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Lecture 13Power Flow
Professor Tom OverbyeDepartment of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS
2
Announcements
Be reading Chapter 6, also Chapter 2.4 (Network Equations). HW 5 is 2.38, 6.9, 6.18, 6.30, 6.34, 6.38; do by October 6 but
does not need to be turned in. First exam is October 11 during class. Closed book, closed
notes, one note sheet and calculators allowed. Exam covers thru the end of lecture 13 (today)
An example previous exam (2008) is posted. Note this is exam was given earlier in the semester in 2008 so it did not include power flow, but the 2011 exam will (at least partially)
3
Multi-Variable Example
1
2
2 21 1 2
2 22 1 2 1 2
1 1
1 2
2 2
1 2
xSolve for = such that ( ) 0 where
x
f ( ) 2 8 0
f ( ) 4 0
First symbolically determine the Jacobian
f ( ) f ( )
( ) =f ( ) f ( )
x x
x x x x
x x
x x
x f x
x
x
x x
J xx x
4
Multi-variable Example, cont’d
1 2
1 2 1 2
11 1 2 1
2 1 2 1 2 2
(0)
1(1)
4 2( ) =
2 2
Then
4 2 ( )
2 2 ( )
1Arbitrarily guess
1
1 4 2 5 2.1
1 3 1 3 1.3
x x
x x x x
x x x f
x x x x x f
J x
x
x
x
x
5
Multi-variable Example, cont’d
1(2)
(2)
2.1 8.40 2.60 2.51 1.8284
1.3 5.50 0.50 1.45 1.2122
Each iteration we check ( ) to see if it is below our
specified tolerance
0.1556( )
0.0900
If = 0.2 then we wou
x
f x
f x
ld be done. Otherwise we'd
continue iterating.
6
Possible EHV Overlays for Wind
AEP 2007 Proposed Overlay
7
NR Application to Power Flow
** * *
i1 1
We first need to rewrite complex power equations
as equations with real coefficients
S
These can be derived by defining
Recal
i
n n
i i i ik k i ik kk k
ik ik ik
ji i i i
ik i k
V I V Y V V Y V
Y G jB
V V e V
jl e cos sinj
8
Real Power Balance Equations
* *i
1 1
1
i1
i1
S ( )
(cos sin )( )
Resolving into the real and imaginary parts
P ( cos sin )
Q ( sin cos
ikn n
ji i i ik k i k ik ik
k k
n
i k ik ik ik ikk
n
i k ik ik ik ik Gi Dik
n
i k ik ik ik ik
P jQ V Y V V V e G jB
V V j G jB
V V G B P P
V V G B
)k Gi DiQ Q
9
Newton-Raphson Power Flow
i1
In the Newton-Raphson power flow we use Newton's
method to determine the voltage magnitude and angle
at each bus in the power system.
We need to solve the power balance equations
P ( cosn
i k ik ikk
V V G
i1
sin )
Q ( sin cos )
ik ik Gi Di
n
i k ik ik ik ik Gi Dik
B P P
V V G B Q Q
10
Power Flow Variables
2 2 2
n
2
Assume the slack bus is the first bus (with a fixed
voltage angle/magnitude). We then need to determine
the voltage angle/magnitude at the other buses.
( )
( )
G
n
P P
V
V
x
x f x
2
2 2 2
( )
( )
( )
D
n Gn Dn
G D
n Gn Dn
P
P P P
Q Q Q
Q Q Q
x
x
x
11
N-R Power Flow Solution
( )
( )
( 1) ( ) ( ) 1 ( )
The power flow is solved using the same procedure
discussed last time:
Set 0; make an initial guess of ,
While ( ) Do
( ) ( )
1
End While
v
v
v v v v
v
v v
x x
f x
x x J x f x
12
Power Flow Jacobian Matrix
1 1 1
1 2
2 2 2
1 2
1 2
The most difficult part of the algorithm is determining
and inverting the n by n Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
n
n
n n n
n
f f fx x x
f f fx x x
f f fx x x
J x
x x x
x x x
J x
x x x
13
Power Flow Jacobian Matrix, cont’d
i
i
i1
Jacobian elements are calculated by differentiating
each function, f ( ), with respect to each variable.
