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Sitaram Appari

The document describes modeling a power system network using an admittance matrix formulation. Key points: 1) Branches are modeled as admittances to relate voltage and current. The admittance matrix (Y-bus) is formed with diagonal elements equal to the sum of incident branch admittances and off-diagonals equal to the negative of branch admittances. 2) Kirchhoff's and Ohm's laws are used to write equations relating bus voltages and branch currents. 3) Simplifying assumptions are made to develop the "DC power flow" equations, including ignoring voltage magnitudes and angles and resistance. This leads to a linear relationship between bus voltage angles and real power injections

Engineering
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Power system representation NODE or BUS (substation) BRANCHES (lines or transformers) NETWORK (but unloaded and unsupplied) 1
Power system representation LOAD: Extracts MW out of the node (injects negative MW into the node) GENERATOR: Injects MW into the node NETWORK (loaded and supplied) 2
Power system representation NETWORK (loaded and supplied) Best branch model Approximate branch model 3
Power system representation Approximate branch model Branch resistance Branch inductive reactance Branch capacitive susceptance Ignore resistance, OK because it is much less than reactance. Ignore susceptance, OK because its affect on MW flows very small. Only model reactance, OK for getting branch flows. 4
Power system representation NETWORK (loaded and supplied) Here is what we will model as a network (reactance only) 5
Power system representation The impedance is a complex number zij=rij+jxij. We ignore the resistance: zij=jxij 6 1 2 34 z12 z14 z34 z23 z13
Power system representation Impedance relates voltage drop and current via Ohm’s law: )( 1 ji ij ij VV z I  i j zij Vi Vj Iij Voltage drop (volts) Current(amps) 7
Power system representation Admittance, yij, is the inverse of impedance, zij: )( 1 ji ij ij VV z I  i j yij Vi Vj Iij )( jiijij VVyI  8
Power system representation Label the admittances yij 9 y12 y14 y34 y23 y13 1 2 34
Power system representation I1 Current injections: Ii flowing into bus i from generator or load. Positive if generator; negative if load. 10 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 I1, I4 will be positive. I3 will be negative. I2 will be positive if gen exceeds load, otherwise negative.
Power system representation I1 Voltages: Vi is voltage at bus i. 11 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4
Power system representation I1 Kirchoff’s current law: sum of the currents at any node must be zero. 12 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 1413121 IIII  I14 I13 I12 Note: We assume there are no bus shunts in this system. Bus shunts are capacitive or inductive connections between the bus and the ground. Although most systems have them, they inject only reactive power (no MW) and therefore affect MW flows in the network only very little.
Power system representation I1 Now express each current using Ohm’s law: 13 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 1413121 IIII  I14 I13 I12 )( jiijij VVyI  )()()( 4114311321121 VVyVVyVVyI 
Power system representation I1 Now collect like terms in the voltages: 14 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 I14 I13 I12 )()()( 4114311321121 VVyVVyVVyI  )()()()( 14413312214131211 yVyVyVyyyVI 
Power system representation I1 Repeat for the other four buses: 15 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 I14 I13 I12 )()()()( 14413312214131211 yVyVyVyyyVI  )()()()( 24423324232122112 yVyVyyyVyVI  )()()()( 34434323133223113 yVyyyVyVyVI  )()()()( 43443424134224114 yVyyyVyVyVI 
Power system representation Repeat for the other four buses: 16 )()()()( 14413312214131211 yVyVyVyyyVI  )()()()( 24423324232122112 yVyVyyyVyVI  )()()()( 34434323133223113 yVyyyVyVyVI  )()()()( 43443424134224114 yVyyyVyVyVI  Notes: 1. yij=yji 2. If branch ij does not exist, then yij=0. I1 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 I14 I13 I12
