4.2 Example of [Z Bus ] matrix building algorithm

Transcription

1 4.2 Example of [Z ] matrix building algorithm The single line diagram of a power system is shown in the Fig The line impedances in pu are also given. The step-by-step procedure for [ Z ] matrix formulation is explained as given below: Figure 4.16: Single Line Diagram of the Power System for the example Preliminary Step: The graph of the network and a tree is shown in Fig Elements 1,2,4 and 5 are the tree branches while 3, 6 and 7 are the links. Figure 4.17: Graph and a tree of the network of Fig Step 1: The step-by-step [ Z ] matrix building algorithm starts with element 1, which is a tree branch connected between nodes 1 and the reference node 0 and has an impedance of z 10 j0.10 pu. This is shown in the accompanying figure,fig

2 Figure 4.18: Partial network of Step 1 The resulting [ Z ] matrix is (1) (1) Z (1) [ z 10 ] (1) [ j0.10 ] Step 2: Next, the element 2 connected between node 2 (q 2) and the reference node 0 is selected. This element has an impedance of z 20 j0.10 p.u. As this is the addition of a tree branch it will add a new node 2 to the existing [ Z ] matrix. This addition is illustrated in Fig Figure 4.19: Partial network of Step 2 The new bus impedance matrix is given by : (1) (2) (1) (2) (1) j (1) j0.1 0 Z [ ] [ (2) 0 z 20 (2) 0 j0.10 ] Step 3: Element 3 connected between existing nodes, node 1 (p 1) and node 2 (q 2), having an impedance of z 12 j0.20 p.u. is added to the partial network, as shown in Fig Since this is an addition of a link to the network a two step procedure is to be followed. In the 119

3 Figure 4.20: Partial network of Step 3 first step a new row and column is added to the matrix as given below : (1) (2) (l) (1) j ( Z 12 Z 11 ) (2) 0.0 j0.10 ( Z 22 Z 21 ) ( Z 21 Z 11 ) ( Z 22 Z 12 ) Zll (1) (2) (l) (1) j j0.10 (2) 0.0 j0.10 j0.10 j0.10 j0.10 j0.40 where, Z ll Z 11 + Z 22 2 Z 12 + z 20 j j j0.20 j0.40 p.u. Next this new row and column is eliminated to restore the size of [ Z ] matrix as given below: j [ Z ] [ 0.0 j0.10 ] [ j0.10 ] [ j0.10 j0.10] j0.10 j0.40 Hence, the impedance matrix after the addition of element 3 is found out to be : (1) (2) (1) j0.075 j0.025 [ Z ] [ (2) j0.025 j0.075 ] Step 4: The element 4, which is added next, is connected between an existing node, node 2 (p 2) and a new node, node 3 (q 3). The impedance of this element is z 23 j0.30 p.u. and it is a tree branch hence, a new node, node 3 is added to the partial network. This addition, shown in Fig. 4.21, thus increases the size of [ Z ] to (3 3). 120

4 Figure 4.21: Partial network of Step 4 The new impedance matrix can be calculated as: (1) (2) (3) (1) j0.075 j0.025 Z12 Z (2) j0.025 j Z22 (3) Z21 Z22 Z22 + z 23 (1) (2) (3) (1) j0.075 j0.025 j0.025 (2) j0.025 j j0.075 (3) j0.025 j0.075 j0.375 Step 5: Element 5 is added next to the existing partial network. This is a tree branch connected between an existing node, node 3 (p 3) and a new node, node 4 (q 4). This is illustrated in Fig Since a new node is added to the partial network, the size of [ Z ] increases to (4 4). The impedance of the new element is z 34 j0.15 p.u. The new bus impedance matrix is : (1) (2) (3) (4) (1) j0.075 j0.025 j0.025 Z31 (2) j0.025 j0.075 j0.075 Z32 Z (3) j0.025 j0.075 j0.375 Z33 (4) Z13 Z23 Z33 Z33 + z 34 (1) (2) (3) (4) (1) j0.075 j0.025 j0.025 j0.025 (2) j0.025 j0.075 j0.075 j0.075 (3) j0.025 j0.075 j0.375 j0.375 (4) j0.025 j0.075 j0.375 j0.525 Step 6: Next,the element 6 connected between two existing nodes node 1 (p 1) and node 4 (q 4) is added to the network, as shown in the Fig The impedance of this element is z 23 j0.25 p.u. As this is a link addition, the two step procedure is used. The bus impedance 121

5 Figure 4.22: Partial network of Step 5 Figure 4.23: Partial network of Step 6 matrix is modified by adding a new row and column as given below: (1) j0.075 j0.025 j0.025 j0.025 ( Z 14 Z 11 ) (2) j0.025 j0.075 j0.075 j0.075 ( Z 24 Z 21 ) (3) j0.025 j0.075 j0.375 j0.375 ( Z 34 Z 31 ) (4) j0.025 j0.075 j0.375 j0.525 ( Z 44 Z 41 ) ( Z 41 Z 11 ) ( Z 42 Z 12 ) ( Z 43 Z 13 ) ( Z 44 Z 14 ) Zll 122

