Ee 1352 Monograph Unitiia

14K.L.N. COLLEGE OF INFORMATION TECHNOLOGY Pottapalayam – 630 611, Sivagangai District.

Year : 2008 -2009,

Branch : III B.E. EEE

Semester : 6

Subject : EE 1352 Power System Analysis

Name of the Writer (Collaborator) : R. Ramesh, Professor (EEE), IT Dept. Unit : II

Date of Submission : 22.11.2008

Syllabus : IInd Unit Introduction – Bus Classification – Bus admittance matrix – Solution of nonlinear algebraic equations – Gauss-Seidal method – Newton Raphson method – Fast decoupled method – Flow charts and comparison of the three methods. -------------------------------------------------------------------------------------EE 1352 - POWER SYSTEM ANALYSIS VI semester EEE Branch (2006-2010 Batch) Unit – II – Power Flow Analysis 1

Syllabus : Introduction – Bus Classification – Bus admittance matrix – Solution of non-linear algebraic equations – Gauss-Seidal method – Newton Raphson method – Fast decoupled method – Flow charts and comparison of the three methods. 1.0 Introduction : The load flow studies involve the solution of the power system network under steady-state conditions. The solution is obtained by considering certain inequality constraints imposed on bus voltages and reactive power of generators. The main information obtained from a load-flow study are the magnitude and phase angle of the voltage at each bus and the real and reactive power flowing in each line. The load flow solution also gives the initial conditions of the system when the transient behavior of the system is to be studied. The load flow study of a power system is essential to decide the best operation of existing system and for planning the future expansion of the power system. The load flow solution of power system under balanced load condition is considered for our studies. The steps involved in load flow studies are : 1. Representation of the system by single-line diagram. 2. Determining the impedance diagram using the information in single line diagram. 3. Formulation of network equations. 4. Solution of network equations. 1.1 Types of Buses In a power system the buses are meeting point of various components. The generators will feed energy to buses and loads will draw energy from buses. In the network of a power system the buses become nodes and so a voltage can be specified for each bus. The bus or bus bar is a conductor made of copper or aluminium having negligible resistance. Hence the bus bar will have zero voltage drop when it conducts the rated current. Therefore the buses are considered as points of constant voltage in a power system. Therefore each bus in a power system is associated with four quantities : 1. Real power, P 2. Reactive power, Q 3. Magnitude of voltage, │V│ 4. Phase angle of voltage, δ In a load flow problem, two quantities (out of four) are specified for each bus and the remaining two quantities are obtained by solving the load flow equations. The classification of buses are : i) Load bus or PQ bus ii) Generator bus or voltage controlled bus iii) Slack bus or swing bus or reference bus. Types of buses can be summarized as a table as follows : Bus type Load bus Generator bus

Quantities specified P, Q P, │V│

Quantities to be obtained │V│, δ Q, δ

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Slack bus │V│, δ P, Q P = PG – PD: │V│= Magnitude of bus voltage Q = QG – QD; δ = Phase angle of bus voltage PG, QG – Real and Reactive powers generated by generators connected to the bus respectively. PD, QD – Real and Reactive powers drawn by the loads connected to the bus respectively. Load Bus : The bus called the load bus, when real and reactive components of power are specified for the bus. The load flow equations can be solved to find the magnitude and phase angle of the voltage. In a load bus the voltage is allowed to vary within permissible limits, for example ± 5 %. Generator Bus : The bus is called generator bus, when real power and magnitude of bus voltage are specify for the bus. The load flow equation can be solved to find the reactive power and phase angle of bus voltage. Usually for generator buses reactive power limits will be specified. Slack Bus : The bus is called slack bus, if the magnitude and phase angle of bus voltage are specified for the bus. The slack bus is the reference bus for load flow solution and usually one of the generator buses is selected as the slack bus. Need for slack bus : Basically the power system has only two types of buses and they are load and generator buses. In these buses only power injected by generators and power drawn by loads are specified, but the power loss in transmission lines are not accounted. Total (complex) power loss in transmission line = Sum of complex power of generators – Sum of complex power of loads. The transmission line losses can be estimated only if the real and reactive power of all buses are known. The powers in the buses will be known only after the solving the load flow equations. For these reasons, the real and reactive power of one of the generator bus is not specified and this bus is called slack bus. It is assumed that the slack bus generates the real and reactive power required for transmission line losses. Hence for a slack bus, the magnitude and phase of bus voltage are specified and real and reactive powers are obtained through the load flow solution. ----------------------------------------------------------------------------------------------------------1,2 Node Equations and Bus admittance matrix When the power system is represented by impedance / reactance diagram, it can be considered as a circuit or network. The buses can be treated as nodes and the voltages of all buses (nodes) can be solved by conventional node analysis technique. Let n be the number of major or principal nodes in a circuit or network. (The principal nodes are meeting points of more than two elements). Since the voltage of a node can be measured only with respect to a reference point, one of the nodes is considered as reference node. Now the network will have (N-1) independent voltages. In

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nodal analysis, the independent voltages are solved by writing KCL equation for (N-1) nodes in the circuit. For writing KCL equations, the voltage sources in the circuit should be converted to equivalent current sources. Let V1, V2, V3,……..Vn = Node voltages of nodes 1,2,3,……n respectively. I11, I22, I33,………Inn = Sum of current sources connected (or injecting current) to nodes 1,2,3,……n respectively. Yjj = Sum of admittances connected to node-j Yjk = Negative of sum of admittances connected between node-j and node-k. Note : If the direction of current in current source is towards the node then the source is considered as positive and if it is away from the node then the source is negative. Now the n-number of nodal equations for N-bus system will be in the form shown below (here n=N). Y11V1 + Y12V2 + Y13V3 + ......... + Y1nVn = I 11 Y21V1 + Y22V2 + Y23V3 + ......... + Y2 nVn = I 22 Y31V1 + Y32V2 + Y33V3 + ......... + Y3nVn = I 33 . . . . . . . . Yn1V1 + Yn 2V2 + Yn 3V3 + ......... + YnnVn = I nn The above n-number of equations can be arranged in the matrix form as shown in equation (1). . . Y1n  V1   I 11   Y11 Y12 Y13 Y . . Y2 n  V 2   I 22   21 Y22 Y23 Y31 Y32 Y33 . . Y3n  V3   I 33      =  ....(1) . . . . .  .   .   .  . . . . . .  .   .       . . Ynn  V n   I nn  Yn1 Yn 2 Yn 3 In matrix notation the equation (1) can be written as YV = I . . . . (2) In power system Y matrix is designated as Ybus, and called the admittance matrix. The node voltages are called bus voltages. Therefore the equation (2) can be written as shown in equation (3) Ybus V = I . . . . . (3) Where, Ybus = Bus admittances matrix of order (n x n) V = Bus voltage matrix of order (n x 1) I = Current sources matrix of order (n x 1) n = Number of independent buses in the system.

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 Y11 Y  21 Y31 Bus admittance matrix, Ybus =   .  .  Yn1

Y12 Y22 Y32 . . Yn 2

Y13 Y23 Y33 . . Yn 3

. . . . . .

. . . . . .

Y1n  Y2 n  Y3n  .......(4) .  .   Ynn 

The bus admittance matrix, Ybus is symmetrical around the principal diagonal. The admittances Y11, Y22, Y33,……..Ynn are called self admittances at the buses and the other admittances are called mutual admittances. In general, Yjj = Sum of all admittances connected to bus-j Yjk = Negative of sum of all admittances between bus-j and bus-k. Also Yjk = Ykj Note : If the power system is represented by reactance diagram, then all the elements of the network are inductive susceptances (which are negative). In this case Yjj will be negative and :Yjk will be positive. 1. 3 Solution of bus voltages Consider the node basis matrix equation of N-bus system Ybus V = I . . . . . (5) −1 On premultiplying the equation (5) by Ybus , we get −1 V = Ybus I . . . . (6)

Adjo int ofYbus Deter min antofYbus Let Δ = Determinant of Ybus and Δjj = Cofactor of Yjk −1 We know that, Ybus =

Now Adjoint of Ybus = .  ∆ 11 ∆ 12 ∆ 13 ∆ .  21 ∆ 22 ∆ 23  ∆ 31 ∆ 32 ∆ 33 .  . . .  .  . . . .  .  ∆ n1 ∆ n 2 ∆ n 3

. . . . . .

