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EE/Econ 458 PF Equations

EE/Econ 458 PF Equations. J. McCalley. NODE or BUS (substation). BRANCHES (lines or transformers). Power system representation. NETWORK (but unloaded and unsupplied). GENERATOR: Injects MW into the node. LOAD: Extracts MW out of the node (injects negative MW into the node).

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EE/Econ 458 PF Equations

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  1. EE/Econ 458PF Equations J. McCalley

  2. NODE or BUS (substation) BRANCHES (lines or transformers) Power system representation NETWORK (but unloaded and unsupplied)

  3. GENERATOR: Injects MW into the node LOAD: Extracts MW out of the node (injects negative MW into the node) Power system representation NETWORK (loaded and supplied)

  4. Approximate branch model Power system representation Best branch model NETWORK (loaded and supplied)

  5. Approximate branch model Power system representation 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.

  6. Power system representation Here is what we will model as a network (reactance only) NETWORK (loaded and supplied)

  7. The impedance is a complex number zij=rij+jxij. We ignore the resistance: zij=jxij Power system representation z12 2 1 z13 z14 z23 3 4 z34

  8. Impedance relates voltage drop and current via Ohm’s law: Power system representation Current(amps) Voltage drop (volts) Iij Vi Vj zij i j

  9. Power system representation Admittance, yij, is the inverse of impedance, zij: Iij Vi Vj yij i j

  10. Label the admittances yij Power system representation y12 2 1 y13 y14 y23 3 4 y34

  11. Current injections: Ii flowing into bus i from generator or load. Positive if generator; negative if load. I1, I4 will be positive. I3 will be negative. I2 will be positive if gen exceeds load, otherwise negative. Power system representation I1 I2 y12 2 1 y13 y14 y23 3 4 I3 y34 I4

  12. Voltages: Vi is voltage at bus i. Power system representation I1 I2 y12 2 1 V2 V1 y13 y14 y23 V4 V3 3 4 I3 y34 I4

  13. Kirchoff’s current law: sum of the currents at any node must be zero. 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 I2 y12 I12 2 1 V2 V1 I14 y13 I13 y14 y23 V4 V3 I3 3 4 I4 y34

  14. Now express each current using Ohm’s law: Power system representation I1 I2 y12 2 1 I12 V2 V1 I14 y13 I13 y14 y23 V4 V3 3 4 I3 y34 I4

  15. Now collect like terms in the voltages: Power system representation I1 I2 y12 2 1 I12 V2 V1 I14 y13 I13 y14 y23 V4 V3 3 4 I3 y34 I4

  16. Repeat for the other four buses: Power system representation I1 I2 y12 2 1 I12 V2 V1 I14 y13 I13 y14 y23 V4 V3 3 4 I3 y34 I4

  17. Repeat for the other four buses: Power system representation Notes: 1. yij=yji 2. If branch ij does not exist, then yij=0. I1 I2 y12 2 1 I12 V2 V1 I14 y13 I13 y14 y23 V4 V3 3 4 I3 y34 I4

  18. Write in matrix form: Power system representation Define the Y-bus: Define elements of the Y-bus:

  19. 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. Power system representation

  20. 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: Power system representation where Yij=Gij+jBij. This requires too much EE, so forget about them. Let’s make some assumptions instead. But first, what is θk and θj?

  21. 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

  22. Simplifying assumptions: • No resistance: Yij=jBij • Angle differences across branches, are small: θi-θj: • Sin(θi-θj)= θi-θj • Cos(θi-θj)=1.0 • All voltage magnitudes are 1.0 in the pu system. Power system representation 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. This is the basis for the “DC power flow.”

  23. Example Collect terms in the same variables Repeat procedure for buses 2, 3, 4:

  24. Example Now write in matrix form:

  25. Compare: Example

  26. Example • 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.

  27. 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. Example But we want power flows:

  28. Resulting solution: Example

  29. Resulting solution: Example

  30. Develop B’ matrix: How to solve power flow problems • Get the Y-bus • Remove the “j” from the Y-bus. • Multiply Y-bus by -1. • Remove row 1 and column 1.

  31. Develop equations to compute branch flows: How to solve power flow problems • 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.

  32. How to develop node-arc incidence matrix: • 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

  33. How to solve power flow problems

  34. How to solve power flow problems

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