Power system representation Write in matrix form: 17 Define the Y-bus:                                          4 3 2 1 434241434241 343432313231 242324232121 141312141312 4 3 2 1 V V V V yyyyyy yyyyyy yyyyyy yyyyyy I I I I                  434241434241 343432313231 242324232121 141312141312 yyyyyy yyyyyy yyyyyy yyyyyy Y Define elements of the Y-bus:              44434241 34333231 24232221 14131211 YYYY YYYY YYYY YYYY Y                                      4 3 2 1 44434241 34333231 24232221 14131211 4 3 2 1 V V V V YYYY YYYY YYYY YYYY I I I I
Power system representation Forming the Y-Bus: 1. The matrix is symmetric, i.e., Yij=Yji. 2. A diagonal element Yii is obtained as the sum of admittances for all branches connected to bus i (yik is non-zero only when there exists a physical connection between buses i and k). 3. The off-diagonal elements are the negative of the admittances connecting buses i and j, i.e., Yij=-yji. 18
Power system representation From the previous work, you can derive the power flow equations. These are equations expressing the real and reactive power injections at each bus. If we had modeled branch resistance, we would obtain: 19          N j jkkjjkkjjkk N j jkkjjkkjjkk BGVVQ BGVVP 1 1 )cos()sin( )sin()cos(   This requires too much EE, so forget about them. Let’s make some assumptions instead. But first, what is θk and θj? where Yij=Gij+jBij.
Power system representation 20          N j jkkjjkkjjkk N j jkkjjkkjjkk BGVVQ BGVVP 1 1 )cos()sin( )sin()cos(   θk and θj are the angles of the voltage phasors at each bus. The angle captures the time difference when voltage phasors cross the zero- voltage axis. In the time domain simulation, the red curve crosses before the blue one by an amount of time Δt and so has an angle of θ=ωΔt where ω=2πf and f is frequency of oscillation, 60 Hz for power systems.
Power system representation Simplifying assumptions: 1. No resistance: Yij=jBij 2. Angle differences across branches, are small: θi-θj: • Sin(θi-θj)= θi-θj • Cos(θi-θj)=1.0 3. All voltage magnitudes are 1.0 in the pu system. 21 This is the basis for the “DC power flow.”      N kj j jkkjk BP ,1 )(  Per-unit system: A system where all quantities are normalized to a consistent set of bases. It will result in powers being expressed as a particular number of “100 MVA” quantities. Admittance is also per- unitized.
Example 22      N kj j jkkjk BP ,1 )(  y13 =-j10 y14 =-j10 y34 =-j10 y23 =-j10 y12 =-j10 Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu 414114313113212112 4114311321121 )()()(   BBBBBB BBBP     41431321211413121  BBBBBBP  Collect terms in the same variables Repeat procedure for buses 2, 3, 4:   42432322423211212  BBBBBBP    43433432312321313  BBBBBBP    44342413432421414  BBBBBBP 
Example 23      N kj j jkkjk BP ,1 )(  y13 =-j10 y14 =-j10 y34 =-j10 y23 =-j10 y12 =-j10 Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu Now write in matrix form:                                          4 3 2 1 434241434241 343432313231 242324232121 141312141312 4 3 2 1     BBBBBB BBBBBB BBBBBB BBBBBB P P P P
Example 24 Compare:                                          4 3 2 1 434241434241 343432313231 242324232121 141312141312 4 3 2 1     BBBBBB BBBBBB BBBBBB BBBBBB P P P P              44434241 34333231 24232221 14131211 BBBB BBBB BBBB BBBB jY                  434241434241 343432313231 242324232121 141312141312 bbbbbb bbbbbb bbbbbb bbbbbb j                  2010010 10301010 0102010 10101030 jY                                           4 3 2 1 2010010 10301010 0102010 10101030 1 4 1 2    
Example 25                                           4 3 2 1 2010010 10301010 0102010 10101030 1 4 1 2                                                1 4 1 2 2010010 10301010 0102010 10101030 1 4 3 2 1     But matlab indicates above matrix is singular which means it does not have an inverse. There is a dependency among the four equations, i.e., we can add the bottom three rows and multiply by -1 to get the top row. This dependency occurs because all four angles are not independent; we have to choose one of them as a reference with a fixed value of 0 degrees.