T

∆1n   ∆ 11  ∆ ∆ 2n   12   ∆13 ∆ 3n  = .   .  . .    ∆ nn   ∆ 1n

∆ 21 ∆ 22 ∆ 23 . . ∆ 2n

∆ 31 ∆ 32 ∆ 33 . . ∆ 3n

. . . . . .

. . . . . .

∆ n1  ∆ n 2  ∆ n3   . . . . (7) .  .   ∆ nn 

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V1   I 11  V  I   2  22  V3   I 33  Also, V =   andI =   . . . . (8) .  .  .  .      Vn   I nn  Using equations (7) and (8), the equation (6) can be written as shown in equation (9) . . ∆ n1   I11  V1   ∆11 ∆ 21 ∆ 31 V  ∆ . . ∆ n 2   I 22   2  12 ∆ 22 ∆ 32 V3  . . ∆ n 3   I 33  1  ∆ 13 ∆ 23 ∆ 33   =     . . . (9) . . . . .   .  ∆ . . .  . . . . . .   .        . . ∆ nn   I nn  Vn   ∆1n ∆ 2 n ∆ 3n By matrix multiplication the equation (9) can be expressed as n-number of linear independent equations shown below. 1 V1 = [ ∆11 I 11 + ∆ 21 I 22 + ∆ 31 I 33 + ......... + ∆ n1 I nn ] ∆ 1 V2 = [ ∆12 I 11 + ∆ 22 I 22 + ∆ 32 I 33 + ......... + ∆ n 2 I nn ] ∆ 1 V3 = [ ∆13 I 11 + ∆ 23 I 22 + ∆ 33 I 33 + ......... + ∆ n 3 I nn ] ∆ . . . . . . . . . . . . . . . 1 Vn = [ ∆ 1n I 11 + ∆ 2 n I 22 + ∆ 3n I 33 + ......... + ∆ nn I nn ] ∆ In general the kth bus voltage is given by 1 Vk = [ ∆ 1k I 11 + ∆ 2 k I 22 + ∆ 3k I 33 + ......... + ∆ nk I nn ] ∆ 1 n ∴Vk = ∑ ∆ jk I jj . . . . (10) ∆ j =1 Where, Δ = Determinant of Ybus matrix. Ijj = Sum of current sources injecting current to node-j Δjk = Cofactor o the element Yjk of bus admittance matrix. Note : The equation (10) is Cramer’s rule and this equation can be expressed in another simpler form as shown below. Let Δ = Determinant of Ybus matrix Δk = Determinant of Ybus matrix after replacing kth column by current source vector I.

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∆k ∆ 1.4 Bus or Node elimination by matrix algebra (Kron reduction technique) The buses or nodes which does not have any current sources can be eliminated by matrix manipulation of the standard node equation. Consider the general form of node basis matrix equation. Ybus V = I . . . . . (11) Now the matices in the equation can be partitioned using the guidelines given below. 1. The column matrices V and I are rearranged such that the element associated with buses to be eliminated are in the lower rows of the matrices. 2. The square matrix Ybus is rearranged such that the elements associated with buses to be eliminated are in the last rows and columns of the matrices. 3. The bus admittance matrix is partitioned so that elements identified only with nodes to be eliminated are separated from the other elements by horizontal and vertical lines. Now, kth bus voltage Vk =

Consider a bus admittance matrix of order (5 x 5). Now the matrix equation (11) = 5 can be written as shown in equation (2). Let us assume that the buses 4 and 5 not have any current source and so they can be eliminated. Y12 Y13 Y14 Y15  V1   I 11  || Y22 Y23 | Y24 Y25  V2   I 22      Y32 Y33 | Y34 Y35  V3   I 33     =   . . . . . (12) _ _ | _ _  _   _  Y42 Y43 | Y44 Y45  V4   I 44      Y52 Y53 | Y54 Y55  V5   I 55  Let us partition the matrix equation (12) and define the following submatrices. Y11 Y12 Y13  Y14 Y15  Y41 Y42 Y43  Y    T K =  21 Y22 Y23  L= Y24 Y25  L =   Y51 Y52 Y53  Y31 Y32 Y33  Y34 Y35  for n does Y11 Y  21 Y31  _ Y41  Y51

 I 11  I  I A =  I 22  I X =  44   I 55   I 33  Where Ix = Submatrix composed of the currents entering the buses to be eliminated. Vx = Submatrix composed of the voltages of the buses to be eliminated. K = Submatrix composed of self and mutual admittances identified only with buses to be retained. M = Submatrix composed of self and mutual admittances identified only with buses to be eliminated. L = Submatrix composed of only mutual admittances between buses to be retained and eliminated. LT = Transpose of L Y M =  44 Y54

Y45  Y55 

V1  V A = V2  V3 

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Using the submatrices as defined above, the matrix equation (12) can be written as shown in equation (13).  K L  V A   I A   T    =   . . . . (13)  L M  V B   I X  By matrix multiplication the equation (13) can be written as, KVA + LVX = IA . . . . (14) LT VA+ MVX = IX . . . .(15) From equation (15) we get, MVX = IX - LTVA . . . . .(16) Here all the elements of the submatrix IX are zero, because the buses to be eliminated does not have any current source. MVx = -LTVA . . . . . (17) On premultiplying equation (17) by M-1 we get, Vx = -M-1LTVA . . . . . . (18) On substituting for VX from equation (18) in equation (14) we get, KVA – LM-1LTVA = IA

[

]

∴ K − LM −1 LT V A = I A Ybus,newVA = IA . . . . . .(19) Where Ybus,new = K − LM −1 LT . . . . (20) The new bus admittance matrix Ybus,new used to reconstruct the circuit with the unwanted buses eliminated. The matrix partitioning method discussed above is a general procedure and it is suitable for computer solutions. When large number of buses are to be eliminated, then the size of matrix M will be large and finding M-1 will be tedious. The process of bus elimination can be simplified if one bus is eliminated at a time. In this case the matrix M will have only one element and M-1 is the reciprocal of this element. Here, the bus to be eliminated must be the highest numbered bus and renumbering may be required. This can be achieved by row interchange and column interchange. When bus-p has to be eliminated in a system with n-independent buses, the pth row is interchanged with nth row and pth column is interchanged with nth column in the bus admittance matrix. Consider a bus admittance matrix of order (n x n) in which the nth bus has to be eliminated.

Ybus

 Y11  Y 21   .  . =  Y( n −1)1   −  Y n1 

Y12 Y22 . . Y( n −1) 2 − Yn 2

. . . . . − .

. . . . . − .

Y1( n −1) Y2 ( n −1) . . Y( n −1)( n −1) − Yn ( n −1)

| | | | | | |

     .      . . . . . (21) Y1n Y2 n . . Y( n −1) n − Ynn

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Let us partition the bus admittance matrix as shown in equation (21). submatrices are given below.  Y1n  Y12 . . Y1( n −1)   Y11  Y   Y   2n  Y . . Y 21 22 2 ( n −1)    .  . . . . Y3( n −1) ; L =  K =  .     . . . . .    .   Yn1  Y . . Y   n 2 ( n − 1 )( n − 1 )   Y( n −1) n  1 LT = Yn1 Yn 2 . . . Yn ( n −1) ; M = Ynn ; M −1 = Ynn From equation (20) we get, Ybus,new = K – L M-1 LT . . . . . (22) On substituting the submatrices in equation (22) we get, Ybus,new =

[

The

]

 Y11  Y 21   .   .  Yn1 

Y12 Y22 . .

. . . .

. . . .

Yn 2

.

.

  Y1n    Y    2n   1   −  .    Yn1 Yn 2 . Yn ( n −1)     Ynn  .    Y( n −1)( n −1)  Y( n −1) n  Y1( n −1) Y2( n −1) Y3( n −1) .

