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LOAD FLOW ANALYSIS
Newton-Raphson
1
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NEWTON-RAPHSON POWER FLOWSOLUTION
N-R method is mathematically superior to theGauss-Seidel.
Less prone to divergence with ill-conditionedproblems.
For large system it is more efficient andpractical.
2
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3
n
i j jiji V Y I
Expressing this equation in polar form gives
)( jijn
i j jiji V Y I
Complex power at bus i is
iiii I V jQ P *
)(1
jij j
n
jijiiii V Y V jQ P
Current entering bus i is given by
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4
)(1
jij j
n
jijiiii V Y V jQ P
Separating real and imaginary parts
)cos(1
jiijij j
n
jii Y V V P
)sin(1
jiijij j
n
jii Y V V Q
We have two equations for each load bus (P and Q)and one equation for each voltage controlled bus (P).
2 equationsper loadbus
1 equationper voltagecontrol bus
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5
)cos(1
jiijij j
n
jii Y V V P )sin(
1 jiijij j
n
jii Y V V Q
Expanding these equations in Taylors series and neglecting all higherorder terms results in the following set of linear equations .
)(
)(2
)(
)(
2
)(
2
)(
)(2
2
)(2
)(
2
)(
)(2
2
)(2
)(
2
)(
)(2
2
)(2
)(
2
)(
)(2
2
)(2
)(
)(2
)(
)(
2
k n
k
k n
k
n
k n
k n
n
k k n
k n
k n
n
k k
n
k n
k n
n
k k n
k n
k n
n
k k
k n
k
k n
k
V
V
V Q
V Q
V
Q
V
Q
V P
V P
V P
V P
P P
P P
Q
Q P
P
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7
)(
)(2
)(
)(2
)(
2
)(
)(2
2
)(2
)(
2
)(
)(2
2
)(2
)(
2
)(
)(2
2
)(2
)(
2
)(
)(2
2
)(2
)(
)(2
)(
)(2
k n
k
k n
k
n
k n
k n
n
k k n
k
n
k
n
n
k k
n
k n
k n
n
k k n
k n
k n
n
k k
k n
k
k n
k
V
V
V Q
V Q
V Q
V Q
V P
V P
V P
V P
P P
P P
Q
Q P
P
The Jacobian matrix gives the linearized relationship between smallchanges in voltage angle i(k) and voltage magnitude Vi(k) withthe small changes in real and reactive power P i(k) and Q i(k) .
Elements of the Jacobian matrix are the partial derivatives
of P and Q, evaluated at i(k)
and Vi(k)
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8
Elements of the Jacobian matrix are the partial derivatives of the P iand Q i equations, evaluated at i(k) and voltage magnitude Vi(k) .
)cos(1
jiijij jn
jii Y V V P
)sin(1
jiijij j
n
jii Y V V Q
In short form
V J J
J J
Q
P
43
21
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9
V J J J J
Q P
43
21
For voltage-controlled buses the voltage magnitudes are known
if m buses are voltage-
controlled , m equationsinvolving Q and V andcorresponding columnsof the Jacobian matrixare eliminated
n-1 real power constraints
and n-1-m reactive powercontraints
Jacobian matrix is of order(2n-2-m)(2n-2-m)
J2 =(n-1)(n-1-m)J1= (n-1)(n-1)
J3 =(n-1-m)(n-1) J4 =(n-1-m)(n-1-m)
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10
Diagonal and off-diagonal elements of J 1:
)sin( jiijij jn
i ji
i
i Y V V P
i jY V V P
jiijij ji j
i )sin(
)cos(1
jiijij j
n
jii Y V V P
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11
Diagonal and off-diagonal elements of J 2:
)cos(cos2 jiijijn
i j jiiiii
i
i
Y V Y V V P
i jY V V
P jiijiji
j
i )cos(
)cos(1
jiijij j
n
jii Y V V P
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12
Diagonal and off-diagonal elements of J 3:
)cos( jiijij jn
i ji
i
i Y V V Q
i jY V V Q
jiijij ji j
i )cos(
)sin(1
jiijij j
n
jii Y V V Q
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13
Diagonal and off-diagonal elements of J 4:
)sin(sin2 jiijijn
i j jiiiii
i
i Y V Y V V Q
i jY V V Q jiijiji
j
i )sin(
)sin(1
jiijij j
n
jii Y V V Q
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14
The terms Pi(k) and Q i(k) are the difference betweenthe scheduled and calculated values (power residuals)
k ii
k i P P P
k ii
k i QQQ
The new estimates for bus voltages
k i
k i
k i
1
k i
k i
k i V V V
1
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Procedure for power flow solution byN-R method
For load buses: Pi and Q i are specified Set voltage magnitude and equal to slack bus values
15
0.1)0(iV 0.0)0(
i
Calculate P i(k), Q i(k) , Pi(k) and Q i(k) using the equations:
k ii
k i P P P )cos(
1 jiijij j
n
jii Y V V P
k ii
k i QQQ )sin(
1 jiijij j
n
jii Y V V Q
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Procedure for power flow solution byN-R method
For voltage-controlled buses: Calculate P i(k) and Pi(k) using
16
k ii
k i P P P
)cos(1
jiijij j
n
jii Y V V P
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17
The Jacobian matrix (J 1, J2, J3 and J 4)
)sin(1
jiijij j
n
ji
i
i Y V V P
i jY V V P
jiijij ji j
i )sin(
J1
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18
The Jacobian matrix (J 1, J2, J3 and J 4)
)sin(1
jiijij j
n
ji
i
i Y V V P
i jY V V P
jiijij ji j
i )sin(
)cos(cos21
jiijijij
n
j jiiiii
i
i Y V V Y V V P
i jY V V P
jiijiji j
i )cos(
J1
J2
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19
Calculate the Jacobian matrix (J 3 and J 4)
)sin(sin21
jiijij
n
j jiiiii
i
i Y V Y V V Q
i jY V V Q jiijiji
j
i )sin(
)cos(1
jiijij j
n
ji
i
i Y V V Q
i jY V Q
jiijiji j
i )cos(
J3
J4
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20
V J J
J J
Q
P
43
21
Solve the linear simultaneous equation
Compute new voltage magnitudes and phase angles
k i
k i
k i
1 k i
k i
k i V V V
1
Continue until the residuals P i(k) and Q i(k) are less than the
specified accuracy
)(
)(
k i
k i
Q
P
Q
P
J J
J J V
1
43
21
)()(
)()(
k i
schi
k i
k i
schi
k i
QQQ
P P P
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Consider a three bus power system1
2
3
200MW
bus to bus Series Z (R+jX) (pu) Series Y (G-jB) (pu)
1-2 0.02+j0.04 10-j20
1-3 0.01+j0.03 10-j30
2-3 0.0125+j0.025 16-j32
Line data Base 100 MVA
21
400 MW
250 MvarSlack busV1 = 1.05 0
V3 = 1.04
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12
3
200MW
22
400 MW
250 MvarSlack bus
V1 = 1.05 0
V3 = 1.04
10-j20
10-j30 16-j32
622632163010
321652262010
301020105020
j j j
j j j
j j j
Y bus Convert to polar
4902.6723095.675651.11677709.354349.10862278.31
5651.11677709.354349.6313777.585651.11636068.224349.10862278.315651.11636068.221986.6885165.53
bu sY
Ybus matrix
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1 2
3
200 MW
400 MW
250 MvarSlack busV1 = 1.05 0
V3 = 1.04
10-j20
10-j30 16-j32
)cos(1 jiijij j
n
jii Y V V P
)cos( 2223
122 j j j j
j
Y V V P
)cos()cos(
)cos(
3223233222222222
122121122
Y V V Y V V
Y V V P
For bus 2
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)cos(1
jiijij j
n
jii Y V V P
)cos( 3333
133 j j j j
j
Y V V P
)cos()cos(
)cos(
3333333323323223
133131133
Y V V Y V V
Y V V P
For bus 3
1 2
3
200 MW
400 MW
250 MvarSlack busV1 = 1.05 0
V3 = 1.04
10-j20
10-j30 16-j32
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26
Determine elements of Jacobian matrix
)cos()cos(
)cos(
3223233222222222
122121122
Y V V Y V V
Y V V P
)cos(cos
)cos(
3223233222222
2
122121122
Y V V Y V