Example 26                                            1 4 1 2 2010010 10301010 0102010 10101030 1 4 3 2 1     Eliminate one of the equations and one of the variables by setting the variable to zero. We choose to eliminate the first equation and set the first variable θ1=0 degrees.                                                   025.0 15.0 025.0 1 4 1 20100 103010 01020 1 4 3 2    But we want power flows: )( jkkjkj BP   25.0)025.00(10)( 211212  BP 5.1)15.00(10)( 311313  BP 25.0)025.00(10)( 411414  BP 25.1)15.0025.0(10)( 322323  BP 25.1)025.015.0(10)( 433434  BP
Example 27 Resulting solution: P13=1.5 P14 =0.25 P43 =1.25 P23 =1.25 P12=0.25  Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu
Example 28 Resulting solution: P13=1.5 P14 =0.25 P43 =1.25 P23 =1.25 P12=0.25  Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu
How to solve power flow problems 29 Develop B’ matrix:                                          4 3 2 1 434241434241 343432313231 242324232121 141312141312 4 3 2 1     BBBBBB BBBBBB BBBBBB BBBBBB P P P P              44434241 34333231 24232221 14131211 BBBB BBBB BBBB BBBB jY                  434241434241 343432313231 242324232121 141312141312 bbbbbb bbbbbb bbbbbb bbbbbb j 1. Get the Y-bus 2. Remove the “j” from the Y-bus. 3. Multiply Y-bus by -1. 4. Remove row 1 and column 1.                  2010010 10301010 0102010 10101030 jY               20100 103010 01020 'B
How to solve power flow problems 30 Develop equations to compute branch flows:  )( ADPB where: • PB is the vector of branch flows. It has dimension of M x 1. Branches are ordered arbitrarily, but whatever order is chosen must also be used in D and A. • θ is (as before) the vector of nodal phase angles for buses 2,…N • D is an M x M matrix having non-diagonal elements of zeros; the diagonal element in position row k, column k contains the negative of the susceptance of the kth branch. • A is the M x N-1 node-arc incidence matrix. It is also called the adjacency matrix, or the connection matrix. Its development requires a few comments.
How to solve power flow problems 31 How to develop node-arc incidence matrix:  )( ADPB • number of rows equal to the number of branches (arcs) and a number of columns equal to the number of nodes. • Element (k,j) of A is 1 if the kth branch begins at node j, -1 if the kth branch terminates at node j, and 0 otherwise. • A branch is said to “begin” at node j if the power flowing across branch k is defined positive for a direction from node j to the other node. • A branch is said to “terminate” at node j if the power flowing across branch k is defined positive for a direction to node j from the other node. • Note that matrix A is of dimension M x N-1, i.e., it has only N-1 columns. This is because we do not form a column with the reference bus, in order to conform to the vector θ, which is of dimension (N-1) x 1. This works because the angle being excluded, θ1, is zero.