[

Ybus,new=

 Y11  Y  21  .   .  Yn1 

Y12 Y22

. .

. .

. . Yn 2

. . .

. . .

Y1( n − 1)   Y1nYn1 Y2( n − 1)   1  Y2nYn1 Y3( n − 1)  − .  Y  .  nn   Y( n − 1) nYn1 Y( n − 1)(n − 1) 

Y1nYn 2

.

Y2 nYn 2 . Y( n − 1) nYn 2

. . .

Y1nYn( n − 1)   Y2nYn ( n − 1)  .   Y( n − 1) nYn( n − 1) 

. . . . (23) From equation (23) the element Yjk (i.e. the element in row-k and column-j) of the resulting (n-1) x (n-1) new bus admittance matrix is given by Y jnYnk Y jk ,new = Y jk − . . . . (24) Ynn For j = 1,2,3,….(n-1) and k = 1,2,3,……(n-1) Where Yjk, Yjn, Ynk and Ynn are elements of original or given bus admittance matrix. The equation (24) can be used to compute the elements of new bus admittance matrix directly from the elements of original bus admittance matrix. The following procedure can be used to compute an element of bus admittance matrix using equation (24). Step 1 : Consider an element Yjk of original bus admittance matrix. Step 2 : Get the product of last element of row-j and last element of column-k.

9

]

Step 3 : Divide the product obtained in step-2 by Ynn of original bus admittance matrix. Step 4 : On subtracting the resultant value obtained in step-3 from Yjk of original bus admittance matrix we get, the new value of Yjk. Note : Since bus admittance matrix is symmetrical we can take Yjk,new = Ykj,new and can avoid m(m-1)/2 calculations for a new bus admittance matrix of order (m x m) where m = (n-1). --------------------------------------------------------------------------------------------------------1.6 BUS IMPEDANCE MATRIX The impedance matrix is given by Zbus = Ybus-1 Since the bus admittance matrix is symmetrical, the bus impedance matrix is also symmetrical around the principal diagonal. In bus impedance matrix, the elements on the main diagonal are called driving point impedance of the buses or nodes and the offdiagonal elements are called the transfer impedances of the buses or nodes. The bus impedance matrix is very useful in fault analysis or calculations. The bus impedance matrix can be determined by 1) Taking the inverse of the bus admittance matrix formed. 2) Bus building algorithm. Bus building algorithm Let us denote the original Zbus of a system with n-number of independent buses as Zorig. When a branch of impedance Zb is added to the system the Zorig. Gets modified. The branch impedance Zb can be added to the original system in the following four different ways. Case 1 : Adding a branch of impedance Zb from a new-p to the reference bus. Case 2: Adding a branch of impedance Zb from a new-p to an existing bus-q. Case 3 : Adding a branch of impedance Zb from an existing bus-1 to the reference bus. Case 4 : Adding a branch of impedance Zb between two existing buses h and q. Case 1 : Adding Zb from a new bus-p to the reference bus. Consider a n-bus system as shown in fig.a. Let us add a bus-p through an impedance Zb to the reference bus. The addition of a bus will increase the order of the bus impedance matrix by one. In this case the elements of (n+1)th column and row all zeros except the diagonal. The diagonal element is the added branch impedance Zb. The elements of original Zbus matrix are not altered. The new bus impedance matrix will be as shown in equation (25).

Z bus ,new

  Z orig  =  − −  0 0

| 0 | 0  | 0  . . . . (25)  − | − 0 | Z b 

Case 2 : Adding Zb from a new bus–p to existing bus-q

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Consider a n-bus system as shown in fig.b in which a new bus-p is added through an impedance Zb to an existing bus-q. The addition of a bus will increase the order of the bus impedance matrix by one. In this the element of (n+1)th column are the elements of qth column and elements of (n+1)th row are the elements of qth row. The diagonal element is given by sum of Z qq and Zb. The elements of original Zbus matrix are not altered. The new bus impedance matrix will be as shown in equation (26).

    Z bus , new =    −   Z q1

. . . . (26)

Zbus − Zq2

− .

− Z qq

| | | | | |

Z1q  Z 2q  .   .  −   Z qq + Z b 

Case 3 : Adding Zb from an existing bus-q to the reference bus Consider a n-bus system as shown in fig.c in which an impedance Z b is added from an existing bus-q to the reference bus. Let us consider as if he impedance Zb is connected from a new bus-p and existing bus-q. Now it will be an addition as that of case-2. Then we can short-circuit the bus-q to reference bus. This is equivalent to eliminating (n+1)th bus (i.e. bus-p in this case) and so the bus impedance matrix has to be modified by eliminating (n+1)th row and (n+1)th column. The reduced bus impedance matrix can be formed by a procedure similar to that of bus elimination in bus admittance matrix. This reduced bus impedance matrix is the actual new bus

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impedance matrix. Every element of actual new bus impedance matrix can be determined using the equation (27). Z jk ,act = Z jk −

Z j ( n +1) Z ( n +1) k

. . . . (27) Z ( n +1)( n +1) Note : 1. Zjk,act is the impedance corresponding to row-j and column-k of actual new bus impedance matrix. 2. Zjk, Z(n+1)k, Zj(n+1), Z(n+1)(n+1) are impedance of new bus impedance matrix of order (n+1). 3. Since bus impedance matrix is symmetrical Zjk, act = Zkj, act Case 4 : Adding Zb between two existing buses h and q Consider a n-bus system shown in fig. d, in which an impedance Zb is added between two existing buses h and q. In this case the bus impedance matrix is formed as shown in equation (28). Here the elements of (n+1)th column is the difference between the elements of column-h and column-q. The elements of (n+1)th row is the difference between the elements of row-h and row-q. The diagonal element is given by equation (28).

Z bus , new

    =   −   Z h1 − Z q1

Zbus − Z he − q 2

− .

− Z hn − Z qn

| | | | | |

Z 1h − Z 1q  Z 2 h − Z 2 q  .   . . . . (28) Z nh − Z nq  −   Z qq + Z b 

Z ( n +1)( n +1) = Z b + Z hh + Z hh + Z qq − 2 Z hq . . . . . (29) Since the modification does not involve addition of new bus, the order of new bus impedance matrix has to be reduced to n x n by eliminating the (n+1) th column and (n+1)th row. This reduced bus impedance matrix is the actual new bus impedance matrix. Every element of the actual new bus impedance matrix can be determined using equation (27) which is also given below for reference. Z j ( n +1) Z ( n +1) k Z jk ,act = Z jk − Z ( n +1)( n +1) Direct determination of a bus impedance matrix The bus impedance matrix can be directly obtained from impedance or reactance diagram instead of forming Ybus and then 12

inverting it. In direct determination of Zbus first consider an impedance Zb connected between bus-1 to reference bus impedance matrix will have a single element a shown below. Then add each element of impedance or reactance diagram one by one and modify the Zbus in each step. Each modification of Zbus involve any one of the four cases discussed above. -----------------------------------------------------------------------------------------------------------

1.7 Formulation of Load Flow equations using Ybus matrix The load flow equations can be formed using either the mesh or node base equations of a power system. However, from the view point of computer time and memory, the nodal admittance formulation using the nodal voltages as the independent variables is the most economic. The node basis matrix equation of a n-bus system is given by Ybus V = I .... Where Ybus = Bus admittance matrix of order (n x n) V = Bus (node) voltage matrix of order (n x 1) I = Source current matrix of order (n x 1) The equation can be written in the form shown below : . . Y1 p . . Y1n  V1   I 1   Y11 Y12  I  Y . . Y2 p . . Y2 n  V2   2   21 Y22   .   . . . . . . . .  .       . . . . . . .  .   . = . ...  I p  Y p1 Y p 2 . . Y pp . . Y pn  V p       . . . . . . .  .  .   . .   . . . . . . . .  .       . . Ynp . . Ynn  Vn   I n   Yn1 Yn 2 Let Ip = Current injected to bus p Vp = Voltage of bus p From equation the current Ip can be expressed as shown in equation  Ip = Yp1V1+Yp2V2+…….+YppVp+….+YpnVn p −1

∴ I p = ∑ Y pqVq + Y ppV p + q =1

n

∑Y

q = p +1

pq

Vq …..

