Midterm2-PracticeSolutions

Midterm2-PracticeSolutions - PRACTICE MIDTERM EXAM 2 SPRING...

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PRACTICE MIDTERM EXAM 2 SPRING 2011 1. Swing Equation Swing Equationt:     .. 60 0.5 3 me e f PP H P  Pre-fault: 2 0.2 0.8 2 0.6 eq t L X X X pu 1.1 1.0 sin 1.83 sin ee eq eq E V E V XX     .. 3  00 15.83 o During-fault: 0.4 0.33 0.8 eq VV  ; ' 0.47 X ' ' 1.1 0.52 0.7 e EV P X   .. 0.52 sin 3 Post-fault: ' 1 eq t L X X X
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' ' 1.1 1.0 sin 1.1sin 1 e eq EV P X   .. 60 0.5 3   0 11 27.04
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2. Transient Stability Pre_fault& Post_fault P epre = P epost = | E a || V | X eq sin δ X eq = 0 . 2 + 0 . 8 = 1 pu P epre = P epost = 1 . 15 × 1 1 sinδ = 1 . 15 sinδ power angle: P epre = P epost = P m 0 . 6 = 1 . 15 × sinδ 0 δ 0 = 0 . 55 rad = 31 . 45 δ max = π - δ 0 = π - 0 . 55 = 2 . 59 rad P_fault P e = 0 ˆ δ cc δ 0 ( P m - P efault ) = ˆ δ max δ cc ( P epost - P m ) ˆ δ cc 0 . 55 (0 . 6 - 0) = ˆ 2 . 59 δ cc (1 . 15 sinδ - P m ) 0 . 6 × ( δ cc - 0 . 55) = - 1 . 15 cosδ | 2 . 59 δ cc - 0 . 6 × (2 . 59 - δ cc ) 0 . 6 × (2 . 59 - 0 . 55) = - 1 . 15 cos 2 . 59 + 1 . 15 cosδ cc 0 . 24 = 1 . 15 cosδ cc δ cc = 1 . 36 rad = 77 . 95 ¨ δ = πf H ( P m - P efault ) ¨ δ = πf H ( P m - 0) δ ( t ) = 1 2 πf H P m t 2 + δ 0 δ cc = 1 2 πf H P m t 2 cc + δ 0 t cc = s ( δ cc - δ 0 ) × 2 H πfP m t cc = r (1 . 36 - 0 . 55) × 2 × 2 π × 60 × 0 . 6 = 0 . 169 1
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3. Synchronous Machine Modeling a)
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Midterm2-PracticeSolutions - PRACTICE MIDTERM EXAM 2 SPRING...

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