Homework3

# Homework3 - Homework#3 10.19.4 Randall Robert 9771 Problem...

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Homework #3 Randall Robert 10.19.4 9771 Problem 1. Distribution voltage RMS value measured at unloaded bus is E = 2.4kV, its short circuit power ia 1MVA and the reactance to resistance ratio ξ = 2. Calculate the load voltage RMS value V, power loss in the supply and draw diagrams of the CRMS values when the bus supplies a load that at the voltage RMS value V = 2.4kV has the active power P = 50kW and the reactive powers: Parameters: Source E = 2.4 kV S sc = 1MVA ξ s = 2 Load V RMS = 2.4kV P = 50kW Reactive Powers 1. Q = 50kVAr 2. Q = 0 3. Q = -50kVAr 1. Q = 50kVAr S = P+jQ S = VI*, I* = S/V, I* = (50k+j50kVAr)/2.4kV I* = 70.7k∟45°A I* = 29.5∟45°A V/I = Z L = 2.4kV/(29.5∟45°A) Z L = 81.4∟45°Ω = 57.6 + j57.6

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Z s = E²/Ssc = (2.4kV²)/(1MVA) Z s = 5.76Ω R s = Z s /(√1+ξ²) = 5.76Ω/(√5) R s = 2.58Ω X s = ξR x = 5.15Ω Therefore Z x = R s + jX s = 5.76∟63.4°Ω I = E/(ZL + ZS) = 2.4kV/(63.4 + j60.18) = 2.4kV/(87.4∟43.5°) I = 27.5(∟-43.5°)A V = E – Izs = 24kV – (27.5(∟-43.5°)A)( 5.76(∟63.4°)Ω)
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## This note was uploaded on 02/06/2012 for the course EE 3410 taught by Professor Staff during the Fall '08 term at LSU.

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Homework3 - Homework#3 10.19.4 Randall Robert 9771 Problem...

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