How to solve power flow problems 32 5 1 4 3 2 Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu numberbranch 5 4 3 2 1 01-0 11-0 01-1 001- 1-00 A 432 numbernode                                            100000 010000 001000 000100 000010 D                                           4 3 2 100000 010000 001000 000100 000010    01-0 11-0 01-1 001- 1-00  )( ADPB
How to solve power flow problems 33                                                                             3 43 32 2 4 3 43 32 2 4 5 4 3 2 1 10 )(10 )(10 10 10 100000 010000 001000 000100 000010           B B B B B P P P P P                         025.0 15.0 025.0 4 3 2                                     5.1 25.1 25.1 25.0 25.0 5 4 3 2 1 B B B B B P P P P P P13=1.5 P14 =0.25 P43 =1.25 P23 =1.25 P12=0.25 Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu

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INCIDENCE MATRIX

  • 1. Power system representation NODE or BUS (substation) BRANCHES (lines or transformers) NETWORK (but unloaded and unsupplied) 1
  • 2. Power system representation LOAD: Extracts MW out of the node (injects negative MW into the node) GENERATOR: Injects MW into the node NETWORK (loaded and supplied) 2
  • 3. Power system representation NETWORK (loaded and supplied) Best branch model Approximate branch model 3
  • 4. Power system representation Approximate branch model Branch resistance Branch inductive reactance Branch capacitive susceptance Ignore resistance, OK because it is much less than reactance. Ignore susceptance, OK because its affect on MW flows very small. Only model reactance, OK for getting branch flows. 4
  • 5. Power system representation NETWORK (loaded and supplied) Here is what we will model as a network (reactance only) 5
  • 6. Power system representation The impedance is a complex number zij=rij+jxij. We ignore the resistance: zij=jxij 6 1 2 34 z12 z14 z34 z23 z13
  • 7. Power system representation Impedance relates voltage drop and current via Ohm’s law: )( 1 ji ij ij VV z I  i j zij Vi Vj Iij Voltage drop (volts) Current(amps) 7
  • 8. Power system representation Admittance, yij, is the inverse of impedance, zij: )( 1 ji ij ij VV z I  i j yij Vi Vj Iij )( jiijij VVyI  8
  • 9. Power system representation Label the admittances yij 9 y12 y14 y34 y23 y13 1 2 34
  • 10. Power system representation I1 Current injections: Ii flowing into bus i from generator or load. Positive if generator; negative if load. 10 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 I1, I4 will be positive. I3 will be negative. I2 will be positive if gen exceeds load, otherwise negative.
  • 11. Power system representation I1 Voltages: Vi is voltage at bus i. 11 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4
  • 12. Power system representation I1 Kirchoff’s current law: sum of the currents at any node must be zero. 12 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 1413121 IIII  I14 I13 I12 Note: We assume there are no bus shunts in this system. Bus shunts are capacitive or inductive connections between the bus and the ground. Although most systems have them, they inject only reactive power (no MW) and therefore affect MW flows in the network only very little.
  • 13. Power system representation I1 Now express each current using Ohm’s law: 13 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 1413121 IIII  I14 I13 I12 )( jiijij VVyI  )()()( 4114311321121 VVyVVyVVyI 
  • 14. Power system representation I1 Now collect like terms in the voltages: 14 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 I14 I13 I12 )()()( 4114311321121 VVyVVyVVyI  )()()()( 14413312214131211 yVyVyVyyyVI 
  • 15. Power system representation I1 Repeat for the other four buses: 15 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 I14 I13 I12 )()()()( 14413312214131211 yVyVyVyyyVI  )()()()( 24423324232122112 yVyVyyyVyVI  )()()()( 34434323133223113 yVyyyVyVyVI  )()()()( 43443424134224114 yVyyyVyVyVI 
  • 16. Power system representation Repeat for the other four buses: 16 )()()()( 14413312214131211 yVyVyVyyyVI  )()()()( 24423324232122112 yVyVyyyVyVI  )()()()( 34434323133223113 yVyyyVyVyVI  )()()()( 43443424134224114 yVyyyVyVyVI  Notes: 1. yij=yji 2. If branch ij does not exist, then yij=0. I1 y12 y14 y34 y23 y13 1 2 34 I2 I4 I3 V1 V2 V3 V4 I14 I13 I12
  • 17. Power system representation Write in matrix form: 17 Define the Y-bus:                                          4 3 2 1 434241434241 343432313231 242324232121 141312141312 4 3 2 1 V V V V yyyyyy yyyyyy yyyyyy yyyyyy I I I I                  434241434241 343432313231 242324232121 141312141312 yyyyyy yyyyyy yyyyyy yyyyyy Y Define elements of the Y-bus:              44434241 34333231 24232221 14131211 YYYY YYYY YYYY YYYY Y                                      4 3 2 1 44434241 34333231 24232221 14131211 4 3 2 1 V V V V YYYY YYYY YYYY YYYY I I I I
  • 18. Power system representation Forming the Y-Bus: 1. The matrix is symmetric, i.e., Yij=Yji. 2. A diagonal element Yii is obtained as the sum of admittances for all branches connected to bus i (yik is non-zero only when there exists a physical connection between buses i and k). 3. The off-diagonal elements are the negative of the admittances connecting buses i and j, i.e., Yij=-yji. 18
  • 19. Power system representation From the previous work, you can derive the power flow equations. These are equations expressing the real and reactive power injections at each bus. If we had modeled branch resistance, we would obtain: 19          N j jkkjjkkjjkk N j jkkjjkkjjkk BGVVQ BGVVP 1 1 )cos()sin( )sin()cos(   This requires too much EE, so forget about them. Let’s make some assumptions instead. But first, what is θk and θj? where Yij=Gij+jBij.
