Part IV Modern Rectifiers and Power System Harmonics Chapter 16 Power and Harmonics in Nonsinusoidal Systems Chapter 17 Line-Commutated Rectifiers Chapter 18 Pulse-Width Modulated Rectifiers 1 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Chapter 16 Power And Harmonics in Nonsinusoidal Systems 16.1. Average power in terms of Fourier series 16.2. RMS value of a waveform 16.3. Power factor THD Distortion and Displacement factors 16.4. Power phasors in sinusoidal systems 16.5. Harmonic currents in three-phase systems 16.6. AC line current harmonic standards 2 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
16.1. Average power Observe transmission of energy through surface S i(t) + + v(t) Source Load – – Surface S ∞ Σ V n cos n ω t – ϕ n Express voltage v ( t ) = V 0 + relate energy and current as n = 1 transmission to ∞ Σ I n cos n ω t – θ n Fourier series: i ( t ) = I 0 + harmonics n = 1 3 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Energy transmittted to load, per cycle T W cycle = v ( t ) i ( t ) dt 0 This is related to average power as follows: P av = W cycle T = 1 v ( t ) i ( t ) dt T T 0 Investigate influence of harmonics on average power: substitute Fourier series T ∞ ∞ Σ Σ P av = 1 V n cos n ω t – ϕ n I n cos n ω t – θ n V 0 + I 0 + dt T n = 1 n = 1 0 4 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Evaluation of integral Orthogonality of harmonics: Integrals of cross-product terms are zero if n ≠ m 0 T V n cos n ω t – ϕ n I m cos m ω t – θ m = dt V n I n cos ϕ n – θ n if n = m 0 2 Expression for average power becomes ∞ Σ V n I n cos ϕ n – θ n P av = V 0 I 0 + 2 n = 1 So net energy is transmitted to the load only when the Fourier series of v(t) and i(t) contain terms at the same frequency. For example, if the voltage and current both contain third harmonic, then they lead to the average power V 3 I 3 cos ϕ 3 – θ 3 2 5 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Example 1 v(t) 1 i(t) Voltage: fundamental 0.5 only Current: third 0 harmonic only -0.5 -1 1 p(t) = v(t) i(t) Power: zero average 0.5 P av = 0 0 -0.5 -1 6 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Example 2 1 v(t), i(t) Voltage: third 0.5 harmonic only Current: third 0 harmonic only, in phase with voltage -0.5 -1 p(t) = v(t) i(t) 1 Power: nonzero average P av = 0.5 0.5 0 -0.5 -1 7 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Example 3 Fourier series: v ( t ) = 1.2 cos ( ω t ) + 0.33 cos (3 ω t ) + 0.2 cos (5 ω t ) i ( t ) = 0.6 cos ( ω t + 30 ° ) + 0.1 cos (5 ω t + 45 ° ) + 0.1 cos (7 ω t + 60 ° ) Average power calculation: P av = (1.2)(0.6) cos (30 ° ) + (0.2)(0.1) cos (45 ° ) = 0.32 2 2 8 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Example 3 v(t) 1.0 Voltage: 1st, 3rd, 5th 0.5 Current: 1st, 5th, 7th i(t) 0.0 -0.5 -1.0 p(t) = v(t) i(t) 0.6 Power: net energy is transmitted at 0.4 P av = 0.32 fundamental and fifth harmonic 0.2 frequencies 0.0 -0.2 9 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
16.2. Root-mean-square (RMS) value of a waveform, in terms of Fourier series T 1 (rms value) = v 2 ( t ) dt T 0 Insert Fourier series. Again, cross-multiplication terms have zero average. Result is ∞ 2 Σ V n 2 + (rms value) = V 0 2 n = 1 • Similar expression for current • Harmonics always increase rms value • Harmonics do not necessarily increase average power • Increased rms values mean increased losses 10 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
16.3. Power factor For efficient transmission of energy from a source to a load, it is desired to maximize average power, while minimizing rms current and voltage (and hence minimizing losses). Power factor is a figure of merit that measures how efficiently energy is transmitted. It is defined as (average power) power factor = (rms voltage) (rms current) Power factor always lies between zero and one. 11 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
