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A power system has 200 busses including 15 generator buses. For the load flow analysis using the NewtonRaphson method in polar coordinates, the size of the Jacobian is
Total number of buses (n) = 200
Number of generator buses = 15
One generator bus acts as slack bus.
Size of Jacobian = (2 × 200)  2  14
= 384
The Z_{bus} of a system is
If a 3 phase fault occurs at BUS – 2, the p.u. fault current in each phase is – (in pu)
Three phase fault current is given by
A single line diagram of a power system is shown in the figure. The per unit values of line impedances are given. The sum of diagonal elements of Y_{BUS} matrix is
Z_{12} = j0.18 pu ⇒ y_{12} = j10 pu
Z_{13} = j0.25 pu ⇒ y_{13} = j4 pu
Z_{23} = j0.2 pu ⇒ y_{23} = j5 pu
Y_{11} = y_{12} + y_{13} = j14 pu
Y_{22} = y_{21} + y_{23} = j15 pu
Y_{33} = y_{31} + y_{32} = j9 pu
Sum of diagonal elements = j14 – j15 – j9
= j38 pu
The power system network is having a 60 bus system. The YbusYbus of this power system network is having 70% of sparse. The minimum number of transmission lines that exist in the bus system___
Total elements in Y_{bus} = 60 × 60 = 3600
Non – zero = 0.3 × 3600 = 1080
Diagonal = 60
Mutual elements = 660
Transmission lines =1020/2=510
The single line diagram of a power system network having 3 buses and 4 lines is shown in the figure below. The line data is provided in the associated table. The values of the (2, 2) and (2,1) elements of the bus admittance matrix (Y_{BUS}) are
From the table
Impedance of line 1 = 0.01 + j0.01
Admittance of line 1 =
Impedance of line 2 = 0.01 + j0.08
Admittance of line 2 =
Resultant admittance between the bus1 and bus2
y_{21} = y_{12} = 0.99 – j9.9 + 1.538 – j 12.3
= 2.528 – j22.2
Impedance of line 3 = 0.01 + j0.08
Admittance of line 3 =
y_{13} = 1.538 – j12.3
Impedance of line 4 = 0.02 + j0.14
Admittance of line 4 =
Charging susceptance at bus2
2y_{c2} = sum of charging susceptance of line 1, 2 and
= 2.528  j22.2 + j1.85 + 1  j7
= 3.528 – j 27.35
Y_{BUS}(2,1) =  y_{21}
=  2.528 + j22.2
For a Z bus system Z_{11} = j0.25,Z_{12} = j0.02,Z_{13} = j0.05,z_{14} = j0.04
There are 2 generators at bus 1 and 3 and their subtransient reactance was induced while calculating Z bus. A threephase fault occurs at bus 1. The magnitude of pu current supplied by generator 3 whose subtransient reactance is j0.05 pu is___ (in pu)
The bus admittance matrix for the network shown in the figure is
After converting the given impedances into admittances,
y_{11} = j0.5  j5  j5 = j10.5
y_{22} = j0.5  j2.5  j2.5 = j5 = j8.0
y_{33} = j0.5  j5  j10  j2.5 = j18.0
y_{44} = j5  j10  j5 = j20.0
y_{12} = y_{21} = 0
y_{13} = y_{31} = j5.0
y_{14} = y_{41} = j5.0
y_{23} = y_{32} = j2.5
y_{24} = y_{42} = j5.0
y_{34} = y_{43} = j10.0
Two generators G_{1} and G_{2} are connected with cable having a reactance of j3 PU and the load demand at two buses are SD_{1} = 20 + j20 PU and SD_{2} = 25 + j2.5 PU the total reactive power in PU at the generating station G_{1} when δ = 20° is ______ PU
S = V_{1}I* = 1∠20° × .116∠10 = 0.116∠10
S_{1 }= SD_{1} + s
= 20 + 20j + .114 + 0.020j
S_{1} = 20.114 + 20.02 j
Total power = Active power + Reactive power
∴ Reactive power = 20.02 PU
A generator is connected to an infinite bus through a double circuit line as shown
The admittance matrix Y is given by
The bus impedance matrix of a 4 – bus power system network is given by
An element having an impedance of j 0.12 pu is connected b/n ref bus and bus 2. Calculate new value of Z_{22}.
22 docs274 tests

22 docs274 tests
