Page 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 ¯ I BUS (4.33) In equation (4.33), ¯ V BUS is (m× 1) bus voltage vector ¯ I 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. Page 2 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 ¯ I BUS (4.33) In equation (4.33), ¯ V BUS is (m× 1) bus voltage vector ¯ I 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. 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 ¯ Z 12 ? ¯ Z 1p ? ¯ Z 1m ¯ Z 1q ¯ Z 21 ¯ Z 22 ? ¯ Z 2p ? ¯ Z 2m ¯ Z 2q ? ¯ Z p1 ¯ Z p2 ? ¯ Z pp ? ¯ Z pm ¯ Z pq ? ¯ Z m1 ¯ Z m2 ? ¯ Z mp ? ¯ Z mm ¯ Z mq ¯ Z q1 ¯ Z q2 ? ¯ Z qp ? ¯ Z qm ¯ Z qq ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ¯ I 1 ¯ I 2 ? ¯ I p ? ¯ I m ¯ I q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4.34) The network is assumed to contain bilateral passive elements and hence, ¯ Z qi = ¯ 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 ati th node and calculate the voltage atq th node with respect to reference, as shown in Fig. 4.27. 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, Page 3 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 ¯ I BUS (4.33) In equation (4.33), ¯ V BUS is (m× 1) bus voltage vector ¯ I 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. 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 ¯ Z 12 ? ¯ Z 1p ? ¯ Z 1m ¯ Z 1q ¯ Z 21 ¯ Z 22 ? ¯ Z 2p ? ¯ Z 2m ¯ Z 2q ? ¯ Z p1 ¯ Z p2 ? ¯ Z pp ? ¯ Z pm ¯ Z pq ? ¯ Z m1 ¯ Z m2 ? ¯ Z mp ? ¯ Z mm ¯ Z mq ¯ Z q1 ¯ Z q2 ? ¯ Z qp ? ¯ Z qm ¯ Z qq ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ¯ I 1 ¯ I 2 ? ¯ I p ? ¯ I m ¯ I q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4.34) The network is assumed to contain bilateral passive elements and hence, ¯ Z qi = ¯ 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 ati th node and calculate the voltage atq th node with respect to reference, as shown in Fig. 4.27. 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, ¯ V 1 = ¯ Z 1i ¯ I i ¯ V 2 = ¯ Z 2i ¯ I i ? ¯ V p = ¯ Z pi ¯ I i ? ¯ V m = ¯ Z mi I i ¯ V q = ¯ Z qi ¯ I i ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4.35) 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, ¯ i pq ¯ i ?s = ¯ y pq,pq ¯ y pq,?s ¯ y ?s,pq ¯ y ?s,?s ¯ v pq ¯ v ?s (4.37) Where, ¯ i pq = the current through the added element ‘p-q’. ¯ i ?s = (m× 1) current vector of the elements of the partial network. ¯ v ?s = (m× 1) voltage vector of the elements of the partial network. ¯ y pq,pq = Self-admittance of the added element. ¯ y pq,?s = (m× 1) vector of mutual admittances between the added element ‘p-q’ and the elements ‘?-s’ of the partial network. ¯ y ?s,?s = (m×m) primitive admittance matrix of the partial network. ¯ y ?s,pq = ¯ y pq,?s T The diagonal elements of primitive impedance matrix [¯ z] are the self impedance of the individual elements while the o?-diagonal elements are mutual impedances between the elements. The inverse of primitive impedance matrix is the primitive admittance matrix [¯ y]. This can be explained with the help of an illustrative example. A single line diagram of a power system is shown in Fig. 4.28. 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 ofj0.10. The primitive impedance matrix for the network can be written as, ¯ z= ? ? ? ? ? ? ? ? ? ? ? ? ? j0.60 0 0 0 0 0 j0.50 0 0 0 0 0 j0.50 0 0 0 0 0 j0.25 j1.0 0 0 0 j1.0 j0.20 ? ? ? ? ? ? ? ? ? ? ? ? ? Page 4 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 ¯ I BUS (4.33) In equation (4.33), ¯ V BUS is (m× 1) bus voltage vector ¯ I 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. 