For example, if f ( ) is the bus i real power equation
f ( ) ( cos sin )n
i k ik ik ik ik Gik
x V V G B P P
x
x
i
1
i
f ( )( sin cos )
f ( )( sin cos ) ( )
Di
n
i k ik ik ik iki k
k i
i j ik ik ik ikj
xV V G B
xV V G B j i
14
Two Bus Newton-Raphson Example
Line Z = 0.1j
One Two 1.000 pu 1.000 pu
200 MW 100 MVR
0 MW 0 MVR
For the two bus power system shown below, use the Newton-Raphson power flow to determine the voltage magnitude and angle at bus two. Assumethat bus one is the slack and SBase = 100 MVA.
2
2
10 10
10 10busj j
V j j
x Y
15
Two Bus Example, cont’d
i1
i1
2 1 2
22 1 2 2
General power balance equations
P ( cos sin )
Q ( sin cos )
Bus two power balance equations
(10sin ) 2.0 0
( 10cos ) (10) 1.0 0
n
i k ik ik ik ik Gi Dik
n
i k ik ik ik ik Gi Dik
V V G B P P
V V G B Q Q
V V
V V V
16
Two Bus Example, cont’d
2 2 2
22 2 2 2
2 2
2 2
2 2
2 2
2 2 2
2 2 2 2
P ( ) (10sin ) 2.0 0
( ) ( 10cos ) (10) 1.0 0
Now calculate the power flow Jacobian
P ( ) P ( )
( )Q ( ) Q ( )
10 cos 10sin
10 sin 10cos 20
V
Q V V
VJ
V
V
V V
x
x
x x
xx x
17
Two Bus Example, First Iteration
(0)
2 2(0)2
2 2 2
2 2 2(0)
2 2 2 2
(1)
0Set 0, guess
1
Calculate
(10sin ) 2.0 2.0f( )
1.0( 10cos ) (10) 1.0
10 cos 10sin 10 0( )
10 sin 10cos 20 0 10
0 10 0Solve
1 0 10
v
V
V V
V
V V
x
x
J x
x1 2.0 0.2
1.0 0.9
18
Two Bus Example, Next Iterations
(1)2
(1)
1(2)
0.9(10sin( 0.2)) 2.0 0.212f( )
0.2790.9( 10cos( 0.2)) 0.9 10 1.0
8.82 1.986( )
1.788 8.199
0.2 8.82 1.986 0.212 0.233
0.9 1.788 8.199 0.279 0.8586
f(
x
J x
x
(2) (3)
(3)2
0.0145 0.236)
0.0190 0.8554
0.0000906f( ) Done! V 0.8554 13.52
0.0001175
x x
x
19
Two Bus Solved Values
Line Z = 0.1j
One Two 1.000 pu 0.855 pu
200 MW 100 MVR
200.0 MW168.3 MVR
-13.522 Deg
200.0 MW 168.3 MVR
-200.0 MW-100.0 MVR
Once the voltage angle and magnitude at bus 2 are known we can calculate all the other system values,such as the line flows and the generator reactive power output
20
Two Bus Case Low Voltage Solution
(0)
2 2(0)2
2 2 2
This case actually has two solutions! The second
"low voltage" is found by using a low initial guess.