Y V V P
)sin()sin( 32232332122121122
2
Y V V Y V V P
)sin( 322323323
2
Y V V P
)cos(cos2
)cos(
322323322222
12212112
2
Y V Y V
Y V V P
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27
Determine elements of Jacobian matrix
)cos()cos(
)cos(
3333333323323223
133121133
Y V V Y V V
Y V V P
33332323323223133131133
coscos
)cos(
Y V Y V V
Y V V P
)sin( 233232232
3
Y V V P
)sin()sin( 23323223133131133
3
Y V V Y V V P
)cos( 23323232
3 Y V V P
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)sin()sin(
)sin(
3223233222222222
122121122
Y V V Y V V
Y V V Q
)sin(sin
)sin(
3223233222222
2
122121122
Y V V Y V
Y V V Q
)cos()cos( 32232332122121122
2
Y V V Y V V Q
)cos( 322323323
2
Y V V Q
)sin(sin2
)sin(
322323322222
12212112
2
Y V Y V
Y V V Q
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1 2
3200MW
29
400 MW
250 MvarSlack busV1 = 1.05 0
V3 = 1.04
10-j20
10-j30 16-j32
5.20.4100
2504002 j jS 0.2
100200
3 P
0.102 V 0.002 0.003
Initial values
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1 2
32.00
30
4 pu
2.5 puSlack busV1 = 1.05 0 V3 = 1.04 0.102 V
0.002 0.003
4902.6723095.675651.11677709.354349.10862278.31
5651.11677709.354349.6313777.585651.11636068.22
4349.10862278.315651.11636068.221986.6885165.53
bu sY
14.1
)cos(cos
)cos(
03
0223233
022222
202
10
2212110
20
2
Y V V Y V
Y V V P
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1 2
3
2.00
31
4 pu
2.5 puSlack busV1 = 1.05 0 V3 = 1.04 0.102 V
0.002 0.003
14.102 P 8600.2)14.1(0.40220
2 P P P
5616.003 P 4384.1)5616.0(0.20
330
3 P P P
28.202 Q 2200.0)28.2(5.2
022
02 QQQ
V J J
J J
Q
P
43
21
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32
Evaluating the Jacobian MatrixV J J
J J
Q
P
43
21
)sin()sin( 32232332122121122
2
Y V V Y V V P
4902.6723095.675651.11677709.354349.10862278.31
5651.11677709.354349.6313777.585651.11636068.22
4349.10862278.315651.11636068.221986.6885165.53
bu sY
V1 = 1.05 0
V3 = 1.04
0.102 V
0.00
2
0.00
3
1.0
1.0528.54
0
2
2
P
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33
Evaluating the Jacobian MatrixV J J
J J
Q
P
43
21
28.33)sin( 322323323
2
Y V V P
86.24
)cos(cos2
)cos(
322323322222
12212112
2
Y V Y V
Y V V P
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34
Evaluating the Jacobian MatrixV J J
J J
Q
P
43
21
28.33)sin( 233232232
3
Y V V P
04.66
)sin()sin( 23323223133131133
3
Y V V Y V V P
16)cos( 23323232
3 Y V V P
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35
Evaluating the Jacobian MatrixV J J
J J
Q
P
43
21
14.27
)cos()cos( 32232332122121122
2
Y V V Y V V Q
04.66)cos( 322323323
2
Y V V Q
72.49
)sin(sin2
)sin(
322323322222
12212112
2
Y V Y V
Y V V Q
JJJJP
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36
72.4964.1614.27
64.1604.6628.33
86.2428.3328.54
43
21
J J
J J
28.542
2
P 28.33
3
2
P 86.24
2
2
V P
28.332
3
P 04.66
3
3
P 162
3
V P
14.272
2
Q04.66
3
2
Q 72.492
2
V Q
?43
21
J J
J J V J J
J J
Q
P
43
21
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37
V J J
J J
Q
P
43
21
Solve the linear simultaneous equation
Compute new voltage magnitudes and phase angles
k i
k i
k i
1 k i
k i
k i V V V
1
Continue until the residuals P i(k) and Q i(k) are less than thespecified accuracy
)(
)(
k i
k i
Q
P
Q
P
J J
J J V
1
43
21
)()(
)()(
k i
schi
k i
k i
schi
k i
QQQ
P P P
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38
V J J
J J
Q
P
43
21
02
03
02
72.4964.1614.27
64.1604.6628.33
86.2428.3328.54
22.0
4384.1
86.2
V
