4.3 Z BUS formation considering mutual coupling between elements

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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.

1 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 current relation for the partial network, shown in Fig. 4.26, can be expressed as: Figure 4.26: Partial Network with m-buses [ V BUS ] = [ Z m BUS ] [Ī BUS ] (4.33) In equation (4.33), V BUS is (m 1) bus voltage vector Ī BUS is (m 1) bus current vector Z m BUS is (m m) bus impedance matrix The new added element p-q may be a branch or may be a link as discussed in the previous algorithm. 128

4.3.1 Addition of a branch to this partial network: The performance equation of the network with an added branch p-q is V 1 V 2 V p V m V q = Z 11 Z12 Z 1p Z 1m Z1q Z 21 Z22 Z 2p Z 2m Z2q Z p1 Zp2 Z

2 4.3.1 Addition of a branch to this partial network: The performance equation of the network with an added branch p-q is V 1 V 2 V p V m V q = Z 11 Z12 Z 1p Z 1m Z1q Z 21 Z22 Z 2p Z 2m Z2q Z p1 Zp2 Z pp Z pm Zpq Z m1 Zm2 Z mp Z mm Zmq Z q1 Zq2 Z qp Z qm Zqq Ī 1 Ī2 Ī p Ī mīq (4.34) The network is assumed to contain bilateral passive elements and hence, Zqi = Z iq for i = 1, 2, m, i q. The added branch p-q is assumed to be mutually coupled with one or more elements of the partial network. To determine element Z qi, inject a current at i th node and calculate the voltage at q th node with respect to reference, as shown in Fig Figure 4.27: Calculation of Z qi for the addition of branch Calculation of Z qi As all other bus currents are zero, bus voltages can be written as, 129

3 V 1 = Z 1i Ī i V 2 = Z 2i Ī i V p = Z pi Ī i (4.35) V m = Z mi I i V q = Z qi Ī i Also from Fig. 4.27, V p and V q can be related as, V q = V p v pq (4.36) Where v pq is the voltage across the added element p q. Also, the currents in the elements of the network can be related to the voltages across the elements as, [īpq ȳ ρσ,pq ī ρσ ] = [ ȳpq,pq ȳ pq,ρσ ] [ v pq ] (4.37) ȳ ρσ,ρσ v ρσ Where, ī pq = the current through the added element p q. ī ρσ = (m 1) current vector of the elements of the partial network. v ρσ = (m 1) voltage vector of the elements of the partial network. = Self-admittance of the added element. ȳ pq,ρσ = (m 1) vector of mutual admittances between the added element p q and the elements ρ σ of the partial network. ȳ ρσ,ρσ = (m m) primitive admittance matrix of the partial network. ȳ ρσ,pq = [ȳ pq,ρσ ] T The diagonal elements of primitive impedance matrix [ z] are the self impedance of the individual elements while the off-diagonal elements are mutual impedances between the elements. The inverse of primitive impedance matrix is the primitive admittance matrix [ȳ]. This can be explained with the help of an illustrative example. A single line diagram of a power system is shown in Fig The self-impedances of lines are written by the side of the line and are in p.u.. The two lines between nodes 1 and 3 are mutually coupled with a mutual impedance of j0.10. The primitive impedance matrix for the network can be written as, j j [ z] = 0 0 j j0.25 j j1.0 j

4 Figure 4.28: Sample Power System The inverse of [ z] is [ȳ], the primitive admittance matrix. j j [ȳ] = 0 0 j j5.0 j j2.5 j6.25 The current ī pq in the added branch p q equal to zero as node q is open. ī pq = 0 (4.38) The voltage v pq, however, is not zero as the added branch is mutually coupled to one or more elements of the partial network. Thus, the voltage across other elements of the network can be expressed as, v ρσ = V ρ V σ (4.39) where V ρ and V σ are the voltages of the nodes of the partial network. With ī pq = 0 from equation (4.37) one can write, v pq + ȳ pq,ρσ v ρσ = 0 Hence, v pq = ȳpq,ρσ v ρσ (4.40) 131

5 Substituting v ρσ from equation (4.39) and v pq equation (4.36) in equation (4.40) one gets, V q = V p + ȳpq,ρσ( V ρ V σ ) Substitution of Īi = 1 pu in equation (4.35) results in V p, V q, V ρ and V σ being replaced by their corresponding impedances and hence, Z qi = Z pi + ȳpq,ρσ( Z ρi Z σi ) (4.41) i = 1, 2,, m, i q For calculating the self impedance Z qq, a current Īq = 1 p.u. is injected into q th node with all other currents equal to zero as shown in Fig Then the voltages of the nodes are calculated from equation (4.35), as V 1 = Z 1q Ī q V 2 = Z 2q Ī q V p = Z pq Ī q (4.42) V m = Z mq I q V q = Z qq Ī q With Īq = 1 p.u., Zqq Can be calculated directly by calculating V q. also, V q = V p v pq (4.43) and ī pq = Īq = 1 (4.44) Hence, from equation (4.37) one gets ī pq = 1 = v pq + ȳ pq,ρσ v ρσ (4.45) And thus v pq can be written as, Substituting v pq and v ρσ, the above equation can be rewritten as, v pq = 1 + ȳ pq,ρσ v ρσ (4.46) V q = V p ȳ pq,ρσ( V ρ V σ ) y pq,pq (4.47) 132