0Set 0, guess
0.25
Calculate
(10sin ) 2.0f( )
( 10cos ) (10) 1.0
v
V
V V
x
x
2 2 2(0)
2 2 2 2
2
0.875
10 cos 10sin 2.5 0( )
10 sin 10cos 20 0 5
V
V V
J x
21
Low Voltage Solution, cont'd
1(1)
(2) (2) (3)
0 2.5 0 2 0.8Solve
0.25 0 5 0.875 0.075
1.462 1.42 0.921( )
0.534 0.2336 0.220
x
f x x x
Line Z = 0.1j
One Two 1.000 pu 0.261 pu
200 MW 100 MVR
200.0 MW831.7 MVR
-49.914 Deg
200.0 MW 831.7 MVR
-200.0 MW-100.0 MVR
Low voltage solution
22
Two Bus Region of Convergence
Slide shows the region of convergence for different initialguesses of bus 2 angle (x-axis) and magnitude (y-axis)
Red regionconvergesto the highvoltage solution,while the yellow regionconvergesto the lowvoltage solution
23
PV Buses
Since the voltage magnitude at PV buses is fixed there is no need to explicitly include these voltages in x or write the reactive power balance equations– the reactive power output of the generator varies to
maintain the fixed terminal voltage (within limits)– optionally these variations/equations can be included by
just writing the explicit voltage constraint for the generator bus
|Vi | – Vi setpoint = 0
24
Three Bus PV Case Example
Line Z = 0.1j
Line Z = 0.1j Line Z = 0.1j
One Two 1.000 pu 0.941 pu
200 MW 100 MVR
170.0 MW 68.2 MVR
-7.469 Deg
Three 1.000 pu
30 MW 63 MVR
2 2 2 2
3 3 3 3
2 2 2
For this three bus case we have
( )
( ) ( ) 0
V ( )
G D
G D
D
P P P
P P P
Q Q
x
x f x x
x
25
400 MVA15 kV
400 MVA15/345 kV
T1
T2800 MVA345/15 kV
800 MVA15 kV
520 MVA
80 MW40 Mvar
280 Mvar 800 MW
Line 3 345 kV
Lin
e 2
Lin
e 1345 kV
100 mi345 kV 200 mi
50 mi
1 4 3
2
5
Single-line diagram
The N-R Power Flow: 5-bus Example
26
Bus Type
V
per unit
degrees
PG
per
unit
QG
per
unit
PL
per
unit
QL
per
unit
QGmax
per
unit
QGmin
per
unit
1 Swing 1.0 0 0 0
2 Load 0 0 8.0 2.8
3 Constant voltage
1.05 5.2 0.8 0.4 4.0 -2.8
4 Load 0 0 0 0
5 Load 0 0 0 0
Table 1. Bus input data
Bus-to-Bus
R’
per unit
X’
per unit
G’
per unit
B’
per unit
Maximum
MVA
per unit
2-4 0.0090 0.100 0 1.72 12.0
2-5 0.0045 0.050 0 0.88 12.0
4-5 0.00225 0.025 0 0.44 12.0
Table 2. Line input data
The N-R Power Flow: 5-bus Example
27
Bus-to-Bus
R
per
unit
X
per
unit
Gc
per
unit
Bm
per
unit
Maximum
MVA
per unit
Maximum
TAP
Setting
per unit
1-5 0.00150 0.02 0 0 6.0 —
3-4 0.00075 0.01 0 0 10.0 —
Table 3. Transformer input data
Bus Input Data Unknowns
1 V1 = 1.0, 1 = 0 P1, Q1
2 P2 = PG2-PL2 = -8
Q2 = QG2-QL2 = -2.8
V2, 2
3 V3 = 1.05
P3 = PG3-PL3 = 4.4
Q3, 3
4 P4 = 0, Q4 = 0 V4, 4
5 P5 = 0, Q5 = 0 V5, 5
Table 4. Input data and unknowns
The N-R Power Flow: 5-bus Example
28
Time to Close the Hood: Let the Computer Do the Math! (Ybus Shown)
29
Ybus Details
02321 YY
unitperjjjXR
Y 91964.989276.01.0009.0
11'24
'24
24
unitperjjjXR
Y 83932.1978552.105.00045.0
11'25
'25
25
22
11 '25
'24
'25
'25
'24
'24
22
Bj
Bj
jXRjXRY
2
88.0
2
72.1)83932.1978552.1()91964.989276.0( jjjj
unitperj 624.845847.284590.2867828.2
Elements of Ybus connected to bus 2
30
Here are the Initial Bus Mismatches
31
And the Initial Power Flow Jacobian
32
])0()0(cos[){0()()0( 211212122222 VYVPxPPP])0()0(cos[]cos[ 233232322222 VYVY
])0()0(cos[ 2442424 VY]})0()0(cos[ 2552525 VY
)624.84cos()0.1(5847.28{0.10.8 )143.95cos()0.1(95972.9 )}143.95cos()0.1(9159.19
unitper99972.7)1089.2(0.8 4
])0()0(sin[)0()0()0(1 2442424224 VYVJ
]143.95sin[)0.1)(95972.9)(0.1( unitper91964.9
And the Hand Calculation Details!
