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ECE4762011_Lect11

# ECE4762011_Lect11 - ECE 476 POWER SYSTEM ANALYSIS Lecture...

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Lecture 11 Ybus and Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

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2 Announcements Be reading Chapter 6. HW 4 is 3.4, 3.10, 3.14, 3.19, 3.23, 3.60; due September 29 in class. First exam is October 11 during class. Closed book, closed notes, one note sheet and calculators allowed
3 Multiple Solution Example 3 The dc system shown below has two solutions: where the 18 watt load is a resistive load 2 2 Load Load Load The equation we're solving is 9 volts I 18 watts 1 +R One solution is R 2 Other solution is R 0.5 Load Load R R = = ÷ = Ω = What is the maximum P Load ?

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4 Bus Admittance Matrix or Y bus First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Y bus . The Y bus gives the relationships between all the bus current injections, I , and all the bus voltages, V , I = Y bus V The Y bus is developed by applying KCL at each bus in the system to relate the bus current injections, the bus voltages, and the branch impedances and admittances
5 Y bus Example Determine the bus admittance matrix for the network shown below, assuming the current injection at each bus i is I i = I Gi - I Di where I Gi is the current injection into the bus from the generator and I Di is the current flowing into the load

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Y bus Example, cont’d 1 1 1 1 2 1 3 1 12 13 1 1 2 1 3 j 1 2 3 2 21 23 24 1 2 3 4 By KCL at bus 1 we have 1 ( ) ( ) (with Y ) ( ) Similarly ( ) G D A B A B j A B A B A A C D C D I I I V V V V I I I Z Z I V V Y V V Y Z Y Y V
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ECE4762011_Lect11 - ECE 476 POWER SYSTEM ANALYSIS Lecture...

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