Mathcad - 03-01

Mathcad - 03-01 - I n 185.54amp = I n 3 I 3 2 I 9 2 I 15 2...

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b. Based on the fundamental component, the total harmonic distortion is THD 1 I 3 2 I 5 2 + I 7 2 + I 9 2 + I 11 2 + I 13 2 + I 15 2 + I 17 2 + I 1 = THD 1 0.59 = or 59% NOTE: THD F = THD 1 If instead, we use the rms current to find THD, we find THD 2 I 3 2 I 5 2 + I 7 2 + I 9 2 + I 11 2 + I 13 2 + I 15 2 + I 17 2 + I rms = THD 2 0.508 = or 50.8% NOTE: THD R = THD 2 c. The displacement power factor is determined by the real power used and the apparent power calculated from the fundamental components of the voltage and current P 10 kW = S 1 V 1 I 1 = S 1 13.924 kVA = F pdisp P S 1 = F pdisp 0.718 = lagging Problem 3-1: Solution homework kVA kW The rms magnitudes of the harmonic components of the current are given as follows: I 1 118.0 amp = I 3 60.0 amp = I 5 20.0 amp = I 7 19.0 amp = I 9 12.0 amp = I 11 12.0 amp = I 13 9.0 amp = I 15 9.0 amp = I 17 6.0 amp = also V 1 118 volt = a. The rms of the current is the square root of the sum of the squares of the rms magnitudes of the components: I rms I 1 2 I 3 2 + I 5 2 + I 7 2 + I 9 2 + I 11 2 + I 13 2 + I 15 2 + I 17 2 + = I rms 137.007 amp =
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Note that the neutral current is larger than the phase current for this load!
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Unformatted text preview: I n 185.54amp = I n 3 I 3 2 I 9 2 + I 15 2 + ⋅ = f. In a three phase system, the neutral will carry the third, ninth, and fifteenth harmonics which add together from the three phases. RMSing them together yields: F pdist 0.861 = F pdist F ptot F pdisp = from equation 3.17 we could also note that the distortion power factor is given by the ratio of the total power factor to the displacement power factor F pdist 0.861 = F pdist 1 1 THD F 100 2 + = (in percent) THD F 100 THD 1 ⋅ = e. We can find the distortion power factor from equation 3.19 lagging F ptot 0.619 = F ptot P S = from equation 3.15: S 16.167kVA = S V 1 I rms ⋅ = NOTE: There is only one component for the voltage d. To find the total power factor, we know P and can find S...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

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Mathcad - 03-01 - I n 185.54amp = I n 3 I 3 2 I 9 2 I 15 2...

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