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1. In the two-bus system shown in Figure 1, bus 1 is a slack bus with 01
01.0V pu. A
load of 100 MW and 50 MVAR is taken from bus 2. The line impedance is
z 12 = 0.12 + j 0.16 pu on a base of 100 MVA. Using Newton-Raphson method, obtain
the voltage magnitude and phase angle of bus 2. Start with an initial estimate of)0(
2V = 1.0 pu and (0)
2δ = 0. Perform two iterations.
Figure 1
Y =
j43 j43 j43 j43
Y
55
55 and θ =
00
00
53.13126.87
126.8753.13
In this problem 0.5QI
1.0PI
2
2
and unknown quantities =
2
2
V
δ
With initial value0
1 01.0V ;0
2 01.0V
iP =
ii
2
iGV +
N
in
1n
iV
nV
niY cos (
niθ +
nδ -
iδ )
iQ =
ii
2
iBV
N
in
1n
iV
nV
niY sin (
niθ +
nδ -
iδ )
2P = )δδθ(cosYVVGV
2112121222
2
2
= (1 x 1 x 3 ) + (1 x 1 x 0)0(126.87cos5 0 x ) = 0 pu
22
2
22BVQ
2V
1V
21Y sin (
21θ +
1δ -
2δ )
= -(1 x 1 x -4) - (1 x 1 x 5 x )00(126.87sin0
= 0 pu
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1.001.0PPIΔP222
pu
0.500.5QQIΔQ222 pu
ii
Hii
2
iiBVQ
iiN iP + ii2
i GV
ii
M i
P -ii
2
iGV
ii
L i
Qii
2
iBV
22
2
2222BVQH 0 - ( 1 x 1 x -4 ) = 4
22
N 2
P +22
2
2GV = 0 + (1 x 1 x 3) = 3
22
M 2
P -22
2
2GV = 0 - (1 x 1 x 3) = -3
22
2
2222BVQL 0 - ( 1 x 1 x -4 ) = 4
43
34
2
2
2
V
VΔ
Δδ
=
0.5
1.0
2
2
2
V
VΔ
Δδ
=
0.2
0.1
0)1(
2.73rad.0.10.1)0δ 5(
pu0.80.2)1.0V)1(
2 (
0
101.0V
0
2.730.8V 5
This completes the first iteration.
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Second iteration:
2P = )δδθ(cosYVVGV
2112121222
2
2
= (0.8 x 0.8 x 3 ) + (0.8 x 1 x 5.73))(-0(126.87cos50
x ) = -0.7875 pu
22
2
22BVQ
2V
1V
21Y sin (
21θ +
1δ -
2δ )
= -(0.8 x 0.8 x -4) - (0.8 x 1 x 5 x )5.73))(0(126.87sin0
= -0.3844 pu
0.2125-0.7875)1.0PPIΔP222
( pu
0.11560.3844)(0.5QQIΔQ222
pu
22
2
2222BVQH -(-0.38440) - (0.8 x 0.8 x -4 ) = 2.9444
22
N 2
P +22
2
2GV = -0.7875 + (0.8 x 0.8 x 3) = 1.1325
22
M 2
P -22
2
2GV = -0.7875 - (0.8 x 0.8 x 3) = -2.7075
22
2
2222BVQL -0.3844 - ( 0.8 x 0.8 x -4 ) = 2.1756
2.17562.7075
1.13252.9444
2
2
2
V
VΔ
Δδ
=
0.1156
0.2125
2
2
2
V
VΔ
Δδ
=
0.0967
0.0350
2
VΔ - 0.0967 x 0.8 = -0.0773
0)(2
27.73rad.0.1350.035)(0.1-δ
pu0.72270.0773)0.8V)2(
2 (
0
101.0V
0
27.730.7227V
This completes the second iteration.
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2. Figure 2 shows the one-line diagram of a simple three-bus power system with
generation at buses 1 and 2. The voltage at bus 1 is 01
01.0V per unit. Voltage
magnitude at bus 2 is fixed at 1.05 pu with a real power generation of 400 MW. A
load consisting of 500 MW and 400 MVAR is taken from bus 3. Line admittances
are marked in per unit on a 100 MVA base. For the purpose of hand calculations,
line resistances and line charging susceptances are neglected. UsingNewton-Raphson method, start with the initial estimates of
)0(
2V = 1.05 + j 0 and
)0(
3V = 1.0 + j 0, and keeping
2V = 1.05 pu, determine the phasor values of
2V and
3V . Perform two iterations.
