2 Power system representation BRANCHES (lines or transformers)NODE or BUS(substation)NETWORK(but unloaded and unsupplied)
3 Power system representation GENERATOR: Injects MW into the nodeLOAD: Extracts MW out of the node (injects negative MW into the node)NETWORK(loaded and supplied)
4 Power system representation Approximate branch modelBest branch modelNETWORK(loaded and supplied)
5 Power system representation Approximate branch modelBranch resistanceBranch inductive reactanceBranch capacitive susceptanceIgnore 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 Power system representation The impedance is a complex number zij=rij+jxij.We ignore the resistance: zij=jxijz12z14z34z23z131234
8 Power system representation Impedance relates voltage drop and current via Ohm’s law:Current(amps)Voltage drop (volts)IijViVjzijij
9 Power system representation Admittance, yij, is the inverse of impedance, zij:IijViVjyijij
10 Power system representation Label the admittances yijy12y14y34y23y131234
11 Power system representation 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.I1I2y12y14y34y23y131234I3I4
12 Power system representation Voltages: Vi is voltage at bus i.I1I2y12y14y34y23y131234V1V2V4V3I3I4
13 Power system representation 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.I1I2y12y14y34y23y131234I12V1V2I14I13V4V3I3I4
14 Power system representation Now express each current using Ohm’s law:I1I2y12y14y34y23y131234I12V1V2I14I13V4V3I3I4
15 Power system representation Now collect like terms in the voltages:I1I2y12y14y34y23y131234I12V1V2I14I13V4V3I3I4
16 Power system representation Repeat for the other four buses:I1I2y12y14y34y23y131234I12V1V2I14I13V4V3I3I4
17 Power system representation Repeat for the other four buses:Notes:1. yij=yji2. If branch ij does not exist, then yij=0.I1I2y12y14y34y23y131234I12V1V2I14I13V4V3I3I4
18 Power system representation Write in matrix form:Define the Y-bus:Define elements of the Y-bus:
19 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.
20 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: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 Power system representation 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 Δtand so has an angle of θ=ωΔt where ω=2πf and f is frequency of oscillation, 60 Hz for power systems.
22 Power system representation Simplifying assumptions:No resistance: Yij=jBijAngle differences across branches, are small: θi-θj:Sin(θi-θj)= θi-θjCos(θi-θj)=1.0All voltage magnitudes are 1.0 in the pu system.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:
26 ExampleBut 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 ExampleEliminate 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.But we want power flows:
30 How to solve power flow problems Develop B’ matrix:Get the Y-busRemove the “j” from the Y-bus.Multiply Y-bus by -1.Remove row 1 and column 1.
31 How to solve power flow problems Develop equations to compute branch flows: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,…ND 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 solve power flow problems 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.