Let Sp = Complex power of bus-p Pp = Real power of bus-p Qp = Reactive power of bus-p Now Sp = Pp + j Qp Also Sp = Vp Ip* 13

Therefore, Vp Ip* = Pp + j Qp . . . . . .  The load flow problem can be handled more conveniently by use of Ip rather than Ip* Therefore take conjugate of equation  (Vp Ip*)* = (Pp + j Qp)* Vp* Ip = Pp - j Qp Pp − jQ p ∴Ip = ... V p* On equating the equation and we get, p −1 n Pp − jQ p Y V + Y V + Y V = .... ∑ ∑ pq q pq q pp p V p* q =1 q = p +1 Method-1 : Gauss or Gauss-Seidal method The equation can be written as shown below n Pp − jQ p p −1 Y ppV p = − Y V − ∑ ∑ Y pqVq pq q V p* q =1 q = p +1 n  1  Pp − jQ p p −1 ∴V p = − Y V − Y V   . . . . . ∑ ∑ pq q pq q Y pp  V p* q =1 q = p +1 

  n 1  Pp − jQ p ∴V p = − ∑ Y pqVq  . . . .   Y pp  V p* q =1   q≠ p The equation is called the load flow equation for Gauss or Gauss-Seidal method. Method – 2 : Newton - Raphson method The equation can be written as shown below n

Pp − jQ p = V p* ∑ Y pqVq . . . . q =1

Let, Vp = ep + j fp . . . (10) Vq = eq + j fq . . . (11) Ypq = Gpq + j Bpq . . . (12) Where, ep, fp = Real and imaginary part of Vp respectively eq, fq = Real and imaginary part of Vq respectively Gpq, Bpq = Conductance & susceptance of the admittance Ypq respectively. On substituting for Vp, Vq and Ypq from equation (10), (11) and (12) we get, n

Pp − jQ p = (e p + jf p ) * ∑ (G pq − jB pq )(eq + jf q ) q =1

14

n

= (e p − jf p )∑ (eq G pq − jeq B pq + jf q G pq − j 2 f q B pq ) q =1 n

[

]

[

]

= (e p − jf p )∑ (eq G pq + f q B pq ) + j ( f q G pq − eq B pq ) q =1

n

= ∑ (e p − jf p ) (eq G pq + f q B pq ) + j ( f q G pq − eq B pq ) q =1

n

[

= ∑ e p (eq G pq + f q B pq ) + je p ( f q G pq − eq B pq ) − jf p (eq G pq + f q B pq ) − j 2 f p ( f q G pq − eq B pq ) q =1

= ∑ [e n

q =1

(eq G pq + f q B pq ) + f p ( f q G pq − eq B pq

p

)] − j ∑ [ f n

q =1

p

(eq G pq + f q B pq ) − e p ( f q G pq − eq B pq )

. . .(13)   On separating the real and imaginary parts of equation (13) we get, n

[

]

[

]

Pp = ∑ e p (eq G pq + f q B pq ) + f p ( f q G pq − eq B pq ) . . . . . (14) q =1 n

Q p = ∑ f p (eq G pq + f q B pq ) − e p ( f q G pq − eq B pq ) . . . . . (15) q =1

From equation we get, Vp

2

= e 2p + f p2 . . . . . . . (16)

The equations (14) and (15) are called load flow equations of Newton - Raphson method. ----------------------------------------------------------------------------------------1.3 LOAD FLOW SOLUTION BY GAUSS – SEIDAL METHOD The Gauss – Seidal method is an iterative algorithm for solving a set of non-linear load flow equations. The non-linear load flow equations are given by equation , when p = 1,2,. . . . n and this equation is n  1  Pp − jQ p p −1 Vp = − Y V − Y pqVq  . . . . . (17)  ∑ ∑ pq q * Y pp  V p q =1 q = p +1  where p = 1,2,3,……..n The variables in these equations are the node voltages V1, V2, V3,……Vn. In Gauss-Seidal method an initial value of voltages are assumed and they are denoted as V10, V20, V30,……Vn0. On substituting these initial values in equation (17) and by taking p=1, the revised value of bus-1 voltage V11 is computed. The revised value of bus voltage is V11 is replaced for initial value V10 and the revised bus-2 voltage V21 is computed. Now replace the V11 for V10 and V21 for V20 and perform the calculation for bus-3 and so on. The process of computing all the bus voltages as explained above is called one iteration. The iterative process is then repeated till the bus voltage converges within prescribed accuracy. The convergence of bus voltage is quite sensitive to the initial

15

]

]

values assumed. Based on practical experience it is easier to get a set of initial voltages very close to final solution.  n 1  Pp − jQ p p −1 k +1 k  V pk +1 = − Y V − Y V ∑ ∑ pq p  . . . . (18) pq q Y pp  V pk * q =1 q = p +1   k th Where, Vi = k iteration value of bus voltage Vi Vik+1 = (k+1)th iteration value of bus voltage Vi In equation (18), to compute the (k+1)th iteration values of bus-p voltage , the (k+1)th iteration values of voltages are used for all buses less than p and kth iteration values are used for all buses greater than or equal to p. The equation (18) is applicable for load bus, since in load bus changes in both magnitude and phase of voltages are allowed. But in generator bus the magnitude of voltage remains constant and so the equation (18) is used to calculate the phase of the voltage.

( )

It is important to note that the slack bus is a reference bus and so its voltage will not change. Therefore in each iteration the slack bus voltage is not modified. For a generator bus, the reactive power is not specified. Therefore in order to calculate the phase of bus voltage of a generator bus using equation (18) we have to estimate the reactive power from the bus voltages and admittances as shown below : From equation (6) we get, p −1

∑ Y pqVq + Y ppV p + q =1

n

Pp − jQ p

q = p +1

V p*

∑ Y pqVq =

n  p −1  ∴ Pp − jQ p = V ∑ Y pqVq + ∑ Y pqVq  . . . . (19) q= p  q =1  From equation (19) the equation for complex power in bus-p during (k+1)th iteration can be obtained as shown in equation (20) * p

k +1 p

P

− jQ

k +1 p

n  p −1  k +1 = (V ) ∑ Y pqV p + ∑ Y pqV pk  . . . . (20) q= p  q =1  k * p

The reactive power of bus-p during (k+1)th iteration is given by imaginary part of equation (20). Reactive power of bus-p during (k+1)th iteration is given by n   p −1  Q pk +1 = (−1) x Im aginary (V pk ) * ∑ Y pqV pk +1 + ∑ Y pqV pk   . . . (21) q= p   q =1   Also, for generator buses a lower and upper limit for reactive powers will be specified. In each iteration, the reactive power of generator bus is calculated using equation (21) and then checked with specified limits. If it violates the specified limit, then the reactive power of the bus is equated to the limits violated and it is treated as load bus. If it does not violate the limits then the bus is treated as generator bus. ----------------------------------------------------------------------------------------1.4 Computation of slack bus power and line flows

16

The slack bus power can be calculated after computing the bus voltage upto the specified accuracy. The equation (19) can be used the slack bus power. Here, bus-p is a slack bus. n  Pp − jQ p = V p* ∑ Y pqVqk  . . . . (22)  q =1  The line flows are the power fed by the buses into various lines and they are calculated as shown below : Consider a line connecting bus-p and bus-q as shown in fig. 1 below. Usually the transmission line is connected to buses using transformers at its ends. The π equivalent of the transmission line with transformers at its end consist of a series admittance Ypq and shunt admittances y’pq as shown in fig. below .

