4.1.3 Addition of a link between an existing node and the reference node (Case 3):

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4.1.3 Addition of a link between an existing node and the reference node (Case 3): When an element is connected between an existing node and the reference, it creates a loop and thus, the addition of

1 4.1.3 Addition of a link between an existing node and the reference node (Case 3): When an element is connected between an existing node and the reference, it creates a loop and thus, the addition of this element is equivalent to the addition of a link. This will not generate any new node and the size of modified Z Bus matrix remains unchanged. However, all the elements are modified and need to be recalculated. Let the added element,with an impedance of z qo, be connected between an existing node q and the reference node 0 as shown in the Fig Ī l is the current Figure 4.12: Addition of a link between an existing node q and the reference through the link as shown in the Fig This current modifies the current injected into q th from Īq to Īq Īl. The modified network equations can be written as, bus V 1 V q V m = Z 11 Ī 1 + Z 12 Ī Z 1q (Īq Īl) + + Z 1m Ī m = Z q1 Ī 1 + Z q2 Ī Z qq (Īq Īl) + + Z qm Ī m = Z m1 Ī 1 + Z m2 Ī Z mq (Īq Īl) + + Z mm Ī m (4.17) Also V q = z qo Ī l (4.18) Substuting V q from the equation (4.17) into the equation (4.18) one can write Z q1 Ī 1 + Z q2 Ī Z qq (Īq Īl) + + Z qm Ī m = z qo Ī l 110

2 or 0 = Z q1 Ī 1 Z q2 Ī 2 Z qq Ī q + Z qm Ī m + ( Z qq + z qo )Īl (4.19) Equations equation (4.17) and equation (4.19) together form the set of (m + 1) simultaneous network equations which can be expressed in matrix form as: V 1 V 2 V q = V m 0 Z 11 Z12 Z 1q Z 1m Z 1q Z 21 Z22 Z 2q Z 2m Z 2q Z q1 Zq2 Z qq Z qm Z qq Z m1 Zm2 Z mq Z mm Z mq Z q1 Z q2 Z qq Z qm Zqq + z qo Ī 1 Ī2 Ī q Ī m Ī l (4.20) The link current Īl has to be eliminated and hence, the last row and column of modified Z Bus matrix have to be eliminated. The partitioned matrix relation of equation (4.20) can be written in compact form as: [ V m ] Bus (m) (1) (m) Z m Bus [ Z] = [ (1) [ Z] T Z ] ll 0 [ Z] = [ Z 1q Z 2q Z qq Z mm ] T Ī m Bus Ī l (4.21) From equation (4.20) one can write 0 = [ Z] T [Ī m Bus] + Ī l or, Ī l = [ Z] T [Ī m Bus ] (4.22) From equation (4.21) one can also write, [ V m Bus] = [ Z m Bus][Ī Bus ] + [ Z]Īl (4.23) Substituting Īl from equation (4.22) into equation (4.23) one obtains, [ V m Bus] = [[ Z m Bus] [ Z][ Z] T ] [Ī m Bus] (4.24) 111

Hence, [ V m Bus] = [ Z Bus ][Ī m Bus] (4.25) where [ Z Bus ] = [[ Z m Bus] [ Z][ Z] T ] (4.26) It is worth observing that the [ Z Bus ] matrix is an m m matrix i.e. the size of the [ Z Bus ] matrix does not increase when a link is added to the partial network of m buses as no new node is created.

3 Hence, [ V m Bus] = [ Z Bus ][Ī m Bus] (4.25) where [ Z Bus ] = [[ Z m Bus] [ Z][ Z] T ] (4.26) It is worth observing that the [ Z Bus ] matrix is an m m matrix i.e. the size of the [ Z Bus ] matrix does not increase when a link is added to the partial network of m buses as no new node is created Addition of a link between two existing nodes (Case 4): Let an element with impedance z pq be connected between two existing nodes p and q. This is an addition of a link as it forms a loop encompassing nodes p and q as shown in Fig Let Īl Figure 4.13: Addition of a link between two existing nodes p and q be the current through the link as shown in the Fig This link current changes the injected current at p th node from Īp to (Īp Īl), while the injected current at node q th is modified from Īq to (Īq + Īl). The modified network equations can be written as: 112

4 V 1 = Z 11 Ī 1 + Z 12 Ī Z 1p (Īp Īl) + + Z 1q (Īq + Īl) + + Z 1m Ī m V p = Z p1 Ī 1 + Z p2 Ī Z pp (Īp Īl) + + Z pq (Īq + Īl) + + Z pm Ī m V q = Z q1 Ī 1 + Z q2 Ī Z qp (Īp Īl) + + Z qq (Īq + Īl) + + Z qm Ī m (4.27) V m = Z m1 Ī 1 + Z m2 Ī Z mp (Īp Īl) + + Z mq (Īq + Īl) + + Z mm Ī m Also from Fig. 4.13, the relation between V p and V q in terms of Īl and z pq can be written as V p V q = z pq Ī l (4.28) or Substituting V p and V q is obtained: 0 = V p + V q + z pq Ī l (4.29) from the equation (4.27) into the equation (4.29) in the following relation 0 =( Z q1 Z p1 )Ī1 + ( Z q2 Z p2 )Ī2 + + ( Z qp Z pp )Īp + ( Z qq Z pq )Īq + + ( Z qm Z pm )Īm + ( Z pp + Z qq 2 Z pq + z pq )Īl (4.30) Equations (4.27) and (4.30) form a set of (m + 1) simultaneous equation which can be written in matrix form as: V 1 V 2 V q = V m 0 Z 11 Z12 Z 1q Z 1m Z1l Z 21 Z22 Z 2q Z 2m Z2l Z q1 Zq2 Z qq Z qm Zql Z m1 Zm2 Z mq Z mm Zml Z l1 Zl2 Z lq Z lm Zll Where, Z 1l = Z l1 = ( Z 1q Z 1p ) ; Z2l = Z l2 = ( Z 2q Z 2p ) Z ql = Z lq = ( Z qq Z qp ) ; Zml = Z lm = ( Z mq Z mp ) Ī 1 Ī2 Ī q Ī m Ī l (4.31) also and 113