Figure 2
Y =
000
000
000
904090209020
902090609040
902090409060
Y
402020
206040
204060
and θ =
000
000
000
909090
909090
909090
In this problem
4QI
5PI
4PI
3
3
2
and unknown quantities =
3
3
2
V
δ
δ
With flat start0
101.0V ;
0
201.05V ;
0
301.0V
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ii
2
iiii
iiiiii
ii
2
iiii
BVQL
PM;PN
BVQH
)δδ(sinYVVM
)δδ(sinYVVN
)δδ(cosYVVH
i j ji ji ji
i j ji ji ji
i j ji ji ji
iP =
ii
2
iGV +
N
in
1n
iV
nV
niY cos (
niθ +
nδ -
iδ )
iQ =
ii
2
iBV
N
in
1n
iV
nV
niY sin (
niθ +
nδ -
iδ )
2P = )δδθ(cosYVV)δδθ(cosYVVGV
233232322112121222
2
2
= 0 + 0))90(cosx20x1x(1.05))90(cosx40x1x1.05(00
3P = )δδθ(cosYVV)δδθ(cosYVVGV
323232233113131333
2
3
= 0 + 0))90(cosx20x1.05x1())90(cosx20x1x(100
22
2
22BVQ
2V
1V )δδθ(sinY
211212)δδ(θsinYVV
232332
23
= 60)-x1.05x(1.05- ))(90sinx40x1x(1.050
))(90sinx20x1x(1.05-0 = 3.15
33
2
33BVQ
3V
1V )δδθ(sinY
311313)δδ(θsinYVV
3232323
2
= 40)-x1x(1- ))(90sinx20x1x(10
))(90sinx20x1.05x(1-0 = -1
404PPIΔP222
505PPIΔP333
31)(4QQIΔQ333
Sinceii
G are zero and ji
θ are0
90
22
2
2222BVQH -3.15 + ( 1.05 x 1.05 x 60 ) = 63
33
2
3333BVQH = 1 + ( 1 x 1 x 40 ) = 41
0PM;0PN333333
33
2
3333BVQL -1 + ( 1 x 1 x 40 ) = 39
32
H2
V3
V )δδ(cosY2332
= 211x20x1x1.05 and23
H = -21
32
N2
V3
V 0)δδ(sinY2332
; 32
M3
V2
V 0)δδ(sinY3223
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3900
04121
02163
3
3
3
2
V
VΔ
Δδ
Δδ
=
3
5
4
3
3
3
2
V
VΔ
Δδ
Δδ
=
0.0769-
0.1078-
0.0275
0)1(
21.5782rad.0.02750.02750δ
0)1(
36.179rad.0.10780.10780δ
0.92310.07691.0V)1(
3
0
101.0V
0
21.581.05V
0
36.180.9231V
This completes the first iteration.
After second iteration :
0)2(
21.61rad.0.02810.00060.0275δ
0)2(
3
6.898rad.0.12040.01260.1078δ
0.90560.01750.9231V)2(
3 pu
Thus at the end of second iteration :
0
101.0V
0
21.611.05V
0
36.90.9056V
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3. Referring to question 2 above, perform 2 iterations in computing :
a) Power flow using Decouple Power Flow method.
b) Power flow using Fast Decouple Power Flow method.
Figure 2
a) Power flow using Decouple Power Flow method.
4121
2163
3
2
Δδ
Δδ
=
5
4
0)1(
2
1.5756rad.0.02750.02750δ 0)1(
36.1765rad.0.10780.10780δ
0
101.0V
0
21.581.05V
0
36.181.0V
33
2
33BVQ
3V
1V )δδθ(sinY
311313)δδ(θsinYVV
3232323
2
= 40)-x1x(1- )))6.18(0(90insx20x1x(100
))6.18)(1.58(90insx20x1.05x(1- 00 = -0.6915
3.3085-0.6915)(4QQIΔQ333
33
2
3333BVQL -0.6915 - ( 1 x 1 x -40 ) = 39.3085
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ii
2
iiii
ii
2
iiii
BVQL
BVQH
)δδ(cosYVVHi j ji ji ji
39.3085
3
3
V
VΔ = 3.3085
3VΔ = - 3.3085 x 1.0 / 39.3085 = - 0.0842
0.91580.08421.0V)1(
3
After first iteration :
0
101.0V
0
21.581.05V
0
36.180.9158V
Second iteration:
2P = )δδθ(cosYVV)δδθ(cosYVVGV
233232322112121222
2
2
= 0 +
3.7548))90(cosx20x0.9158x(1.05))90(cosx40x1x1.05(00
000
58.118.658.1
3P = )δδθ(cosYVV)δδθ(cosYVVGV
323232233113131333
2
3
= 0 +
4.5685))90(cosx20x1.05x())90(cosx20x1x(0.915800
000
18.658.19158.018.6
22
2
22BVQ
2V
1V )δδθ(sinY
211212)δδ(θsinYVV
232332
23
=
