Transmission Line Power Loss and High Voltage Benefits

1. The Power Loss Equation

As electricity flows through a conductor, some energy is inevitably converted into heat due to the conductor's electrical resistance. This is the power loss we need to minimise.

The key to understanding and calculating this loss is a fundamental formula. The following reading will introduce this equation and its components.

As the reading explains, the power loss ($P_{loss}$) in a conductor is given by:

$P_{loss}=I^{2}R$

Where:

• $P_{loss}$ is the power lost as heat, measured in Watts (W).
• $I$ is the current flowing through the conductor, measured in Amperes (A).
• $R$ is the resistance of the conductor, measured in Ohms (Ω).

The most important thing to notice here is the term $I^2$. The power loss is not just proportional to the current; it's proportional to the square of the current. This means that if you double the current, you quadruple the power loss. This non-linear relationship is the entire reason high-voltage transmission is necessary.

2. Why High Voltage is the Solution

So, if high current is the problem, how do we reduce it? The answer lies in the relationship between power, voltage, and current. The total power ($P$) being transmitted is the product of voltage ($V$) and current ($I$):

$P=VI$

For a power station producing a certain amount of power, this value of $P$ is fixed. Therefore, to deliver that power, we have a choice: we can use a low voltage and a high current, or a high voltage and a low current.

The following video provides an excellent demonstration of why the second option is the only viable one for an efficient grid.

The video makes the point powerfully:

• At low voltage, the current required is enormous (6,667 A), leading to catastrophic power losses (530 MW) that are far greater than the power being generated in the first place.
• At high voltage, the current is much smaller (17 A), resulting in manageable power losses (3.3 kW), which is only a tiny fraction of the transmitted power.

This is the core principle of efficient power transmission: use transformers to step the voltage up as high as possible at the start of the transmission line, which in turn reduces the current. This minimised current then flows through the lines, and the $I^{2}R$ losses are kept to a minimum. At the other end, another transformer steps the voltage back down for local distribution.

3. A Practical Calculation

Let's work through a more realistic UK-based example together.

Imagine a power station generating 600 MW of power. The power is sent to a city over a long transmission line with a total resistance of 4 Ω.

We will calculate the power loss for two scenarios.

Scenario A: Transmitting at a "low" voltage of 33,000 V (33 kV)

1. Calculate the current (I):
   We rearrange the power formula $P=VI$ to find $I=P/V$.
   $I=\frac{600,000,000\text{ W}}{33,000\text{ V}}\approx 18,182\text{ A}$

2. Calculate the power loss ($P_{loss}$):
   Now we use the power loss formula, $P_{loss}=I^{2}R$.
   $P_{loss}=(18,182\text{ A}{)}^2\times 4\Omega \approx 1,322,320,000\text{ W}$
   This is 1,322 MW. The loss is more than double the power we were trying to send! The transmission line would melt.

Scenario B: Transmitting at a high voltage of 400,000 V (400 kV)

1. Calculate the current (I):
   $I=\frac{600,000,000\text{ W}}{400,000\text{ V}}=1,500\text{ A}$
   Notice how much smaller the current is compared to Scenario A.

2. Calculate the power loss ($P_{loss}$):
   $P_{loss}=(1,500\text{ A}{)}^2\times 4\Omega =9,000,000\text{ W}$
   This is 9 MW.

Comparing the Results

• In Scenario A (33 kV), we lost 1,322 MW trying to send 600 MW. This is completely unworkable.
• In Scenario B (400 kV), we lost only 9 MW. The efficiency of the transmission is:
  $\text{Efficiency}=\frac{\text{Power Delivered}}{\text{Power Generated}}=\frac{600\text{ MW}-9\text{ MW}}{600\text{ MW}}=\frac{591}{600}\approx 98.5\%$

This calculation clearly shows why the UK national grid uses extremely high voltages like 400 kV for long-distance power transmission.

Self-Assessment

To check your understanding, consider this question:

If you keep the amount of power transmitted and the resistance of the line constant, but you double the transmission voltage, what happens to the power loss?

1. It is halved.
2. It is quartered.
3. It stays the same.

For more practice, you can review the worked examples in the document "Section 3.2.5 – Power Transmission" (resource ID fd307), which provides several clear, step-by-step problems.

Conclusion

This lesson has focused on one of the most fundamental principles of electrical engineering and grid design.

Key Takeaways:

• Power loss in transmission lines is caused by the line's resistance and is calculated with the formula $P_{loss}=I^{2}R$.
• Because the loss is proportional to the square of the current, reducing the current is the most effective way to improve efficiency.
• For a fixed amount of power ($P=VI$), increasing the transmission voltage ($V$) proportionally decreases the current ($I$).
• This is why we use very high voltages (like 400 kV and 275 kV) for bulk power transmission—it dramatically reduces energy waste.

Preview of the Next Lesson

We've just established why it's so important to step voltage up for transmission and down for use. In the next lesson, we will look at the facilities where this happens. We will address the learning outcome: "Explain the functions of transmission substations, including voltage transformation and power routing." We'll explore the key equipment inside these critical nodes of the grid, starting with the power transformer.