ELEC4612 Power System Analysis Power Flow Analysis

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1 ELEC462 Power Sstem Analsis Power Flow Analsis Dr Jaashri Ravishankar

2 Busbars The meeting point of various components of a PS is called bus. The bus or busbar is a conductor made up of copper or aluminium having negligible resistance. A busbar has zero voltage drop when it conducts the rated current. Thus, the buses are considered as points of constant voltage in a power sstem. 2

3 Node Equations When the PS is represented b impedance diagram, it can be considered as a circuit or network. The buses can be treated as nodes and the voltages of all buses can be solved b conventional nodal analsis. Nodal analsis is based on KCL and hence analsis is easier if impedances are converted to admittances. The admittance between buses and 3

4 Admittance Matrix (Y bus ) Draw the impedance diagram from the OLD. Obtain the admittances of the individual elements and convert the impedance diagram to admittance diagram, b replacing voltage sources with current sources. Appl KCL to all independent nodes (buses). Finall, arrange the node equations in matrix form. [I bus ]=[Y bus ][V bus ] 4

5 5

6 Extending this for a n-bus sstem, Or I bus is the vector of injected bus currents (external sources). The current is positive when flowing towards the bus and negative if flowing awa from the bus. Y bus is the bus admittance matrix. 6

7 Y bus b inspection Can be used onl if there is no mutual coupling between the lines.. Obtain the admittances of the individual elements 2. Diagonal elements (self-admittance or driving point admittance) is the sum of admittances of all elements connect to a node, including admittance between the node and ground (node 0) 3. Off-diagonal elements (mutual or transfer admittance) is equal to the negative of the admittance between the nodes 7

8 Example 5. 8

9 Impact of Shunt Branches Shunt admittances are added to the diagonal elements corresponding to the nodes at which the are connected. The off-diagonal elements are unaffected. Shunt admittances are seen when the transmission line is modelled as a PI- or T- network. 9

10 Sstems with Mutual Coupling Invert the primitive impedance matrices of the network branches to obtain the corresponding primitive admittance matrices. Two mutuall coupled branches have a 2 x 2 matrix, threemutuallcoupledbrancheshavea3x3matrix, and so on. Multipl the elements of each primitive admittance matrix b the 2 X 2 building block matrix. Take care to label from dotted to undotted terminal. Combine b adding together, those elements with identical row and column labels to obtain the Ybus matrix. 0

11 Primitive admittance matrix for three mutuall coupled branches

12 Building Block Matrix 2

13 Example 5.2 Two branches having impedances equal to 0.25 pu are coupled through mutual impedance Z M = 0.5 pu as shown. Find the Y bus matrix for the mutuall coupled branches. Solution 3

14 Example 5.3 Consider that onl the two branches 3 and 2 3 in the circuit of Figure below are mutuall coupled as indicated b the dots beside them and that their mutual impedance is j0.5 per unit. Determine the circuit Y bus. Values in the figure are impedances in pu. j0.5 4

15 5

16 Tap Changing Transformer V V 2 Bus t Bus 2 I t is the ratio of the per-unit voltage on the left-hand-side of the ideal transformer to the per-unit voltage on the right-hand-side of the ideal transformer. is the transformer admittance. This means that the secondar side of the ideal transformer has voltage (/t )V and current of t I. V V 2 V /t Bus t Bus 2 I We can express the current t I using Ohm s Law: V t I V t 2 () t I I 2 I 2 6

17 Dividing through b t and expanding the right-hand-side ields: I V V 2 2 t t (2) Now express the current I 2 : Bus (3) I V I 2 V2 V2 V t t We can re-write eqs. (2) and (3) in matrix form as: I I 2 2 t t t V V 2 The equivalent circuit for Y bus calculation now becomes, (4) V V 2 V /t t t I I 2 Bus 2 I I 2 /t V V 2 7

18 8 Comparing, we have, & Solving the above for result in: t t t Y bus 2 /t 2 I I 2 V V 2 t t 2 t & t t t t 2 t t t /t 2 t t t t I I 2 V V

19 Example 5.4 (a) Obtain the admittance matrix for the network shown. The values given are admittances. (b)recalculate the admittance matrix b considering an off-nominal tap transformer between buses 3 and 4 with t =.02. -j4 3 4 I 2-j3 2 2-j4 j0. I 2 2-j5 j0.2 j0.3 I 3 j0.4 I 4 9

