Assignment_4_Solution - Assignment #4 Chapter 4 Fall 2010-...

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Assignment #4 – Chapter 4 Fall 2010- EE 362Q /EE 394.9 Power Quality Prof. S. Santoso (instructor) Problem 4.14 Power input to the motor is given by, P = 3000 x 746= 2.238 MW Efficiency is given as 90% and power factor is 0.8 for the motor. Hence motor MVA rating is given by, 2.238 2.8257 0.9 0.8 MVA m S == × Note: There are two possible ways to solve this problem: a) Using circuit analysis and b) Using short-circuit capacity. The motor MVA rating is needed if one uses circuit analysis (e.g. the circuit in Fig. 4.23) for solving the problem. The following solution uses short-circuit capacity and hence does not require the knowledge of the motor MVA rating. Either solution method leads to the same result. The transformer short circuit capacity is given by, MVA 150 08 . 0 12 = = SC TR MVA Short circuit capacity at bus ‘A’ is 40 MVA Hence short circuit capacity at bus ‘B’ can be calculated as follows: SC A SC TR SC B MVA MVA MVA 1 1 1 + = 40 1 150 1 1 + = SC B MVA =31.58 MVA If only one motor was started, then the voltage sag at starting is 15% or V min = 0.85 p.u.
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This note was uploaded on 11/21/2010 for the course EE 362Q taught by Professor Santoso during the Spring '10 term at University of Texas at Austin.

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Assignment_4_Solution - Assignment #4 Chapter 4 Fall 2010-...

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