please explain from where came the method of The bus voltage changes ( in the re
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please explain from where came the method of The bus voltage changes ( in the red collar) ??
EXAMPLE 4.1 The one line diagram of a simple three-bus power system is shown in the Figure 4.1. Each generator is represented by an emf behind the transient reactance. All impedances are expressed in per unit in a common 100 MVA base, and for simplicity, resistances are neglected. The following assumption are made: shunt capacitances are neglected and the system is considered on noo load. all generators are running at their rated voltage and rated frequency with their emf in phase Determine the fault current, the bus voltages and the line currents during the fault when a balanced three-phase fault with a fault impedance 2, 0.16 p.u. fault occurs on a) bus 3 b) bus 2 EET308 POWER SYSTEMANALYSIS SOLUTION (a) Fault on bus 3 j0.2 j0.4 j0.8 j0.4 j0.4 Impedance network with fault at bus 3 Zf j0.16 EET3 08 POWER SYSTEM ANALYSIS EXAMPLE 4.1 j0.8 j0.4 Figure 4. EET308 POWER SYSTEM ANALYSIS 10 SOLUTION Thevenin's theorem states that the changes in the network voltages caused by the added branch (the fault impedance) is equivalent to those caused by the added voltage Vs(0) with a other sources short-circuited All prefault voltages are assumed to be equal to 1.0 per unit Vi(0) V2(0) V3(0) 1.0 pu EET308 POWER SYSTEM ANALYSIS j0.2 j0.4 one-line diagram of a simple power system j0.4 j0.8 j0.4 j0.4 13(F) Thevenin's equivalent networkExplanation / Answer
When the fault is at bus 3:
It is mentioned in the yellow box that ‘Thevenin’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 Thevenin’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 Thevenin’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 Thevenin’s equivalent voltage across bus 3. Remaining all explanation is same as the fault at bus 3 only. (** Here the Thevenin’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 aslo.
When the fault is at bus 1:
Here at V3 we had taken -0.5 as the Thevenin’s voltage and V2 and V1 are taken as zero. (** here the Thevenin’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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