Fig.1 π equivalent of a transmission line for evaluating line flows

In fig.1,

I pq = I pq1 + I pq 2 = (V p − Vq )Y pq + V p

Y pq' 2

. . . (23)

Complex power injected by bus-p in line pq - Spq is given by,  Y pq'  * * S pq = Ppq − jQ pq = V p I pq = V p (V p − Vq )Y pq + V p  . . . (24) 2   In fig.2, I qp = I qp1 + I qp 2 = (Vq − V p )Y pq + Vq

Y pq'

. . . . (25) 2 Complex power injected by bus-q in line pq - Spq is given by,  Y pq'  S qp = Pqp − jQqp = Vq* I qp = Vq* (Vq − V p )Y pq + Vq  . . . (26) 2   Power loss in the transmission line pq is Spq, loss = Spq – Sqp . . . .(27) --------------------------------------------------------------------------------------------1.4 Procedure for load flow solution by Gauss – Seidal method Step 1 : Assume a flat voltage profile 1 + j0 for all buses (nodes) except in the slack bus. The voltage of slack bus is the specified voltage and it is not modified in any iteration. Step 2 : Assume a suitable value of ε called convergence criterion. Here ε is a specified change in bus voltage that is used to compare the actual change in bus voltage between kth and (k+1)th iteration.

17

Step 3 : Set iteration count k = 0 and assumed profile of the buses are denoted as V 10, V20, V30,……Vn0 except slack bus. Step 4 : Set bus count p = 1. Step 5 : Check for slack bus. If it is a slack bus then go to step-1, or otherwise go to next step. Step 6 : Check for generator bus. If it is a generator bus go to next step or otherwise (i.e. if it is load bus) go to step 8. k Step 7 : Temporarily set V p = V p and phase of Vpk as the kth iteration value if the spec

bus-p is a generator bus where V

p p spec

is the specified magnitude of voltage for bus-p.

Then calculate the reactive power of the generator bus using the following equation : n   p −1  Q pk +1 = (−1) x Im aginary (V pk ) * ∑ Y pqV pk +1 + ∑ Y pqV pk   q= p  q =1    The calculated reactive power may be within specified limits or it may violate the limits. If the calculated reactive power is within the specified limits then consider the bus k +1 as generator bus and set Q p = Q p ,cal for this iteration and go to step-8. If the calculated reactive power violates the specified limit for reactive power then treat this bus as a load bus. The magnitude of the reactive power at this bus will correspond to the limit it has violated. k +1 i.e. if Q p ,cal ≤ Q p ,min thenQ p = Q p ,min k +1 (or) Q p ,cal ≥ Q p ,max thenQ p = Q p ,max Since the bus is treated as load bus, take actual value of Vpk for (k+1)th iteration. i.e. │Vpk│need not be replaced by │Vp│spec when the generator bus is treated as load bus. Then go to step-9. Step 8 : For generator bus the magnitude of voltage does not change and so for all iterations the magnitude of bus voltage is the specified value. The phase of the bus voltage can be calculated as shown below.  n 1  Pp − jQ p p −1 k +1 k +1 k  V p ,temp = − ∑ Y pqVq − ∑ Y pqV p  Y pp  V pk *  q =1 q = p +1   k +1  Im ag . partofV p ,temp  δ pk +1 = tan −1   k +1  Re alpartofV p ,temp  Now, the (k+1)th iteration voltage of the generator bus is given by V pk +1 = V p ∠δ pk +1

( )

spec

where │Vp│spec = magnitude of specified bus voltage. After calculating Vpk+1 for generator bus go to step-11. Step 9 : For the load bus the (k+1)th iteration value of load bus-p voltage, Vpk+1 can be calculated using the following equation.  n 1  Pp − jQ p p −1 k +1 k  V pk +1 = − Y V − Y V ∑ ∑ pq p  pq q Y pp  V pk * q =1 q = p +1  

( )

18

Step 10 : An acceleration factor, α can be used for faster convergence. If acceleration factor is specified then modify the (k+1)th iteration value of bus-p voltage using the following equation. 1 V pk,+acc = V pk + α (V pk +1 − V pk ) Then set, Vpk+1 = Vp,acck+1 . Note : A suitable value of α for a particular system can be obtained by running trial load flows. For most of the systems the recommended value of α is 1.6. Step 11 : Calculate the change in the bus-p voltage, using the relation, ∆V pk +1 = V pk +1 − V pk Step 12 : Repeat the steps 4 to 11 until all the bus voltages have been calculated. For this increment the bus count by 1 and go to step-5, until the bus count is n. Step 13 : Find out the largest of the absolute value of the change in voltage. i.e. find the largest among │ΔV1k+1│,│ΔV2k+1│,│ΔV3k+1│,………..│ΔVnk+1│. Let this largest change be │ΔVmax│. Check whether this largest change │ΔVmax│ is less than the prescribed tolerance ε. If │ΔVmax│ is less than ε then move to next step. Otherwise increment the iteration count and go to step-4. Step 14 : Calculate the line flows and slack bus power using the bus (node) voltages. A computer program can be developed using the procedure given above. A flowchart for the computational procedure have been presented in fig.2. Note : For pure load bus the real and reactive power are considered as negative. For pure generator buses the real and reactive power are always considered as positive, even if the bus is treated as load bus. In buses having generators and loads connected to it, either the net power will be specified or the generated and load powers will be individually specified. The specified net power is the difference between generator & load power and so it can be used as such in calculation without changing the sign. When the generator and load power are individually specified, then the power used for calculation is the value obtained by subtracting load power from generator power (i.e. Sp = ΣSG – ΣSL and Sp = Pp + jQp).

19

Fig. 2 Flowchart for load flow solution by Gauss - Seidal method

20

2. NEWTON - RAPHSON METHOD OF LOAD FLOW ANALYSIS : 21

The Newton – Raphson method of load flow analysis is an iterative method which approximates the set of non-linear simultaneous equations to a set of linear simultaneous (load flow) equations using Taylor’s series expansion and the terms are limited to first order approximation. The load flow equations of Newton-Raphson method [equation (1) to (3)] are non-linear equations in terms of real and imaginary part of bus voltages. n

[

]

[

]

Pp = ∑ e p (eq G pq + f q B pq ) + f p ( f q G pq − eq B pq ) . . . . (1) q =1 n

Q p = ∑ f p (eq G pq + f q B pq ) − e p ( f q G pq − eq B pq ) . . . . (2) q =1

Vp

2

= e 2p + f p2 . . . . (3)

where, ep = Real part of Vp (Voltage of bus-p) fq = Imaginary part of Vp (Voltage of bus-p) Gpq, Bpq = Conductance & susceptance of admittance Ypq respectively. The voltage of a slack bus will be known quantity in a power system and so it need not be solved. For load buses Pp and Qp will be specified and we have to solve Vp. For a generator bus pp and │Vp│ will be specified and we have to solve Qp and phase of Vp. In order to calculate Vp of load bus and Qp & phase of Vp for generator bus we have to first calculate the real and imaginary part of bus voltages. Therefore the unknown quantities are real part (ep) of bus voltages and imaginary part (fp) of bus voltages. For a n-bus system, the bus-1 is considered as slack bus and the real and imaginary part of remaining (n-1) bus voltages have to be solved. Hence we have 2(n-1) variables to be solved and they are e2, e3, …….en, f2, f3,…….fn. Thus to solve the problem for 2(n-1) variables we need to solve 2(n-1) set of equations. ----------------------------------------------------------------------------------------------------------2.1 Mathematical background for Newton - Raphson method of analysis Let x1, x2, ……….xn be a set of unknown variables and y1, y2,……..yn be a set of specified quantities. Now the specified quantities can be expressed as a non-linear function of unknown variables as shown below. y1 = f1(x1, x2,…..,xn)| y2 = f2(x1, x2,…..,xn)| . . │ . . │. . . . (1) . . │ yn = fn(x1, x2,…..,xn)⌡ Let us assume an approximate initial solution x01,x02,…..,x0n for the above equations. The prefix zero refers to zeroth iteration in the process of solving the above non-linear equations. 22

