Advantages and Limitations of High Transmission Voltage

Electron Electronics & Electrical Digital Electronics

Advantages of High Transmission Voltage

Electric power is transmitted at very high voltages due to some technical and economic reasons which are described as follows −

1. Reduces the Volume of Conductor Material

Consider the electric power being transmitting through the three-phase three-wire transmission system.

Let,

P = Power transmitted (in Watts)
V = Line voltage (in Volts)
$\mathrm{cos}\:\phi$ = Load power factor
R = Resistance per conductor (in ohms)
$\mathit{\rho}$ = Resistivity of conductor material
l = length of transmission line (in meters)
a = cross sectional area of conductor

Therefore, the load current is given by,

$$\mathrm{\mathit{I}\:=\:\frac{\mathit{P}}{\sqrt{3}\mathit{V}\mathrm{cos\:\phi }}}$$

And the resistance per conductor is

$$\mathrm{\mathit{R}\:=\:\rho \:\frac{\mathit{l}}{\mathit{a}}}$$

Thus, the total power loss in the transmission line is

$$\mathrm{\mathit{W}\:=\:3\mathit{I^{\mathrm{2}}}\mathit{R}\:=\:3\mathrm{\left( \frac{\mathit{P}}{\sqrt{3}cos\:\phi} \right )^{\mathrm{2}}}\:\times \:\mathrm{\left(\rho \:\frac{\mathit{l}}{\mathit{a}} \right )}\:=\:\frac{\mathit{\rho lP^{\mathrm{2}}}}{\mathit{aV^{\mathrm{2}}\mathrm{cos^{\mathrm{2}}\:\phi }}}}$$

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

As there are three conductors, the total volume of conductor material required is given by,

$$\mathrm{\mathrm{Total\: volume\: of\: conductor \:material}\:=\:3\:\times \:\mathit{a}\:\times \:\mathit{l}\:=\:3\:\times \:\mathrm{\left (\frac{\mathit{P^{\mathrm{2}}\rho \mathit{l}}}{\mathit{WV^{\mathrm{2}}\:\mathrm{cos^{2}}\:\phi }} \right )\:\times \:\mathit{l}}}$$

$$\mathrm{\therefore \mathrm{Volume \:of\: conductor \:material}\:=\:\frac{3\mathit{P^{\mathrm{2}}\rho \mathit{l^{\mathrm{2}}}}}{\mathit{WV^{\mathrm{2}}\:\mathrm{cos^{2}}\:\phi }}\:\:\:\cdot \cdot\cdot \mathrm{\left ( 1 \right )}}$$

From equation (1), it is clear that for the given values of P, $\rho$, l and W, the volume of conductor material required is inversely proportional to the square of transmission voltage and load power factor. Therefore, if the power is transmitted at high voltage, then lesser is the conductor material required.

2. Decreases Percentage Line Drop

The voltage drop in the transmission line is given by,

$$\mathrm{\mathrm{Line \:drop}\:=\:\mathit{IR}\:=\:\mathit{I}\:\times \:\mathrm{\left ( \rho \frac{\mathit{l}}{\mathit{a}} \right )}}$$

Let J is the current density of the conductor, then

$$\mathrm{\mathit{a}\:=\:\frac{\mathit{I}}{\mathit{J}}}$$

$$\mathrm{\Rightarrow \mathrm{Line\: drop}\:=\:\mathit{I}\:\times \mathrm{\left [ \rho \:\frac{\mathit{l}}{\mathrm{\left ( \frac{\mathit{I}}{J} \right )}} \right ]}\:=\:\rho \mathit{Jl}}$$

$$\mathrm{\therefore \mathrm{Percentage\: line\: drop}\:=\:\frac{\mathit{\rho Jl}}{\mathit{V}}\:\times \:100\%\:\:\:\cdot \cdot \cdot \mathrm{\left ( 2 \right )}}$$

From equation (2), it is clear that the percentage line drop is inversely proportional to the transmission voltage. Therefore, the percentage line drop decreases when the transmission voltage increases.

3. Increases Transmission Efficiency

The input power to the transmission line is given by,

$$\mathrm{\mathit{P_{\mathit{in}}}\:=\:\mathit{P}\:+\:\mathrm{Total \:power\: loss}\:=\:\mathit{P}\:+\:\frac{\mathit{\rho lP^{\mathrm{2}}}}{\mathit{aV^{\mathrm{2}}\mathrm{cos^{\mathrm{2}}\:\phi }}}}$$

$$\mathrm{\Rightarrow \mathit{P_{\mathit{in}}}\:=\:\mathit{P}\:+\:\frac{\rho \mathit{lP}^{\mathrm{2}}\:\times \:\mathit{J}}{\mathit{V^{\mathrm{2}}\mathrm{cos^{\mathrm{2}}\:\phi }\:\times \:\mathit{I}}}\:=\:\mathit{P}\:+\:\mathrm{\left ( \frac{\rho \mathit{lP^{\mathrm{2}}J}}{\mathit{V^{\mathrm{2}}\mathrm{cos^{2}}\:\phi }}\:\times \:\frac{1}{\mathit{I}} \right )}}$$

$$\mathrm{\Rightarrow \mathit{P_{\mathit{in}}}\:=\:\mathit{P}\:+\:\mathrm{\left [ \mathrm{\left ( \frac{\rho \mathit{l}\mathit{P}^{\mathrm{2}}\mathit{J}}{\mathit{V^{\mathrm{2}}}\mathrm{cos}^{\mathrm{2}}\:\phi}\right )}\:\times \:\mathrm{\left(\frac{\sqrt{3}\mathit{V}\mathrm{cos}\:\phi}{\mathit{P}} \right )} \right ]}}$$

$$\mathrm{\Rightarrow \mathit{P_{\mathit{in}}}\:=\:\mathit{P}\:+\:\frac{\sqrt{3}\rho \mathit{lPJ}}{\mathit{V}\mathrm{cos\:\phi }}\:=\:\mathit{P}\mathrm{\left(1\:+\: \frac{\sqrt{3}\rho \mathit{lJ}}{\mathit{V}\mathrm{cos\:\phi }} \right )}}$$

Since the transmission efficiency is defined as,

$$\mathrm{\eta \:=\:\frac{\mathrm{Output\: power}}{\mathrm{Input\:power}}\:=\:\frac{\mathit{P}}{\mathit{P}\mathrm{\left(1\:+\:\frac{\sqrt{3}\rho \mathit{lJ}}{\mathit{V}\mathrm{cos\:\phi }} \right)}}\:=\:\frac{1}{\mathrm{\left(1\:+\:\frac{\sqrt{3}\rho \mathit{lJ}}{\mathit{V}\mathrm{cos\:\phi }} \right )}}}$$

By using Binomial theorem, we get,

$$\mathrm{\eta \:\cong \:\mathrm{\left ( 1\:-\:\frac{\sqrt{3}\rho \mathit{lJ}}{\mathit{V}\mathrm{cos\:\phi }} \right )}\:\:\:\cdot \cdot \cdot \mathrm{\left ( 3 \right )}}$$

Since $\rho$, l and J are constants, therefore the transmission efficiency increases when the transmission voltage increases.

Limitations of High Transmission Voltage

The limitations of the high transmission voltage in AC transmission system are as follows −

High transmission voltage increases the cost of insulting the conductors.
High voltage also increases the cost of electrical equipment such as transformers, switches and circuit breakers, etc.

Therefore, there is also a limit to the high transmission voltage which can be economically employed in transmission of electric power.

Manish Kumar Saini

Updated on 15-Feb-2022 07:51:20

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