power part 6 fall 2011.pptpeople.ece.umn.edu/class/ee0001/pefall2011/power part 6... · 2011. 10....
TRANSCRIPT
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EE1PEEE Refresher Class
Power Part 6
notes by T. Ernst
EE1 – Power Part 6 Notes Fall 2011, Page 1
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EE1 – Power Part 6 Notes
• Sequence Impedances• T-Lines• Rotating Machines• Transformers
• Short Circuit Calculations• 3-phase faults• Phase-phase faults• Phase-ground faults• Phase-phase-ground faults
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EE1 – Power Part 6 Notes
Symmetrical components
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EE1 – Power Part 6 Notes
Sequence impedances
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EE1 – Power Part 6 Notes
T-line sequence impedances
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EE1 – Power Part 6 Notes
Rotating machine sequence impedances
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EE1 – Power Part 6 Notes
Transformer sequence impedances
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EE1 – Power Part 6 Notes
Transformer zero sequence impedance: Connection dependent
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EE1 – Power Part 6 Notes
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EE1 – Power Part 6 Notes
Neutral impedances
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EE1 – Power Part 6 Notes
Delta – GrdY Transformers
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EE1 – Power Part 6 Notes
Three winding transformers
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EE1 – Power Part 6 Notes
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EE1 – Power Part 6 Notes
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EE1 – Power Part 6 Notes
Fault Calculations and Superposition Theory
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EE1 – Power Part 6 Notes
Load current versus fault current
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EE1 – Power Part 6 Notes
3-Phase Faults
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EE1 – Power Part 6 Notes
3-Phase Faults
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EE1 – Power Part 6 Notes
Phase-phase faults
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EE1 – Power Part 6 Notes
Phase-phase Faults
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EE1 – Power Part 6 Notes
Phase-ground Faults
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EE1 – Power Part 6 Notes
Phase-ground Faults
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EE1 – Power Part 6 Notes
Phase-ground Faults
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EE1 – Power Part 6 Notes
Phase-phase-ground Faults
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EE1 – Power Part 6 Notes
Phase-phase-ground Faults
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EE1 – Power Part 6 Notes
Example ProblemFeeder from Large Utility System
480V Main Bus
23,000 – 480 GY Volt
5700/6000/6500 KVA
Z = j8.7%
GENHydro Generator
1000 KVA, 460V
Impedance data (see next page)
Remote 480V Bus
Calculate all types of faults here
0.1 Miles
500 KCMil Aluminum
PVC Conduit
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EE1 – Power Part 6 Notes
XD = synchronous impedance = j1.25 pu
XD’ = transient impedance = j0.30 pu
XD’’ = sub-transient impedance = j0.30 pu
X2 = negative sequence impedance = j0.48 pu
X0 = zero sequence impedance = j0.19 pu
all at generator base of: 1000 KVA, 460 V
Generator Z
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EE1 – Power Part 6 Notes
Solution approach:1. Define system pu bases2. Calculate all component impedances on system
base3. Draw positive, negative and zero sequence
networks4. Reduce sequence networks to equivalent sequence
impedances5. Calculate each type of fault
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EE1 – Power Part 6 Notes
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EE1 – Power Part 6 Notes
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EE1 – Power Part 6 Notes
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