volume of conductor material required in two wire DC system (say K) is given by,

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

$$\mathrm{\mathit{\therefore K\mathrm{\, =\, }\frac{\mathrm{4}P^{\mathrm{2}}\rho l^{\mathrm{2}}}{WV_{m}} }\; \; \; ...\left ( 1 \right )}$$

Conductor Material Required in Two-Wire DC System with Mid-Point Earthed

Consider a two wire DC system with mid-point earthed as shown in Figure-2.

In this system, the maximum voltage between any conductor and the earth is 𝑉_𝑚, therefore the maximum voltage between two conductors is 2𝑉_𝑚. Therefore, the load current is given by,

$$\mathrm{\mathit{I_{L\mathrm{2}}\mathrm{\, =\, }\frac{P}{\mathrm{2}V_{m}}}}$$

Let a₂ is the area of cross section of the conductor and R₂ be the resistance of each line conductor, then

$$\mathrm{\mathit{R_{\mathrm{2}}\mathrm{\, =\, }\frac{\rho l}{a_{\mathrm{2}}}}}$$

Therefore, the total lines losses are given by,

$$\mathrm{\mathit{W\mathrm{\, =\, }\mathrm{2}\,I_{L\mathrm{2}}^{\mathrm{2}}\,R_{\mathrm{2}}\mathrm{\, =\, }\mathrm{2}\times \left ( \frac{P}{\mathrm{2}V_{m}} \right )^{\mathrm{2}}\times \frac{\rho l}{a_{\mathrm{2}}}\mathrm{\, =\, }\frac{P^{\mathrm{2}}\rho l}{\mathrm{2}V_{m}^{\mathrm{2}}a_{\mathrm{2}}}}}$$

⇒ Area of cross section,

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

Hence, the volume of conductor material required in the two wire DC system with mid-point earthed (let 𝐾₁) is given by,

$$\mathrm{\mathit{K_{\mathrm{1}}\mathrm{\, =\, }\mathrm{2}\times a_{\mathrm{2}}\times l\mathrm{\, =\, }\frac{P^{\mathrm{2}}\rho l^{\mathrm{2}}}{V_{m}^{\mathrm{2}}W}\; \; \: \cdot \cdot \cdot \left ( \mathrm{2} \right )}}$$

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

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

$$\mathrm{\mathit{\therefore K_{\mathrm{1}}\mathrm{\, =\, }\frac{K}{\mathrm{4}} }\; \; \; \cdot \cdot \cdot \left ( 3 \right )}$$

Thus, the volume of conductor material required in a two-wire DC system with mid-point earthed is one-fourth of that required in a two-wire DC system with one conductor earthed.

Conductor Material Required in Three-Wire DC System

Consider a three-wire DC system in which two outers and a neutral wire be earthed at the generator end as shown in Figure-3.

If a balanced load is connected to the system, no current will flow in the neutral wire. Then, for the balanced load, the load current is given by,

$$\mathrm{\mathit{I_{L\mathrm{3}}\mathrm{\, =\, }\frac{P}{\mathrm{2}V_{m}}}}$$

If a₃ is the area of cross section of each outer wire, then the resistance R3 of each outer line conductor is given by,

$$\mathrm{\mathit{R_{\mathrm{3}}\mathrm{\, =\, }\frac{\rho l}{a_{\mathrm{3}}}}}$$

Thus, the total line losses are given by,

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

⇒ Area of cross section, $\mathrm{\mathit{a_{\mathrm{3}}\mathrm{\, =\, }\frac{P^{\mathrm{2}}\rho l}{\mathrm{2}V_{m}^{\mathrm{2}}W}}}$

Also, assuming the area of cross-section of the neutral wire to be half that of the outer line wires. Then, the volume of the conductor material required in the three-wire DC system (say K₃) is given by,

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

$$\mathrm{\mathit{\therefore K_{\mathrm{3}}\mathrm{\, =\, }\frac{\mathrm{1.25}P^{\mathrm{2}}\rho l^{\mathrm{2}}}{V_{m}^{\mathrm{2}}W }\; \; \; \cdot \cdot \cdot \left ( \mathrm{4} \right ) }}$$

Comparing equations (1) & (4), we get,

$$\mathrm{\mathit{ \frac{K_{\mathrm{3}}}{K}\mathrm{\, =\, }\frac{\left ( \frac{\mathrm{1.25}P^{\mathrm{2}}\rho l^{\mathrm{2}}}{V_{m}^{\mathrm{2}}W } \right )}{\left ( \mathrm{\mathit{\frac{\mathrm{4}P^{\mathrm{2}}\rho l^{\mathrm{2}}}{WV_{m}}}} \right )}}\mathrm{\, =\, }\frac{5}{16}}$$

$$\mathrm{\mathit{\therefore K_{\mathrm{3}}\mathrm{\, =\, }}\frac{5}{16}\times \mathit{K}\; \; \; \cdot \cdot \cdot \left ( 5 \right )}$$

Hence, from the equation (5) it is clear that the volume of conductor material required in the three-wire DC system is (5/16)^th of that required in a two-wire DC system with one conductor earthed.