chapter 6. power flow (load flow) analysis...
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Chapter 6. Power Flow (Load Flow) Analysis (전력조류 해석)
- Steady-state circuital analysis of a power system in normal operation.
- Find SIV ,, of all circuital elements (Generator, load, line).
- 발전기에서 생산된 전력은 송전 및 배전 선로를 통해 수용가에게 전송되어
소비되는데 이러한 전력의 흐름을 전력 조류 또는 조류라고 함.
- Backbone of power system analysis and design.
- Nonlinear static equations → Numerical technique
Linear static (algebraic) equations
3x+4y=2 3 4 x 2
5x+6y=4 5 6 y 4
Nonlinear static equations
3x2 +4y=2 # of variable = # of equation
2xy+ y =3
6.1 Gauss-Seidel Method of nonlinear algebraic equations.
one variable nonlinear equation
)(0)( xgxxf =⇒=
*)(*0*)( xgxxf =⇒=
Assume initial 0x
?)( 0101 ε<−⇒= xxxgx
?)( 1212 ε<−⇒= xxxgx
?)( 2323 ε<−⇒= xxxgx
!)( 11 Yesxxxgx kkkk ε<−⇒= ++
MM
→ kk xxx ≈= +1*
Update rule : )(1 kk xgx =+
0x1x
)( 01 xgx =
)( 12 xgx =
2x
)( 23 xgx =
3x 4x*xx
)(xgy =
y xy =
0x3x
)( 01 xgx = )( 12 xgx =
)( 23 xgx =
2x 1xx
)(xgy =
y xy =
1)'(1 <<− xg
)(xg
3x 1x 0x 2x
Ex 6.2 (Page 196)
?*0496)( 23 ==−+−= xxxxxf
)3)(1(3)94(3)'( 2 −−=+−= xxxxxf 3,1'=x
1 3 4
4,1* =x
1 3
04961)1( =−+−=f
44275427)3( −=−+−=f
By Gauss – Seidel 20 =x
)(xgx = 469 23 ++−= xxx
)(94
96
92
3
xgxxx =++−=
22222.2944
96
98)2()( 0 =++−== gxg
5173.2)22222.2( =g
8966.2)5173.2( =g
3376.3
7398.3
9568.3
9988.3
M
Fig 6.3 Page 197
n-variable with n-equations
T
nxxxX ],,,[ 21 L=
=
==
0)(
0)(0)(
2
1
Xf
XfXf
n
M 0
)(
)()(
)( 2
1
=
=
Xf
XfXf
XF
n
M
=
==
)(
)()(
22
11
Xgx
XgxXgx
nn
M
==
)(
)()(
)(),( 2
1
Xg
XgXg
XGXGX
n
M
Update rule
)(1 kk XGX =+
=
=
=
=
+
+
+
+
)(
)(
)(
)(
)1(
3)1(3
2)1(2
1)1(1
knkn
kk
kk
kk
Xgx
Xgx
Xgx
Xgx
M 조금이라도 더 좋은 값으로 표현하기 위해서
→ Updated values are used immediately in subsequent equation.
Stop condition
)( 11 kkk XXX −∆∆ ++ → ε≤∆ +1kX
Euclidean norm : 2/1
1
2)1(1 ])([∑
=++ ∆=∆
n
ikik xX
Sup norm : )1(1 max ++ ∆=∆ kiik xX
Vector form
Vector form
EX)
