E&CE 261: Power Systems 
                                                                                                                            
                      STEPS TO SOLVE POWER FLOW ANALYSIS: FOR DUMMIES 
                       
                          1.  Represent the system by its one line diagram 
                              The point of this is to just identify all the buses in the system and see how all the 
                              impedances relate between them.  Label all the buses and write all the data that 
                              has been given.  Normally, this diagram is already given. 
                               
                              Example 
                               
                          2.  Convert all quantities to Per Unit 
                              Often, the values that we are given are not in per unit with respect to one common 
                              base value, so we need to find all the parameters that we are given with respect to 
                              one common base value.  This base value we compute with respect to is normally 
                              explicitly specified, but if not we can assume one and move on. 
                           
                              Example Cont. 
                           
                              If we consider the above example, with V  and V  already given in per unit values and 
                              S     = 100 MVA                              1       3
                                base
                              In Bus 2: The base is: -(400 + j250)/100 = -4 – j2.5 pu    
 P  = - 4  Q  = - 2.5 
                                                                                              2        2
                              In Bus 3: The base us: +(200)/100 = 2 pu                   
 P  = 2 
                                                                                              3
                          3.  Draw the Impedance Diagram 
                              Now that all the values have been expressed in terms of one common per unit 
                              base, we can represent the power system with inductors. 
                               
                                  -    A Generator is represented with a source and an inductor, X  
                                                                                                          L
                                           o   The value of X  will always be given, but not in terms of the 
                                                                 L 
                                               common per unit base value – we need to calculate the base value 
                                               in step 2 
                                  -    Transmission Lines are represented by an impedance – we need to 
                                       calculate the base value in step 2 
                                  -    A Motor is represented with a source and an inductor, XM 
                                           o   The value of X  will always be given, but not in terms of the 
                                                                 M 
                                               common per unit base value – we need to calculate the base value 
                                               in step 2 
                                  -    Transformers are represented by an inductor – we need to calculate the 
                                       base value in step 2 
                               
                           
                           
                           
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                                                                                E&CE 261: Power Systems 
                                                                                                         
                         Example Cont. 
                          
                         In the example above we are not given enough data to find this step (We don’t know the 
                         voltage power, rated voltage, and inductance/impdedance) 
                          
                      4.  Obtain the Y  matrix. 
                                     bus 
                         We now need to find the relationships between all the bus lines.  We need to 
                         calculate the self admittance and mutual admittance.  For mutual admittance we 
                         multiply by negative 1. 
                          
                         Example Cont. 
                                          -1             -1
                         Y  = (0.02 + j0.04) + (0.01 + j0.03)  = 20 –j50  
                           11                   -1
                         Y  = Y  = - [0.02 + j0.04]  = -10 + j20 
                           12   21              -1
                         Y  = Y  = - [0.01 + j0.03]  = -10 + 30j 
                           13   31        -1                -1
                         Y  = (0.02 + j0.04)  + (0.0125 + j0.025)  = 26 – j52 
                           22                      -1
                         Y  = Y  = - [0.0125 + j0.025]  = -16 + j32 
                           23   32        -1                -1
                         Y  = (0.01 + j0.03)  + (0.0125 + j0.025)  = 26 - j62 
                           33
                         Note: 
                         Y  =  - [Y  + Y ] 
                           11     12   13
                         Y   = - [Y  + Y ] 
                           22     12   23
                         Y   = - [Y  + Y ] 
                           33     23   13
                         Putting this all together we get: 
                          
                          
                          
                          
                          
                          
                         IMPORTANT STEP: It is very useful to covert these values to polar form (|V |, θθ ): 
                                                                                                  θθ
                                                                                                ij ij
                       
                       
                       
                       
                       
                         Note: Angles are in radians for this example, but for consistency use degrees. 
                       
                       
                      5.  Classify the buses as follows: 
                          (Delta is the voltage angle) 
                           
                                 Bus Type               Given Parameters          Unknown Parameters 
                                 Slack Bus                     V, δδ                      P, Q 
                                                                 δδ
                              Generator Bus                   P, |V|                      Q, δδ 
                                                                                             δδ
                                 Load Bus                      P, Q                       V, δδ 
                                                                                            δδ
                    
                   
                   
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                                                                                                                                                                        E&CE 261: Power Systems 
                                                                                                                                                                                                                            
                                              6.  Start answering the missing variables, by assumptions (unless it is specified 
                                                     otherwise): 
                                               
                                                             a.  Slack, assume nothing 
                                                             b. Generator, assume δδ = 0 
                                                                                                           δδ
                                                             c.  Load, assume V = 1 pu, δδ = 0 
                                                                                                                    δδ
                                       
                                       
                                                     Example Cont. 
                                               
                                                                                                                                                                          Given                   Required to 
                                        Bus Number                             Type                          Given                       Unknown                    Parameters                   Approximate 
                                                                                                                                                                         to Use 
                                                   1 Slack                                                                                  P , Q                                                           - 
                                                                                                             V , δ                             1      1                   V , δ  
                                                                                                               1      1                                                      1     1
                                                   2 Load P, Q                                                                             |V |, δ                        P , Q                        |V |, δ  
                                                                                                               2      2                        2      2                     2      2                      2       2
                                                   3 Voltage P, |V|                                                                         Q, δ                         P , |V |                          δ  
                                                                                                              3       3                        3     3                      3      3                         3
                                       
                                      Assume (for now) that: 
                                      |V | = 1 pu 
                                          2
                                      δ  = 0
                                         2        
                                      δ  = 0 
                                        3
                                              7.  Find approximations for the Real and Reactive Power that we are given, using the 
                                                     assumed and given values for voltage/angles/admittance.  Find the difference in 
                                                     this with the value that was actually given. 
                                       
                                                     Example Cont. 
                                       
                                                     We now need equations for P , Q , and P : 
                                                                                                            2      2              3
                                       
                                       
                                       
                                       
                                       
                                       
                                       
                                       
                                       
                                       
                                       
                                       
                                                     We know all these parameters so we can solve for the first approximation of P P  and 
                                                     Q                                                                                                                                               2,    3,
                                                        2
                                                     We find: 
                                                     P = -1.14 
                                                        2 
                                                     P = 0.5616 
                                                        3 
                                                     Q2 = -2.28 
                                                      
                                                     Since we know P , Q , and P , we can find ∆P , ∆Q , and ∆P : 
                                                                                     2      2              3                              2         2                3
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                                                                               E&CE 261: Power Systems 
                                                                                                       
                         ∆value = Given Value – Approximated Value 
                   
                   
                   
                   
                   
                   
                   
                   
                   
                   
                   
                      8.  Write the Jacobian Matrix for the first iteration of the Newton Raphson Method. 
                         This is in the form: 
                           [∆∆values] = [Jacobian Matrix] * [∆∆ for Unknown Parameters] 
                                        ∆∆                          ∆∆
                          
                         Example Cont. 
                         So in this case we know ∆P , ∆Q , and ∆P and need to find the Jacobian partial 
                                                 2   2       3 
                         derivatives for the unknown values: δ , δ , |V |, 
                                                         2  3   2
                         So this means the Jacobian matrix is a 3x3 matrix, so we need to find 9 partial 
                         derivatives. 
                         We can do this as follows: 
                          
                           
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
                          
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