EEE471HW2solp2

# EEE471HW2solp2 - 15MVA Tertiary (3) Δ connected, 2.3 kV,...

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Problem 3.34 Model: 3 – 1 phase transformers: 450 MVA, 20:288.7kV, x = 0.1 pu a) HV: Grounded wye LV: delta b) HV: Grounded wye LV: wye Δ 500:20 kV 1350 MVA X = 0.1 pu 500 kV j 0.1 pu 20 kV 1 : 30 j e Y 500:34.6 kV 1350 MVA X = 0.1 pu 500 kV j 0.1 pu 34.6 kV

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Problem 3.41 Element Z pu1 MVA B1 kV B1 MVA B2 kV B2 Z pu2 * G1 j.2 750 18 100 20 j.0216 G2 j.2 750 18 100 20 j.0216 M j.2 1500 20 100 20 j.0133 T1,T2,T3,T4 j.1 750 500 100 500 j.0133 T5 j.1 1500 500 100 500 j.0067 L45 40Ώ = = 2500 2500 ) 500 ( 2 B Z j.016 L36 25Ώ Z B = 2500Ώ j.01 L37 25Ώ Z B = 2500Ώ j.01 * 2 2 1 1 2 1 2 = B B B B pu pu kV kV MVA MVA Z Z Δ 1 2 3 4 5 6 7 1 2 m Δ Δ Δ j 40Ώ j25Ώ j25Ώ T1 T2 T3 T4 T5 Y
3.41 (continued) Problem 3.51 3-winding Transformer, 3-phase Primary (1) Y connected, 66kV, 20MVA Secondary (2) Y connected, 13.2kV,

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Unformatted text preview: 15MVA Tertiary (3) Δ connected, 2.3 kV, 5MVA X 12 = 0.08 pu on 20MVA, 66kV bases X 13 = 0.10 pu on 20MVA, 66kV bases X 23 = 0.09 pu on 15MVA, 13.2kV bases On 20MVA, 66kV base X 12 = 0.08 pu, X 13 = 0.10 pu pu X 12 . ) 09 (. 15 20 23 = = pu X 03 . 2 12 . 1 . 08 . 1 =-+ = pu X 05 . 2 1 . 12 . 08 . 2 =-+ = pu X 07 . 2 08 . 12 . 1 . 3 =-+ = j.0133 j.016 j.0133 j.0216 j.0133 j.01 j.01 j.0133 j.0216 j.0067 j.0133 n1 n1 n1 E g1 E m E g2 3.51 (continued) Loads: Secondary load: 12MW at 13.2kV, pu S V R u 67 . 1 20 12 1 2 2 = = = Tertiary load: 5MW at 2.3kV pu R u 4 20 5 1 3 = = rgf:2/9/09 j.03 pu j.05 pu j.07 pu 1.67 pu 4 pu P S T n1...
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## This note was uploaded on 02/21/2010 for the course EEE ??? taught by Professor Farmer during the Spring '10 term at ASU.

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EEE471HW2solp2 - 15MVA Tertiary (3) Δ connected, 2.3 kV,...

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