=+−+=
=−−+=
0110),(
0510),(
23
21212
213
1211
xxxxxf
xxxxxf
TxxX ],[ 21= ,
TX ]6.0,6.0[0 =
Sol) With initial value
−=
=→
04.4)6.0,6.0(616.0)6.0,6.0(
2
1
ff
, TXF )04.4,616.0()( 0 −=
=++=
=++−=
),(101
101
101
),(21
101
101
2123
212
21123
11
xxgxxx
xxgxxx
Update rule: )(1 kk XGX =+
==
==
1754.0)6.0,5384.0(
5384.0)6.0,6.0(
2)1(2
1)1(1
gx
gx
)1754.06.0,5384.06.0max( −−=∆X
4246.0)4246.0,0616.0max( ==
==
==
1507.0)1754.0,5019.0(
5019.0)1754.0,5384.0(
2)2(2
1)2(1
gx
gx
)1507.01754.0,5019.05384.0max( −−=∆X
0365.0)0247.0,0365.0max( ==
0176.0)1507.0,5019.0(0053.0)1507.0,5019.0(
2
1
=−=
ff
==
==
1506.0)1507.0,5024.0(
5024.0)1507.0,5019.0(
2)3(2
1)3(1
gx
gx
)1506.01507.0,5024.05019.0max( −−=∆X
0005.0)0001.0,0005.0max( ==
0191.0)1506.0,5024.0(0002.0)1506.0,5024.0(
2
1
==
ff
-> TXF )0191.0,0002.0()( 3 =
6.2 Newton – Raphson Method of nonlinear algebraic equations.
<Linearization of Nonlinear function>
Taylor Series )(xf at ax =
∑∞
=−=
0
)()(
!)()(
n
nn
axn
afxf
L+′′′−+′′−
+′−+= )(!3
)()(!2
)()()()(32
afaxafaxafaxaf
),( yxf at ),( oo yx
...),()(),()(),(),( tohyxyfyyyx
xfxxyxfyxf +
∂∂
−+∂∂
−+= oooooooo
Ex1) 22)( 2 ++= xxxf at 1=x
L+′′′−+′′−
+′−+= )1(6
)1()1(2
)1()1()1()1()(32
fxfxfxfxf
4|22)1( 1=+=′ =xxf
2|2)1( 1==′′ =xf
0)1( =′′′f
2
2)1(4)1(5
2
⋅−
+⋅−+=xx
222 ++= xx → 고차 미분 Term 이 Zero 가 아니면 무한대항까지!
Ex2) 222)( 2 =++= xatxxxf
...)2('''6
)3()2(''2
)2()2(')2()2()(22
+−
+−
+−+= fxfxfxfxf
10244)2( =++=f
6|22)2(' 2 =+= =xxf
2|2)2('' 2 == =xf
xx
xx2)2(6)2(10 −
+−+=
222 ++= xx
First order approximation with Taylor expansion
)(')()()( afaxafxf −+≅
oooo yxoyxooo yfyy
xfxxyxfyxf ,, |)(|)(),(),(
∂∂
−+∂∂
−+≅
Ex3) 222)( 2 =++= xatxxxf
)2(')2()2()( fxfxf −+=
26 −= x → Tangential line
Ex4) 322)( 2 =++= xatxxxf
8)3(17)3(')3()3()( −+=−+= xfxfxf
78 −= x
EX5) )2,1(22),( 22 atyxyxyxf ++=
)2,1()2,1( )2()1()2,1(),(yfy
xfxfyxf
∂∂
−+∂∂
−+=
1044222122)2,1( 2 =++=+⋅⋅+=f
8221424 )2,1()2,1( =⋅+⋅=+=∂∂ yx
xf
6221222 )2,1()2,1( =⋅+⋅=+=∂∂ yxyf
)2(6)1(810 −+−+= yx
> 가장 빨리 증가하는 방향 (접평면에 수직)
1068 −+= yx →Tangential plane
- linearization of a nonlinear function
비선형 함수의 선형화에 적용 !
< Newton-Raphson method>
One variable nonlinear equation
cxf =)( c : constant
Assume initial x0 and linearization at x0
cxfxxxfxf =′−+≅ )()()()( 000
)]([)(
1)( 00
0 xfcxf
xx −′
=−
100
0 )]([)(
1 xxfcxf
xx ∆−′
+= , )( 0xfc − : error
similarly,
)]([)(
11
112 xfc
xfxx −
′+=
update rule
)()(
1k
kkk xf
xfcxx′
−+=+
0x1xx
2x*x
c
y
cf) limitation of Newton-Raphson
0x3x1xx
2x*x
→ Initial point needs to be near the solution!