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 ¯ Z 12 ? ¯ Z 1p ? ¯ Z 1m ¯ Z 1q ¯ Z 21 ¯ Z 22 ? ¯ Z 2p ? ¯ Z 2m ¯ Z 2q ? ¯ Z p1 ¯ Z p2 ? ¯ Z pp ? ¯ Z pm ¯ Z pq ? ¯ Z m1 ¯ Z m2 ? ¯ Z mp ? ¯ Z mm ¯ Z mq ¯ Z q1 ¯ Z q2 ? ¯ Z qp ? ¯ Z qm ¯ Z qq ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ¯ I 1 ¯ I 2 ? ¯ I p ? ¯ I m ¯ I q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4.34) The network is assumed to contain bilateral passive elements and hence, ¯ Z qi = ¯ 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 ati th node and calculate the voltage atq th node with respect to reference, as shown in Fig. 4.27. 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, ¯ V 1 = ¯ Z 1i ¯ I i ¯ V 2 = ¯ Z 2i ¯ I i ? ¯ V p = ¯ Z pi ¯ I i ? ¯ V m = ¯ Z mi I i ¯ V q = ¯ Z qi ¯ I i ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4.35) 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, ¯ i pq ¯ i ?s = ¯ y pq,pq ¯ y pq,?s ¯ y ?s,pq ¯ y ?s,?s ¯ v pq ¯ v ?s (4.37) Where, ¯ i pq = the current through the added element ‘p-q’. ¯ i ?s = (m× 1) current vector of the elements of the partial network. ¯ v ?s = (m× 1) voltage vector of the elements of the partial network. ¯ y pq,pq = Self-admittance of the added element. ¯ y pq,?s = (m× 1) vector of mutual admittances between the added element ‘p-q’ and the elements ‘?-s’ of the partial network. ¯ y ?s,?s = (m×m) primitive admittance matrix of the partial network. ¯ y ?s,pq = ¯ y pq,?s T The diagonal elements of primitive impedance matrix [¯ z] are the self impedance of the individual elements while the o?-diagonal elements are mutual impedances between the elements. The inverse of primitive impedance matrix is the primitive admittance matrix [¯ y]. This can be explained with the help of an illustrative example. A single line diagram of a power system is shown in Fig. 4.28. 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 ofj0.10. The primitive impedance matrix for the network can be written as, ¯ z= ? ? ? ? ? ? ? ? ? ? ? ? ? j0.60 0 0 0 0 0 j0.50 0 0 0 0 0 j0.50 0 0 0 0 0 j0.25 j1.0 0 0 0 j1.0 j0.20 ? ? ? ? ? ? ? ? ? ? ? ? ? Figure 4.28: Sample Power System The inverse of ¯ z is ¯ y, the primitive admittance matrix. ¯ y= ? ? ? ? ? ? ? ? ? ? ? ? ? -j1.67 0 0 0 0 0 -j2.0 0 0 0 0 0 -j2.0 0 0 0 0 0 -j5.0 j2.5 0 0 0 j2.5 -j6.25 ? ? ? ? ? ? ? ? ? ? ? ? ? The current ¯ i pq in the added branch ‘p-q’ equal to zero as node ‘q’ is open. ¯ i pq = 0 (4.38) Thevoltage ¯ v pq , however, isnotzeroastheaddedbranchismutuallycoupledtooneormoreelements of the partial network. Thus, the voltage across other elements of the network can be expressed as, ¯ v ?s = ¯ V ? - ¯ V s (4.39) where ¯ V ? and ¯ V s are the voltages of the nodes of the partial network. With ¯ i pq = 0 from equation (4.37) one can write, ¯ y pq,pq ¯ v pq + ¯ y pq,?s ¯ v ?s = 0 Hence, ¯ v pq =- ¯ y pq,?s ¯ v ?s ¯ y pq,pq (4.40) Page 5 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 ¯ I BUS (4.33) In equation (4.33), ¯ V BUS is (m× 1) bus voltage vector ¯ I 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. 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 ¯ Z 12 ? ¯ Z 1p ? ¯ Z 1m ¯ Z 1q ¯ Z 21 ¯ Z 22 ? ¯ Z 2p ? ¯ Z 2m ¯ Z 2q ? ¯ Z p1 ¯ Z p2 ? ¯ Z pp ? ¯ Z pm ¯ Z pq ? ¯ Z m1 ¯ Z m2 ? ¯ Z mp ? ¯ Z mm ¯ Z mq ¯ Z q1 ¯ Z q2 ? ¯ Z qp ? ¯ Z qm ¯ Z qq ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ¯ I 1 ¯ I 2 ? ¯ I p ? ¯ I m ¯ I q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4.34) The network is assumed to contain bilateral passive elements and hence, ¯ Z qi = ¯ 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 ati th node and calculate the voltage atq th node with respect to reference, as shown in Fig. 4.27. 