33
Five Bus Power System Solved
slack
One
Two
ThreeFourFiveA
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.000 pu 0.974 pu
0.834 pu
1.019 pu
1.050 pu 0.000 Deg -4.548 Deg
-22.406 Deg
-2.834 Deg
-0.597 Deg
395 MW
114 Mvar
520 MW
337 Mvar
800 MW 280 Mvar
80 MW 40 Mvar
34
37 Bus Example Design Case
slack
Metropolis Light and Power Electric Design Case 2SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
A
MVA
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHI MKO69
ROGER69
UI UC69
PETE69HI SKY69
TI M69
TI M138
TI M345
PAI 69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
J O138 J O345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.03 pu
1.02 pu
1.03 pu
1.03 pu
1.01 pu
1.00 pu1.01 pu
1.00 pu
1.02 pu
1.01 pu
1.00 pu
1.01 pu
1.01 pu
1.01 pu
1.01 pu
1.02 pu
1.00 pu
1.00 pu
1.02 pu
0.99 pu
0.99 pu
1.00 pu
1.02 pu
1.00 pu1.01 pu
1.01 pu
1.00 pu 1.00 pu
1.01 pu
1.02 pu 1.02 pu
1.02 pu 1.03 pu
A
MVA
1.02 pu
A
MVA
A
MVA
LYNN138
A
MVA
1.02 pu
A
MVA
1.00 pu
A
MVA
System Losses: 10.70 MW
220 MW 52 Mvar
12 MW 3 Mvar
20 MW 12 Mvar
124 MW 45 Mvar
37 MW
13 Mvar
12 MW 5 Mvar
150 MW 0 Mvar
56 MW
13 Mvar
15 MW 5 Mvar
14 MW
2 Mvar
38 MW 3 Mvar
45 MW 0 Mvar
25 MW 36 Mvar
36 MW 10 Mvar
10 MW 5 Mvar
22 MW 15 Mvar
60 MW 12 Mvar
20 MW 28 Mvar
23 MW 7 Mvar
33 MW 13 Mvar
15.9 Mvar 18 MW 5 Mvar
58 MW 40 Mvar
60 MW 19 Mvar
14.2 Mvar
25 MW 10 Mvar
20 MW 3 Mvar
23 MW 6 Mvar 14 MW
3 Mvar
4.9 Mvar
7.3 Mvar
12.8 Mvar
28.9 Mvar
7.4 Mvar
0.0 Mvar
55 MW 25 Mvar
39 MW 13 Mvar
150 MW 0 Mvar
17 MW 3 Mvar
16 MW -14 Mvar
14 MW 4 Mvar
KYLE69A
MVA
35
Good Power System Operation
• Good power system operation requires that there be no reliability violations for either the current condition or in the event of statistically likely contingencies• Reliability requires as a minimum that there be no transmission
line/transformer limit violations and that bus voltages be within acceptable limits (perhaps 0.95 to 1.08)
• Example contingencies are the loss of any single device. This is known as n-1 reliability.
• North American Electric Reliability Corporation now has legal authority to enforce reliability standards (and there are now lots of them). See http://www.nerc.com for details (click on Standards)
36
Looking at the Impact of Line Outages
slack
Metropolis Light and Power Electric Design Case 2SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
A
MVA
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHI MKO69
ROGER69
UI UC69
PETE69HI SKY69
TI M69
TI M138
TI M345
PAI 69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
J O138 J O345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.03 pu
1.02 pu
1.03 pu
1.03 pu
1.01 pu
1.00 pu1.01 pu
1.00 pu
1.02 pu
1.01 pu
1.00 pu
1.01 pu
1.01 pu
1.01 pu
1.01 pu
1.02 pu
1.01 pu
1.00 pu
1.02 pu
0.90 pu
0.90 pu
0.94 pu
1.01 pu
0.99 pu1.00 pu
1.00 pu
1.00 pu 1.00 pu
1.01 pu
1.01 pu 1.02 pu
1.02 pu 1.03 pu
A
MVA
1.02 pu
A
MVA
A
MVA
LYNN138
A
MVA
1.02 pu
A
MVA
1.00 pu
A
MVA
System Losses: 17.61 MW
227 MW 43 Mvar
12 MW 3 Mvar
20 MW 12 Mvar
124 MW 45 Mvar
37 MW
13 Mvar
12 MW 5 Mvar
150 MW 4 Mvar
56 MW
13 Mvar
15 MW 5 Mvar
14 MW
2 Mvar
38 MW 9 Mvar
