Presentation on theme: "EE/Econ 458 PF Equations J. McCalley 1. Power system representation NODE or BUS (substation) BRANCHES (lines or transformers) NETWORK (but unloaded and."— Presentation transcript:
EE/Econ 458 PF Equations J. McCalley 1
Power system representation NODE or BUS (substation) BRANCHES (lines or transformers) NETWORK (but unloaded and unsupplied) 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) 3
Power system representation NETWORK (loaded and supplied) Best branch model Approximate branch model 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. 5
Power system representation NETWORK (loaded and supplied) Here is what we will model as a network (reactance only) 6
Power system representation The impedance is a complex number z ij =r ij +jx ij. We ignore the resistance: z ij =jx ij z 12 z 14 z 34 z 23 z 13
Power system representation Impedance relates voltage drop and current via Ohms law: i j z ij ViVi VjVj I ij Voltage drop (volts) Current(amps) 8
Power system representation Admittance, y ij, is the inverse of impedance, z ij : i j y ij ViVi VjVj I ij 9
Power system representation Label the admittances y ij 10 y 12 y 14 y 34 y 23 y
Power system representation I1I1 Current injections: I i flowing into bus i from generator or load. Positive if generator; negative if load. 11 y 12 y 14 y 34 y 23 y I2I2 I4I4 I3I3 I 1, I 4 will be positive. I 3 will be negative. I 2 will be positive if gen exceeds load, otherwise negative.
Power system representation I1I1 Voltages: V i is voltage at bus i. 12 y 12 y 14 y 34 y 23 y I2I2 I4I4 I3I3 V1V1 V2V2 V3V3 V4V4
Power system representation I1I1 Kirchoffs current law: sum of the currents at any node must be zero. 13 y 12 y 14 y 34 y 23 y I2I2 I4I4 I3I3 V1V1 V2V2 V3V3 V4V4 I 14 I 13 I 12 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 I1I1 Now express each current using Ohms law: 14 y 12 y 14 y 34 y 23 y I2I2 I4I4 I3I3 V1V1 V2V2 V3V3 V4V4 I 14 I 13 I 12
Power system representation I1I1 Now collect like terms in the voltages: 15 y 12 y 14 y 34 y 23 y I2I2 I4I4 I3I3 V1V1 V2V2 V3V3 V4V4 I 14 I 13 I 12
Power system representation I1I1 Repeat for the other four buses: 16 y 12 y 14 y 34 y 23 y I2I2 I4I4 I3I3 V1V1 V2V2 V3V3 V4V4 I 14 I 13 I 12
Power system representation Repeat for the other four buses: 17 Notes: 1. y ij =y ji 2. If branch ij does not exist, then y ij =0. I1I1 y 12 y 14 y 34 y 23 y I2I2 I4I4 I3I3 V1V1 V2V2 V3V3 V4V4 I 14 I 13 I 12
Power system representation Write in matrix form: 18 Define the Y-bus: Define elements of the Y-bus:
Power system representation Forming the Y-Bus: 1. The matrix is symmetric, i.e., Y ij =Y ji. 2. A diagonal element Y ii is obtained as the sum of admittances for all branches connected to bus i (y ik 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., Y ij =-y ji. 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: 20 This requires too much EE, so forget about them. Lets make some assumptions instead. But first, what is θ k and θ j ? where Y ij =G ij +jB ij.
Power system representation 21 θ k and θ j 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: Y ij =jB ij 2.Angle differences across branches, are small: θ i -θ j : Sin(θ i -θ j )= θ i -θ j Cos(θ i -θ j )=1.0 3.All voltage magnitudes are 1.0 in the pu system. 22 This is the basis for the DC power flow. 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 23 Collect terms in the same variables Repeat procedure for buses 2, 3, 4:
Example 24 Now write in matrix form:
Example 25 Compare:
Example 26 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 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 θ 1 =0 degrees. But we want power flows:
Example 28 Resulting solution:
Example 29 Resulting solution:
How to solve power flow problems 30 Develop B matrix: 1.Get the Y-bus 2.Remove the j from the Y-bus. 3.Multiply Y-bus by Remove row 1 and column 1.
How to solve power flow problems 31 Develop equations to compute branch flows: where: P B 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 k th 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 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 k th branch begins at node j, -1 if the k th 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, θ 1, is zero.