By P. Bellomo, J. Sebek
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The second harmonic is twice the fundamental frequency, the third harmonic is 3 X the fundamental frequency, etc. 015 Section 3. 02 Trigonometric form of the Fourier Series T 1 a 0 = ∫ f ( t )dt T0 T 2 a k = ∫ f ( t )cos kω0tdt T0 T 2 b k = ∫ f ( t )sin kω0tdt T0 Complex form from Euler e j x = cos x + j sin x T 2 − jkω 0 t ck = ∫ f ( t )e dt T0 June 21 – July 2, 2004 Section 3. 02 Even function symmetry f ( t ) = f ( -t ) No sine terms Has DC component if no half-wave symmetry Odd function symmetry f ( t ) = - f ( -t ) No cosine terms No DC component Half-wave symmetry f ( t ) = - f ( t -1/2T) Have sines and cosines but only odd harmonics No DC component June 21 – July 2, 2004 Section 3.
73 V C C VC-A VC V LL = 3 Vφ and I L = I φ IC VA-B S 3φ = 3 Vφ I φ S 3φ = June 21 – July 2, 2004 3 V LL I L IB 30O IA VB V B-C Section 3. 73 I A ,I V V ,I VLL = V VLL = V B V ,I VLL = V C V LL = Vφ and I L = 3 Iφ S 3φ = 3 Vφ I φ S 3φ = June 21 – July 2, 2004 3 V LL I L Section 3. Power Line Considerations 36 Three Phase Quantities For W ye S 3φ = 3 V LL I L For Delta S3φ = 3 VLL I L VA − B =|VA − B | e j0 VB − C =|VB − C | e− j120 VC − A =|VC − A | e− j240 June 21 – July 2, 2004 Section 3. 025 • p (t) = vA (t)*iA (t) + vB (t)*iB (t) + vC (t)*iC (t) • 3 times the single phase power with only 3 conductors, not 6 • For balanced load, p (t) is constant June 21 – July 2, 2004 Section 3.
5 A 138 kV ( 138 kV ) 2 Z= = 1900 Ω 10,000 kVA Section 3. 56 pu* 145 A = 113 A = 226 A Section 3. 05 pu I = 1 pu June 21 – July 2, 2004 Section 3. Power Line Considerations 48 Homework Problem #1 Referring to the one-line diagram below, determine the line currents in the: A. Generator June 21 – July 2, 2004 B. Transmission Line C. M1 Section 3. Power Line Considerations D. M2 49 Harmonics, Complex Waveforms and Fourier Series • Non-sinusoidal waves are complex and are composed of sine and cosine harmonics • The harmonics are integral multiples of the fundamental frequency (1st harmonic) of the wave.