45 MW 0 Mvar
25 MW 36 Mvar
36 MW 10 Mvar
10 MW 5 Mvar
22 MW 15 Mvar
60 MW 12 Mvar
20 MW 40 Mvar
23 MW 7 Mvar
33 MW 13 Mvar
16.0 Mvar 18 MW 5 Mvar
58 MW 40 Mvar
60 MW 19 Mvar
11.6 Mvar
25 MW 10 Mvar
20 MW 3 Mvar
23 MW 6 Mvar 14 MW
3 Mvar
4.9 Mvar
7.2 Mvar
12.8 Mvar
28.9 Mvar
7.3 Mvar
0.0 Mvar
55 MW 32 Mvar
39 MW 13 Mvar
150 MW 4 Mvar
17 MW 3 Mvar
16 MW -14 Mvar
14 MW 4 Mvar
KYLE69A
MVA
80%A
MVA
135%A
MVA
110%A
MVA
Opening one line (Tim69-Hannah69) causes an overload. This would not be allowed
37
Contingency Analysis
Contingencyanalysis providesan automaticway of lookingat all the statisticallylikely contingencies. Inthis example thecontingency setIs all the single line/transformeroutages
38
Power Flow And Design
• One common usage of the power flow is to determine how the system should be modified to remove contingencies problems or serve new load• In an operational context this requires working with the
existing electric grid• In a planning context additions to the grid can be considered
• In the next example we look at how to remove the existing contingency violations while serving new load.
39
An Unreliable Solution
slack
Metropolis Light and Power Electric Design Case 2SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
A
MVA
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHI MKO69
ROGER69
UI UC69
PETE69HI SKY69
TI M69
TI M138
TI M345
PAI 69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
J O138 J O345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.02 pu
1.01 pu
1.02 pu
1.03 pu
1.01 pu
1.00 pu1.01 pu
1.00 pu
1.02 pu
1.01 pu
1.00 pu
1.01 pu
1.01 pu
1.01 pu
1.01 pu
1.02 pu
0.99 pu
1.00 pu
1.02 pu
0.97 pu
0.97 pu
0.99 pu
1.02 pu
1.00 pu1.01 pu
1.01 pu
1.00 pu 1.00 pu
1.01 pu
1.02 pu 1.02 pu
1.02 pu 1.03 pu
A
MVA
1.02 pu
A
MVA
A
MVA
LYNN138
A
MVA
1.02 pu
A
MVA
1.00 pu
A
MVA
System Losses: 14.49 MW
269 MW 67 Mvar
12 MW 3 Mvar
20 MW 12 Mvar
124 MW 45 Mvar
37 MW
13 Mvar
12 MW 5 Mvar
150 MW 1 Mvar
56 MW
13 Mvar
15 MW 5 Mvar
14 MW
2 Mvar
38 MW 4 Mvar
45 MW 0 Mvar
25 MW 36 Mvar
36 MW 10 Mvar
10 MW 5 Mvar
22 MW 15 Mvar
60 MW 12 Mvar
20 MW 40 Mvar
23 MW 7 Mvar
33 MW 13 Mvar
15.9 Mvar 18 MW 5 Mvar
58 MW 40 Mvar
60 MW 19 Mvar
13.6 Mvar
25 MW 10 Mvar
20 MW 3 Mvar
23 MW 6 Mvar 14 MW
3 Mvar
4.9 Mvar
7.3 Mvar
12.8 Mvar
28.9 Mvar
7.4 Mvar
0.0 Mvar
55 MW 28 Mvar
39 MW 13 Mvar
150 MW 1 Mvar
17 MW 3 Mvar
16 MW -14 Mvar
14 MW 4 Mvar
KYLE69A
MVA
96%A
MVA
Case now has nine separate contingencies with reliability violations
40
A Reliable Solution
slack
Metropolis Light and Power Electric Design Case 2SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
A
MVA
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHI MKO69
ROGER69
UI UC69
PETE69HI SKY69
TI M69
TI M138
TI M345
PAI 69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
J O138 J O345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.03 pu
1.01 pu
1.02 pu
1.03 pu
1.01 pu
1.00 pu1.01 pu
1.00 pu
1.02 pu
1.01 pu
1.00 pu
1.01 pu
1.01 pu
1.01 pu
1.01 pu
1.02 pu
1.00 pu
0.99 pu
1.02 pu
0.99 pu
0.99 pu
1.00 pu
1.02 pu
1.00 pu1.01 pu
1.01 pu
1.00 pu 1.00 pu
1.01 pu
1.02 pu 1.02 pu
1.02 pu 1.03 pu
A
MVA
1.02 pu
A