16.3.1. Linear resistive load, nonsinusoidal voltage Then current harmonics are in phase with, and proportional to, voltage harmonics. All harmonics result in transmission of energy to load, and unity power factor occurs. I n = V n θ n = ϕ n so cos ( θ n – ϕ n ) = 1 R ∞ 2 Σ V n 2 + (rms voltage) = V 0 2 n = 1 ∞ 2 2 ∞ 2 Σ Σ I n V 0 V n 2 + (rms current) = = R 2 + I 0 2 2 R 2 n = 1 n = 1 = 1 R (rms voltage) ∞ Σ V n I n cos ( ϕ n – θ n ) P av = V 0 I 0 + 2 n = 1 12 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
16.3.2. Nonlinear dynamical load, sinusoidal voltage With a sinusoidal voltage, current harmonics do not lead to average power. However, current harmonics do increase the rms current, and hence they decrease the power factor. P av = V 1 I 1 cos ( ϕ 1 – θ 1 ) 2 ∞ 2 Σ I n 2 + (rms current) = I 0 2 n = 1 I 1 2 cos ( ϕ 1 – θ 1 ) (power factor) = ∞ 2 ∑ I n 2 + I 0 2 n = 1 = (distortion factor) (displacement factor) 13 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Distortion factor Defined only for sinusoidal voltage. I 1 = (rms fundamental current) 2 (distortion factor) = ∞ (rms current) 2 ∑ I n 2 + I 0 2 n = 1 Related to Total Harmonic Distortion (THD): ∞ Σ 2 I n n = 2 (THD) = I 1 1 (distortion factor) = 1 + (THD) 2 14 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Distortion factor vs. THD 100% Distortion factor 90% 80% 70% 0% 20% 40% 60% 80% 100% THD 15 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Peak detection rectifier example Conventional single- phase peak detection rectifier 100% 100% 91% percent of fundamental Harmonic amplitude, THD = 136% Typical ac line 80% Distortion factor = 59% 73% current spectrum 60% 52% 40% 32% 19% 15% 15% 13% 9% 20% 0% 1 3 5 7 9 11 13 15 17 19 Harmonic number 16 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Maximum power obtainable from 120V 15A wall outlet with peak detection rectifier (ac voltage) (derated breaker current) (power factor) (rectifier efficiency) = (120 V) (80% of 15 A) (0.55) (0.98) = 776 W at unity power factor (ac voltage) (derated breaker current) (power factor) (rectifier efficiency) = (120 V) (80% of 15 A) (0.99) (0.93) = 1325 W 17 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
16.4. Power phasors in sinusoidal systems Apparent power is the product of the rms voltage and rms current It is easily measured —simply the product of voltmeter and ammeter readings Unit of apparent power is the volt-ampere, or VA Many elements, such as transformers, are rated according to the VA that they can supply So power factor is the ratio of average power to apparent power With sinusoidal waveforms (no harmonics), we can also define the real power P reactive power Q complex power S If the voltage and current are represented by phasors V and I , then S = VI * = P + jQ with I* = complex conjugate of I , j = square root of –1 . The magnitude of S is the apparent power (VA). The real part of S is the average power P (watts). The imaginary part of S is the reactive power Q (reactive volt-amperes, or VARs). 18 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
Example: power phasor diagram Imaginary The phase angle between axis S = VI * the voltage and current, or Q (ϕ 1 – θ 1 ), coincides with the || S || = V rms I rms angle of S . The power factor is power factor = P = cos ϕ 1 – θ 1 ϕ 1 – θ 1 S ϕ 1 θ 1 P Real axis ϕ 1 – θ 1 In this purely sinusoidal case, the distortion factor is unity, and the V power factor coincides with the displacement factor. I 19 Fundamentals of Power Electronics Chapter 16: Power and Harmonics in Nonsinusoidal Systems
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