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QUESTION: 1

A power system has 200 busses including 15 generator buses. For the load flow analysis using the Newton-Raphson method in polar coordinates, the size of the Jacobian is

Solution:

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

*Answer can only contain numeric values

QUESTION: 2

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)

Solution:

Three phase fault current is given by

QUESTION: 3

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

Solution:

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

*Answer can only contain numeric values

QUESTION: 4

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___

Solution:

Total elements in Y_{bus} = 60 × 60 = 3600

Non – zero = 0.3 × 3600 = 1080

Diagonal = 60

Mutual elements = 660

Transmission lines =1020/2=510

QUESTION: 5

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

Solution:

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

*Answer can only contain numeric values

QUESTION: 6

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 sub-transient reactance was induced while calculating Z bus. A three-phase fault occurs at bus 1. The magnitude of pu current supplied by generator 3 whose sub-transient reactance is j0.05 pu is___ (in pu)

Solution:

QUESTION: 7

The bus admittance matrix for the network shown in the figure is

Solution:

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

QUESTION: 8

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

Solution:

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

QUESTION: 9

A generator is connected to an infinite bus through a double circuit line as shown

The admittance matrix Y is given by

Solution:

QUESTION: 10

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}.

Solution:

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