Homework2

# Homework2 - Homework#2 10.10.04 Problem 1 Parameters LOAD...

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Homework #2 Randall Robert 10.10.04 9771 Problem 1. Parameters: LOAD; V RMS = 480V P = 10kW pF = 0.5 or 0.9 SOURCE; S SC = 0.25[MVA] X s /R s = ξ = 2 Variable of interest is V RMS of the source which is noted E, and the loss of active power ΔP s in the supply. First we see for λ = 0.5 Using λ=P/VI we get; I = 10kW/(480V(0.5)) I = 41.67A Using E-V = Z s I 1 where Z s = √(R s ²+X s ²) and ξ = √(S sc Z s ) we get the equation; S sc Z s = X s ²I² + 2Z s( I 1) * V + V² 1736.4Z s ² - 209997 Z s + 230400 = 0 Z s = 119.8 Using R s = Z s /(√(5)) we get; R s = 119.8/(√(5)) R s = 53.6 Using X s = 2R s we get; X s = 2*(53.6) X s = 107.2 Using E = √ ((Ssc*X s (√((1/ξ²)+1))) we get; E = 5.474[kV]

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Next we see for λ = 0.9 Using λ=P/VI we get; I = 10kW/(480V(0.9)) I = 23.15A Using E-V = Z s I 1 where Z s = √(R s ²+X s ²) and ξ = √(S sc Z s ) we get the equation; S sc Z s = X s ²I² + 2Z s( I 1) * V + V² 535.9Z s ² - 227776 Z s + 230400 = 0 Z s = 424 Using R s = Z s /(√(5)) we get; R s = 424/(√(5)) R s = 189.6 Using X s = 2R s we get; X s = 2*(189.6) X s = 379.2 Using E = √ ((Ssc*X s (√((1/ξ²)+1))) we get; E = 10.295[kV]
Problem 2. Parameters: LOAD; V RMS = 2.4kV P = 25kW pF = 0.5 SOURCE; S SC = 0.5[MVA]

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