Mathcad - 05-07

Mathcad - 05-07 - of the components I rms I 1 2 I 3 2 I 5 2...

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I 11pu 0.088 = I 3pu I 3 I rms = I 3pu 0.438 = I 13pu I 13 I rms = I 13pu 0.066 = I 5pu I 5 I rms = I 5pu 0.146 = I 15pu I 15 I rms = I 15pu 0.066 = I 7pu I 7 I rms = I 7pu 0.139 = I 17pu I 17 I rms = I 17pu 0.044 = I 9pu I 9 I rms = I 9pu 0.088 = then from the definition of K-factor (equation 5-39) K 1 2 I 1pu 2 3 2 I 3pu 2 + 5 2 I 5pu 2 + 7 2 I 7pu 2 + 9 2 I 9pu 2 + 11 2 I 11pu 2 + 13 2 I 13pu 2 + 15 2 I 15pu 2 + 17 2 + = K 7.747 = Problem 5-7: Solution 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 = The rms of the current is the square root of the sum of the squares of the rms magnitudes
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Unformatted text preview: 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 137.007amp = To find the K-factor we need the ratio of each component of current to the RMS current I 1pu I 1 I rms = I 1pu 0.861 = I 11pu I 11 I rms = I 17pu 2 ⋅...
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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 - 05-07 - of the components I rms I 1 2 I 3 2 I 5 2...

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