Question

A HVDC line, made of copper with resistivity (or specific resistance) \(\rho=1.7 \times 10^{-8}\) \([\Omega \mathrm{m}]\) has a length \(l=800[\mathrm{~km}]\) and a diameter of \(2 r=22[\mathrm{~cm}]\). It is delivering 1000 [MW] of electricity at a grid point with a voltage of \(1000[\mathrm{kV}]\). Calculate (a) the electric resistance of the line; (b) the electrical current; (c) the heat loss in the cable, both in absolute value and in percentage of the transmitted power; (d) the voltage difference \(\Delta V\) between the beginning and the end of the line. Note: in practice, the resistance of the cable is higher and losses are in the order of 3 to \(2.5 \% / 1000[\mathrm{~km}]\).

Short answer

The resistance is 0.35789 Ω, current is 1000 A, heat loss is 357.89 kW (0.0358% of power), and voltage drop is 357.89 V.

Step-by-step solution

Calculate the Cross-Sectional Area of the Cable

First, we need the cross-sectional area \(A\) of the cable, as the wire is cylindrical. The diameter \(d\) of the wire is 22 cm, so the radius \(r\) is half of that: \(11 \, \text{cm} = 0.11 \, \text{m}\). The cross-sectional area is given by the formula \(A = \pi r^2\). Hence, \(A = \pi \times (0.11)^2 = 0.038 \text{ m}^2\).

Calculate the Resistance of the Line

The formula to calculate resistance \(R\) is given by \(R = \frac{\rho l}{A}\), where \(\rho\) is the resistivity, \(l\) the length of the wire, and \(A\) the cross-sectional area. Substituting the given values: \(R = \frac{1.7 \times 10^{-8} \times 800,000}{0.038}\). After calculation, \(R \approx 0.35789 \, \Omega\).

Calculate the Electrical Current

The power \(P\) delivered is 1000 MW = \(1000 \times 10^6 \, \text{W}\). The voltage \(V\) is 1000 kV = \(1000 \times 10^3 \, \text{V}\). Using the formula \(P = IV\), where \(I\) is the current, \(I = \frac{P}{V} = \frac{1000 \times 10^6}{1000 \times 10^3}\). Thus, \(I = 1000 \, \text{A}\).

Calculate Heat Loss in the Cable

The heat loss \(P_{\text{loss}}\) due to resistance is calculated using the formula \(P_{\text{loss}} = I^2 R\). Substituting for \(I\) and \(R\): \(P_{\text{loss}} = (1000)^2 \times 0.35789 = 357,890 \, \text{W} = 357.89 \, \text{kW}\).

Calculate Heat Loss Percentage

To find the percentage, we take the heat loss and divide it by the total power transmitted. Thus percentage loss = \(\frac{357.89}{1000,000} \times 100\% \). Hence, the percentage loss is approximately \( 0.0358\% \).

Calculate the Voltage Drop Across the Line

The voltage drop \(\Delta V\) can be calculated using Ohm's law: \(\Delta V = IR\), substituting \(I = 1000 \, A\) and \(R = 0.35789 \, \Omega\). Thus, \(\Delta V = 1000 \times 0.35789 = 357.89 \, \text{V}\).

Key concepts

Resistivity

Resistivity is a fundamental property of materials that indicates how much they resist the flow of electric current. It is often denoted by the Greek letter \( \rho \) and measured in ohm-meters \([\Omega \cdot m]\). A material with low resistivity allows electric charge to flow more easily through it, making it a good conductor, like copper. Resistors and components in circuits use materials with higher resistivity to hinder current flow intentionally.
Resistivity depends on several factors, including the material type and temperature. For example, the resistivity of copper typically used in electrical systems is around \(1.7 \times 10^{-8} \Omega m\). This low resistivity contributes to its widespread use in electrical wiring, allowing efficient electricity transport over vast distances, such as in HVDC lines. When designing anything electrical, choosing materials with appropriate resistivities ensures efficiency and minimizes energy loss.

Power Loss

Power loss in electrical systems often occurs due to resistive components, turning some electrical energy into heat. Calculating power loss is essential to ensure systems operate efficiently and safely. The formula for power loss due to resistance is given by \( P_{\text{loss}} = I^2R \), where \( I \) is the current, and \( R \) is the resistance of the circuit or component.

• A higher current or increased resistance results in greater power loss.
• Reducing resistance or optimizing current flow can lower power losses, conserving energy.

Power loss also affects system performance. In HVDC lines delivering massive energies across cities, minimizing losses by choosing materials with low resistivity is crucial. Understanding how power loss works help engineers save energy and preserve the environment by reducing the carbon footprint associated with power generation.

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering, describing the relationship between voltage, current, and resistance in a circuit. It can be expressed as \( V = IR \), where \( V \) is the voltage across the component, \( I \) is the current flowing through it, and \( R \) is the resistance. This relationship helps calculate unknown values when the other two are provided, making it a cornerstone in circuit analysis and design.

• Ohm's Law helps determine how much voltage is needed to drive a certain current through a specific resistance in a device or network.
• Understanding this law aids in predicting how changes in one quantity affect the others, crucial for troubleshooting and managing circuit components.

By applying Ohm's Law, we can compute the voltage drop across transmission lines, helping engineers ensure voltage at delivery points remains consistent for end-users.

Cross-Sectional Area

In electrical conductors, the cross-sectional area plays a significant role in determining resistance. A conductor's cross-sectional area is the surface area of a slice through the conductor perpendicular to the current's flow, typically measured in square meters \(m^2\). Using a larger cross-sectional area decreases resistance, allowing currents to pass more freely.
To calculate the cross-sectional area of a cylindrical wire like our copper HVDC line, use the formula \( A = \pi r^2 \), where \( r \) is the radius. With a diameter of 22 cm, the radius is half, or 11 cm (0.11 m). This calculation results in \( A = \pi \times (0.11)^2 \approx 0.038 \text{ m}^2\).

• Larger cross-sectional areas lead to lower resistance and less heat, beneficial for high-power applications.
• Designers adjust this parameter to optimize performance based on application requirements.

Understanding cross-sectional area helps engineers choose the right wire size to avoid overheating and ensure structures support the desired current flows efficiently.