Question

A high-voltage direct current (HVDC) transmission line carries electrical power a distance of \(643.1 \mathrm{~km}\). The line transmits \(7935 \mathrm{MW}\) of power at a potential difference of \(1.177 \mathrm{MV}\). If the HVDC line consists of one copper wire of diameter \(2.353 \mathrm{~cm}\), what fraction of the power is lost in transmission?

Short answer

Answer: 1.434%

Step-by-step solution

Determine the current transmitted by the wire.

We can find the current transmitted by the wire by dividing the transmitted power by the potential difference. The provided information states that the transmission line transmits 7935 MW of power at a potential difference of 1.177 MV. I = \frac{Power}{Potential \, Difference} = \frac{7935 \times 10^6 \mathrm{W}}{1.177 × 10^6 \mathrm{V}} = 6740.88 \mathrm{A}

Calculate the resistance of the copper wire.

To calculate the resistance, we first need to find the resistivity (ρ) of the copper wire. The resistivity of copper is approximately 1.68 × 10^{-8} Ωm. Next, we need to find the cross-sectional area (A) of the wire using its diameter. The diameter of the copper wire is given as 2.353 cm, or 0.02353 m. A = \frac{π d^2}{4} = \frac{π \times (0.02353 \mathrm{m})^2}{4} = 4.342 × 10^{-4} \mathrm{m^2} Now, we can find the resistance (R) using the formula: R = \frac{ρL}{A} where L is the length of the wire, which is 643.1 km or 643100 m. R = \frac{1.68 × 10^{-8} Ωm \times 643100 \mathrm{m}}{4.342 × 10^{-4} \mathrm{m^2}} = 0.0250 \mathrm{Ω}

Calculate the power loss due to the resistance.

Now that we have the current (I) and resistance (R) of the transmission line, we can find the power loss (P_loss) using the formula: P_loss = I^2R = (6740.88 \mathrm{A})^2 \times 0.0250 \mathrm{Ω} = 1.138 × 10^8 \mathrm{W} = 113.8 \mathrm{MW}

Find the fraction of the power lost in transmission.

Finally, we can determine the fraction of power lost in the transmission by dividing the power loss (P_loss) by the transmitted power (P_trans). Fraction of power lost = \frac{P_{loss}}{P_{trans}} = \frac{113.8 \times 10^6 \mathrm{W}}{7935 \times 10^6 \mathrm{W}} = 0.01434 Multiplying this by 100 gives the percentage of power loss: Percentage power loss = 0.01434 \times 100 = 1.434 % Thus, 1.434% of the power is lost during the transmission through the HVDC line.

Key concepts

HVDC (High-Voltage Direct Current)

High-Voltage Direct Current (HVDC) is a technology used to transmit electricity over long distances. It is an efficient way of transferring large amounts of power over vast distances because it minimizes power losses when compared to alternating current (AC) transmission systems. HVDC systems utilize direct current (DC), which means the direction of the flow of electric charge is constant.

Unlike AC, which oscillates back and forth, DC provides a steady current and voltage, which simplifies the design of transformers and allows for consistent power transmission. HVDC is particularly useful in underwater and underground cables, where AC systems would face significant challenges due to capacitance and the need for thicker cables.

Moreover, HVDC allows for asynchronous interconnection of grids, facilitating power exchange between different regions without the need for frequency synchronization. This feature contributes to the stabilization of the grid and provides a more reliable power supply.

Electrical Resistance

Electrical resistance is a fundamental concept in the field of electricity, which represents the opposition to the flow of electric current through a conductor. The resistance of a conductor depends on its material, length, and cross-sectional area. Materials with low resistivity, such as copper or aluminum, are commonly used in electrical wiring to reduce resistance and hence, energy losses.

When electric current passes through a conductor with resistance, energy is lost in the form of heat due to the collisions between the charge carriers (electrons) and the atoms within the conductor. This is known as the Joule heating effect. The resistance (\( R \) ) of a wire can be calculated with the formula: \[ R = \frac{\rho L}{A} \[ where \( \rho \) is the resistivity of the material, \( L \) is the length of the conductor, and \( A \) is the cross-sectional area.

The resistivity of copper, which is low compared to most other materials, makes it an ideal conductor for electrical cables. However, even with a good conductor like copper, resistance can still cause a significant amount of power to be lost over long distances, which is why high voltages are used in power transmission to reduce these losses.

Power Loss in Transmission

Power loss during transmission is a key concern for the efficiency of electrical power systems. One of the main causes of power loss is the inherent resistance of the materials used to make power lines. As electrical current travels along these lines, energy is dissipated in the form of heat — a phenomenon explained by Joule’s law. This law states that the power loss (\( P_{loss} \) ) due to resistance can be expressed by the equation: \[ P_{loss} = I^2R \[ where \( I \) is the current flowing through the conductor and \( R \) is its resistance.

In the context of the exercise, calculating the power loss involves determining the current and the resistance of the transmission line, which requires knowledge of the wire’s dimensions and material properties. The power loss calculation exemplifies how electrical engineering principles are used to address efficiency challenges in real-world systems.

Reducing power loss is crucial, as it directly impacts the cost of electricity and also has environmental implications. One of the strategies to reduce these losses is by using HVDC for long-distance power transmission, as it reduces resistive losses when compared to equivalent AC transmission systems. The exercise concludes that a fraction of the transmitted power is inevitably lost, but HVDC technology ensures that this loss remains as low as possible, thereby maximizing efficiency and reducing costs in power transmission networks.