  • 20. Power system representation 20          N j jkkjjkkjjkk N j jkkjjkkjjkk BGVVQ BGVVP 1 1 )cos()sin( )sin()cos(   θk and θj are the angles of the voltage phasors at each bus. The angle captures the time difference when voltage phasors cross the zero- voltage axis. In the time domain simulation, the red curve crosses before the blue one by an amount of time Δt and so has an angle of θ=ωΔt where ω=2πf and f is frequency of oscillation, 60 Hz for power systems.
  • 21. Power system representation Simplifying assumptions: 1. No resistance: Yij=jBij 2. Angle differences across branches, are small: θi-θj: • Sin(θi-θj)= θi-θj • Cos(θi-θj)=1.0 3. All voltage magnitudes are 1.0 in the pu system. 21 This is the basis for the “DC power flow.”      N kj j jkkjk BP ,1 )(  Per-unit system: A system where all quantities are normalized to a consistent set of bases. It will result in powers being expressed as a particular number of “100 MVA” quantities. Admittance is also per- unitized.
  • 22. Example 22      N kj j jkkjk BP ,1 )(  y13 =-j10 y14 =-j10 y34 =-j10 y23 =-j10 y12 =-j10 Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu 414114313113212112 4114311321121 )()()(   BBBBBB BBBP     41431321211413121  BBBBBBP  Collect terms in the same variables Repeat procedure for buses 2, 3, 4:   42432322423211212  BBBBBBP    43433432312321313  BBBBBBP    44342413432421414  BBBBBBP 
  • 23. Example 23      N kj j jkkjk BP ,1 )(  y13 =-j10 y14 =-j10 y34 =-j10 y23 =-j10 y12 =-j10 Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu Now write in matrix form:                                          4 3 2 1 434241434241 343432313231 242324232121 141312141312 4 3 2 1     BBBBBB BBBBBB BBBBBB BBBBBB P P P P
  • 24. Example 24 Compare:                                          4 3 2 1 434241434241 343432313231 242324232121 141312141312 4 3 2 1     BBBBBB BBBBBB BBBBBB BBBBBB P P P P              44434241 34333231 24232221 14131211 BBBB BBBB BBBB BBBB jY                  434241434241 343432313231 242324232121 141312141312 bbbbbb bbbbbb bbbbbb bbbbbb j                  2010010 10301010 0102010 10101030 jY                                           4 3 2 1 2010010 10301010 0102010 10101030 1 4 1 2    
  • 25. Example 25                                           4 3 2 1 2010010 10301010 0102010 10101030 1 4 1 2                                                1 4 1 2 2010010 10301010 0102010 10101030 1 4 3 2 1     But matlab indicates above matrix is singular which means it does not have an inverse. There is a dependency among the four equations, i.e., we can add the bottom three rows and multiply by -1 to get the top row. This dependency occurs because all four angles are not independent; we have to choose one of them as a reference with a fixed value of 0 degrees.
