A transmission line that has a resistance per unit length of 4.50 ×10^-1 Ω/ m is to be used to t

A transmission line that has a resistance per unit length of...

A transmission line that has a resistance per unit length of $4.50 \times 10^{-1} \Omega / \mathrm{m}$ is to be used to transmit 5.00 MW over 400 miles $\left(6.44 \times 10^{5} \mathrm{m}\right)$ The output voltage of the generator is 4.50 kV (rms). (a) What is the line loss if a transformer is used to step up the voltage to 500. kV (rms)? (b) What fraction of the input power is lost to the line under these circumstances? (c) What difficulties would be encountered on attempting to transmit the 5.00 MW at the generator voltage of 4.50 kV (rms)?

Key Concepts

High Voltage Power Transmission Efficiency

High voltage power transmission improves efficiency by reducing current, thereby lowering resistive losses along the transmission line. This allows for the delivery of large amounts of power over long distances with minimal energy lost as heat, making it a key principle in modern power grids.

Transformer Step-Up Voltage

Using a transformer to step up voltage is fundamental in power transmission systems. By increasing the transmission voltage, the current required to deliver a certain power level is reduced, which in turn minimizes I²R losses. This strategy improves the overall efficiency of the power system.

Challenges of Low Voltage Transmission

Transmitting power at low voltage necessitates high current to deliver the same amount of power, which leads to excessive resistive losses and potential overheating of the transmission lines. This not only decreases efficiency but also poses significant engineering and safety challenges, requiring larger conductors and more robust cooling solutions.

Transmission Line Resistance

Transmission line resistance is a measure of how much a conductor opposes the flow of electric current per unit length. It is crucial in power transmission because energy is lost as heat, with the loss being proportional to the square of the current flowing through the line and the resistance of the conductor.

Resistive Power Loss (I²R Losses)

Resistive power losses, often referred to as I²R losses, occur when electric current flows through a conductor with resistance. Because the power loss is proportional to the current squared, even small increases in current can lead to large increases in energy dissipated as heat, which is critical in long distance power transmission.

Transcript

00:01 So the resistance in the line, r subline would be equaling 4 .50 times 10 to the negative 4th ums per meter multiplied by 6 .44 times 10 to the 5th meters.

00:23 This is giving us 290 ums.

00:27 We now know that here for part a, the average power transmitted would be equaling the rms voltage voltage.

00:46 Multiplied by the rms current.

00:51 Now we can then solve for the rms current.

00:55 This would be equaling the average power transmitted divided by the rms voltage and so we can say that the rms current is equaling 5 .00 times 10 to the 6th watts and this would be divided by 500 times 10 to the third volts and this is giving us 10 .0 amps.

01:29 And so we can then see that the average power loss would be equalling rms squared multiplied rms current squared multiplied by the resistance in the line.

01:44 And so this would be 10 .0 amps squared multiplied by 2090 ums and this is giving us 2 .90 times 10 to the 4th watts this would be our final answer for part a for part b then we can say the power input to the line would be power average input this would be equalling the rms current squared rather my apologies this would be equaling the average power transmitted plus the average power lost.

02:36 And so this would be equaling 5 .00 times 10 to the 6 watts plus 2 .90 times 10 to the 4th watts.

02:50 And so this is going to be equaling 5 .03 times 10 to the 6 watts.

03:00 And so now the fraction of input power lost during transmission would be, we can say, some fraction is going to be equaling the average power loss divided by the average power input.

03:20 So this would be 2 .90 times 10 to the fourth watts divided by 5 .03 times 10 to the 4 watts divided by 5 .03 times 10 to the 6 watts and this is giving us 0 .00577 or we can say 0 .577%.

03:47 So this would be our final answer for part b.

03:52 For part c, we know that it is impossible to deliver the needed power at the generator voltage of 4 .5, 4 .50 kilovolts...

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