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

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

2 Overview and HW # 8 Chapter 2.4 Chapter 6.4 Chapter Homework: Special Problem 1 and 2 (see handout)

3 Overview and HW # 8 Homework: Special problem 1 Solve for the Y bus of the given system

4 Overview and HW # 8 Homework: Special problem 2 Solve for the Y bus of the given system (note that the right image shows the equivalent mutual inductance)

5 HW # 8 Solutions Special Problem 1 Y bus = j19.98 j10 j10 j10 j19.98 j10 j10 j10 j19.98

6 HW # 8 Solutions Special Problem 2 Y bus v = I where Y bus = j16.75 j11.75 j2.5 j2.5 j11.75 j19.5 j2.5 j5 j2.5 j2.5 j5.8 0 j2.5 j5 0 j8.3 v = V 1 V 2 V 3 V 4 I =

7 Development: Models can be solved using either nodal (bus) analysis or port analysis Nodal analysis much easier to use than loop equations Nodal analysis used by computer programs For ease of calculations, admittance is used (Y) Ohm s Law: V=IR Complex: V=IZ Y = 1 Z I = YV (Siemens) Y is symmetric Y = Y T only have to solve for either the upper or lower triangular part of the matrix!!

8 Simple Example Solve for voltages (x i i = 1.. 6) using KCL and develop the impedance matrix

9 Chapter 2.4: Review of Nodal Analysis Node 3: 1 x 2 3 6V = x 1 x (x 5 4 x 3 ) x x 3 0.2x 4 = 3V System of network equations: 2x 1 x 2 x 3 = 0 x 1 1.5x x 4 = 0 x x 3 0.2x 4 = 3 0.5x 2 0.2x x 4 = 0 Ax = B

10 Chapter 2.4: Review of Nodal Analysis Similarly, can solve for the currents such that: Ax = B such that we have ZI = V (note Z is real here) x 1 x 2 = 0 6

11 Outline of steps Step 1: Define a reference bus (node), then define all voltages with respect to the reference bus. Step 2a: Source transform any voltage sources in parallel to an equivalent current sources. Step 2b: Convert all impedance values (Z) to admittance values (Y). Step 3: Write the equations found from nodal analysis in a matrix format Step 4: Solve the problem using direct or indirect methods.

12 Simple transmission network model (values in admittance) Solve using 1) a two port network 2) again using nodal analysis

13 Two port method: Y 11 = I 1 V 1 V 2 =0 = 0.1j 0.3j V 1 V 1 = 0.4j Y 12 = I 1 V 2 V 1 =0 = 0.1j V 2 V 2 Y 21 = Y 12 = 0.1j = 0.1j Y 22 = I 2 V 2 V 1 =0 = 0.1j 0.2j V 2 V 2 = 0.3j Y = Y 11 Y 12 Y 21 Y 22 = 0.4j 0.1j 0.1j 0.3j

14 Nodal analysis method: I 1 = 0.1j 0.3j V 1 0.1j V 2 I 2 = 0.1j V j 0.2j V 2 Matrix format: YV=I YV = 0.4j 0.1j 0.1j 0.3j V 1 V 2 = I 1 I2 Hence both solutions yield the same answer and are valid

15 Example from 2.4: Solve the model using nodal analysis:

16 Step 1 and step 2:

17 Step 3: I 1 = 10j V j V 10 V 20 = j3 j10 V 10 j3 V 20

18 Step 3: I 2 + 3j V 10 V 20 + j V 30 V j V 30 V 20 = jv 20 I 2 = j3 + j j2 j V 20 + j3 V 10 + j 2j V 30

19 Step 3: I 3 = j V 30 V j V 30 V j V 30 = j 2j V 20 + j 2j 4j V 30

20 Step 3: Combining equations j3 j10 V 10 j3 V 20 = I 1 j3 + j j2 j V 20 + j3 V 10 + j 2j V 30 = I 2 j 2j V 20 + j 2j 4j V 30 = I 3 j V 10 V 20 V 30 = I 1 I 2 I 3 Y V I

21 A Wedeward level example: Solve the following 6 bus

22 Circuit equivalent model ( using the transmission line π equivalent circuit)

23 Given parameters: Line Data: Name From Bus To Bus Z' (p.u.) Y'/2 (p.u.) TL j 0.16j TL j 0.78j TL j 0.78j TL j 0.16j Transformer Data: Name From Bus To Bus Zeq (p.u.) Ye (p.u.) T j 0 T j 0 T j 0

24 Bus 1: I 1 = 1 Z eq1 V 1 V 3 = Y eq1 V 1 Y eq1 V 3

25 Bus 2: I 2 = 1 Z eq2 V 2 V 4 = Y eq2 V 2 Y eq2 V 4

26 Bus 3: 0 = 1 V Z 1 V 3 + Y e1 V 3 + Y 2 V eq V Z 3 V 5 + Y 1 V V Z 3 V Y 4 2 V Z 4 V 3 V 4 = Y eq1 V 1 + Y eq1 + Y e1 + Y 2 + Y Y 1 + Y Y 4 + Y 2 4 V 3 Y 1 + Y 4 V 4 Y 2 V 5

