Figure 1 shows the network diagram of a 132kV power system. The network has three buses and two generating units. The loads of (l.O+jO.35) pu and (3.2+J1.2) pu are connected at bus 2 and 3...

Solution:
Representing the impedances of the bus system
BUS CODE

IMPEDANCE
1-2 0+j0.18
2-3 0+j0.32
1-3 0+j0.24
Schedule of generation and load as given is:
BUS CODE

ASSUMED VOLTAGE GENERATION LOAD
MW MVAR MW MVAR
1-2 1.042+j0 0 0 0 0
2-3 1+j0 1.2 0 1 j0.35
1-3` 1+j0 0 0 3.2 j1.2
Calculating bus admittance matrix:
BUS CODE IMPEDANCE

ADMITTANCE

1-2 j0.18 -5.556j
2-3 j0.32 -3.125j
1-3 j0.24 -4.1667j
Forming bus admittance matrix
ADMITTANCE CODE

ADMITTANCE FORMULA ADMITTANCE VALUE
-9.722j
5.556j
4.1667j
5.556j
-8.681j
3.125j
4.1667j
3.125j
-7.2917j
Hence matrix can be written as:
[



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[



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Now development of load flow equations:
For any bus:


∑
Now
Hence separating real and imaginary parts:
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Now from nodal admittance matrix we get:
B E f


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Now we have f1=0, f2=0, f3=0
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Now we have f1=0, f2=0, f3=0
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Now we have f1=0, f2=0, f3=0
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Now we have f1=0, f2=0, f3=0
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For iteration 2:


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∑


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Hence we get:

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