Lecture Methods of Electric power systems analysis - Lesson 4: Power flow provide students with knowledge about using the Ybus; solving for bus currents; solving for bus voltage; complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudes; power balance equations; early power flow system size; . | ECEN 615 Methods of Electric Power Systems Analysis Lecture 4 Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering Texas A amp M University overbye@ Announcements Start reading Chapters 1 to 3 from the book mostly background material Homework 1 is assigned today. It is due on Thursday September 3 Public website is https ecen-615-fall-2020 1 Using the Ybus If the voltages are known then we can solve for the current injections Ybus V I If the current injections are known then we can solve for the voltages 1 Ybus I V Zbus I where Z bus is the bus impedance matrix However this requires that Ybus not be singular note it will be singular if there are no shunt connections 2 Solving for Bus Currents For example in previous case assume V j Then 12 12 j16 j 12 j16 12 j j Therefore the power injected at bus 1 is S1 V1I1 j j S2 V2 I 2 j j j 3 Solving for Bus Voltages For example in previous case assume I Then 1 12 12 j16 j 12 j16 12 Therefore the power injected is S1 V1I1 j 5 j S2 V2 I 2 j 4 Power Flow Analysis When analyzing power systems we know neither the complex bus voltages nor the complex current injections Rather we know the complex power being consumed by the load and the power being injected by the generators plus their voltage magnitudes Therefore we can not directly use the Ybus equations but rather must use the power balance equations 5 Power Balance Equations From KCL we know at each bus i in an n bus system the current injection I i must be equal to the current that flows into the network n I i I Gi I Di Iik k 1 Since I Ybus V we also know n I i I Gi I Di YikVk k 1 The network power injection is then Si Vi I i 6 Power Balance Equations cont d n n Si Vi I i Vi YikVk Vi YikVk k 1 k 1 This is an equation