6 Figure 4.29: Calculation of Z qq for the addition of a branch With Īq = 1 p.u., from equation (4.35) V p, V q, V ρ and V σ can be replaced by respective transfer impedances, Z qq = Z pq ȳ pq,ρσ( Z ρq Z σq ) y pq,pq (4.48) Addition of a link to this partial network: If the added element p q is a link, then a fictitious node l is created by connecting an ideal voltage in series with the added element, as shown in Fig The value of the source voltage is selected such that the current Īl through the added link is zero. If ē l is the voltage of node l with respect to node q and Īl is the current injected into node l from node q. The performance equation of the partial network with the added link p l and ideal series voltage source e l is, V 1 Z Z1p... Z1m Z1l V p = Z p1... Zpp... Zpm Zpl V m Z m1... Zmp... Zmm Zml ē l Z l1... Zlp... Zlm Zll Ī 1 Ī p Ī mīl (4.49) 133

7 Figure 4.30: Calculation of Z li for the addition of a link Here, Zli represents the transfer impedance relating the current, I i, injected into the i th bus and the voltage of the added source ē l, connected between nodes l and q. They are conceptually different from the elements Z ij of Z BUS matrix which relate the current injected into the j th bus and the voltage of the i th bus with repect to the reference. As ē l = V l V q, The elements Z li, 1 = 1, 2, m, i l, of the added row and column, can be calculated by injecting a current Īi into the i th node and determining the voltage of l th node with repect to q th node. Hence, Z li = ēl Ī i, Īk = 0, k = 1, 2, m, k l Also, ē l = V p V q v pl (4.50) The current ī pl through the link can be written as, ī pl = ȳ pl,pl v pl + ȳ pl,pσ v ρσ (4.51) Since the current through the link is zero, ī pl = ī pq = 0 Hence, v pl = ȳpl,ρσ v ρσ ȳ pl,pl (4.52) 134

Since the voltage source is ideal source, one can write, ȳ pl,ρσ = ȳ pq,ρσ and So, ȳ pl,pl = (4.53) v pl = ȳpq,ρσ v ρσ = ȳpq,ρσ( V ρ V σ ) (4.54) With I i = 1 p.u., substituting V p, Vq, Vp and V σ from equation (4.

8 Since the voltage source is ideal source, one can write, ȳ pl,ρσ = ȳ pq,ρσ and So, ȳ pl,pl = (4.53) v pl = ȳpq,ρσ v ρσ = ȳpq,ρσ( V ρ V σ ) (4.54) With I i = 1 p.u., substituting V p, Vq, Vp and V σ from equation (4.35) and v pl from equation (4.52) in equation (4.50) one gets Z li = Z pi Z qi + ȳpq,ρσ( Z pi Z σi ) (4.55) i = 1, 2, m, i l To calculate Z ll, a current is injected at the l th node with respect to node q, as shown in Fig As all other node currents are zero, the node voltages can be written as, Figure 4.31: Calculation of Z ll for the addition of a link V k = Z kl Ī l, k = 1, 2, m 135

9 ē l = Z ll Ī l (4.56) With Īl = 1 p.u., Z ll can be directly computed by calculating ē l. The current in the element p l is ī pl = Īl = 1 p.u. (4.57) From equation (4.37) one gets ī pl = ȳ pl,pl v pl + ȳ pl,pσ v pσ = 1 Further as ȳ pl,pl = and ȳ pl,pσ = ȳ pq,pσ, hence v pl can be expressed as Substituting v pl from equation (4.50), one can write v pl = 1 + ȳ pq,pσ v pσ (4.58) ē l = V p V q ȳ pq,pσ( V p V σ ) With Īl = 1 p.u., substituting V p, V q, V ρ, V σ and ē l from equation (4.56), Zll is obtained as Z ll = Z pi Z qi ȳ pq,ρσ( Z pl Z σl ) (4.59) i = 1, 2, m, i l In the case of link addition the additional row and column corresponding to fictitious node l are to be eliminated. For this the fictitious series voltage source ē l is short circuited. From equation (4.49) the bus voltages can be written in compact from as [ V m Bus] = [ Z m Bus][Ī m Bus] + [ Z][Ī l ] (4.60) Where, [ Z] = [ Z1l Z2l Z ml ] T is an (m 1) vector comprising of the entries of the column added to the Z m Bus matrix [Ī m ], [ Bus V Bus m ] = (m 1) bus current and voltage vectors respectively, of the partial network before the addition of element p l. [ Z m ] = (m m) [ Z Bus Bus ] matrix of the partial network before the addition of element p l. Ī l = current injected in the link. Also, [ē l ] = [ Z] T [Ī Bus ] + Z ll Ī l = 0 (4.61) 136

10 On substituting Īl from equation (4.61) into equation (4.60) [ V Bus ] can be written as, [ V Bus ] = [ Z m Bus] [ Z. Z T Z ll ][ĪBus] (4.62) Hence, the final [ Z Bus ] which is (m m) in size can be written as [ Z Final Bus ] = [ Z m Bus] [ Z. ZT ] (4.63) Z ll An example illustrating the [ Z Bus ] building procedure will be discussed in the next lecture. 137

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

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