Two variables nonlinear equations
1( , )f x y C=
2( , )g x y C= with initial 0 0( , )x y
Linearization at 0 0( , )x y
0, 0 0, 00 0 0 ( ) 0 ( ) 1( , ) ( , ) ( ) ( )x y x yf ff x y f x y x x y y Cx y
∂ ∂≅ + − + − =
∂ ∂
0, 0 0, 00 0 0 ( ) 0 ( ) 2( , ) ( , ) ( ) ( )x y x yg gg x y g x y x x y y Cx y
∂ ∂≅ + − + − =
∂ ∂
0, 0 0, 00 ( ) 0 ( ) 1 0 0( ) ( ) ( , )x y x yf fx x y y C f x yx y
∂ ∂− + − = −
∂ ∂
0, 0 0, 00 ( ) 0 ( ) 1 0 0( ) ( ) ( , )x y x yg gx x y y C g x yx y
∂ ∂− + − = −
∂ ∂
x∆ y∆
−
=
∆∆
∂∂
∂∂
∂∂
∂∂
−
),(),(
2
1
)0,0()0,0(
)0,0()0,0(
kk
kk
yxyx
yxyx
yxgcyxfc
yx
yg
xg
yf
xf
= Jacobian matrix at 0 0( , )x y
0 0( , )J x y=
−
=
∆∆ −−
),(),(
),(002
00100
1
yxgCyxfC
yxJyx
−
+
=
−−
),(),(
),(002
001
00
1
0
0
1
1
yxgCyxfC
yxJyx
yx
update rule
−
+
=
−−
+
+
),(),(
),(2
11
1
1
kk
kkkk
k
k
k
k
yxgcyxfc
yxJyx
yx
n-variables & n-equations
( )nxxxX ⋅⋅⋅⋅= 21 [ ]0,20100 ,)( nxxxXCXF ⋅⋅⋅⋅==
=•••
==
nn CXf
CXfCXf
)(
)()(
22
11
•••
=
)(
)()(
)(
2
1
Xf
XfXf
XF
n
,
•••
=
nC
CC
C
2
1
update rule
)()(
)()(1
11
CXFXJX
XFCXJXX
kkk
kkkk
−−=
−+=−
−+
where,
kXXn
nnn
n
n
k
xf
xf
xf
xf
xf
xf
xf
xf
xf
XJ
=
∂∂
⋅⋅⋅⋅⋅∂∂
∂∂
⋅⋅⋅⋅⋅∂∂
⋅⋅⋅⋅⋅∂∂
∂∂
∂∂
⋅⋅⋅⋅⋅∂∂
∂∂
=
21
2
2
2
1
2
1
2
1
1
1
)(MMM
Stop condition
)( 11 kkk XXX −∆∆ ++ → ε≤∆ +1kX
Euclidean norm : 2/1
1
2)1(1 ])([∑
=++ ∆=∆
n
ikik xX
Sup norm : )1(1 max ++ ∆=∆ kiik xX
P. 205 Ex 6.5 6.6
Discussion) Why use the 1+∆ kX as a stop condition instead of CXF k −+ )( 1 ?
Summary
Gauss – Seidel
· Derivatives are not used → Relatively simple & fast calculation
· Hard to converge & slow convergence
Newton-Raphson
· Calculation of derivative & (n×n) matrix inversion
→ long time and not correct!
· Good convergence & fast convergence
→ sparse matrix : A matrix has a few non zero elements.
· sparse matrix technique with Newton- Raphson.
