Please answer PROBLEM 3
Problem \#3 (25 pts) Validate your answers by using a software solver (MatPower or PowerWorld) to solve for Problem 1. Problem 1: (100 pts) Using Example 6.9 or your text, and modifying the system as a lossless transmission line system, compute the following; 1. Formulate the Newton Raphson Power Flow problem (20 pts): a. Define the knowns/unknowns (5 pts) b. Mismatch equations ( 5pts) c. Jacobian Formulation (5 pts) d. Iterative Update equations ( 5pts) 2. Perform a single iteration of the NR method and show (45 pts): a. Mismatch Equations ( 15pts) b. Jacobian matrix (15 pts) c. Iterative update Equations ( 15pts). 3. Using the same system, compare the first iteration of the NR Power Flow computation to the Decoupled Power Flow solution, showing (15pts): a. Decoupled PF Jacobian (5 pts) b. Decoupled PF Iteration update equation (5 pts) c. Discuss the differences between the Decoupled Power Flow and your answer in Item 2.c. (5 pts) 4. Using the same system, compare the first iteration of the NR power flow computation to the Fast Decoupled Power Flow solution, showing (20 pts): a. Fast Decoupled B' and B" matrices (10 pts) b. Fast Decoupled Iteration Update equation (10 pts) Power flow input data and Ybus Figure 6.2 shows a single-line diagram of a five-bus power system. Input data are given in Tables 6.1,6.2, and 6.3. As shown in Table 6.1, bus 1, to which a generator is connected, is the swing bus. Bus 3 , to which a generator and a load are connected, is a voltage-controlled bus. Buses 2,4 , and 5 are load buses. Note that the loads at buses 2 and 3 are inductive since Q2=QL2=2.8 and QL3= 0.4 are negative. For each bus k, determine which of the variables Vk,k,Pk, and Qk are input data and which are unknowns. Also, compute the elements of the second row of Ybus. Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall leaming experience. Cengage Leaming reserves the right to remove additional content at any time if subsequent rights restrictions require it. Section 6.4 The Power Flow Problem Bus input data for Example 6.9 Sbsst=100 MVA, Vbsse=15kV at buses 1,3 , and 345kV at buses 2,4,5 TABLE 6.2 Line input data for Example 6.9 Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, seanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Chapter 6 | Power Flows TABLE 6.3 Transformer input data for Example 6.9 TABLE 6.4 Input data and unknowns for Example 6.9 The input data and unknowns are listed in Table 6.4. For bus 1, the swing bus, P1 and Q1 are unknowns. For bus 3 , a voltage-controlled bus, Q3 and 3 are unknowns. For buses 2,4 , and 5 , load buses, V2,V4,V5 and 2,4,5 are unknowns. The elements of Ybus are computed from (6.4.2). Since buses 1 and 3 are not directly connected to bus 2 , Y21=Y23=0 Using (6.4.2), Y24=R24+jX241=0.009+j0.11=0.89276+j9.91964perunit=9.95972/95.143perunitY25=R25+jX251=0.0045+j0.051=1.78552+j19.83932perunit=19.9195/95.143perunitY22=R24+jX241+R25+jX251+j2B24+j2B25 Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Section 6.4 The Power Flow Problem Screen for Example 6.9 =(0.89276j9.91964)+(1.78552j19.83932)+j21.72+j20.88=2.67828j28.4590=28.584784.624perunit where half of the shunt admittance of each line connected to bus 2 is included in Y22 (the other half is located at the other ends of these lines). This five-bus power system is modeled in PowerWorld Simulator case Example 6_9 (see Figure 6.3). To view the input data, first click on the Edit Mode button (on the far left-hand side of the ribbon) to switch into the Edit mode (the Edit mode is used for modifying system parameters). Then by selecting the Case Information tab, you can view tabular displays showing the various parameters for (Continued) Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the elhook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Chapter 6 | Power Flows the system. For example, use Network, Buses to view the parameters for each bus, and Network, Lines and Transformers to view the parameters for the transmission lines and transformers. Fields shown in blue on the screen can be directly changed simply by typing over them, and those shown in green can be toggled by clicking on them. Note that the values shown on these displays match the per unit values from Tables 6.1 to 6.3, except the power values are shown in actual MW/Mvar units. The elements of Ybus also can be displayed by selecting Solution Details, Ybus. Since the Ybus entries are derived from other system parameters, they cannot be changed directly. Notice that several of the entries are blank, indicating that there is no transmission line or transformer directly connecting these two buses (a blank entry is equivalent to zero). For larger networks, most of the elements of the Ybus are zero since any single bus usually only has a few incident lines (such sparse matrices are considered in Section 6.8). The elements of the Ybus can be saved in a Matlab compatible format by first right-clicking within the Ybus matrix to display the local menu, and then selecting Save Ybus in Matlab Format from the local menu. Finally, notice that no flows are shown on the oneline because the nonlinear power flow equations have not yet been solved, and the solution of these equations are covered next. Using Ybus, the nodal equations for a power system network are written as I=YbusV where I is the N vector of source currents injected into each bus and V is the N vector of bus voltages. For bus k, the k th equation in (6.4.3) is
