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Chapter 24: Problem 52

You're working on a new high-speed rail system. It uses \(6000-\) horsepower electric locomotives, getting power from a single overhead wire with resistance \(15 \mathrm{m} \Omega / \mathrm{km},\) at \(25 \mathrm{kV}\) potential relative to the track. Current returns through the track, whose resistance is negligible. Energy-efficiency standards call for no more than \(3 \%\) power loss in the wire. How far from the power plant can the train go and still meet this standard?

Short Answer

Expert verified
The train can travel up to approximately \(282.32 \, \text{km}\) from the power plant and still meet the energy-efficiency standard.

Step by step solution

01

Convert Horsepower to Watts

First, it is essential to convert the power from horsepower to watts. The conversion factor is 1 horsepower = 746 watts. So, \(6000 \cdot 746 = 4476000 \, \text{W}\).
02

Find the Current

Next, calculate the current being used by the train by using the power equation, \(P = IV\), where \(P\) is power, \(I\) is current, and \(V\) is voltage. Rearranging for \(I\): \(I = \frac{P}{V} = \frac{4476000}{25000} = 179.04 \, \text{A}\).
03

Find Maximum Allowable Power Loss

The question states that the power loss shouldn't exceed 3% of the total power. So, the maximum power loss is \(0.03 \times 4476000 = 134280 \, \text{W}\).
04

Calculate the Maximum Distance

Now, use the power loss formula \(P_{\text{loss}} = I^2 \cdot R \cdot L\). Rearranging it for \(L\): \(L = \frac{P_{\text{loss}}}{I^2 \cdot R} = \frac{134280}{179.04^2 \cdot 15 \times 10^{-3}} = 282.32 \, \text{km}\).
05

Interpret the Result

The result of 282.32 km means that the train can travel up to that distance from the power plant and still comply with the energy-efficiency standards stipulated.

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