Lecture 19

# 050 09044 j 02973 j 03 100 158 the motors internal

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 02/09/2014 for the course ECE 5042 taught by Professor Degroat during the Fall '13 term at Ohio State.

Ask a homework question - tutors are online