6.3. Y-Bus Matrix (Bus Admittance Matrix)
용어: Power System
= External Source + Transmission Network(Power Network)
- External Source: 발전기와 부하
- Transmission Network(Power Network): 송전선로로 이루어진 network부분
Transmission Network
ExternalSource
ExternalSource
ExternalSource
iI : 모선전류(Bus current)
: External Source (발전기/부하)에서 Transmission network으로 i- 모선을
통하여 유입되는 전류
- (발전기 혹은 부하)만 존재하는 모선
: (발전기 혹은 부하)에서 모선을 통하여 공급되는 전류
- 발전기와 부하가 동시에 존재하는 모선 : (발전기 전류+부하 전류)
- 발전기와 부하가 모두 없는 모선 : 모선전류는 zero.
ex1) a five-bus power system
2L 5L2G
4I
1G 4G
13I
12I
2I 5I
1I 1
2
3 4
5
11 GI I= 12 13( )I I = +
2 22 G LI I I= + 21 23( )I I = +
3 0I = 32 35 51( )I I I = + +
44 GI I= 45( )I =
55 LI I= 53 54( )I I = +
Bus Matrix Equation
- A matrix equation which represents the bus current as a function of bus
voltage.
- Ohm’ s Law of entire power system.
- Nodal equation of a power system from KCL
BUSBUSBUS VYI = ,
where,
=
=
n
BUS
n
BUS
V
VV
V
I
II
IMM2
1
2
1
, ,
n: number of bus
YBUS: Bus Admittance Matrix (n*n)
ex2) A 2-bus system with a short line
+
-
2V1V +
-
1I 2I][ΩjX
−
−=
−
−=
2
1
matrixAdmittanceBus
2
1
2
111
11
11
11
VV
Xj
Xj
Xj
Xj
VV
jXjX
jXjXII
BUSY44 344 21
ex3) A two bus system with a medium or long transmission line
1G2L
1V
2G
2V① ②
회로적으로 표현
ay
G1 G2
1I 2I1V
L2 by by
2V
211222
212111
)()()()(
VyyVyVVyVyIVyVyyVVyVyI
baaab
abaab
++−=−+=
−+=−+=
+−
−+=
2
1
2
1
VV
yyyyyy
II
bab
aba
BUSY=
)(1
)(1
122
211
VVjX
I
VVjX
I
−=
−=
EX4) A three bus system
1G 2G
3L
① ②
③
ay
byby
βy
βy
αy
G1
G2
L3
1I 1V 2V
3V
2I
3I
KCL at bus ①,②,③
)()( 3112111 VVyVyVVyVyI ab −++−+= αβ
321)( VyVyVyyyy aba αβα −−+++=
211222 )()( VyyVyVVyVyI baaab ++−=−+=
211333 )()( VyyVyVVyVyI βαααβ ++−=−+=
+−+−
−−+++=
3
2
1
3
2
1
)(00)(
)(
VVV
yyyyyy
yyyyyy
III
baa
aba
βαα
αβα
BusY of n-bus power system
=
nnnnn
n
n
n V
VV
YYY
YYYYYY
I
II
M
L
MLMM
L
L
M2
1
21
22221
11211
2
1
niniii VYVYVYI +++= L2211 ∑=
=n
jjijVY
1
iiY : Self-admittance of i-bus (diagonal)
sum of all admittance connected to i-bus
ijY : Mutual-admittance of i-j bus (off-diagonal)
Negative admittance between i-j bus
→ BusY is a symmetric matrix ( ijY = jiY )
→ ijY = jiY = 0 for no connection between i-j bus
(← impedance 는 ∞)
→ BusY has the information of network topology & transmission line parameters
→ Y-Bus matrix 와 모든 모선전압을 알면 임의의 모선 전류 표현 가능.