Let Δx01, Δx02,….., Δx0n are the corrections required for x01, x02,…..,x0n respectively for the next better solution. Now the non-linear equations can be expressed as shown below, i.e. they can be expressed as functions of modified variables, x 01 + Δx01, x02 + Δx02,…..,x0n + Δx0n . y1 = f1(x01 + Δx01, x02 + Δx02,….., x0n + Δx0n ) | y2 = f2(x01 + Δx01, x02 + Δx02,….., x0n + Δx0n ) | . . │ . . │. . . . (2) . . │ 0 0 0 0 0 0 yn = fn(x 1 + Δx 1, x 2 + Δx 2,….., x n + Δx n )⌡ The above equations are linearised about the initial guess using Taylor’s series expansion. The linearised equations with second order and higher derivatives neglected are given below. ∂f ∂f 1 ∂f + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n ∂f ∂f ∂f y 2 = f 2 ( x10 , x 20 ,........., x n0 ) + ∆x10 1 + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n . . . . . . ∂f ∂f ∂f y n = f n ( x10 , x 20 ,........., x n0 ) + ∆x10 1 + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n The equation (3) can be rearranged as shown in equation (4) ∂f ∂f ∂f y1 − f 1 ( x10 , x 20 ,........., x n0 ) = ∆x10 1 + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n ∂f ∂f ∂f y 2 − f 2 ( x10 , x 20 ,........., x n0 ) = ∆x10 1 + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n . . . . . . ∂f ∂f ∂f y n − f n ( x10 , x 20 ,........., x n0 ) = ∆x10 1 + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n y1 = f 1 ( x10 , x 20 ,........., x n0 ) + ∆x10

. . . . . (3)

. . . . . (4)

0 0 0 Let y1 − f 1 ( x1 , x 2 ,........., x n ) = ∆y1

y 2 − f 2 ( x10 , x 20 ,........., x n0 ) = ∆y 2 . . . . . . 0 0 y n − f n ( x1 , x 2 ,........., x n0 ) = ∆y n Now the equation (4) can be written as

23

∂f ∂f 1 ∂f + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n ∂f ∂f ∂f ∆y 2 = ∆x10 1 + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n . . . . . . . . . (5) . . ∂f ∂ f ∂ f ∆y n = ∆x10 1 + ∆x 20 2 + ........ + ∆x n0 n ∂x1 ∂x 2 ∂x n The equation (5) can be arranged in the matrix form as shown in equation (6) ∂f1   ∂f 1 ∂f 1 . . .  ∂x ∂x 2 ∂x n   1  ∂f 2   ∂f 2 ∂f 2 . . . ∆x10   ∆y1   ∂x  ∂ x ∂ x 2 n  ∆y   1   ∆x 29    2  . . . . . .    .   .  ..   . . . . . . (6)  =  .   . . . . . .  .   .    .      0  ∆y n   . . . . . .  ∆x n    ∂f n   ∂f n ∂f n . . .  ∂x1 ∂x 2 ∂x n  In matrix (vector) notation, the equation (6) can be expressed as B = JC . . . . (7) ∂f1   ∂f1 ∂f1 . . .  ∂x ∂x ∂xn  2  1  ∂f 2 ∂f 2 ∂f 2   . . . ∆x10   ∆y1   ∂x1 ∂x2  ∂ x n  9 ∆y    ∆x2   2  . . . . . .   .   .   .. ; C =   Where B =  ;J=   .    .  . . . . . .    .   .     0    . . . . . .  ∆yn  ∆xn     ∂f ∂f n ∂f n   n  . . . ∂xn   ∂x1 ∂x2 Here J is the first derivative matrix and it is called Jacobian matrix. The elements of Jacobian matrix are obtained by evaluating the first derivative at the assumed solution. The B matrix is called residual column vector. The elements of matrix B are the difference between the specified quantities and calculated quantities at the assumed ∆y1 = ∆x10

24

solution. With the elements of J and B are known, the elements of matrix C are obtained by solving the matrix equations (7). The C matrix is called increment solution vector. 0 0 0 The solution of the matrix equation (7) gives ∆x1 , ∆x 2 ,........∆x n . The next better solution is obtained as follows. x11 = x10 + ∆x10 x 12 = x 20 + ∆x 20 . . . . . . . . (8) . . x 1n = x n0 + ∆x n0 With the new solution equation (8) the process is repeated to find next new solution. The iterative process is terminated if any one of the following condition is satisfied. i) The largest (magnitude of the) element in the B matrix is less than a prespecified value. ii) The largest (magnitude of the) element in the C matrix is less than a prespecified value. In this method the initial values assumed should be close to actual solution. Otherwise, there are chances of the solution diverging rather than converging and it may not be possible to get a solution. In power system problems, the choice of initial values may not pose a problem because, a flat voltage profile, i.e. Vp = 1 + j0 for p=1,2,….,n except the slack bus has been found to be satisfactory for almost all practical systems. -----------------------------------------------------------------------------------------------------2.2 Applying Newton – Raphson technique to load flow problem Consider a system with n-buses. The bus-1 is usually selected as slack bus. The other buses (i.e. bus-2 to bus-n) can be either generator bus or load bus. For load buses the specified quantities are Pp and Qp. For generator buses the specified quantities are Pp and |Vp|. 2.2.1 Formation of 2(n-1) equation for a system with n buses : Case (i) When all the (n-1) buses are load buses : In this case, bus-1 is slack bus and bus-2 to bus-n are load buses. Let P2, P3,.. . . . .Pn be the specified real powers and Q2, Q3,. . . . . Qn be the specified reactive powers of (n-1) load buses. The unknown variables are ral part of bus voltages e2, e3,…….en and reactive part of bus voltages f2,f3,…..fn. Now the matrix equation B = JC for this power system problem will be in the form shown in equation (9)

25

. . . (9) Case (ii) When the system has both load and generator buses In this case also the bus-1 is slack bus. Let buses 2,3,4,……m, be load buses and the remaining buses the generator buses. Here buses are numbered such that the first bus is slack bus, bus-2 to bus-m are load buses and bus-(m+1) to bus-n are generator buses. Let P2, P3,……….Pn be the specified real powers of (n-1) buses. Let Q2, Q3,….. Qm be the specified reactive powers of load buses. Let |Vm+1|,|Vm+2|,……|Vn| be the specified magnitude of voltages of generator buses. The unknown variables are real part of bus voltages e2, e3,……en and imaginary part of bus voltages f2, f3,…….fn. Now the matrix B = JC for this case will be in the form shown in equation (10)

26

. . . (10) 2.3 Computing the elements of matrices : The elements of Jacobian matrix (J) can be derived from the three load flow equations (1) to (3). The elements of the matrices are obtained by partially differentiating the load flow equations with respect to a unknown variable and then evaluating the first derivaties using the solution of previous iteration. For first iteration the initial assumed values are used as the solution of previous iteration. The initial values are ep0 and fp0 for p=1,2,3,……n. The element of the residual column (B) is the difference between the specified value of the quantity and the calculated value of the quantity using the solution of previous iteration. Let Pp,spec, Qp,spec and |Vp|spec be the specified quantities at the bus-p. For the initial solution the values of Pp0, Qp0 and |Vp0| can be calculated using the load flow equations. Now for the first iteration, ΔPp = Pp,spec – Pp0 ΔQp = Qp,spec – Qp0 After calculating the elements of Jacobian matrix (J) and residual column vector (B), the elements of increment voltage vector (C) can be calculated by using any standard technique. Now the next better solution will be e1p = e 0p + ∆e 0p f p1 = f p0 + ∆f p0 These values of voltages will be used in the next iteration. This process will be repeated and in general the new better estimates for bus voltages will be e kp+1 = e kp + ∆e kp 27

f pk +1 = f pk + ∆f pk The process is repeated till the magnitude of the largest element in the residual column vector is less than the prescribed value. 2.3.1 Procedure for load flow solution by Newton – Raphson method : Step 1 : Assume a flat voltage profile 1 + j0 for all buses (nodes) except the slack bus. The voltage of the slack bus is the specified voltage and it is not modified in any iteration. Step 2 : Assume a suitable value of ε called convergence criterion. Hence ε is a specified change in the residue that is used to compare the actual residues (ΔP and ΔQ or ΔV) at the end of each iteration. Step 3 : Set iteration count k = 0, and assumed voltage profile of the buses are denoted as V10, V20, V30,…….Vn0 except slack bus. The real part of the voltage are denoted as e p0 and the reactive part as fp0 for p = 1,2,3,…….n. Step 4 : Set bus count p = 1. Step 5 : Check for slack bus. If it is a slack bus then go to step-13, otherwise go to next step. Step 6 : Calculate the real and reactive power of bus-p using the following equation. n

