Final answer:
Transmission lines, used for electrical power distribution, include overhead, underground, and submarine cables, each suited to specific applications. High voltages help reduce losses, as demonstrated by the calculation where the loss is 0.25% for a 100 MW power transmission at 200 kV over a line with 1 ohm resistance.
Step-by-step explanation:
Types of Transmission Lines and Their Limits
- There are various types of transmission lines used for transporting electrical power, including overhead lines, underground cables, and submarine cables. Each type has unique applications and physical constraints. Overhead lines are most common for long-distance, high-voltage power transmission, because they are cost-effective and easy to maintain. Underground and submarine cables are used in dense urban areas or underwater crossings where overhead lines are not feasible. Superconducting transmission lines represent an emerging technology with the potential to transfer power without loss, although they are still under development due to material challenges.
Limitations on Power Transmission
- The maximum power that transmission lines can handle is limited by several factors. One primary limitation is the current capacity of the line, which is related to the conductor size and material. As current increases, so does resistive heating, which can cause the conductive material to deteriorate or sag, leading to reduced efficiency and possible failure. To mitigate resistance, high voltages are used for long-distance transmission. However, insulation and clearance requirements also limit the maximum voltage that can be safely used.
Calculations for Power Transmission
- To illustrate the concept, let's consider a scenario where 100 MW of power needs to be transmitted at 200 kV. The current needed for this transmission can be found using the formula I = P/V, where I is the current, P is the power, and V is the voltage. Therefore, the current required is I = 100 MW / 200 kV = 500 A. If the transmission lines have a resistance of 1 ohm (R = 1 Ω), the power dissipated due to resistance (Ploss = I2R) would be Ploss = (500 A)2(1 Ω) = 250 kW. The percentage of power lost in the transmission process can be calculated as (Ploss / P) * 100%, which in this case is (250 kW / 100 MW) * 100% = 0.25%.