Exam3Solutions-Fall07

# Exam3Solutions-Fall07 - P FW P core = P IN 2.5 10 5 × = f...

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Here's what I used to generate the plot: V o 500 = I Fo 3.1 = V OC I F ( 29 V o 1 e I F I Fo - - = I SC I F ( 29 1200 8.5 I F = I F 0 0.1 , 10 .. = 0 1 2 3 4 5 6 7 8 9 10 0 50 100 150 200 250 300 350 400 450 500 0 500 1000 400 455 V OC I F ( 29 I SC I F ( 29 7 5 I F This problem is essentially Homework Problem 5.2 with the numbers changed a bit.

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Exam 3 Solution: Given: R A 0 = X S j 0.1 = PF 0.8 = P 300000 = V L 400 = 1) The field current is read directly from the graph: I F 5 = 0 1 2 3 4 5 6 7 8 9 10 0 50 100 150 200 250 300 350 400 450 500 0 500 1000 400 449.027 V OC I F ( 29 I SC I F ( 29 7 5 I F 2) Internally generated voltage: I A P 3 V L e j - acos PF ( ) = I A 433.013 = 180 π arg I A ( 29 36.87 - = V φ V L 3 = V φ 230.94 = E A V φ R A I A + X S I A + = E A 259.246 = δ arg E A ( 29 = 180 π δ 7.679 = 3) For E A as in 2: V TOC 3 E A = V TOC 449.027 = From the graph, I F 7 = 4, 5) Power and Torque: P OUT P PF = P OUT 2.4 10 5 × =
or 3 V φ E A X S sin δ ( 29 2.4 10 5 × = P CU 3 I A ( 29 2 R A = P CU 0 = P FW 5000 = P core 5000 = P IN P OUT P CU
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Unformatted text preview: + P FW + P core + = P IN 2.5 10 5 × = f sync 60 = ϖ sync 2 π ⋅ f sync ⋅ = τ applied P IN ϖ sync = τ applied 663.146 = 6) Efficiency η P OUT P IN 100 ⋅ = η 96 = 7) Voltage Regulation: VR E A V φ-V φ 100 ⋅ = VR 12.257 = 8) Voltage regulation for unity PF E AUPF V φ R A I A ⋅ + X S I A ⋅ + = E AUPF 234.965 = VR UPF E AUPF V φ-V φ 100 ⋅ = VR UPF 1.743 = 9) Torque angle: δ arg E AUPF ( 29 = 180 π δ ⋅ 10.62 = P SSL 3 V φ ⋅ E A ⋅ X S = P SSL 1.796 10 6 × = 10) Static Stability Limit:...
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## This note was uploaded on 07/28/2011 for the course EEL 3211 taught by Professor Staff during the Summer '08 term at University of Florida.

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Exam3Solutions-Fall07 - P FW P core = P IN 2.5 10 5 × = f...

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