Study Notes for Electrotechnology 2A - Power in AC Circuits

UNIT 2: Power in AC Circuits

MODULE CONTENTS

• Power in resistive AC circuit
• Power in inductive AC circuit
• Power in capacitive AC circuit
• Power in series and parallel AC circuit (RL, RC and RLC)
• Power/Load Triangle Concept and Power Factor (Apparent, Reactive, and Active)
• Power Factor Correction
• Maximum Power Transfer Theorem
• Power Measurement
• Energy Billing

Power in AC Circuits

• Definition: The rate of flow of energy through any given point in a circuit.
• Power Delivery: In AC circuits, power is delivered to resistive and reactive loads.
• Real Power: The portion of power averaged over a complete cycle of the AC waveform results in net transfer of energy in one direction. It is known as real power or active power, which results in energy dissipation in the load.
• Reactive Power: The rate of energy flowing into and out of reactive loads, associated with the reactive components in AC circuits.
• Conclusion: In a ‘mixed’ AC circuit, there is a mix of active and reactive powers contributing to the overall power factor.

Power in Resistive AC Circuits

• Case Overview: Purely resistive AC circuit with an AC voltage source, resulting current due to resistance R (Ohms).
• Phase Shift: No phase shift exists between Voltage (V) and Current (I) due to pure resistance.

Power in Purely Resistive AC Circuits

• Power Calculation: Power is calculated as:
    - Instantaneous Power: $P(t) = V(t) imes I(t)$
    - Given voltage, the resulting current is: $I_{rms} = \frac{V_{rms}}{R}$
• Average Power: Average power is calculated using the trig double angle formula through integration over one cycle:**
    - Average Power: $P_{avg} = V_{rms} imes I_{rms}$

Voltage and Current - Inductors and Capacitors

• Operational Characteristics:
    - In a capacitor, current leads voltage.
    - In an inductor, current lags voltage.

Power Delivered to Purely Reactive Components

• Ideal Reactive Devices: Average power absorbed by ideal reactive devices is zero due to a 90° phase difference between voltage and current.
• Power Flow: Power oscillates into and out of the network with no net power loss; instantaneous power is zero at specific instants during the cycle.

Purely Inductive AC Circuits

• Current-Lagging: In an ideal inductor, current lags voltage by 90°
• Power Calculation:
    - Formulate the instantaneous power; integrate to find average power, which results in zero power.

Purely Capacitive AC Circuits

• Current-Leading: In an ideal capacitor, current leads voltage by 90°
• Power Calculation:
    - Similar to inductors, the average calculated power results in zero power.

Power in RC and RL AC Circuits

• Combination of R, L, and C: When combining reactive components with resistance, the current will flow as modeled by the respective phase shifts (positive for RC, negative for RL).
• Instantaneous Power: Different formulations yield components that add up through phase shifts.
• Average Power Calculation: The average power over one cycle can be derived and will also include contributions from real and imaginary components due to phase differences.

Power Triangle and Power Factor

• Concept Definition: Power is expressed in real and imaginary parts:
    - Active Power (P)
    - Reactive Power (Q)
    - Apparent Power (S): Magnitude of complex power illustrated as a triangle.
• Power Factor:
    - Defined as the ratio $(PF = \frac{P}{S})$; dimensionless.
    - Lagging power factor for inductive circuits and leading for capacitive circuits.

Complex Power

• Complex Power Definition: Vector representation of power defined as $S = P + jQ$.
• Magnitude of Complex Power: Calculated as $|S| = ext{Magnitude}(S)$.
• Calculation Method: When calculating complex power from voltage and current phasors, the use of the complex conjugate is crucial:
    - $S = VI^*$
    - $|S| ext{∠} \frac{ heta_v - heta_I}{180}$

Complex Power and Apparent Power Example

• Given: Load $Z = 5 ext{∠} 90° ext{Ω}$, Voltage $E = 50 ext{∠} 0° ext{V}$.
• Calculations:
    - Current: $I = \frac{E}{Z} = \frac{50 ext{∠} 0°}{5 ext{∠} 90°} = 10 ext{∠} -90°$.
    - Complex Power: $S = EI^* = 50 ext{∠} 0° imes 10 ext{∠} 90° = 500 ext{∠} 90° ext{VA}$.
• Note: Angle of complex power corresponds to impedance.

Tutorials and Assessment

Tutorial 1

1. Problem: 1000 Hz, 50V source delivers 300 W to a load with current of 8 A; find the impedance.
      - Answer: C = 38.5 µF
2. Problem: Load $Z = 10 ext{∠} 36.9° ext{Ω}$; V = 30 ext{∠} 0°; find complex power and related values.
      - Answer: 90 ∠ 36.9° VA, 0.8 lagging, 72 W, 54 VAR
3. Problem: Calculate resistance and inductance of an inductor coil connected to a supply of 230 V at 50 Hz, with current of 5.0 A and dissipating 750 W.

Power Factor Correction Example

• Given: Average power of 1 kW; different reactive situations to find power factor:
    - (a) Q = 0 VAR; PF = 1 (b) Q = 500 VAR; PF = 0.894 lagging (c) Q = 200 VAR; PF = 0.981 leading

Power Factor Correction

• Purpose: To improve power factor for reduced running costs due to lower current flow and decreased losses (I²R losses).
• Utility Tariffs: Encourage higher power factors through penalties for low readings.
• Correction Method: Connect capacitors in parallel with inductive loads to mitigate lagging power factor.

Power Factor Correction Example

• Transformer Problem: 300 kVA transformer with a lagging pf of 0.70 brought to 0.90:
• Calculate required capacitor rating (answer: 112.5 kVAR).

Power Measurement

• Wattmeter Definition: Measures average power absorbed by a load using a current coil and a voltage coil.
    - Configuration: Current coil in series (low impedance) and voltage coil in parallel (high impedance).

Energy Billing

• Cost Analysis: Low power factor loads are expensive; utilities categorize customers based on residential, commercial, industrial tariffs, accounting for fixed costs and variable energy consumption charges.
• Structure: Monthly energy bill includes:
    - Fixed cost and variable rate in kWh.

Energy Billing Examples

1. Company Example: 200,000 kWh at varying charges resulting in a total bill of R19,500.
2. Load Example: 300 kW load at 13 kV, calculating average costs, including power factor penalties based on above-mentioned computations.

Tutorials

1. Calculate monthly bill for paper mill based on given consumption.
2. Determine bill for an 800 kW induction furnace operated under specified conditions.