27 Bus 4 (similarly): 0 = Y eq2 V 2 + Y 4 + Y 4 + Y Y 1 + Y Y 3 + Y 2 e2 + Y eq2 V 4 Y 1 + Y 4 V 3 Y 3 V 5 Bus 5: 0 = Y 2 V 3 Y 3 V 4 + Y 2 + Y 2 + Y 2 eq3 + Y 3 + Y 3 2 Y eq3 V 6 Bus 6: I 6 = Y eq3 V 5 + Y eq3 + Y e3 V 6

28 Formation of the Admittance Matrix: Note that: Y kk = admittances connected directly to k th bus Y kn = admittances connected between buses k and n (k n)

29 Admittance matrix formation via inspection: Single-line four bus Equiv. Reactance diagram

30 Find Y bus via inspection of the per-unit diagram (converted to current sources): Y c + Y d + Y f Y d Y c Y f Y d Y b + Y d + Y e Y b Y e Y c Y b Y a + Y b + Y c 0 Y f Y e 0 Y e + Y f + Y g V = I Y kk = admittances connected directly to k th bus Y kn = admittances connected between buses k and n (k n)

31 Find Y bus via inspection of the per-unit diagram (converted to current sources): j j8 j4 j2.5 j8 j j4 j5 j4 j4 j j2.5 j5 0 j V = I Note: book should have Y e = j5.0

32 Find Y bus via inspection of the per-unit diagram (converted to current sources): j j8 j4 j2.5 j8 j j4 j5 j4 j4 j j2.5 j5 0 j V = I Note: book should have Y e = j5.0

33 Find Y bus via inspection of the per-unit diagram (converted to current sources): j14.5 j8 j4 j2.5 j8 j17 j4 j5 j4 j4 j8.8 0 j2.5 j5 0 j8.3 V 1 V 2 V 3 V 4 = Note: book should have Y e = j5.0

34 Chapter 6.4: Power Flow and Bus Type Four bus variables per bus: V k the voltage magnitude δ k the phase angle P k the net real power, where P k = P Generator,k + P Load,k Q k the net reactive power, where where Q k = Q Generator,k + Q Load,k Bus k

35 Chapter 6.4: Power Flow and Bus Type Bus Types: 1) Swing Bus The only reference bus (notated as k = 1 for V k δ k ) Input (known) data: V 1 δ 1 = p. u. (typically) Computed variables: P 1 and Q 1 2) Load (PQ) Bus The most common type Input (known) data: P k and Q k Computed variables: V k and δ k If no generation: P k = P load, k; if the load is purely inductive: Q k = Q Load,k 3) Voltage controlled (PV) bus Input (known) data: P k and V k Computed variables: Q k and δ k

36 Chapter 6.4: Power Flow and Bus Type Example 6.9 Bus 1 Swing (reference) bus Buses 2, 4, 5 load bus Bus 3 PV bus

37 Chapter 6.4: Power Flow and Bus Type Example 6.9 continued: Y bus is symmetric, so solve for upper triangular Buses 1 and 5 connected Y 11 and Y 15 0, else Y 1i = 0 Buses 2, 4, and connected Y 22, Y 24, and Y 25 0 else Y 2,i+1 = 0 Buses 3 and 4 connected Y 33 and Y 33 0, else Y 3,i+2 = 0 Etc

38 Chapter 6.4: Power Flow and Bus Type Example 6.9 continued: Y bus is symmetric, so solve for upper triangular Y bus upper = Since Y is symmetric Y Y 15 Y 22 0 Y 24 Y 25 Y 33 Y 34 0 Y 44 Y 45 Y 55 Y bus = Y Y 15 0 Y 22 0 Y 24 Y Y 33 Y Y 24 Y 34 Y 44 Y 45 Y 15 Y 25 0 Y 45 Y 55

39 Chapter 6.4: Power Flow and Bus Type Example 6.9 continued: Y bus is symmetric, so solve for upper triangular Y 11 = 1 Z 11 = Y 22 = 1 R 11 +jx 11 = 1 + j B j B 25 R 24 +jx 24 2 R 25 +jx j0.02 = j49.72; Y 15 = Y 11 = 2.68 j28.6 Y 24 = 1 1 R 24 +jx 24 = j0.1 = j9.9 Y km = Admittances Y 25 = 1 1 R 25 +jx 25 = j0.05 = j19.9 Y 33 = 1 R 34 +jx 34 = j0.01 = 7.49 j99.4 Y 34 = Y 33 = j99.4 Y 44 = 1 Z 24 + j B Z 45 + j B Z 34 = 11.9 j147.96

40 Chapter 6.4: Power Flow and Bus Type Example 6.9 continued: Y 45 = j B 45 R 45 +jx 45 2 = j0.1 = j39.5 Y km = Admit. Y 55 = j B j B 25 Z 15 Z 45 2 Z 25 2 = 9.0 j108.6 Final matrix (using symmetric property of Y bus ): Y bus = 3.7 j j j j j j j j j j j j j j j108.6

41 Chapter 6: Solving Network Equation Solutions to the matrices: Direct method Gaussian Elimination LU Decomposition Indirect methods Jacobi Gauss-Seidel Newton-Raphson

ELEC4612 Power System Analysis Power Flow Analysis

ELEC4612 Power System Analysis Power Flow Analysis ELEC462 Power Sstem Analsis Power Flow Analsis Dr Jaashri Ravishankar jaashri.ravishankar@unsw.edu.au Busbars The meeting point of various components of a PS is called bus. The bus or busbar is a conductor