Question

The one-line diagram of a three-bus power system is shown in Figure 4. All impedances are
expressed in per unit on a common MVA base. All resistances and shunt capacitances are
neglected. Information on each component in this system is given below:
• Each generator is represented by an emf
behind the sub-transient reactance of j0.045
and their neutrals are connected to the
ground.
• Line 1-2 has reactance of j0.88
• Line 2-3 has reactance of j0.65
• Line 1-3 has reactance of j0.4
• The generators are operating on no load at
their rated voltage (1.02 per unit) with their
emf in phase.
a) Construct the bus impedance matrix by adding one impedance element at a time and
using the matrix elimination technique, when needed.
b) Consider, a three-phase fault, which occur sequentially at bus 1, 2 and 3 through a fault
impedance of Zf=j0.20 per unit. Find the bus location at which fault current in per unit is
the most severe.
c) Determine the healthy bus voltages for the fault location at bus 3.
A second line is built between Bus 1 and Bus 2 with an impedance of j0.88.
d) Repeat Questions (a), (b) and (c) for the new power systems.
e) Compare the bus matrix in this case with the original bus matrix. What is the implication
of introducing the second transmission line between bus 1 and bus 2?
Figure 4

Figure 4

Answer 1

Solution According to given information, bus network will be as shown in adjacent figure 30.045€ 30.0458 90.9 j 0.65 (a) Bus impedance matrise (Zeus) construction Step I Choosing bus 1 es reference bus Z bus? [0.045 30.045 130.045) Step 2 Adding 2 bus a Step 3 Adding - Lars 2 branch impedance between bus ① and new bus 2 (30.045 30.0457 jours & – 0.88 j0.045 jo.95] impedance between bus 2 and nere bus ③ 80.045 20.045 0.045] 30.045 j0.925 20.825 3 ₃0.045 0.925 1.575] 30.045 & 7 20.65 Step 4 Adding impedance between bus ③ and existing bus ① 20.045 Zunz 2,, + Z33 - 22,3 1.5W + Zu 10-045 30.045 20.045 10.925 ; 0.925 L;0.045 0.925 j1.575 Loj 0.88 j 1.53 07 0.88 j 1.53 jtni (0.045 + 1.575 = 4-2X0.045 + 0.48 1 By Matrise 3.1.93 elimination was a (20.045 120.045 120.045 j0.045 0.925 j1.425 10.045) 0.925 7 1.575 ] - jah! 181.88 To 0.88 1.53) 1.43 1.53 20.0458 10.88 0 + Te Zeur a (10.045 10.045 j0.045 20.045 jo.0457 0.524 20.227 50.227 10.362 ] 20.4 20.65 S

Steb 5 Adding inkedance between existing bus ② and reference bus Zbus 120.045 20.045 20.045 L; 0.045 20.045 20.045 20.045 30.524 30.227 36.524 10.227 2 0.362 ; 0.227 jo.524 j0.227 2 0.569] 0.045 & 7 30.045 & 7 3 of 20.88 10 0.4 20.65 Materin elimination 0.045 Zbers 2 jo.045 20.045 10.045 20.045 0.524 j0.227 20.045 j 0.227) : 0.362 0.524 045 0.524 0.227 0.569 0.227 Zrous ? ( 20.0414 2 0.0036 j. 20270) j0.0036 5 0.69414 j0.0179 L; 0.0270 jo.0179 j0.2714 | . (6) Fault corrent at bus l with Zr2 20.20 and Ef 21.02, IF ² e 2,14 Zr 1.02 (0.0414 +0.20 III) 2 4.225 units Fault inorent at bus 2 with similar conditions, Tral n za roolihat 0..) → tr 4.225 unite Fault norrent at leus 3 with similar conditions, EL Ifal - - 33+2 1.02 (0.2714 +0.2) 2.2 - TFS) 2 2.164 units Fal - 2.10 Se fault at laus and lens And are more severe than I ll IF 2) > Ifa) are equally severe. fault at bus ③

be) For fault at bus , We E - 713 If = 1.02- (0.0270) (2.164) [vila 0.9616 units 1₂1 2 - 2 12 JFZ 1.02 - (0.0036)(4.225) (1₂) 2 1.0048 units (d) between For second bus of impedance j0.88 connected bus 1 and bus & figure becomes as shown: as 312 2 ; (0.88 110.88) 2;D.44 30.04s 04 10044 204587 . 0 jo.4 30.65 Similar steps can be followed as in las, (b) and (1) to obtain required result using adjacent figure instead of given figure.