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39
026548.0
44221.0
5934.2
0
2
03
02
V
The new bus voltages in the first iteration
97345.0)026548.0(1
44221.0)44221.0(0
5934.2)5934.2(0
12
13
12
V
k i
k i
k i
1
k i
k
i
k
i V V V 1
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40
For the second iteration
12
13
12
103589.48402838.17538577.28
379086.15656383.65981642.32
302567.21765618.31724675.51
050914.0
021715.0
099218.0
V
001767.0
05641.0
10023.0
02
13
12
V
The new bus voltages in the 2nd iteration
971684.0)001767.0(97345.0
49862.0)05641.0(44221.069363.2)10023.0(5934.2
22
23
2
2
V
h h d
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41
For the third iteration
22
23
22
954870.47396932.17548205.28
351628.15597585.65933865.32
1474477.21693866.31596701.51
000143.0
000038.0
000216.0
V
0000044.0
00013751.0
00217724.0
22
23
22
V
The new bus voltages in the 3rd iteration
97168.0
49876.06963.2
32
33
3
2
V
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42
The solution is assumed to converge after 3 iterationswith a maximum power mismatched of 2.510 -4
4988.004.1
696.297168.0
3
2
V
V
puQ
pu P puQ
4085.1
1842.24617.1
1
1
3
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Exercise
In a two bus system, bus 1 is a slack bus with V 1=1.0 0pu. A load of 100 MW and 50 Mvar is taken from bus 2.
The line impedance is z 12 = 0.12+j0.16 pu on a base of100 MVA.
Using Gauss-Seidel method determine V 2. Continueuntil converge. How many iteration?.
Repeat using Newton-Raphson. Perform untilconvergence. How many iteration?
Discuss/compare between the two solutions. Explain how S 1 and real and reactive power loss in the
line can be calculated.
43
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Solution with N-R
44
V1=1.0 0
12
50 Mvar
100 MW
z12 = 0.12+j0.16
y12 = 3-j4 y12 = ?
Ybus = ?
13.53587.1265
87.126513.535busY
V1 1 0 0
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45
V1=1.0 0
1 2
50 Mvar
100 MWy12 = 3-j4
Power flow equations in polar forms:
)cos(1
jiijij j
n
jii Y V V P
)sin(1
jiijij j
n
jii Y V V Q
V1=1 0 0
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46
V1=1.0 0
1 2
50 Mvar
100 MWy12 = 3-j4
)cos(1
jiijij j
n
jii Y V V P
13.53587.1265
87.126513.535busY
)13.53cos(5)87.126cos(5
)13.53cos(5)87.126cos(5
)cos()cos(
)cos(
2
21212
221212
2222222212212112
2221
22
V V V
V V V V
Y V V Y V V
Y V V P j j j jn
j
At bus 2
V1=1 0 0
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47
V1=1.0 0
1 2
50 Mvar
100 MWy12 = 3-j4
)sin(1 jiijij j
n
jii Y V V Q
)13.53sin(5)87.126sin(5
)13.53cos(5)87.126sin(5
)sin()sin(
)sin(
2
21212
221212
2222222212212112
2221
22
V V V
V V V V
Y V V Y V V
Y V V Q j j j jn
j
13.53587.1265
87.126513.535busY
At bus 2
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48
Partial derivatives of P 2 w.r.t. 2 and V2
)13.53cos(5)87.126cos(5 2
212122 V V V P
)87.126sin(5 12122
2
V V P
)13.53cos(52)87.126cos(5 21212
2 V V V P
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49
Partial derivatives Q 2 w.r.t. 2 and V2
)13.53sin(5)87.126sin(5 2
212122 V V V Q
)87.126cos(5 12122
2
V V Q
)13.53sin(10)87.126sin(5 21212
2 V V V Q
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50
Expressed Load in p.u. 5.00.1100
501002 j
jS
0.102 V 0.002 Slack bus voltage is V 1 = 1.0 0 pu.Initial estimates :
033
)13.53cos(510087.126cos511
)13.53cos(5)87.126cos(5
2
2021
021
02
02 V V V P
100.10220
2 P P P
Initial values?
P2 = ?; Q 2 =?