20 Network Solution using Y bus If the voltages are known, the current injections can be solved using, If the currents are known, the voltages can be solved using, This can be solved using Cramer s rule: Voltage of the k th bus, 20

21 Example 5.5 Solve the node voltages and in the network shown. The voltages and impedances are in pu. 2

22 Power Sstem under Stead State Successful power sstem operation under stead-state conditions requires: Generation supplies load demand plus losses. Bus voltage magnitudes are close to rated values. Generators operate within specified real and reactive power limits. Transmission lines and transformers are not overloaded. Power flow analsis is the tool for investigating these requirements. 22

23 Power Flow Analsis (PFA) Stead state analsis of a power sstem Information obtained comprises voltage magnitudes of each bus voltage phase angle of each bus real power flow reactive power flow Power loss in the sstem This information is essential for continuous monitoring of the current state of the sstem and for future sstem expansion. 23

24 Conventional Network Analsis? Conventional circuit analsis is based on node voltage / loop current method. We write simultaneous linear equations (algebraic) o Nodal KCL equations YV = I o Loop KVL equations IZ = V Given values: Voltage / current of sources & impedances In a PS, input data for loads normall are given in terms of power, not impedance. Generators are considered as power sources, not voltage or current sources. 24

25 Vs 25

26 The solution of power flow becomes simple when onl two buses are involved. However, it still needs trail and error method, when voltages and impedances are unknown. The solution is therefore not unique. In normal power sstem there are large number of buses and the solution becomes quite complex, tedious and time consuming. Hence there is no choice but to use iterative solutions. 26

27 Power Flow Equations To find all bus voltages (magnitude & angle) Consider a tpical bus (bus i) i 27

28 PFA Solution Steps. Represent sstem b OLD. 2. Obtain the impedance diagram in per unit. 3. Find admittance matrix. 4. Formulate network equations. 5. Solve iterativel. 6. Following are assumed. The loads are constant and the are defined b their real and reactive power consumption. Generator terminal voltages are tightl regulated and therefore are constant. 28

29 Tpes of Buses Bus Tpe Load bus Generator bus Slack bus Quantities Specified P, Q Remarks Voltage allowed to var 5% P, V Q limits are specified V, Reference bus; one of the generator buses 29

30 Need for Slack bus To account for line losses In a PS, complex S gen =S load +S loss S loss can be estimated onl if P and Q at all buses are known The powers in the buses will be known onl after solving the power flow equations Thus, P and Q of the slack bus are not specified It is assumed that the generator connected to this bus will suppl the line losses 30

31 Iterative Methods. Gauss-Seidel 2. Newton-Raphson 3. Fast Decoupled The objective is to find,, and at all the PS buses. All these methods start with an initial assumption of unknown values for each bus. These are then updated at each iteration. The process continues till errors between the quantities in successive iteration reduces below a pre-specified value. 3

32 Summar Power flow stud is essential to decide the operation of the existing sstem and for planning the future sstem expansion. This is a stead state analsis. The information obtained are magnitude & phase angle of voltages, P & Q flowing in each line and line losses. For this analsis, the buses are classified as load bus, generator bus and slack bus. For large sstems, power flow is done using iterative methods like GS, NR and FD methods. Iterative methods are required because the power flow equations are non-linear algebraic equations. 32

33 Exercise 5.. Suppose that mutual coupling exists pairwise between branches 3 and 2 3, and also between branches 2 3 and 2 5 of Example 3.3. The mutual impedance between the former and latter pairs of branches are j0.5 and j0. per unit, respectivel. Find Y bus for the overall circuit. Hint: Primitive admittances for mutual branches is calculated b inverting the primitive impedance matrix as a single entit (3 x 3 matrix). Answer: 33

34 2. Find out the Y bus matrix of the power sstem shown. Data for this sstem is given in the table. Answer: 3. Determine the 4 x 4 bus admittance matrix and write nodal equations in matrix format for the circuit shown in Figure. Do not solve the equations. Answer: 34

EE 581 Power Systems. Admittance Matrix: Development, Direct and Iterative solutions

EE 581 Power Systems. Admittance Matrix: Development, Direct and Iterative solutions EE 581 Power Systems Admittance Matrix: Development, Direct and Iterative solutions Overview and HW # 8 Chapter 2.4 Chapter 6.4 Chapter 6.1-6.3 Homework: Special Problem 1 and 2 (see handout) Overview