Lecture 32 Review: AC power analysis Power factor correction

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Presentation transcript:

Lecture 32 Review: AC power analysis Power factor correction Average power, complex power, power triangles RMS values Power factor Power factor correction Related educational modules: Section 2.9.1, 2.9.2

AC power analysis Average power: Average power in terms of RMS (or effective) values: Complex power:

Power triangles Complex power (rectangular form): Real (average) and reactive power: Presented graphically:

Power factor (pf) Power factor: Load impedance:

Example 1 For the circuit below, determine: (a) the complex power delivered by the source (b) the average power delivered by the source

Outline problem on previous slide: 1. find equivalent impedance 2. find source current 3. complex power = VI*/2 4. Average power = (Vm*Im/2)*cos(thetav-thetai)

(a) Determine the complex power delivered by the source

(b) Determine the average power delivered by the source

Effect of pf on power delivery If v - i  0, we have some reactive power that is not consumed by the load The current provided to the load is higher than necessary Results in additional power dissipated during delivery Power companies don’t like this!

Power factor correction Power companies may require that users maintain a minimum power factor e.g. pf > 0.9 Most large loads are inductive in nature e.g. inductive motors Power factor correction may be necessary The approach must be inexpensive & simple to implement Adding a capacitor in parallel with the inductive load will increase the power factor

Power factor correction – continued We have an inductive load with some power factor cos1: The power triangle is shown below:

Power factor correction – continued again We can increase the power factor by adding a capacitor in parallel with the load: The power triangle then becomes:

Example 2 – power factor correction For the circuit below if (a) Determine the power factor (b) Re-design the circuit so that pf = 1

Example 2 – Determine pf

Example 2 – Re-design circuit so that pf = 1