Complex power of i-bus : iS
)()(
*
LiGiLiGiLiGi
iiiii
QQjPPSSjQPIVS
+++=+=
+==
- Supplying power from i-bus to the network
- Sum of i-bus Generator & Load
- Zero for a bus has neither Generator nor load
1S in a two – bus system
*
22111112
1
*11
*11111 )()( VYVYVVYVIVjQPS
jjj +===+= ∑
=
ijijijijij
jiijiii
YjBGY
VV
θ
δδδδ
∠=+=
−∆∠= ,
∑∑
∑
∑
==
=
=
∠−=−∠−=
−∠−∠=
∠+∠=
2
11111
2
11111
2
11111
2
1
*1111
)()()(
)(
)(
jjjjj
jjjjj
jjjjj
jjjjj
jBGVVjBGVV
VjBGV
VjBGV
δδδ
δδ
δδ
∑
∑
∑
=
=
=
−+
+=
+−=
2
111111
2
111111
2
111111
cossin
sincos
sincos)(
jjjjjj
jjjjjj
jjjjjj
BGVVj
BGVV
jjBGVV
δδ
δδ
δδ
∑
∑
=
=
−=
+=
2
1111111
2
1111111
)cossin(
)sincos(
jjjjjj
jjjjjj
BGVVQ
BGVVP
δδ
δδ
Similarly,
∑
∑
=
=
−=
+=
2
1222222
2
1222222
)cossin(
)sincos(
jjjjjj
jjjjjj
BGVVQ
BGVVP
δδ
δδ
→ Note : 1 2 0S S+ ≠ (∵transmission line)
iS in n-bus power system (Power Balance Equation)
( )
( )1
1
cos sin1, 2, 3, ,
sin cos
n
i i j ij ij ij ijj
n
i i j ij ij ij ijj
P VV G Bi n
Q VV G B
δ δ
δ δ
=
=
= +
= ⋅ ⋅ ⋅
= −
∑
∑
-> Power of External Source (발전기+부하) of i-Bus
= Power to the Network through i-Bus
→ 1
0n
ii
S=
≠∑
cf) Power balance equation with ijijij YY θ∠=
∑=
===−=n
jjjiiiiiiiii VYVIVIVjQPS
1
***** )()(
∑∑
∑
==
=
−∠=+−∠=
∠∠−∠=
n
jijijjiji
n
jjiijjiji
n
jjjijijii
VYVVYV
VYV
11
1
)()( δθδδθ
δθδ
∑∑
∑
==
=
−+−=
−∠=
n
jijijjiji
n
jijijjiji
n
jijijjiji
VYVjVYV
VYV
11
1
)sin()cos(
)(
δθδθ
δθ
−=−−=
−=−=
∑∑
∑∑
==
==n
jijijjiji
n
jijijjijii
n
jijijjiji
n
jijijjijii
VYVVYVQ
VYVVYVP
11
11
)sin()sin(
)cos()cos(
θδδθ
θδδθ ni ,,2,1 L=
Ex1) An elementary DC circuit.
+
-
+
-
][1 Ω
][5 V ][4 V
][1 Ω
][1 Ω
V
I1 I2
V 4[V] 5[V]
1[A] 2[A]3[A]
Sol)
Nodal: 11
51
4 VVV=
−+
−
][339,29 VVVV ===−
Mesh:
+=++=+=++=
21212
21211
2)(52)(4
IIIIIIIIII
2,1 21 == II
Supply power:
][41*4][4 WP V ==
][102*5][5 WP V ==
][9)3(*3][1 WP shunt −=−=Ω
][1)1(*)34(][1 WP left −=−−=Ω
][4)2(*)35(][1 WP right −=−−=Ω
0419104 =−−−+=∑ iP
Ex2) Understanding of bus matrix eq. and power eq.
ex1) as a three bus DC power system (not a formal terminology)
with given ?&,]5,3,4[ == iBUST
BUS PIV
][1 S Bus1 Bus2 Bus3
LoadL2
][1 SV1 V2 V3
I1 I2 I3
−−−
−=
=
110121
011
333231
232221
131211
YYYYYYYYY
YBUS
From Bus matrix equation
−=
+−−+−
−=
=
−−−
−=∗=
23
1
53564
34
534
110121
011
BUSBUSBUS VYI
From Power balance equation
V I + -
∑∑==
⋅+⋅=+=n
jjiji
n
jijijijjijii GVVBGVVP
11)001()sincos( δδ
in DC ,0( =−= jiij δδδ ,0=ijθ )0=ijB
∑∑==
−=−=n
jijji
n
jijijijji YVVYVV
11)00cos()cos( θδ
∑∑==
==n
jijji
n
jijji YVVGVV
11
131112113
11111111 YVVYVVYVVYVVP
jjj ++== ∑
=
][4216054)1(34144 W=−=⋅⋅+−⋅⋅+⋅⋅=
33322221123
1222 YVYVYVVYVVP
jjj ++== ∑
=
][9)3(3)1(5)2(3)1(43 W−=−⋅=−⋅+⋅+−⋅=
33332231133
1333 YVYVYVVYVVP
jjj +++== ∑
=
][10)1(5)1(3033 W=+−⋅+⋅=
0321 ≠++⇒ PPP
transmission loss
][1)34(1)( 222112 WVVyPLEFT =−=−=
][4)35(1)( 223223 WVVyPRIGHT =−=−=
0411094321 =−−+−=++++⇒ RL PPPPP
(Conclusion)
* BUSY → Network Topology + line parameters
* Given BUSY & BUSV → iI , iS of all buses
→ Circuital analysis completed
* Finding out BUSV gives the complete circuital analysis.