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51
Load 5.00.12 jS
0.102 V 0.002 Slack bus voltage is V 1 = 1.0 0 pu.Initial estimates :
044
)13.53sin(510087.126sin511
)13.53sin(5)87.126sin(5
2
20
21
0
21
0
2
0
2 V V V Q
5.0050.00
22
0
2
QQQ
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52
The elements of Jacobian matrix
)87.126sin(5 12122
2 V V P
)13.53cos(52)87.126cos(5 21212
2 V V V P
4)87.126sin(5112
201
P J
3)13.53cos(512)87.126cos(152
22
V P J
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53
)87.126cos(5 12122
2
V V Q
)13.53sin(10)87.126sin(5 21212
2 V V V Q
3)87.126cos(5112
203
Q J
4)13.53sin(110)87.126sin(152
204 V
Q J
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54
The set of linear equations in the first iteration becomes
02
02
43
34
5.0
0.1
V
02
02
1
5.0
0.1
43
34V
Solution of the matrix gives:
2.0
10.00
2
02
V
8.0)2.0(1
7296.510.0)10.0(01
2
12
V
radian
16.012.0
12.016.0
43
34 1
F h d i i
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55
7875.092.17075.2
)13.53cos(58.0)7296.5(87.126cos518.0
)13.53cos(5)87.126cos(5
2
2121
121
12
12 V V V P
2125.0)7875.0(0.11221
2 P P P
For the second iteration
3844.056.29444.2
)13.53sin(58.0)7296.5(87.126sin518.0
)13.53sin(5)87.126sin(5
2
2121
121
12
12 V V V Q
1156.0)3844.0(50.012212
QQQ
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56
The set of linear equations in the 2nd iteration becomes
12
12
7195.27075.24157.19444.2
1156.02125.0
V
Solution of the matrix gives:
0773.0
0350.01
2
12
V
7227.0)0773.0(8.0
135.0)0350.0(10.02
2
22
V
radian
CONTINUE WITH THE 3RD ITERATION!Good Luck
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Example of Exam Question
a) In the Gauss Seidel method there is termcalled acceleration factor. Explainacceleration factor briefly.
(5 marks)
57
E l f E Q i
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Example of Exam Questionb) For the power system shown, the bus admittance matrix Y bus
in per unit is given by:
58
5.45.15.15.031
5.15.05.75.262316293
j j j
j j j j j j
For each bus k ,specify the bustype, anddeterminewhich of thevariables V k , k ,P k and Q k are
input data andwhich areunknowns
(2 marks)
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The mismatch
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equation invector-matrixform of the
Newton-Raphsonpower flowmethod
(3 marks)
V J J J J
Q P
43
21 General equation
The mismatch
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equation invector-matrixform of the
Newton-Raphsonpower flowmethod
(3 marks)
2
3
2
2
2
3
2
2
2
2
3
3
3
2
3
2
2
3
2
2
2
2
3
2
V
V QQQV P P P V P P P
Q P
P
E l f E Q ti
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Example of Exam Question
62
5.45.15.15.0315.15.05.75.262
316293
j j j j j j
j j jAssume an initial
estimate of voltageV 2 = 1.0/0 and therotor angle 3 = 0 ,calculate the busreal and reactivepower mismatchesto be used in thefirst iteration of theNewton-Raphson
power flowmethod.