[

]

[

]

Pp = ∑ e p (eq G pq + f q B pq ) + f p ( f q G pq − eq B pq ) q =1 n

Q p = ∑ f p (eq G pq + f q B pq ) − e p ( f q G pq − eq B pq ) q =1

Step 7 : Calculate the change in real power. k k Change in real power, ∆ p = Pp ,spec − Pp where Pp , spec = Specified real power of bus-p Step 8 : Check for generator bus. If it is a generator bus then go to next step (step-9) or otherwise (i.e. if it is a load bus) go to step-12. Step 9 : Check for reactive power limit violation of generator buses. For this compare the calculated reactive power Qpk with specified limits. If the limit is violated go to step-11, otherwise go to next step. Step 10 : If the calculated reactive power is within the specified limits then consider this bus as generator bus. Now calculate the voltage residue (change in voltage) using the following equation. ∆V pk Where V p

spec

2

= Vp

2 spec

− V pk

2

= specified voltage magnitude for generator bus. Then go to step-13.

Step 11: If the reactive power limit is violated then treat this bus as a load bus. Now the specified reactive power for this bus will correspond to the limit violated. k i.e. if Q p ≤ Q p ,min thenQ p ,spec = Q p ,min k i.e. if Q p ≥ Q p ,max thenQ p ,spec = Q p ,max Step 12 : Calculate the change in reactive power for load bus (or for the generator bus treated as load bus).

28

k k Change in reactive power, ∆Q p = Q p ,spec − Q p Step 13 : Repeat steps 5 to 12 until all resides (change in P and Q or V) are calculated. For this increment the bus count by 1 and go to step 6, until the bus count is n. Step 14 : Determine the largest of the absolute value of the residue (i.e. find the largest 2

among ∆Ppk , ∆Q pk or ∆V pk . Let this largest change by ΔE. Step 15 : Compare ΔE and ε then go to step 20. If ΔE < ε then go to step 20. If ΔE > ε then go to next step (step 16) Step 16 : Determine the elements of Jacobian matrix (J) by partially differentiating the load flow equations and evaluating the equations using the kth iteration values. k k Step 17 : Calculate the increments in real and reactive part of voltages, ∆e p and ∆f p by solving the matrix equation B = JC [The elements of B matrix are ∆Ppk , ∆Q pk or ∆V pk

2

k k calculated in the previous steps The elements of C matrix are ∆e p and ∆f p which are voltage increments to be solved from the matrix equation B = JC]. Step 18 : Calculate the new bus voltages as shown below e kp+1 = e kp + ∆e kp ; p = 1,2,3, ………n except slack bus.

f pk +1 = f pk + ∆f pk ; V

k +1 p

= (e

k +1 2 p

) +(f

p = 1,2,3,……….n except slack bus. ) andδ

k +1 2 p

k +1 p

 e kp +1  = tan  k +1  f   p  −1

V pk +1 = V pk +1 ∠δ pk +1 Step 19 : Advance the iteration count, i.e. k = k+1 and to step-4. Step 20 : Calculate the line flows.

29

Fig. 2.3 Flowchaart for load flow solution by Newton – Raphson method

30

2.3.1 COMPARISION OF G.S. AND N.R. METHOD OF LOAD FLOW STUDIES 1. For G.S. method the variables are expressed in rectangular coordinates whereas in N.R. method, they are expressed in polar coordinates. If rectangular coordinates are used for N.R. method then memory requirement will be more. 2. The number of mathematical operations per iteration will be lesser in G.S. method. 3. The G.S. method has linear convergence characteristics whereas the N.R. method has quadratic convergence characteristics. Hence N.R. method converges faster than G.S. method. 4. In G.S. method number of iterations increases with number of buses but in N.R. method the number of iteration remains constant and it does not depend on the size of the system. 5. In G.S. method convergence is affected by the choice of slack bus and the presence of series capacitors but the N.R. method is less sensitive to these factors. 6. The N.R. method needs only 3 to 5 iterations to reach an acceptable solution for a large system. But G.S. method requires large number of iterations (30 or more) for same level of accuracy. Advantages of G.S. method 1. Calculations are simple and so the programming task is faster. 2. The memory requirement is less. 3. Useful for small size system. Disadvantages of G.S. method 1. Require large number of iteration to reach convergence. 2. Not suitable for larger systems. 3. Convergence time increases with size of the system. Advantages of N.R. method 1. The N.R. method is faster, more reliable and the results are accurate. 2. Requires less number of iterations for convergence. 3. The number of iteration are independent of the size of the system (number of buses). 4. Suitable for large size systems. Disadvantages of N.R. method 1. The programming logic is more complex than G.S method 2. The memory requirement is more. 3. Number of calculations per iterations are higher than G.S. method. ------------------------------------------------------------------------------------------------------FAST DECOUPLED METHOD

31

3.1 Representation of Transformers : 1.1 Fixed tap setting transformer In a system having different voltage levels, two buses are separated by a transformer operating with either a nominal or an off nominal turns ratio. In the case of a nominal turns ratio, the transformer is represented by its leakage admittance.

1.2 Off nominal turns ratio transformer If an off nominal turns ratio is present, the transformer can be represented by its leakage admittance in series with an ideal autotransformer as shown.

Representation of off nominal turns ratio transformer Here the leakage admittance y of the transformer has been referred to the base value associated with bus q, the non-tap side of the transformer and the tap side bus is connected directly with the ideal transformer whose turns ratio α = (1 ± t) where t is the per unit tap setting. The current flow in the transformer in the direction from ‘x’ to ‘q’ is given by I x = ( E x − E q ) y pq Since the autotransformer is an ideal one, there is no power loss in the transformer. *

*

∴E p I p = ExI x *

*

1 Ix =  Ix * a Ep 1 Since α is real, α* = α, = Ix a  1 E p Ip =  − E q  y pq a a  y pq y pq = 2 Ep − E q . . . . . (1) a a y pq y pq From which Y pp = 2 andY pq = − a a Similarly the current flowing from q to x is Ip =

Ex

32

I q = ( E q − E x ) y pq

In general

 Ep   y pq =  E q −  a   y pq = y pq E q − E p . . . . . . (2) a I q = Yqq E q + Yqp E p

From which

Yqq = y pq andYqp = −

y pq

a Since Ypq = Yqp, we can suggest a π equivalent of the above transformer which is shown in fig. below.

Fig. π equivalent circuit of off nominal turns ratio transformer From the π equivalent circuit, the self-admittance at any bus is the sum of the admittances connected at that particular bus and the shunt admittance to ground. To find the shunt admittance to ground at bus p. Ypp = Ypq + yqo y pq y pq = + y po a a2 y pq y pq y po = 2 − a a y pq  1  y po = − 1 . . . . . . (3) a  a  At bus q, Yqq = Yqp + y qo y pq =

y pq a

+ y qo

y qo = y pq −

y pq

a  1 y qo = y pq 1 −   a Thus the elements are marked in the π equivalent circuit. When the off nominal turns ratio is represented at bus p for a transformer connecting buses ‘p’ and ‘q’, the self admittance at bus p is

33

Y pp = y p1 + y p 2 + ........ +

y pq a

+ ...... +

= y p1 + y p 2 + ..... + The mutual admittance from p to q is Y pq =

11   − 1 y pq + .......... + y pn aa 

y pq a2

+ ..... + y pn

− y pq a

The self admittance at bus q is

= y q1 + y q 2

y pq

 1 + ...... + y qp 1 −  + y qn a  a + ...... + y pq + ..... + y qn

Yqq = y q1 + y q 2 + ....... +

and is left unchanged. Only on the tapped side, the admittance value is to be divided by a2 and the untapped side is left as such. 1.3 Power flow through phase shifting transformer A phase shifter regulates the flow of active power by varying its phase angle Φ. In load flow studies, it can be represented by an admittance in series with an ideal auto transformer having a complex turns ratio as shown in figure below.

