EEE471Ch7

# EEE471Ch7 - Symmetrical Faults Chapter 7 1 Faults Shunt...

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1 Symmetrical Faults Chapter 7

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2 Faults Shunt faults: Three phase a b c Line to line Line to ground 2 Line to ground b a c a b c a b c g
3 Faults Series faults One open phase: a b c 2 open phases a b c Increased phase impedance Z a b c

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4 Why Study Faults? Determine currents and voltages in the system under fault conditions Use information to set protective devices Determine withstand capability that system equipment must have: Insulating level Fault current capability of circuit breakers: Maximum momentary current Interrupting current
5 Symmetrical Faults AC R L ) ( 2 ) ( α ϖ + = t VSin t e α t=0 i(t) Fault at t = 0 2 V

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6 Symmetrical Faults For a short circuit at generator terminals at t=0 and generator initially open circuited: dt di L Ri t e + = ) ( dt di L Ri t VSin + = + ) ( 2 α ϖ by using Laplace transforms i(t) can be found ( L is considered constant)
7 Symmetrical Faults ] / ) ( ) ( [ 2 ) ( T t e Sin t Sin Z V t i - - - - + = θ α ϖ 2 2 2 2 ) ( X R L R Z + = + = R X Tan R L Tan 1 1 - - = = Where: R X R L T = = Time Constant ] / ) ( ) ( [ 2 ) ( T t e Sin t Sin ac I t i - - - - + = θ Where: I ac = ac RMS fault current at t=0 (Examples) Note that for a 3- phase system α will be different for each phase. Therefore, DC offset will be different for each phase

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8 EXAMPLE FOR α = θ = 90 O 0 1 2 3 4 5 6 7 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0 .2 0 .4 0 .6 0 .8 1 VOLTAGE 0 1 2 3 4 5 6 7 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0 .2 0 .4 0 .6 0 .8 1 I 90 O I DC = 0 0 1 2 3 4 5 6 7 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0 .2 0 .4 0 .6 0 .8 1 I RMS Fault at t=0 ] / ) ( ) ( [ 2 ) ( T t e Sin t Sin ac I t i - - - - + = θ α θ ϖ 0 0 0 0
9 Example θ = 90 o and α = 0 0 1 2 3 4 5 6 7 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0 .2 0 .4 0 .6 0 .8 1 Voltage I AC I DC I RMS 0 1 2 -1 0 1 Fault at t=0 0 ] / ) ( ) ( [ 2 ) ( T t e Sin t Sin ac I t i - - - - + = θ α θ ϖ

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10 Symmetrical Faults I ac and I dc are independent after t = 0 2 2 dc I ac I RMS I + = T t e aco I dc I - = 2 (max) Substituting: T t e ac I T t e ac I ac I RMS I 2 2 1 2 ) 2 2 ( 2 (max) - + = - + =
11 Asymmetry Factor I RMS (max) = K(τ) I ac Asymmetry Factor = K(τ) r x e K τ π 4 2 1 ) ( - + = Where: τ = number of cycles (Example 7.1)

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12 Example 7.1 Short circuit on System: V=20 kV LN , X = 8 Ω, R = .8 Ω, Fault occurs at a time to produce maximum DC offset Find: a)I AC b) I RMS momentary at cycles 5 . 0 = τ c)I RMS interrupting current a) RMS AC kA I 488 . 2 8 . 8 20 2 2 = + = b) At .5 cycles: 438 . 1 2 1 ) ( 10 / ) 5 (. 4 = + = - π e K I momentary =(1.438)(2.488) =3.577 kA c) At 3 cycles: 023 . 1 2 1 ) ( 10 / ) 3 ( 4 = + = - e K I (interrupting) = (1.023)(2.488) = 2.545.kA
13 AC Decrement In the previous analysis we treated the generator as a constant voltage behind a constant impedance for each phase. The constant inductance is valid for steady state conditions but for transient conditions, the generator inductance is not constant. Recall that for steady state conditions, the

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