  • 26. Example 26                                            1 4 1 2 2010010 10301010 0102010 10101030 1 4 3 2 1     Eliminate one of the equations and one of the variables by setting the variable to zero. We choose to eliminate the first equation and set the first variable θ1=0 degrees.                                                   025.0 15.0 025.0 1 4 1 20100 103010 01020 1 4 3 2    But we want power flows: )( jkkjkj BP   25.0)025.00(10)( 211212  BP 5.1)15.00(10)( 311313  BP 25.0)025.00(10)( 411414  BP 25.1)15.0025.0(10)( 322323  BP 25.1)025.015.0(10)( 433434  BP
  • 27. Example 27 Resulting solution: P13=1.5 P14 =0.25 P43 =1.25 P23 =1.25 P12=0.25  Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu
  • 28. Example 28 Resulting solution: P13=1.5 P14 =0.25 P43 =1.25 P23 =1.25 P12=0.25  Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu
  • 29. How to solve power flow problems 29 Develop B’ matrix:                                          4 3 2 1 434241434241 343432313231 242324232121 141312141312 4 3 2 1     BBBBBB BBBBBB BBBBBB BBBBBB P P P P              44434241 34333231 24232221 14131211 BBBB BBBB BBBB BBBB jY                  434241434241 343432313231 242324232121 141312141312 bbbbbb bbbbbb bbbbbb bbbbbb j 1. Get the Y-bus 2. Remove the “j” from the Y-bus. 3. Multiply Y-bus by -1. 4. Remove row 1 and column 1.                  2010010 10301010 0102010 10101030 jY               20100 103010 01020 'B
  • 30. How to solve power flow problems 30 Develop equations to compute branch flows:  )( ADPB where: • PB is the vector of branch flows. It has dimension of M x 1. Branches are ordered arbitrarily, but whatever order is chosen must also be used in D and A. • θ is (as before) the vector of nodal phase angles for buses 2,…N • D is an M x M matrix having non-diagonal elements of zeros; the diagonal element in position row k, column k contains the negative of the susceptance of the kth branch. • A is the M x N-1 node-arc incidence matrix. It is also called the adjacency matrix, or the connection matrix. Its development requires a few comments.
  • 31. How to solve power flow problems 31 How to develop node-arc incidence matrix:  )( ADPB • number of rows equal to the number of branches (arcs) and a number of columns equal to the number of nodes. • Element (k,j) of A is 1 if the kth branch begins at node j, -1 if the kth branch terminates at node j, and 0 otherwise. • A branch is said to “begin” at node j if the power flowing across branch k is defined positive for a direction from node j to the other node. • A branch is said to “terminate” at node j if the power flowing across branch k is defined positive for a direction to node j from the other node. • Note that matrix A is of dimension M x N-1, i.e., it has only N-1 columns. This is because we do not form a column with the reference bus, in order to conform to the vector θ, which is of dimension (N-1) x 1. This works because the angle being excluded, θ1, is zero.
  • 32. How to solve power flow problems 32 5 1 4 3 2 Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu numberbranch 5 4 3 2 1 01-0 11-0 01-1 001- 1-00 A 432 numbernode                                            100000 010000 001000 000100 000010 D                                           4 3 2 100000 010000 001000 000100 000010    01-0 11-0 01-1 001- 1-00  )( ADPB
  • 33. How to solve power flow problems 33                                                                             3 43 32 2 4 3 43 32 2 4 5 4 3 2 1 10 )(10 )(10 10 10 100000 010000 001000 000100 000010           B B B B B P P P P P                         025.0 15.0 025.0 4 3 2                                     5.1 25.1 25.1 25.0 25.0 5 4 3 2 1 B B B B B P P P P P P13=1.5 P14 =0.25 P43 =1.25 P23 =1.25 P12=0.25 Pg1=2pu Pd3=4pu Pd2=1pu 1 2 34 Pg2=2pu Pg4=1pu