(9 marks)
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V J J
J J
Q
P
43
21
Solve the linear simultaneous equation
Compute new voltage magnitudes and phase angles
k i
k i
k i
1 k i
k i
k i V V V
1
Continue until the residuals P i(k) and Q i(k) are less than thespecified accuracy
)(
)(
k i
k i
Q
P
Q
P
J J
J J V
1
43
21
)()(
)()(
k i
schi
k i
k i
schi
k i
QQQ
P P P
JJP
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V J J
J J
Q
P
43
21
2
3
2
2
2
3
2
2
2
2
3
2
2
3
3
2
3
3
2
2
2
2
3
2
V
V QQQV P
V P
P P
P P
Q P
P
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5.45.15.15.031
5.15.05.75.262
316293
j j j
j j j
j j j
Y bus
565.717134.4435.1085811.1435.1081623.3
435.1085811.1565.719057.7435.1083246.6435.1081623.3435.1083246.6565.714868.9
busY
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)cos(1
jiijij j
n
jii Y V V P
)cos( 2223
122 j j j j
j
Y V V P
)cos()cos(
)cos(
3223233222222222
122121122
Y V V Y V V
Y V V P
For bus 2
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)cos()cos(
)cos(
3223233222222222
122121122
Y V V Y V V
Y V V P
565.717134.4435.1085811.1435.1081623.3
435.1085811.1565.719057.7435.1083246.6
435.1081623.3435.1083246.6565.714868.9
bu sY
)435.108cos(5811.1
)565.71cos(9057.7)435.108cos(3246.6
3232
2212122
V V
V V V P
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)cos()cos(
)cos(
3333333323323223
133131133
Y V V Y V V
Y V V P
565.717134.4435.1085811.1435.1081623.3
435.1085811.1565.719057.7435.1083246.6435.1081623.3435.1083246.6565.714868.9
bu sY
)565.71cos(7134.4
)435.108cos(5811.1)435.108cos(1623.32
3
2313133
V
V V V V P
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)sin()sin(
)sin(
3223233222222222
122121122
Y V V Y V V
Y V V Q
565.717134.4435.1085811.1435.1081623.3
435.1085811.1565.719057.7435.1083246.6435.1081623.3435.1083246.6565.714868.9
bu sY
)435.108sin(5811.1
)565.71sin(9057.7)435.108sin(3246.6
3232
2212122
V V
V V V Q
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)435.108cos(5811.1
)565.71cos(9057.7)435.108cos(3246.6
3232
2212122
V V
V V V P
002.100.1
005.1
23
22
11
V V
V
11.0)00435.108cos(5811.102.10.1
)565.71cos(9057.71)00435.108cos(3246.605.10.1 202 P
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)565.71cos(7134.4)435.108cos(5811.1)435.108cos(1623.3
23
2313133
V V V V V P
002.100.1
005.1
23
22
11
V V
V
02.0
)565.71cos(7134.402.1
)435.108cos(5811.1102.1)435.108cos(1623.305.102.1
)565.71cos(7134.4
)435.108cos(5811.1)435.108cos(1623.3
2
23
0231313
03
V
V V V V P
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)435.108sin(5811.1
)565.71sin(9057.7)435.108sin(3246.6
3232
2212122
V V
V V V Q
002.100.1
005.1
23
22
11
V V
V
33.0
)435.108sin(5811.102.10.1)565.71sin(9057.71)435.108sin(3246.605.10.1
2
2Q
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)()(
)()(
k i
schi
k i
k i
schi
k i
QQQ
P P P
67.0)33.0(0.1
62.0)02.0(6.0
39.1)11.0(5.1
)0(22
)0(2
)0(22
)0(3
)0(22
)0(2
QQQ
P P P
P P P
sch
sch
sch
Additional questions
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Additional questions
74
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j j j j j j
j j jAssume an initialestimate of voltageV 2 = 1.0/0 and therotor angle 3 = 0 ,calculate JacobianMatrix to be used inthe first iteration ofthe Newton-Raphson powerflow method.
(9 marks)
A four bus power system data is shown in Tables Q1a and Q1b. Form a busd itt t i Y d d t i th lt t th d f fi t
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admittance matrix, Y bus , and determine the voltages at the end of firstiteration using Gauss-Seidel method. The reactive power constraint ongenerator 2 is 0.2 Q 2 1.0. Use flat start estimates to start calculation.
(16 marks)
Lines Admittance (per unit)1 2 2 j8.01 3 1 j4.02 3 0.666 j2.6642 4 1 j4.03 - 4 2 j8.0
Bus Generation Load Bus Voltage Remarks
P Q P Q V 1 - - 0.2 0.1 1.06 0 Slack
2 0.5 - - - 1.04 - PV
3 - - 0.4 0.3 - - PQ
4 - - 0.3 0.1 - - PQ
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12382410
82664.14666.3664.2666.041
41664.2666.0664.14666.382
04182123
j j j
j j j j
j j j j
j j j
Y bu s
The admittance bus matrix
Lines Admittance (per unit)1 2 2 j8.01 3 1 j4.0
2 3 0.666 j2.6642 4 1 j4.03 - 4 2 j8.0
Wh i h f B 2? Wh i k ?
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Bus 2 is PV busKnown P and V2 Unknown Q 2 and 2
1108.0
424323222121*0
20
2 V Y V Y V Y V Y V Q
What is the type of Bus 2? What is known?What is unknown?
Write the equation for Q 2
Whats next?
The value of Q2 is less than the minimum specified
Reactive power is fixed at 0.2 (lower limit)
