050 09044 j 02973 j 03 100 158 the motors internal

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Unformatted text preview: 1.103∠7.1° 5! Network Fault Example, cont'd The motor's terminal voltage is then 1.05∠0 - (0.9044 - j 0.2973) × j 0.3 = 1.00∠ − 15.8° The motor's internal voltage is 1.00∠ − 15.8° − (0.9044 - j 0.2973) × j 0.2 = 1.008∠ − 26.6° We can then solve as a linear circuit: 1.103∠7.1° 1.008∠ − 26.6° If = + j 0.15 j 0.5 = 7.353∠ − 82.9° + 2.016∠ − 116.6° = j 9.09 6! Fault Analysis Solution Techniques !  !  Circuit models used during the fault allow the network to be represented as a linear circuit There are two main methods for solving for fault currents: 1.  2.  Direct method: Use prefault conditions to solve for the internal machine voltages; then apply fault and solve directly Superposition: Fault is represented by two opposing voltage sources; solve system by superposition –  first voltage just represents the prefault operating point –  second system only has a single voltage source 7! Superposition Approach Faulted...
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