6 Substituting the values of appropriate [ Z ] intermediate impedance matrix is: matrix elements in the last row and column the (1) j0.075 j0.025 j0.025 j0.025 j0.05 (2) j0.025 j0.075 j0.075 j0.075 j0.05 (3) j0.25 j0.075 j0.375 j0.375 j0.35 (4) j0.025 j0.075 j0.375 j0.525 j0.50 j j0.35 j0.50 j0.80 where, Z ll Z 44 + Z 11 2 Z 14 + z 14 j j j j0.25 j0.80 p.u. The additional row and column l are to be eliminated to restore the impedance matrix size to (m m), and the [ Z ] matrix after the addition of element 6 is calculated as: Hence, j0.05 j0.50 j0.075 j0.025 j0.025 j0.025 j0.35 [ j0.05 j0.05 j0.35 j0.50] j0.025 j0.075 j0.075 j0.075 [ Z ] j0.50 j0.25 j0.075 j0.375 j0.375 j0.80 j0.025 j0.075 j0.375 j0.525 (1) (2) (3) (4) (1) j j j j (2) j j j j Z (3) j j j j (4) j j j j Step 7: Finally the element 7 connected between two existing nodes node 2 (p 2) and node 4 (q 4) is added to the partial network of step 6. The impedance of this element is is z 23 j0.40 pu. This is also a link addition, as shown in Fig and hence the two step precedure will be followed to obtain the [ Z ] matrix. In the first step the is calculated after a row and a 123

7 Figure 4.24: Partial network of Step 7 column are added to the exiting Z as follows: (1) j j j j ( Z 14 Z 12 ) (2) j j j j ( Z 24 Z 22 ) (3) j j j j ( Z 34 Z 32 ) (4) j j j j ( Z 44 Z 42 ) ( Z 41 Z 21 ) ( Z 42 Z 22 ) ( Z 43 Z 23 ) ( Z 44 Z 24 ) Zll Substituing the values of the elements of impedance matrix one gets: (1) j j j j j0.281 (2) j j j j j0.281 (3) j j j j j1031 (4) j j j j j j0.281 j0.281 j j j where, Z ll Z 22 + Z 44 2 Z 24 + z 24 j j j j0.40 j p.u. The additional row and column l are to be eliminated to restore the impedance matrix size to 124

8 (m m), and [ Z ] after the addition of element 7 is calculated as: j j j j j j j [j j j j0.1688] j j j j [ Z ] j j0.469 j j j j j j j j Hence, (1) (2) (3) (4) (1) j j j j (2) j j j j Z (3) j j j j (4) j j j j As can be seen that the final [ Z ] matrix is a (4 4) matrix, as the network has 4 nodes and a reference node. As there are 7 elements is the network, 7 steps are required for the formation of [ Z ] matrix Modifications in the existing [ Z ] : If in an existing network, for which the [ Z ] matrix is known, some modification such as line removal or line impedance alteration is carried out then the [ Z ] matrix can be easily modified without any need of reconstructing the matrix from scratch. As an example, let the Z matrix be the final bus impedance matrix given for the network of Fig Next, let the element 7 connecting nodes 2 and 4 be removed from the network and it is required to find the modified Z. Removal of element 7 is equivalent to setting its impedance z 24 to infinite. This can be obtained by connecting a fictitious element z 24 add in parallel to the existing element z org 24 such that the resultant impedance z 24 result is infinite i.e. or 1 z result 24 1 z org z add z add 24 z org 24 j0.40 p.u. Hence, by adding an element z add 24 j0.4 p.u. in parallel to z org 24 the removal of line between nodes 2 and 4 can be simulated. The new added fictitious element is a link addition between the two nodes, p 2 and q 4 and is shown in Fig Hence, this will require a two-step procedure. The addition of the fictitious element 8, which is a link, will introduce a temporary row and column. 125

9 Figure 4.25: Adding a link to simulate the removal of element 7 The is given as: (1) j j j j ( Z 14 Z 12 ) (2) j j j j ( Z 24 Z 22 ) (3) j j j j ( Z 34 Z 32 ) (4) j j j j ( Z 44 Z 42 ) ( Z 41 Z 21 ) ( Z 42 Z 22 ) ( Z 43 Z 23 ) ( Z 44 Z 24 ) Zll Substituting the appropriate values one gets: (1) j j j j j (2) j j j j j (3) j j j j j (4) j j j j j j j j j j where, Z ll Z 22 + Z 44 2 Z 24 + z add 24 j j j ( j0.40) j p.u. 126

10 The additional row and column is eliminated in the following step: j j j j j j j [j j j j0.1131] j j j j [ Z ] j j j j j j j j j j Thus, the final impedance matrix after the removal of element 7 is : (1) (2) (3) (4) (1) j j j j (2) j j j j Z (3) j j j j (4) j j j j The obtained Z matrix is identical to the Z matrix obtained in step 6 of the previous example, which is the impdance matrix of the network before the addition of element 7. So far we have considered the Z matrix building algorithm without any presence of mutually coupled elements. In the next lecture, we will take into account the presence of mutually coupled elements while forming the Z matrix. 127