<p> Power Flow Examples 1.0 Introduction</p><p>We will solve the same problem in three different ways: - Full Newton - Fast decoupled - DC This problem and the solution by full Newton and the solution by decoupled power flow were adapted the problem in [1].</p><p>2.0 The Problem</p><p>The network to be solved in shown in Fig. 1.</p><p>SG1 10-j20 </p><p>V1=1.050 SD2=400+j250 </p><p>16-j32 10-j30 </p><p>|V3|=1.04 </p><p>SG3=200 MW Fig. 1</p><p>1 In the network, charging capacitance has been neglected. All admittances are represented in pu on a 100 MVA base.</p><p>The Y-bus is given below:  20  j50  10  20  10  j30   Y   10  j20 26  j52  16  j32  10  j30  16  j32 26  j62  We will ignore reactive limits. Stopping criterion is |ΔPk|, |ΔQk|<ε=0.001.</p><p>3.0 Solution by Full Newton</p><p>Bus 1 is the swing bus, bus 2 is a PV bus, and bus 3 is a PQ bus. Therefore, the equations we need are:</p><p>P2  P2 (x)  P2 </p><p>P3  P3 (x)  P3</p><p>Q2  Q2 (x)  Q2</p><p>Using a 100 MVA base, P2=-4.0, P3=2.0, and Q2=-2.5. Then </p><p>2 P2  P2 (x)  4 </p><p>P3  P3 (x)  2 </p><p>Q3  Q3 (x) - 2.5 where: n Pi   Vi Vk Gik cos(i  k )  Bik sin(i  k ) k 1 n Qi   Vi Vk Gik sin(i k )  Bik cos(i k ) k 1 We use a “flat start” for the first iteration:</p><p> 2   0       0   3        V2  1.0 And then the mismatch vector is:</p><p> P   P2   1.14   4.0  2.8600          P   0.5616   2.0   1.4384    3                   Q  Q2    2.28   2.5  0.2200  We then get the Jacobian elements using</p><p>11 Pp (x) J pq   V p Vq G pq sin( p   q )  B pq cos( p   q )  q</p><p>11 Pp (x) 2 J pp   Q p (x)  B pp V p  p</p><p>3 21 Q p (x) J pq   V p Vq G pq cos( p   q )  B pq sin( p   q )  q</p><p>21 Q p (x) 2 J pp   Pp (x)  G pp Vp  p</p><p>12 Pp (x) J pq   Vp G pq cos( p   q )  B pq sin( p   q ) Vq</p><p>12 Pp (x) Pp (x) J pp    G pp V p V p V p</p><p>22 Q p (x) J pq   Vp G pq sin( p  q )  B pq cos( p   q ) Vq</p><p>22 Q p (x) Q p (x) J pp    Bpp Vp  Vp Vp Again, using the “flat start” solution:</p><p>2   0       0   3        V2  1.0 We evaluate the above expressions to get the the Jaobian matrix, resulting in:  54.28  33.28 24.86  Note the off-diagonal submatrices are not all   zero in this case, in contrast to the example J   33.28 66.04  16.64 given in “PowerFlowAlgorithm.doc,” (pg. 10) because here G≠0.  27.14 16.64 49.72  Our update equation is then</p><p>4  P  ( j) ( j )   J  x         Q </p><p> 54.28  33.28 24.86   2   2.8600     33.28 66.04  16.64    1.4384   3          27.14 16.64 49.72 V2   0.2200  Solution to this yields:</p><p>  2   0.045263      0.007718  3        V2   0.026548 The updated solution is then 0  0.045263  0.045263 (1) (0) (0)       x  x  x  0   0.007718   0.007718 1  0.026548  0.97345  For the second iteration, we get</p><p> 51.724675  31.765618 21.302567   2   0.099218     32.981642 65.656383  15.379086    0.021715   3           27.14 16.64 48.103589 V2   0.050914  Solution to this yields:</p><p>  2   0.001795      0.000985  3        V2   0.001767 The updated solution is then</p><p>5  0.045263  0.001795  0.04706 (2) (1) (1)       x  x  x   0.007718   0.000985   0.00870   0.97345   0.001767  0.971684  For the third iteration, we get</p><p> 51.596701  31.693866 21.147447   2   0.000216     32.933865 65.597585  15.379086    0.000038   3           27.14 17.396932 47.954870 V2   0.000143  We note that the mismatch vector satisfies the stopping criterion, but we will go ahead and complete this iteration to obtain:</p><p>  2    0.000038       0.0000024  3        V2   0.0000044 The updated solution is then  0.04706   0.000038    0.04706  (3) (2) (2)       x  x  x   0.00870   0.0000024   0.008705  0.971684   0.0000044  0.971680  So solution is obtained in 3 iterations.