5 = Z qq + Z pp 2 Z pq + z pq. For eliminating the link current Īl equation (4.31) can be written in the compact form as: V m Bus 0 (m) (1) (m) Z m Bus [ = [ Z] (1) [ Z] T Z ] ll Ī m Bus Ī l (4.32) [ Z] = [( Z 1q Z 1p ) ( Z pq Z pp ) ( Z qq Z qp ) ( Z mq Z mp )] T Now using equation (4.26), the [ Z Bus ] matrix, after the elemination of the link current Īl, can be determined. It is worth noting that the [ Z Bus ] is still (m m) in size as no new node has been created. Summarizing the step-by-step procedure for building the [ Z Bus ] as follows: Step 1: Draw the graph of the network and select a tree of the graph. Identify the branches and the links of the graph. A tree of a graph with branches and links is shown in Fig Figure 4.14: Tree of a graph Step 2: Select a branch connected to the reference node to initiate the [ Z Bus ] matrix building process. From Fig. 4.14, it is evident that the first branch selected could be either 1 or 3 or 4 as these are the only branches connected to the reference node. Let the branch 1 be selected as the starting branch and z po be the impedance of the branch then Z (m) Bus = [ (1) (1) z p0 ] Step 3: Pick up another element from the graph. It should either be connected to an existing node or the reference node. Never select an element connected to two new nodes as it will be isolated from the existing partial network and this will result in the elements of [ Z Bus ] matrix becoming infinite. For instance, with reference to Fig. 4.14, in the next step of [ Z Bus ] matrix building process, if 114

6 element 5 is next added to the partial network as shown in Fig. 4.15, the resultant network is disjointed. This is an incorrect choice. The proper choice could be any one of the elements 2 or 3 or 4 or 6. Figure 4.15: Selecting a wrong element in the step-by-step process If the bus impedance matrix of a partial network with m-nodes, [ Z m Bus ], is known, then depending on whether the added (m + 1) th element is a branch or a link, the following steps are to be followed to obtain the new [ Z Bus ] matrix: (a) If the added element is a branch between a new node q and the reference node with an impedance z qo, then the size of the new [ Z Bus ] matrix will increase by one and the new matrix is given as : (1) ( ) (m) (q) (1) Z11 Z 1m 0 Z Bus = (m) Zm1 Z mm 0 (q) z q0 (b) If the added element is a branch between an existing node p and a new node q with an impedance z pq, then a new row and column corresponding to the new node q is added to the existing [ Z m Bus ] matrix. The new matrix is calculated as follows: (1) (p) (m) (q) (1) Z11 Z 1p Z 1m Z1p (p) Zp1 Z pp Z pm Zpp Z Bus = (m) Zm1 Z mp Z mm Zmp (q) Zp1 Z pp Z pm Zpp + z pq 115

7 (c) If the added element is a link between an existing node q and the reference node with an impedance z qo, then no new node is added to the network. A two-step procedure has to be followed to find the new bus impedance matrix. In the first step, a column and a row will be temporarily added to existing [ Z m Bus ] matrix as: Z (temp) Bus = (1) (q) (m) (l) (1) Z11 Z 1q Z 1m Z 1q (q) Zq1 Z qq Z qm Z qq (m) Zm1 Z mq Z mm Z mq (l) Z q1 Z qq Z qm Zll = Z qq + z q0 The additional row and column have to be deleted so that [Z Bus ] matrix is (m m) in size. The elimination process is carried out using [ Z Bus ] = [[ Z m Bus] [ Z][ Z] T ]] [ Z] = [ Z 1q Z 2q Z qq Z mq ] T (d) If the added element is a link between two existing nodes p and q with an impedance z pq, then again the two step procedure as outlined in (step c) is to be followed. The temporary impedance matrix Z (temp) Bus is calculated as: (1) (p) (q) (m) (l) (1) Z11 Z1p Z1q Z1m ( Z 1q Z 1p ) (p) Zp1 Zpp Zpq Zpm ( Z pq Z pp ) (q) Zq1 Zqp Zqq Zqm ( Z qq Z qp ) (m) Zm1 Zmp Zmq Zmm ( Z mq Z mp ) (l) ( Z q1 Z p1 ) ( Z qp Z pp ) ( Z qq Z pq ) ( Z qm Z pm ) Zll = Z pp + Z qq 2 Z pq + z pq 116

8 Next eliminate the added row and column l using the expression: [ Z Bus ] = [[ Z m Bus] [ Z][ Z] T ] [ Z] = [( Z 1q Z 1p ) ( Z pq Z pp ) ( Z qq Z qp ) ( Z mq Z mp )] T Step 4: Repeat Step 3 till all the elements are considered. In the next lecture, we will be looking at an example of [Z Bus ] matrix building algorithm. 117

4.3 Z BUS formation considering mutual coupling between elements

4.3 Z BUS formation considering mutual coupling between elements Assume that the bus impedance matrix [ Z m BUS ] is known for a partial network of m nodes and a reference node. The bus voltage and bus