6.4 Power flow (Load flow, 전력조류)
- steady-state circuital analysis of a power system
- nonlinear equations
load: LLL jQPS +=
generator: fixed GP & fixed GG VV =
← 발전기의 출력은 real power 와 전압의 크기.
Ex3) Given V1 = 4, P2 = -9, P3 = 10 of ex2)
find P1 = ?, V2 = ?, V3 = ?
← 어떤 모선은 전압, 어떤 모선은 전력이 주어짐!
← 단순한 nodal, mesh 로는 풀 수 없다!
][1 S ][1 S41 =V ?2 =V ?3 =V
92 −=P 1033 −=P ?1 =P
sol)
−−−
−=
110121
011
BusY
Finding out BUSV gives the complete circuital analysis.
2 unknown voltage variables (V2, V3) ← 2 power equation at Bus 2 & 3
(power 를 알고 있는 모선에서 식을 만들 수 있다.)
Power equation at bus 2
∑=
==3
122222
jjjYVVIVP
)( 2332222112 YVYVYVV ++=
)1()2()1(49 322 −∗+∗+−∗=− VVV
)42(9 322 −−=− VVV - ①
Power equation at bus 3
∑=
==3
133333
jjjYVVIVP
)( 3333223113 YVYVYVV ++=
)1()1(0410 323 ∗+−∗+∗= VVV
)(10 233 VVV −= - ②
①,② → Numerical technique <V2, V3>=<3,5>
←답을 알고 있기 때문에 확인 가능
T
BUSV ]5,3,4[= & BUSY give the other variables!
∑=
++===3
11331221111111 )(44
jj YVYVYVYVIVP
][405)1(3144 W=⋅+−+⋅=
loss: ][1)34( 2
][1 WR
P l =−
=Ω
][4)35( 2
][2 WR
P r =−
=Ω
Power conservation 0411094321 =−−+−=++++ rl PPPPP
- A circuital problem can be defined with given bus voltage and bus power.
← 어떤 모선은 전압, 어떤 모선은 전력이 주어짐.
-“ Power equations” are used to find unknown voltage variables.
EX4) Nonlinear eq.s does not always have the answer!
Given P1= 3[W] P2= -50[W] P3= 4[W]
Find V1= ? V2= ? V3= ?
sol) → Three power equations & 3 variables give <V1,V2,V3>
→ Power conservation can not hold because power losses are negative.
3 - 50 + 4 + loss < 0 (≠ 0)
rl PPPPP ++++ 321 ≠ 0
→ No answer! (Numerical method is not converged!)
EX5) Nonlinear eq.s does not always have the answer!
Given P1 = 300[W] P2 = -50[W] P3 = 400[W]
Find V1 =? V2 =? V3=?
Sol)
A load 50[W]
Two generator 300+400=700[W]
→ three power equations & 3 variables
power balance gives
line loss = 700-50 = 650[W] !
→ Generally, no answer ! (Numerical method is not converged!)
How to avoid the situation of (EX4) & (EX5) ?
→ All generation power should not be given.
→ At least, one generator power should not be given: “ Slack generator”
→ Reasonable amount of generations considering the loads
Load flow is a Simulation of real situation to feed the loads.
