4.4 Example of [ Z Bus ] matrix formulation in the presence of mutual impedances

Transcription

1 4.4 Example of [ Z matrix formulation in the presence of mutual impedances Consider the network shown in Fig Figure 4.32: The power system for [ Z example A tree for the network is shown in Fig The system data is given in Table 4.1. Figure 4.33: Tree of the network Table 4.1: System data Self Mutual Element no. code Impedance code Impedance p - q z pq,pq (p.u.) r - s z pq,rs (p.u.) 1 0-1(1) j (2) j (2) j j (1) j j j

2 Step 1: The algorithm starts building the [ Z matrix element by element. To initiate the process, start with element 1 connected between nodes p = 0 and q = 1, shown in Fig The [ Z matrix of the partial network is given as, Figure 4.34: Partial network in Step 1 (1) [ Z = (1) [ j0.4 Step 2: Next add element 2 connected between p = 0 and q = 1 which is mutually coupled to the existing element 1, connected between ρ = 0 and σ = 1. This new element is a link as it does not create a new node, the partial network for this step is shown in Fig The augmented [ Z (temp) matrix after the addition of this element, is given by Figure 4.35: Partial network in Step 2 Z (temp) (1) (l) (1) j0.4 = [ Z 1l (l) Zl1 Zll Z l1 = Z 01 Z 11 + ȳ0 1(2),0 1(1)( Z 01 Z 11 ) ȳ 0 1(2),0 1(2) Z ll = Z 0l Z 1l ȳ 0 1(2),0 1(1)( Z 0l Z 1l ) ȳ 0 1(2),0 1(2) where, Z01 and Z 0l are the elements of [ Z matrix associated with the reference node. The primitive impedance matrix [ z for the partial network is 0 1(1) 0 1(2) 0 1(1) j0.4 j0.2 [ z = [ 0 1(2) j0.2 j0.5 The primitive admittance matrix [ȳ for the partial network in nothing but the inverse of primitive 139

3 impedance matrix [ z and is given by 0 1(1) 0 1(2) [ȳ = [ z 1 0 1(1) j3.125 j1.25 = [ 0 1(2) j1.25 j2.5 With Z 01 = 0, since 0 is the reference node, Zl1 is evaluated as Z l1 = j0.4 + j1.25( j0.4) j2.5 = j0.2 = Z 1l Also as Z 0l = 0, since 0 is the reference node, and hence, Zll is calculated as Z ll = j j1.25(j0.2) j2.5 = j0.50 The augmented Z (temp) bus matrix is given as Z (temp) (1) (l) (1) j0.4 j0.2 = [ (l) j0.2 j0.5 The row and column corresponding to the l th row and column corresponding to a link addition,(shown in red in the above matrix), need to be eliminated as the link addition does not create a new node. The [ Z matrix, after the addition of second element to the partial network, is calculated using the following expression [ Z = [ Z Z 1l Zl1 Z ll = j0.4 ( j0.2)( j0.2) j0.5 (1) Z = (1) [ j0.32 Note that the size of Z matrix is still (1 1) as no new node has been added to the partial network as yet. Step 3: Next add element 3, which is connected between the nodes p = 0 and q = 2. This is a branch addition as a new node, node 2 is created. This element is mutually coupled to the existing element 1. Hence, the primitive [ z matrix of the partial network, shown in Fig. 4.36, is 140

4 Figure 4.36: Partial network in Step 3 0 1(1) 0 1(2) (1) j0.4 j0.2 j0.1 [ z = 0 1(2) j0.2 j j0.1 0 j0.5 The primitive [ȳ matrix is calculated as [ z 1 and is equal to 0 1(1) 0 1(2) (1) j3.333 j1.333 j0.667 [ȳ = 0 1(2) j1.333 j2.533 j j0.667 j j2.133 The modified [ Z matrix is expressed as (1) (2) (1) j0.32 [ Z = [ Z 12 (2) Z21 Z22 For this element p = 0 and q = 2 and the set of elements [ ρ σ mutually coupled to this element is [0 1(1) 0 1(2) Z 21 = Z 01 + [ȳ 0 2,0 1(1) ȳ 0 2,0 1(2) [ Z 02 Z 11 Z 01 Z 11 ȳ 0 2,0 2 Z 01 and Z 02 are the transfer impedances associated with the reference node and are equal to zero. Z 21 = [j0.667 j [ j0.32 j0.32 j = j0.06

