# HW11Solution - HW 11 Solution ECE 476 11.19(a Xeq 0.0375...

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HW 11 Solution ECE 476 11.19 (a)        1 0.0375 0.09 0.09 0.0825 400 100 cos ( ) 0 4.0 0 ( ) (1.0)(1.0) eq bus X P I pf V pf     1.0 0 0.0825(4 0 ) 1 0.33 1.053 18.2629 bus eq E E V jX I j j (b) From eq (11.1.18) and eq (11.1.19) in the book, ?𝛿 ? ?? = 𝜔 ? − 𝜔 ?𝑦? ?𝜔 ? ?? = 𝜔 ?𝑦? 2? (? ?,?.? − ? ?,?.?. ? 𝜔 ?𝑦? ?𝛿 ? ?? ) (b-2) From the equation in lecture notes 22 24 (difference is how you define D; normalized with 𝜔 ?𝑦? or not) ?𝛿 ? ?? = 𝜔 ? − 𝜔 ?𝑦? ?𝜔 ? ?? = 𝜔 ?𝑦? 2? (? ?,?.? − ? ?,?.?. − ? ?𝛿 ? ?? ) (c) 𝜔 ?𝑦?. = 2𝜋60 , ? ?,?.? = 4 Before the fault occurs, the equivalent reactance is 𝑋 ?? = ?0.0375 + j0.09//j0.09 = 0.0825 ? 𝑎 = 𝑉 𝑖?? + ?𝑋 ?? ? = 1.0 + ?0.0825 ∗ 4 = 1 + ?0.33 = 1.053∠18.263°

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δ(0) = 18.263° = 0.3187 ??? During the fault, the Thevening reactance and the Thevening voltage are 𝑋 ?ℎ = ?0.0375 + j0.03//j0.09 = 0.06 ?. ?. 𝑉 ?ℎ = 1.0∠0° × 0.03 0.03 + 0.09 = 0.25∠0° ?. ?. ? ?,?.?. = 1.053 × 0.25 0.06 ???𝛿 = 4.3875 ???𝛿 ?. ?. Let 𝑥 1 = 𝛿 and 𝑥 2 = 𝛿 ̇ . Therefore, 𝑥̇ 1 = 𝑥 2 𝑥̇ 2 = 2 × 60𝜋 2 × 20 (4 − 1.053 × 0.25 0.06 ???𝑥 1 0.1 2 × 60𝜋 𝑥 2 ) = 9.4248(4 − 4.3875 ???𝑥 1 − 0.00027𝑥 2 ) x(0) = [ 0.3187 0 ]
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