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, ¯ V 1 = ¯ Z 1i ¯ I i ¯ V 2 = ¯ Z 2i ¯ I i ? ¯ V p = ¯ Z pi ¯ I i ? ¯ V m = ¯ Z mi I i ¯ V q = ¯ Z qi ¯ I i ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4.35) 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, ¯ i pq ¯ i ?s = ¯ y pq,pq ¯ y pq,?s ¯ y ?s,pq ¯ y ?s,?s ¯ v pq ¯ v ?s (4.37) Where, ¯ i pq = the current through the added element ‘p-q’. ¯ i ?s = (m× 1) current vector of the elements of the partial network. ¯ v ?s = (m× 1) voltage vector of the elements of the partial network. ¯ y pq,pq = Self-admittance of the added element. ¯ y pq,?s = (m× 1) vector of mutual admittances between the added element ‘p-q’ and the elements ‘?-s’ of the partial network. ¯ y ?s,?s = (m×m) primitive admittance matrix of the partial network. ¯ y ?s,pq = ¯ y pq,?s T The diagonal elements of primitive impedance matrix [¯ z] are the self impedance of the individual elements while the o?-diagonal elements are mutual impedances between the elements. The inverse of primitive impedance matrix is the primitive admittance matrix [¯ y]. This can be explained with the help of an illustrative example. A single line diagram of a power system is shown in Fig. 4.28. 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 ofj0.10. The primitive impedance matrix for the network can be written as, ¯ z= ? ? ? ? ? ? ? ? ? ? ? ? ? j0.60 0 0 0 0 0 j0.50 0 0 0 0 0 j0.50 0 0 0 0 0 j0.25 j1.0 0 0 0 j1.0 j0.20 ? ? ? ? ? ? ? ? ? ? ? ? ? Figure 4.28: Sample Power System The inverse of ¯ z is ¯ y, the primitive admittance matrix. ¯ y= ? ? ? ? ? ? ? ? ? ? ? ? ? -j1.67 0 0 0 0 0 -j2.0 0 0 0 0 0 -j2.0 0 0 0 0 0 -j5.0 j2.5 0 0 0 j2.5 -j6.25 ? ? ? ? ? ? ? ? ? ? ? ? ? The current ¯ i pq in the added branch ‘p-q’ equal to zero as node ‘q’ is open. ¯ i pq = 0 (4.38) Thevoltage ¯ v pq , however, isnotzeroastheaddedbranchismutuallycoupledtooneormoreelements of the partial network. Thus, the voltage across other elements of the network can be expressed as, ¯ v ?s = ¯ V ? - ¯ V s (4.39) where ¯ V ? and ¯ V s are the voltages of the nodes of the partial network. With ¯ i pq = 0 from equation (4.37) one can write, ¯ y pq,pq ¯ v pq + ¯ y pq,?s ¯ v ?s = 0 Hence, ¯ v pq =- ¯ y pq,?s ¯ v ?s ¯ y pq,pq (4.40) Substituting ¯ v ?s from equation (4.39) and ¯ v pq equation (4.36) in equation (4.40) one gets, ¯ V q = ¯ V p + ¯ y pq,?s ( ¯ V ? - ¯ V s ) ¯ y pq,pq Substitution of ¯ I i = 1 pu in equation (4.35) results in ¯ V p , ¯ V q , ¯ V ? and ¯ V s being replaced by their corresponding impedances and hence, ¯ Z qi = ¯ Z pi + ¯ y pq,?s ( ¯ Z ?i - ¯ Z si ) ¯ y pq,pq (4.41) ?i= 1, 2,?,m, i?q For calculating the self impedance ¯ Z qq , a current ¯ I q = 1 p.u. is injected into q th node with all other currents equal to zero as shown in Fig. 4.29. Then the voltages of the nodes are calculated from equation (4.35) , as ¯ V 1 = ¯ Z 1q ¯ I q ¯ V 2 = ¯ Z 2q ¯ I q ? ¯ V p = ¯ Z pq ¯ I q ? ¯ V m = ¯ Z mq I q ¯ V q = ¯ Z qq ¯ I q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4.42) With ¯ I q = 1 p.u., ¯ Z qq Can be calculated directly by calculating ¯ V q . also, ¯ V q = ¯ V p - ¯ v pq (4.43) and ¯ i pq =- ¯ I q =-1 (4.44) Hence, from equation (4.37) one gets ¯ i pq =-1= ¯ y pq,pq ¯ v pq + ¯ y pq,?s ¯ v ?s (4.45) And thus ¯ v pq can be written as, ¯ v pq =- 1+ ¯ y pq,?s ¯ v ?s ¯ y pq,pq (4.46) Substituting ¯ v pq and ¯ v ?s , the above equation can be rewritten as, ¯ V q = ¯ V p + 1+ ¯ y pq,?s ( ¯ V ? - ¯ V s ) y pq,pq (4.47)Read More

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