Fig. Representation of a phase shifting transformer From the figure, the terminal voltage at x and p are related by Ep = a s + jbs Ex Since there is no power loss in an ideal transformer, *

*

Ep I p = Ex I x

[

]

I p Ep  1 = =  Ix E x  a s + jbs 1 = a s − jbs 1 Ip == a s − jbs I x

  

*

From the figure, I x = E p − E q y pq ∴I p =

1 a s − jbs

 Ep  − E q  y pq   a s + jbs 

34

Ip =

y pq a s2 + bs2

Ep −

y pq a s − jbs

Eq

In general I p = Y pp E p + Y pq E q y pq y pq Y pp = 2 andY pq = − From which 2 a s − jbs a s + bs Similarly, the current from q to s is I q = E q − E s y pq

(

)

 Ep  = E q −  y pq a s + jbs   y pq = y pq E q − Ep a s + jbs In general,

I q = Yqq E q + Yqp E p

From which Yqq = y qq andYqp =

y pq a s + jbs

Since Y pq ≠ Yqp , an equivalent π circuit cannot be formed. When the phase shifting transformer is connected between buses ‘p’ and ‘q’, the self admittance at bus-p is y pq Y pp = y p1 + y p 2 + ...... + 2 + ...... + y pn a s + bs2 The mutual admittance between buses ‘p’ and ‘q’ is given by − y pq Y pq = a s − jbs The self admittance at bus ‘q’ is Yqq = y q1 + y q 2 + ....... + y qp + ...... + y qn The mutual admittance between buses ‘q’ and ‘p’ is given by − y pq y qp = a s + jbs In matrix form (p) (q) (p)  y pq 2 2  as + bs  − y pq (q)   as + jbs

− y pq   a − jbs2   y pq   2 s

Since Y is unsymmetrical, the equivalent circuit representation is not possible.

35

Review Questions (2 marks) 1. Name any two iterative methods need for the solution of load flow problems. 2. How is the swing bus selected in a load flow study ? 3. What are the quantities specified at a generator bus ? 4. What is meant by acceleration factor in load flow solution ? What is its best value ? 6. What are the uses of phase shifting transformer ? 7. What is the important factor on which the real power transfer over line depend? 8. State Gauss-Seidal load flow formula. 9. State Newton-Raphson load flow formula. 10. What are the applications of load flow analysis? 11. Mention the items specified and not specified in a reference bus. 12. Define voltage-controlled bus. 13. State the advantages of Newton-Raphson method over Gauss-Seidal method. 14. Explain the convergence criterion for Gauss-Seidal load flow studies ? 15. What is a Jacobian matrix? 16. What are the advantages of Gauss-Seidal load flow method? 17. In a load flow problem, how various buses are classified ? 18. State the elements of a Jacobian matrix. 19. What are power balance equations in power flow analysis? 20. What is the necessity for slack bus? 21. What is a load bus? 22. Given the simultaneous equations: x1+x2 = 4; 2x1+x2=5; using initial values x10=2 and x20=2. Write down the values for x11 and x21 (first iteration values) using Gauss-Seidal method. 23. Write down the general expression for complex power injected by the source into the ith bus of a power system. 24. What is half line charging admittance? 25. What is meant by a transformer with off-nominal turns ratio? 26. Draw the equivalent circuit of tapped transformer for nodal admittance matrix. 27. Draw the circuit representation of a line with voltage regulating transformer. Review Questions (16 marks) 1.i) State the load flow problem. ii) Discuss the classification of buses. 2. Draw the flow chart for load flow solution by Gauss-Seidal iterative method and explain.

36

3. Derive the static load flow equations of a power system. Explain the Newton-Raphson method of solving the load flow equations. 4. Draw the flow chart for load flow solution by Newton-Raphson method. 5. Compare Gauss-Seidal method and Newton-Raphson method in detail. 6. Explain clearly how the nodal admittance matrix of a system is changed when an off nominal turns ratio transformer is introduced in a line connected between two buses. 7. Discuss the various types of buses and their significance in detail. 8. Derive the static load flow equations using Ybus. 9. Derive Newton-Raphson power flow analysis algorithm and give steps for implementation of this algorithm. 10. Derive the basic equations for the load flow study using Gauss-Seidal method. With respect to this method, explain the following : a. Acceleration factor. b. Convergence criteria c. Handling of PV buses. 11. Draw the representation schemes for a. Phase shifting transformer b. Tap changing transformer. 12. Derive mathematical model of phase shifting transformer to be used in power flow analysis. 13. Carry out one iteration of load flow analysis of the system given below by Gauss-Seidal method. p.u. generation p.u. load Bus Bus │V│ No. Type p.u. P Q P Q 1 Slack 1.02 2 PV 1.0 0.8 3 PQ 1.0 0.4 Line reactance in p.u. are given below : Bus code Impedance 1-2 J0.5 2-3 J0.5 3-1 J0.5 14. Consider the power system with the following data : 1 2 3 _______________ 1 | -j12 j8 j4 Ybus = 2 | j8 -j12 j4 3 | j4 j4 -j8 Generation Load Voltage Bus Bus No. Type P Q P Q │V│ δ 1 Slack 1.0 0° 2 PV 6.0 0 1.04 3 PQ 0 0 4.0 0.5 Assume that bus 2 can supply any amount of reactive power. Assuming flat start, perform the first iteration of power flow analysis using NewtonRaphson method. 15. The bus admittance matrix of a 3 bus power system is 37

Ybus

− j 20  =  j10  j10 

j10 − j15 j5

j10   j5  − j15 

Bus No.

Bus Type

PG

QG

PL

QL

1

PV

2.903 4

-

-

-

2

PQ

-

-

4.008 9 -

1.7915

3 Slack The latest solution is V1 = 1.05 ∠ 6.96° V2 = 0.9338 ∠ -8.8° V3 = 1.0 ∠ 0° Determine the bus voltages at the end of next iteration using (i) GS method (ii) NR method 16. The following data are given for a three bus power system. Bus 1 slack bus V specified = 1.05 ∠ 0° Bus 2 PV bus │V│ specified = 1.02 p.u. PG = 3 p.u. Bus 3 PQ bus PL = 4 p.u. QL = 2 p.u. Line reactances in p.u. are given below: Bus code Impedance Carry out one 1-2 j0.5 iteration of load 2-3 j0.5 flow solution by 3-1 j0.5 GS method. Take Q limits of generator 2 is 0 ≤ Q ≤ 4. 17. Repeat problem No. 16 when the limits of reactive power is specified as 0 ≤ Q ≤ 2.5 18. The load flow data of a four bus system is given in tables. Assume the bus voltage of bus No.3 as 1.04 p.u. and the maximum and minimum reactive power constraints for bus 3 are -0.1 and 1.0 p.u. respectively. Taking bus 1 as slack determine the voltages of all the buses t the end of first iteration starting with flat voltage profile for all buses expect the slack bus using (i) GaussSeidal method and (ii) Newton-Raphson method. Bus Assumed Generation Load No. voltages MW MVAR MW MVAR 1 1.06 + j0.0 0.0 0.0 0.0 0.0 2 1.0 + j0.0 0.0 0.0 0.2 0.1 3 1.06 + j0.0 0.6 0.3 0.4 0.2 4 1.06 + j0.0 0.0 0.0 0.4 0.0

38

Bus code

Impedance

1-2 1-3 2-3 2-4 3-4

0.02 + j0.08 0.06 + j0.24 0.04 + j0.16 0.04 + j0.16 0.01 + j0.04

Line charging admittance y’pq/2 0.0 + j0.04 0.0 + j0.03 0.0 + j0.025 0.0 + j0.025 0.0 + j0.015

19 Explain clearly with flow chart the computational procedure for load flow solution using Gauss-Seidal method when the system contains all types of buses. 20 Explain clearly with flow chart the computational procedure for load flow solution using Gauss-Seidal method when the system contains all types of buses. ----------------------------------------------------------------------------------------------

39