The one-line diagram of a simple three-bus power system is...

The one-line diagram of a simple three-bus power system is shown in the figure below: j0.30 10.05 j0.075 j0.75 1 2 3 j0.45 Each generator is represented by an emf behind the subtransient reactance. All impedances are expressed in per unit on a commom MVA base. All resistance and shunt capacitances are neglected. The generators are operating on lo load at their rated voltage with their emfs in phase. A three-phase fault occurs at bus 3 through a fault impedance of $Z_f = j0.19$ per unit. Using Z matrix approach, obtain the following: A. Y-bus matrix and Z-bus matrix B. Determine the bus voltages C. line currents during fault.

Transcript

00:01 We are considering a line diagram of a simple power system which is shown here.

00:05 Let's say here is g1 which is connected let's say plate 1, plate 2.

00:12 Two coils are present here that is at the value of x of t1 which is equal to 0 .1 which is further connected to xl.

00:21 Xl that is at 0 .25 then again here is plate 3.

00:27 Here is a plate 4 which from here is connected in this ways where the value of x of t2 is equal to 0 .12 which is further connected to z2 which is at x of d2 double dash that is 0 .1 and this is x of d1 double dash that is equal to 0 .1.

00:47 So this is the given figure from here.

00:49 In the first part of the question now this is the circuit.

00:53 If we simplify or basically in the first part of the circuit we are using the trevin theorem to calculate the impedance of the point of the fault that is x of t, x of 33 and xif.

01:06 So here we are considering about its trevin impedance.

01:12 So trevin impedance across the circuit let's say simplified circuit from here 0 .45, 0 .3 inductor of 0 .25 that is grounded.

01:27 From here it is grounded and from here it is grounded.

01:30 So this is z of 33.

01:32 So here we have to find out the value of z33.

01:35 So z33 is equal to 0 .45 which is in parallel to 0 .25.

01:40 So solving the term from here so we are having the value of z33 that is equal to j of 0 .1607 and in the same way the value of z33 becomes equal to j of 0 .1607.

01:55 So this is the value of z33.

01:58 Now if we are considering that we are considering about the trevin equivalent circuit.

02:06 So this circuit from here will be a battery source of kilowatt is connected from here which is connected to a z33 inductor that is j of 0 .1607.

02:18 Here we are getting the current that is if.

02:21 Here 2f that is equal to j of 0 .07 is grounded from here.

02:27 So the value of i of fpu from here is equal to j which is less than 0 that is j of 0 .1607 plus 0 .07.

02:39 So solving the term from here we get the value of if or pu that become equal to minus j of 4 .33pu.

02:46 So this is the answer to the part a of the question...

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