))58.118.690sin(x20x9158.0x05.1())58.190sin(x40x1x05.1( 00000
60)-x1.05x(1.05-
33
2
33BVQ
3V
1V )δδθ(sinY
311313)δδ(θsinYVV
3232323
2
=
.658.190sin(x20x05.1x9158.0())18.690sin(x20x1x9158.0( 0000
40)-x0.9158x(0.9158-
= -3.7177
2452.07548.3 4PPIΔP222
4315.0)5685.4( 5PPIΔP333
0.2823)(4QQIΔQ333
7177.3
Sinceii
G are zero and ji
θ are0
90
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22
2
2222BVQH -5.1103 + ( 1.05 x 1.05 x 60 ) = 61.0397
33
2
3333BVQH = 3.7177 + ( 0.9158 x 0.9158 x 40 ) = 37.2653
32
H2
V3
V )δδ(cosY2332
= -1.05 x 0.9158 x 20 x cos (-6.180-1.58
0)
= -19.0557
23H = -19.0557
37.2653.0557
.055761.0397
19
19
3
2
Δδ
Δδ
=
0.4315
0.2452
3
2
Δδ
Δδ
=
0.0113
0.0005
0)2(
21.60rad.0.02800.00050.0275δ
0)2(
36.82rad.0.11910.01130.1078δ
0
101.0V
0
21.05V 6.1
0
30.9158V 85.6
33
2
33BVQ
3V
1V )δδθ(sinY
311313)δδ(θsinYVV
3232323
2
=
8.66.190sin(x20x05.1x9158.0())85.690sin(x20x1x9158.0(
0000
40)-x0.9158x(0.9158- = -3.6607
0.3393-.6607)(4QQIΔQ333
3
33
2
3333BVQL -3.6607 + ( 0.9158 x 0.9158 x 40 ) = 29.8869
29.8869
3
3
V
VΔ = 0.3393
3VΔ = - 0.3393 x 0.9158 / 29.8869 = - 0.0104
0.90540.01040V)1(
3 9158.
Thus at the end of second iteration, bus voltages are
0
101.0V
0
21.V 6.105
0
36.850.9054V
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B =
b) Power flow using Fast Decouple Power Flow method.
40
60
60
20203
20402
20401
321
The constant matrices are
40203
2602
32
0
Initial solution is
0
3
0
2
0
1
01.0V
01.05V
01.0V
4ΔP2 ; 5ΔP
3
5V
ΔP;3
V
ΔP
3
3
2
2 8095.
V
ΔPΔδB
'
40 20
2060
3
2
Δδ
Δδ
=
5
3.8095
On solving the above
3
2
Δδ
Δδ
=
0.1119
0.0262
0)1(
2rad.0.02620.02620δ 50.1
0)1(
341rad.0.11190.11190δ .6
0
3
0
2
0
1
411.0V
1.05V
01.0V
.6
50.1
'
B and "
B 40
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33
2
33BVQ
3V
1V )δδθ(sinY
311313)δδ(θsinYVV
3232323
2
= ))41.65.190sin(x20x05.1x1())41.690sin(x20x1x1( 00000
40)-x1x(1-
= -0.6752
3.3248-)(4QQIΔQ 333 6752.0
V
ΔQVΔB
"
40 3
VΔ = - 3.3248 i.e.3
VΔ = - 0.0831
0.91690.08311.0V)1(
3
At the end of first iteration, bus voltages are
0
3
0
2
0
1
410.9169V
1.05V
01.0V
.6
50.1
Second iteration:
2P = )δδθ(cosYVV)δδθ(cosYVVGV
233232322112121222
2
2
= 0 +
3.7492))90(cosx20x0.9169x(1.05))90(cosx40x1x1.05(00
000
50.141.65.1
3P = )δδθ(cosYVV)δδθ(cosYVVGV
323232233113131333
2
3
= 0 +
6971.441.65.19169.041.6 000
))90(cosx20x1.05x())90(cosx20x1x(0.916900
2508.07492.3 4ΔP2
.30295ΔP3
06971.4
3304.02389.0 3
3
2
2
V
ΔP
;V
ΔP
V
ΔPΔδB
'
40 20
2060
3
2
Δδ
Δδ
=
3304.0
0.2389
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3
2
Δδ
Δδ
=
0.0075
0.0015
0)1(
2rad.0.02770.001502620δ 59.1.
0)1(
3 6.84rad.0.11940.00750.1119δ
0
3
0
2
0
1
0.9169V
1.05V
01.0V
84.6
59.1
33
2
33BVQ
3V
1V )δδθ(sinY
311313)δδ(θsinYVV
3232323
2
=
.659.190sin(x20x05.1x9169.0())84.690sin(x20x1x9169.0( 0000
40)-x0.9169x(0.9169-
= -3.6261
0.3739-)(4QQIΔQ333
6261.3
V
ΔQVΔB
"
40 3
VΔ = - 0.4078 i.e.3
VΔ = - 0.0102
0.90670.01020.9169V)1(
3
At the end of second iteration, bus voltages are
0
3
0
2
0
1
0.9067V
1.05V
01.0V
84.6
59.1
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