MVA
A
MVA
LYNN138
A
MVA
1.02 pu
A
MVA
A
MVA
System Losses: 11.66 MW
266 MW 59 Mvar
12 MW 3 Mvar
20 MW 12 Mvar
124 MW 45 Mvar
37 MW
13 Mvar
12 MW 5 Mvar
150 MW 1 Mvar
56 MW
13 Mvar
15 MW 5 Mvar
14 MW
2 Mvar
38 MW 4 Mvar
45 MW 0 Mvar
25 MW 36 Mvar
36 MW 10 Mvar
10 MW 5 Mvar
22 MW 15 Mvar
60 MW 12 Mvar
20 MW 38 Mvar
23 MW 7 Mvar
33 MW 13 Mvar
15.8 Mvar 18 MW 5 Mvar
58 MW 40 Mvar
60 MW 19 Mvar
14.1 Mvar
25 MW 10 Mvar
20 MW 3 Mvar
23 MW 6 Mvar 14 MW
3 Mvar
4.9 Mvar
7.3 Mvar
12.8 Mvar
28.9 Mvar
7.4 Mvar
0.0 Mvar
55 MW 29 Mvar
39 MW 13 Mvar
150 MW 1 Mvar
17 MW 3 Mvar
16 MW -14 Mvar
14 MW 4 Mvar
KYLE69A
MVA
Kyle138A
MVA
Previous case was augmented with the addition of a 138 kV Transmission Line
41
Generation Changes and The Slack Bus
• The power flow is a steady-state analysis tool, so the assumption is total load plus losses is always equal to total generation• Generation mismatch is made up at the slack bus
• When doing generation change power flow studies one always needs to be cognizant of where the generation is being made up• Common options include system slack, distributed across
multiple generators by participation factors or by economics
42
Generation Change Example 1
slack
SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
A
MVA
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHI MKO69
ROGER69
UI UC69
PETE69
HI SKY69
TI M69
TI M138
TI M345
PAI 69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
J O138 J O345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
0.00 pu
-0.01 pu
0.00 pu
0.00 pu
0.00 pu
-0.03 pu-0.01 pu
0.00 pu
0.00 pu
0.00 pu
-0.03 pu
-0.01 pu0.00 pu
0.00 pu
0.00 pu
0.00 pu
0.00 pu
0.00 pu
0.00 pu
-0.002 pu
0.00 pu
0.00 pu
0.00 pu
0.00 pu0.00 pu
0.00 pu
0.00 pu 0.00 pu
0.00 pu
0.00 pu0.00 pu
0.00 pu0.00 pu
A
MVA
-0.01 pu
A
MVA
A
MVA
LYNN138
A
MVA
0.00 pu
A
MVA
0.00 pu
A
MVA
162 MW 35 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
-157 MW -45 Mvar
0 MW
0 Mvar
0 MW 0 Mvar
0 MW 2 Mvar
0 MW
0 Mvar
0 MW 0 Mvar
0 MW
0 Mvar
0 MW 3 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 4 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
-0.1 Mvar 0 MW 0 Mvar
0 MW 0 Mvar 0 MW
0 Mvar
-0.1 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar 0 MW
0 Mvar
-0.1 Mvar
0.0 Mvar
-0.1 Mvar
-0.2 Mvar
0.0 Mvar
0.0 Mvar
0 MW 51 Mvar
0 MW 0 Mvar
0 MW 2 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
Display shows “Difference Flows” between original 37 bus case, and case with a BLT138 generation outage; note all the power change is picked up at the slack
43
Generation Change Example 2
slack
SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
A
MVA
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHI MKO69
ROGER69
UI UC69
PETE69
HI SKY69
TI M69
TI M138
TI M345
PAI 69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
J O138 J O345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
0.00 pu
-0.01 pu
0.00 pu
0.00 pu
0.00 pu
-0.03 pu0.00 pu
0.00 pu
0.00 pu
0.00 pu
-0.03 pu
-0.01 pu-0.01 pu
0.00 pu
0.00 pu
0.00 pu
0.00 pu
0.00 pu
0.00 pu
-0.003 pu
0.00 pu
0.00 pu
0.00 pu
0.00 pu0.00 pu
0.00 pu
0.00 pu 0.00 pu
0.00 pu
0.00 pu0.00 pu
