Hi guys, today we are going to teach you how to model a bus admittance matrix (Y-bus) of a given power system. Modeling and solving of Y-bus matrices is an important part of Power system analysis and design, and is used extensively in diagnosing, solving and finding problems in power systems especially different kind of faults. But before you do any of that stuff you first must know how to make a Y-Bus matrix (short form for Admittance, often denoted in electrical equations as Y, inverse of impedance Z) of a given power system, involving grids, transmission lines transformers and loads. Next we will teach you how to solve a Y- Bus matrix.
Consider following single line diagram of a Power system:
The system consists of 4 (numbered in circles) Buses, 0 bus is a ground or reference bus, a generator with an EMF of 1.25 V (per unit) and an internal impedance of j1.15 is connected to bus no 3.All values here are given in per unit, we will enlist conversion formulas for per unit system at the end) it is essential to convert all given and rated voltages, currents, impedances and admittances to be converted to per unit values before power system analysis. You can see a Junction (a small line) between j1.15 (generator impedance) and j0.1, the junction represents here a step up transformer and j0.1 is its internal impedance, since these are in series we will combine these as j1.25 (j1.15+j0.1). Similarly a motor (load) with an internal emf of 0.85<-45deg and an impedance of j1.15 is connected through a step down transformer of impedance j0.1, at bus no 4. Again we will combine these two into single impedance of j1.25. Bus 3 and 2 are connected through a transmission line of impedance j0.25, you can read the rest of single line diagram in the same manner. Now 4 bus system means you will need a 4X4 matrix. The matrix will be of the form:
Y11 means all admittance connected at node or bus no 1. But in the figure we are given values in impedances, so will convert them to admittances first and then add them. As you can see bus no 1 has impedances j0.25 (Z13, impedance between bus 1 and 3), j0.125 (Z12) and j0.4 (Z14) connected to it, we will individually convert them to admittances and then add them that will be the value of Y11.
So Y13 = 1/Z13 = 1/j0.25 = -j4.0
Similarly Y12 = -j8.0
And Y14 = -j2.5
Now the first entry of Ybus matrix will be
Y11 = Y13+Y12+Y14 = (-j4.0) +(-j8.0)+(-j2.5) = -j14.5
Remember you will first have to see the figure that which admittances are connected to the given bus.
In the same manner you will have to calculate Y22 (some of all admittances connected at bus 2), Y33 and Y44, this deal with diagonal entries of Y-bus matrix.
For off diagonal entries, for example take Y12, you will have to take the the admittance between bus 1 and 2, and then negate it (that is take its negative), here the impedance between bus 1 and 2 is
= j0.125 (Z12), the admittance will be
=-j8.0 (Y12 = 1/j0.125),
But the entry Y12, bus will be –Y12 that is,
Y12, bus = -Y12 = -(-j8.0) = j8.0
Y13, Bus = -Y13 = 1/Z13 = -(1/j0.25) = -(-j4.0) = j4.0
After calculating values for all other entries of Y bus, You will have final version as:
Entries Y43,bus and Y32, Bus are 0 because there is no link between these two busses.
In the next article we will teach you how to solve Y-Bus for Ohm’s law
V=IZ or V=IY-1
To calculate missing bus voltages or current, for power rid analysis or for fault analysis.
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