IB DP Physics: HL复习笔记11.2.5 AC Electrical Power Distribution

AC Electrical Power Distribution

  • Energy losses due to the heating of transmission lines in national power grids are significant
    • This is because the electrical energy is transmitted across long distances from power stations to buildings
  • Inefficiencies in a transformer appear not from just the core, but also in the wires
  • The coils of wire have resistance
    • This causes heat energy to be lost from the current flowing through the coils
    • The larger the current, the greater the amount of heat energy lost
  • In the core, the inefficiencies appear from:
    • Induced eddy currents
    • The reversal of magnetism
    • Poor insulation between the primary and secondary coil
  • Ways to reduce energy loss in a transformer are:
    • Making the core from soft iron or iron alloys to allow easy magnetisation and demagnetisation and reduce hysteresis loss
    • Laminating the core
    • Using thick wires, especially in the secondary coil of step-down transformers
    • Using a core that allows all the flux due to the primary coil to be linked to the secondary coil
  • Power losses from the current are calculated using the equation:

P = I2R

  • Where:
    • P = power (W)
    • I = current (A)
    • R = resistance (Ω)
  • The equation shows that:
    • P ∝ I2
    • This means doubling the current produces four times the power loss
  • Therefore, step-up transformers are used to increase the voltage which decreases the current through transmission lines
    • This reduces the overall heat energy lost in the wires during transmission
  • A step-down transformer is then used to decrease the voltage to that required in homes and buildings

7.10.5-Power-Loss-National-Grid

The use of step-up and step-down transformers in the National Grid

Worked Example

A current of 2500 A is transmitted through 150 km of cables. The resistance of the transmission cable is 0.15 Ω per km.

Calculate the power wasted.

Step 1: List the known quantities

    • Current, I = 2500 A
    • Length of cables, L = 150 km = 150 × 103 m
    • Resistance of the cables, R = 0.15 Ω km-1

Step 2: Write out the power equation

P = I2R

Step 3: Determine the total resistance, R

R = Resistance of the wires × Length of wires

R = 0.15 × 150

Step 4: Substitute values into the power equation

Power lost = (2500)2 × (0.15 × 150) = 141 MW

 

 

 

 

 

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