0.00 pu0.00 pu
A
MVA
0.00 pu
A
MVA
A
MVA
LYNN138
A
MVA
0.00 pu
A
MVA
0.00 pu
A
MVA
0 MW 37 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
-157 MW -45 Mvar
0 MW
0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW
0 Mvar
0 MW 0 Mvar
0 MW
0 Mvar
42 MW -14 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
99 MW -20 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
-0.1 Mvar 0 MW 0 Mvar
0 MW 0 Mvar 0 MW
0 Mvar
-0.1 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar 0 MW
0 Mvar
0.0 Mvar
0.0 Mvar
-0.1 Mvar
-0.2 Mvar
-0.1 Mvar
0.0 Mvar
19 MW 51 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
0 MW 0 Mvar
Display repeats previous case except now the change in generation is picked up by other generators using a participation factor approach
44
Voltage Regulation Example: 37 Buses
Display shows voltage contour of the power system, demo will show the impact of generator voltage set point, reactive power limits, and switched capacitors
slack
SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
A
MVA
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHI MKO69
ROGER69
UI UC69
PETE69
HI SKY69
TI M69
TI M138
TI M345
PAI 69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
J O138 J O345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVAA
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.03 pu
1.01 pu
1.02 pu
1.03 pu
1.01 pu
1.00 pu1.00 pu
0.99 pu
1.02 pu
1.01 pu
1.00 pu
1.01 pu1.01 pu
1.01 pu
1.01 pu
1.02 pu
1.00 pu
1.00 pu
1.02 pu
0.997 pu
0.99 pu
1.00 pu
1.02 pu
1.00 pu1.01 pu
1.00 pu
1.00 pu 1.00 pu
1.01 pu
1.02 pu1.02 pu
1.02 pu 1.03 pu
A
MVA
1.02 pu
A
MVA
A
MVA
LYNN138
A
MVA
1.02 pu
A
MVA
1.00 pu
A
MVA
219 MW 52 Mvar
21 MW 7 Mvar
45 MW 12 Mvar
157 MW 45 Mvar
37 MW
13 Mvar
12 MW 5 Mvar
150 MW 0 Mvar
56 MW
13 Mvar
15 MW 5 Mvar
14 MW
2 Mvar
38 MW 3 Mvar
45 MW 0 Mvar
58 MW 36 Mvar
36 MW 10 Mvar
0 MW 0 Mvar
22 MW 15 Mvar
60 MW 12 Mvar
20 MW 9 Mvar
23 MW 7 Mvar
33 MW 13 Mvar 15.9 Mvar 18 MW
5 Mvar
58 MW 40 Mvar 51 MW
15 Mvar
14.3 Mvar
33 MW
10 Mvar
15 MW 3 Mvar
23 MW 6 Mvar 14 MW
3 Mvar
4.8 Mvar
7.2 Mvar
12.8 Mvar
29.0 Mvar
7.4 Mvar
20.8 Mvar
92 MW 10 Mvar
20 MW 8 Mvar
150 MW 0 Mvar
17 MW 3 Mvar
0 MW 0 Mvar
14 MW 4 Mvar
1.010 pu 0.0 Mvar
System Losses: 11.51 MW
45
Solving Large Power Systems
The most difficult computational task is inverting the Jacobian matrix– inverting a full matrix is an order n3 operation, meaning
the amount of computation increases with the cube of the size size
– this amount of computation can be decreased substantially by recognizing that since the Ybus is a sparse matrix, the Jacobian is also a sparse matrix
– using sparse matrix methods results in a computational order of about n1.5.
– this is a substantial savings when solving systems with tens of thousands of buses
46
Newton-Raphson Power Flow
Advantages– fast convergence as long as initial guess is close to solution– large region of convergence
Disadvantages– each iteration takes much longer than a Gauss-Seidel iteration– more complicated to code, particularly when implementing
sparse matrix algorithms
Newton-Raphson algorithm is very common in power flow analysis