Explanation Given: Value of S i j = S j i ∗ Calculation: The circuit diagram is redrawn as shown in Figure 1. Complex power balance satisfies at each bus. Write the KCL equation at bus 1 . S G 1 − S D 1 − S 13 − ( 0.4 + j 0.2 ) = 0 ...... (1) Here, S G 1 is complex power entering bus 1 , S D 1 and S 13 are the complex powers leaving the bus 1 . Substitute 1 + j 1 for S G 1 and 1 − j 1 for S D 1 in equation (1). 1 + j 1 − ( 1 − j 1 ) − S 13 − ( 0.4 + j 0.2 ) = 0 1 + j 1 − 1 + j 1 − S 13 − ( 0.4 + j 0.2 ) = 0 − 0.4 + j 1.8 − S 13 = 0 Rearrange for S 13 . S 13 = − 0.4 + j 1.8 Substitute S 13 for S j i and S 31 for S i j in given function. S 31 = S 13 ∗ Substitute − 0.4 + j 1.8 for S 13 in above equation. S 31 = ( − 0.4 + j 1.8 ) ∗ = − 0.4 − j 1.8 Write the KCL equation at bus 2 . S G 2 − S D 2 − S 23 − ( − 0.4 + j 0.2 ) = 0 ...... (2) Here, S G 2 is complex power entering bus 2 , S D 2 and S 23 are the complex powers leaving the bus 2 . Substitute 0.5 + j 0.5 for S G 2 and 1 + j 1 for S D 2 in equation (2). 0.5 + j 0.5 − ( 1 + j 1 ) − S 23 − ( − 0

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN: 9781305632134

Author: J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher: Cengage Learning

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Question

Chapter 2, Problem 2.29P

To determine

The value of complex power S13, S31, S23, S32 and SG3.

Expert Solution & Answer

Chapter 2 Solutions

Power System Analysis and Design (MindTap Course List)

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The value of complex power S 13 , S 31 , S 23 , S 32 and S G 3 .

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The value of complex power S13, S31, S23, S32 and SG3.

Chapter 2 Solutions