5 Hence, Z12 = Z 21 = j0.06 Z 22 = Z 02 + Z 21 = The modified [ Z matrix is 1 + [ȳ 0 2,0 1(1) ȳ 0 2,0 1(2) [ Z 02 Z 12 Z 01 Z 12 ȳ 0 2, [j0.667 j [ j0.32 j0.32 j2.133 = j0.48 (1) (2) (1) j0.32 j0.06 [ Z = [ (2) j0.06 j0.48 Step 4: On adding element 4 between p = 2 and q = 3, a new node, node 3 is created. Hence, this is a branch addition and is shown in Fig The modified [ Z matrix can be written as Figure 4.37: Partial network in Step 4 (1) (2) (3) (1) j0.32 j0.06 Z 13 Z = (2) j0.06 j0.48 Z 23 (3) Z31 Z32 Z33 As this element is not mutually coupled to other elements the elements of vector ȳ pq,ρσ are zero. Hence, the new elements of [ Z matrix can be calculated, using the expression given in (4.41), as : Off-diagonal elements Z qi = Z pi i = 1, 2, 3 i q 142

6 Z 31 = Z 21 = j0.06 Z 32 = Z 22 = j0.48 Z 13 = Z 31 = j0.06 Z 23 = Z 32 = j0.48 Diagonal element Using the expression of (4.48) with no mutual coupling, the diagonal element can be written as: Z qq = Z pq + z pq,pq hence, Z 33 = Z 23 + z 23,23 = j j0.4 = j0.88 (1) (2) (3) (1) j0.32 j0.06 j0.06 Z = (2) j0.06 j0.48 j0.48 (3) j0.06 j0.48 j0.88 Step 5: Finally add element 5 between nodes p = 1 and q = 3. This is an addition of a link hence a temporary row and column are added. Fig showns the final network after the addition of this element. The modified Z (temp) matrix can be written as Figure 4.38: The complete network after the addition of link in step 5 Z (temp) = (1) (2) (3) (l) (1) j0.32 j0.06 j0.06 Z 1l (2) j0.06 j0.48 j0.48 Z 2l (3) j0.06 j0.48 j0.88 Z 3l (l) Zl1 Zl2 Zl3 Zll 143

7 Since this element is not mutually coupled to other elements, the new elements of [ Z (temp) matrix can be calculated, using the expression of (4.55), as : Off-diagonal elements Z li = Z pi Z qi i = 1, 2, 3 Z 1l = Z 11 Z 13 = j0.32 j0.06 = j0.26 = Z l1 Z 2l = Z 21 Z 23 = j0.06 j0.48 = j0.42 = Z l2 Diagonal element Z 3l = Z 31 Z 33 = j0.06 j0.88 = j0.82 = Z l3 For calculating the diagonal element, the expression given in (4.59) is used. Hence, Z ll = Z pl Z ql + z pq,pq Z ll = Z 1l Z 3l + z 13,13 = j j j0.6 = j1.68 Hence, the temporary [Z (temp) matrix can be written as [ Z (temp) = (1) (2) (3) (l) (1) j0.32 j0.06 j0.06 j0.26 (2) j0.06 j0.48 j0.06 j0.42 (3) j0.06 j0.48 j0.88 j0.82 (l) j0.26 j0.42 j0.82 j1.68 The l th row and l th column are to be eleminated to restore the size of Z to 3 3. The elimination is done using the relation Z = Z (temp) Z Z T Z ll Z T = [j0.26 j0.42 j0.82 j0.23 j0.32 j0.06 j0.06 j0.42 [ j0.26 j0.42 j0.82 Z = j0.06 j0.48 j0.48 j0.82 j1.68 j0.06 j0.48 j

8 Hence, the final matrix [ Z is (1) (2) (3) (1) j j j [ Z = (2) j j j (3) j j j After this discussion of formulation of [ Z matrix, we are now ready to discuss fault analysis, which we will start from the next lecture. 145