</p><p>4.0 Solution by FDC-nonalternating Recalling that the Y-bus is  20  j50 10  j20 10  j30   Y  10  j20 26  j52 16  j32 10  j30 16  j32 26  j62 </p><p>6 To obtain the B-matrix, we use the imaginary parts of the Y-bus elements only, to get:  50 20 30    B   20  52 32   30 32  62 To obtain the matrix we will use to obtain updates on the angles, we remove the first row and column. Let’s call this B’:  52 32  B     32  62 Since bus 3 is a PV bus, we will not have a reactive power equation for it, and so to obtain the B-matrix on the update equation we need to remove the corresponding column and row. This will be the second row and column of the B’ matrix. We can call the resulting matrix B’’. It is B   52 Recalling from the above example that for the “flat start”</p><p>7  2   0       0   3        V2  1.0 The mismatch vector is:  P    1.14   4.0  2.8600               0.5616   2.0    1.4384    Q    2.28  2.5  0.2200  Then the first iteration of the FDC (non- alternating) is: ( j )  P2   V   2   2.86   P3   ( j )  52 32  2    B       1.0 V3      1.4384   32  62  3 ⋮         1.04   Pn     Vn </p><p>8 ( j )  Q2   V   2   Q3   ( j ) 0.22 B V   V   52V2   3  1.0  ⋮  Qn     Vn </p><p>Solution to these equations yields:</p><p> 2   0.060483       3   0.008989</p><p>V2  0.0042308 The updated solution is then 0   0.060483   0.060483 (1) (0) (0)       x  x  x  0    0.008989    0.008989 1  0.0042308  0.995769  The mismatch vector for the new solution is computed to be</p><p>9  P   0.175895           0.070951     Q   1.579042  And the solution is repeated. The algorithm requires 13 iterations before the stopping criterion on the mismatch vector is satisfied, and the solution obtained is the same as the solution obtained by the full Newton.</p><p>5.0 Solution by FDC-alternating This algorithm is the same as that presented in Section 4.0 except we will update the bus 3 reactive power mismatch vector with the angle solution obtained from the B’ matrix before we use the B’’ matrix to obtain the voltage update. This should result is less iterations.</p><p>6.0 Solution by DC Power Flow </p><p>10 The B’ matrix is used here, with the real power injection according to  B  P  52 32  B     32  62  4 P     2 </p><p> 52 32 2   4         32  623   2 </p><p>2  - 0.0282 - 0.0145 4 - 0.0836          3  - 0.0145 - 0.0236 2  - 0.0109 Need to review the comparison here. Compare to full Newton solution:   0.04706  (3)   x   0.008705  0.971680  To compare the two answers in terms of flows, one can use the approximate DC flow equation, given by:</p><p>Pkj  Bkj (k  j ) where the Bkj elements are given by</p><p>11  50 20 30    B   20  52 32   30 32  62 From the DC approximate solution:</p><p>P12  B12 (1 2 )  20(0  .0836) 1.672</p><p>P13  B13 (1 3 )  30(0  .0109)  0.327</p><p>P23  B23 (2 3 )  32(.0836  .0109)  2.3264 From the Full Newton solution, but using the approximate DC flow equations:</p><p>P12  B12 (1 2 )  20(0  .04706)  0.9412</p><p>P13  B13 (1 3 )  30(0  .008705)  0.2415</p><p>P23  B23 (2 3 )  32(.04706  .008705)  1.22736   0.04706  (3)   x   0.008705  0.971680 </p><p>12 1[] H. Saadat, “Power System Analysis,” 1999, McGraw-Hill.</p>