A one-line diagram of a simple three-bus power system is shown in Figure 1. The

Question

A one-line diagram of a simple three-bus power system is shown in Figure 1. The currents entering buses 1 and 2 are given in Table 1. The line impedances in Table 2 are impedances in per unit on a base of 100 MVA. Use Matlab to calculate the following parameters 1a. Determine the bus admittance matrix 1b. Determine the phasor bus voltage using any of the following a Gauss-Siedel Method b. Newton-Raphson Method 1c. Use Matlab to find the slack bus real and reactive power 1d. Use Matlab to determine the line flows and line losses. Construct a power flow diagram showing the direction of line flow 1e. Explain the steps that was carried out in the load flow analysis. j1.03 Figure 1 Table 111 and 12 Currents 1 in pu)2 (in pu) 1 1.38-j2.72 0.69-j1.36 1.75-i3.5 0.75-1.6 Table 2 Line Impedances (in pu) 1-2 2-3 3-0 0.01 0.02 j0.04 125+j0.0 04 j0.2 03

Explanation / Answer

It is mentioned in the yellow box that ‘The venin’s theorem states that the changes in the network voltages caused by added branch (the fault impedance) is equivalent to those caused by the added voltages V3 (0) with all other sources shorted’ from this statement we can say that if fault impedance is added at any bus or if fault occurs at any bus, the equivalent voltage must be assumed at bus 3 only remaining all voltages are taken as 0.

                And there is also a fact that whenever a fault occurs or whenever a fault impedance is added to the circuit there is increase in bus line voltage and it must added to pre-fault voltage. Here in the RED box they had calculated the change in bus line current due to fault impedance.

                In the first equation V1= 0 – (j0.2) (-j1.2) = -0.24 p.u

                                                        V2= 0 – (j0.4) (-j0.8) = -0.32 p.u

                                                        V3= (j0.16) (-j2) -1 = -0.68 p.u

Here in the above equations for V1 and V2 we had taken zero (0) because it is mentioned that The venin’s voltage is taken only at bus 3 and other are shorted. For a short circuit line voltage is zero so we had subtracted from zero.

When the fault is at bus 2:

                The same is applicable as of above that The venin’s is taken only at bus 3 and other are shorted. But here for V3 we had taken -0.2 as voltage it is the The venin’s equivalent voltage across bus 3. Remaining all explanation is same as the fault at bus 3 only. (** Here the The venin’s voltage at V3 is obtained from the change in voltage due fault impedance at bus V1**) so it is taken as Vth for bus 3, and at bus 3 calculation the current is divided by 2 because the impedances is equal so the flow of current will be halved. So we divided the bus going to bus 3 by 2. The same applicable to bus 1 also.

When the fault is at bus 1:

                Here at V3 we had taken -0.5 as the The venin’s voltage and V2 and V1 are taken as zero. (** here the The venin’s voltage at V3 is obtained from the change in voltage due fault impedance at bus V1**) so it is taken as Vth for bus 3, and at bus 3 calculation the current is divided by 2 because the impedances is equal so the flow of current will be halved. So we divided the bus going to bus 3 by 2.

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