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Chapter 27: Problem 58

A high-voltage transmission line carries \(1000 \mathrm{A}\) starting at \(700 \mathrm{kV}\) for a distance of \(100 \mathrm{mi}\). If the resistance in the wire is \(0.500 \Omega / \mathrm{mi},\) what is the power loss due to resistive losses?

Short Answer

Expert verified
The power loss due to resistive losses is 50 MW.

Step by step solution

01

Find the Total Resistance

The total resistance (R_total) can be found by multiplying the resistance/mile by the total miles. R_total = R_per_mile * total_miles = 0.500 Ohm/mile * 100 miles = 50 Ohms.
02

Calculate the Power Loss

Next, the power loss (P) due to resistive losses can be found by substituting the values of I (current) and R_total (total resistance) into our power loss formula (P = I^2*R). Thus, P = 1000 A^2 * 50 Ohms = 50,000,000 W or 50 MW.
03

Convert Watts to Kilowatts

It is common in power systems to express power in terms of Kilowatts (KW) or Megawatts (MW). As 1W = 1*10^-6 MW, convert the power loss from Watts to Megawatts using this factor. Thus, P = 50,000,000 W * 1 * 10^-6 MW/W = 50 MW.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistive Losses
Resistive losses occur when electrical energy is converted into heat as electricity flows through a conductor. This is often referred to as "I squared R loss," because the power lost in the form of heat is calculated using the formula \( P = I^2 \times R \). Here, \( I \) is the electrical current flowing through the conductor, and \( R \) is the resistance of the conductor. These losses are inevitable in any electrical system, and they increase with the square of the current.
  • The higher the current, the greater the resistive losses.
  • Materials with higher resistance will also lead to bigger energy losses.
Understanding resistive losses is crucial for efficient energy transmission, as they can significantly affect the performance and economics of transmitting electricity over long distances.
High-Voltage Transmission
High-voltage transmission lines are designed to minimize energy losses when power is sent over long distances. They operate by increasing the voltage of the electricity being transmitted, allowing the same power to be carried at a lower current. This approach is beneficial because resistive losses (\( P = I^2 \times R \)) are directly proportional to the square of the current flowing through the line. Reducing the current can effectively minimize these losses.
  • Higher voltage means lower current for the same power transfer.
  • This results in reduced resistive losses.
  • High-voltage transmission lines often carry voltages in the several hundred kilovolt (kV) range.
Thus, by using high-voltage transmission, energy companies can improve efficiency and reduce the overall cost of delivering electricity from power plants to end-users.
Resistance in Conductors
Resistance is a property of a material that describes how difficult it is for an electric current to pass through it. Every conductor, like the wires in power lines, has some resistance. This resistance leads to heating and energy loss when current flows through the conductor.
  • Conductors with lower resistance are preferred for long-distance transmission.
  • The resistance of a wire is influenced by its material, length, and cross-sectional area.
  • Thinner, longer wires have higher resistance than shorter, thicker ones.
The equation to calculate resistance in a wire is \( R = \rho \times \frac{L}{A} \), where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area. Choosing materials with lower resistivity, like copper or aluminum, helps reduce resistive losses.
Electrical Current
Electrical current is the flow of electric charge, often carried by moving electrons in a conductor. It is measured in amperes (A) and is a fundamental concept in understanding power transmission. The flowing current is what enables the delivery of electrical energy from one point to another. However, with the flow of current through a conductor comes resistive losses, a key concern in power transmission.
  • Higher currents lead to more significant resistive losses.
  • Controlling the amount of current can help manage energy efficiency in transmission systems.
  • Current must be balanced with other factors like voltage and resistance to ensure safe and efficient operation of the transmission network.
Engineers often adjust current levels based on the needs of the system, and employing high-voltage systems can allow for lower current levels while maintaining power delivery, thus optimizing the system's efficiency.

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Most popular questions from this chapter

A \(500-\) W heating coil designed to operate from \(110 \mathrm{V}\) is made of Nichrome wire \(0.500 \mathrm{mm}\) in diameter. (a) Assuming that the resistivity of the Nichrome remains constant at its \(20.0^{\circ} \mathrm{C}\) value, find the length of wire used. (b) What If? Now consider the variation of resistivity with temperature. What power will the coil of part (a) actually deliver when it is heated to \(1200^{\circ} \mathrm{C} ?\)

Suppose that a voltage surge produces \(140 \mathrm{V}\) for a moment. By what percentage does the power output of a \(120-\mathrm{V}, 100-\mathrm{W}\) lightbulb increase? Assume that its resistance does not change.

An electric car is designed to run off a bank of \(12.0-\mathrm{V}\) batteries with total energy storage of \(2.00 \times 10^{7} \mathrm{J} .\) (a) If the electric motor draws \(8.00 \mathrm{kW},\) what is the current delivered to the motor? (b) If the electric motor draws \(8.00 \mathrm{kW}\) as the car moves at a steady speed of \(20.0 \mathrm{m} / \mathrm{s}\), how far will the car travel before it is "out of juice"?

Suppose that the current through a conductor decreases exponentially with time according to the equation \(I(t)=I_{0} e^{-t / \tau}\) where \(I_{0}\) is the initial current (at \(t=0),\) and \(\tau\) is a constant having dimensions of time. Consider a fixed observation point within the conductor. (a) How much charge passes this point between \(t=0\) and \(i=\pi^{2}\) (b) How much charge passes this point between \(t=0\) and \(t=10 \tau ?\) (c) What If? How much charge passes this point between \(t=0\) and \(t=\infty ?\)

A more general definition of the temperature coefficient of resistivity is $$\alpha=\frac{1}{\rho} \frac{d \rho}{d T}$$ where \(\rho\) is the resistivity at temperature \(T\). (a) Assuming that \(\alpha\) is constant, show that $$\rho=\rho_{0} e^{\alpha\left(T-T_{0}\right)}$$ where \(\rho_{0}\) is the resistivity at temperature \(T_{0} .\) (b) Using the series expansion \(e^{x} \approx 1+x\) for \(x<<1,\) show that the resistivity is given approximately by the expression \(\rho=\rho_{0}\left[1+\alpha\left(T-T_{0}\right)\right]\) for \(\alpha\left(T-T_{0}\right)<<1\).
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