EX6) Given P1 = 30[W] P2 = -50[W] P3 = ?
Find V1 =? V2 =? V3=10[V]
Slack generator P3 = 20(load) + loss
<Power flow formulation>
- Given BUSY & Load LS , and assumed Generator outputs ( GP & GV )
→ Network condition to feed the loads in the system.
- Firstly, find the voltage of all buses → the other variables.
Type of buses
ⅰ) load bus (85%)
- a bus has no generator.
- given bus power LiLiLi jQPS +=
- find iii VV δ∠=
cf) A bus has neither G nor L: 0=iLS
ⅱ) generator bus (voltage controlled bus,15%))
- a bus with generator.
- given GiP & GiV
- find GiQ & Giδ
cf) A bus has Generator and Load
LiGii SSS +=
= )()( LiiGLiGi QQjPP +++
cf) Note the sign of real power. (usually, positive convention)
ⅲ) slack bus (One bus in a power flow study)
- usually, the largest G bus
- Given °∠= 01slackV
- Find slackslackslack jQPS +=
cf) °∠= 01slackV
← per unit voltage base
← reference phase angle in a AC circuit problem
Given Power balance equation
∑=
+=+=n
jijijijijjiLiGii BGVVPPP
1)sincos( δδ
∑=
−=+=n
jijijijijjiLiGii BGVVQQQ
1)cossin( δδ , ni ,,1L=
Problem formulation for n-bus power system
Given
variables
Unknown
variables
# of unknown
variables
Load Bus P, Q |V|, δ 2nL
Generator Bus P, |V| Q, δ 2nG
Slack Bus |V|=1, δ=0 P, Q 2
총모선수 : 1++= lg nnn
where, nl : # of load bus
ng : (#-1) of G-bus
unknown voltage variables: (2* nl)+ ng
At load bus
∑=
+=n
jijijijijjiLi BGVVP
1)sincos( δδ
∑=
−=n
iijijijijjiLi BGVVQ
1)cossin( δδ i = 1, …, nl
→ 2* nl equations
At G-bus
∑=
+=+=n
jijijijijjiLiGii BGVVPPP
1)sincos( δδ i = 1, …, ng
→ ng equation
∴ )2( gl nn + variables & )2( gl nn + equations
→ All bus voltages BUSV by numerical technique
→ BUSV gives the complete circuital analysis.
Other variables
→ iQ of G-bus
∑=
−=+n
jijijijijjiLiGi BGVVQQ
1)cossin( δδ , gni ,,1L=
→ slackslack jQ P +=slackS
∑=
+=+n
jijijijijjiLiGi BGVVPP
1)sincos( δδ
∑=
−=+n
jijijijijjiLiGi BGVVQQ
1)cossin( δδ
If bus ① is slack
∑=
−+=+n
jjjjjjLslack BGVPP
1111 )sin(cos δδ
∑=
−−=+n
jjjjjjLsalck BGVQQ
1111 cos)sin( δδ
→ SI , of transmission lines → BusBus VY &
Ex) Load flow analysis of following three bus power system
(Find PG1, QG1, δ2, QG2, |V3|, δ3 )
°∠= 011V
111 GGG jQPS +=
22 05.1 δ∠=V
6661.02 =GP
333 δ∠=VV
2244.18653.23 jSL +=
−−
−=
98.1910101098.1910101098.19
jjjjjjjjj
YBUS
Sol) 1++= lg nnn (3 =1+1+1)
Unknown voltage variables : < 332 ,, δδ V >
← )2( gl nn + =(2×1+1)=3
Power Equation
∑∑
∑∑
==
==
−=−=+
=+=+
n
jijijji
n
jijijijijjiLiGi
n
jijijji
n
jijijijijjiLiGi
BVVBGVVQQ
BVVBGVVPP
11
11
cos)cossin(
sin)sincos(
δδδ
δδδ
There power equation with given 332 ,, QPP
· ∑=
=+3
122222 sin
jjjjLG BVVPP δ
232332222222212112 sinsinsin δδδ BVVBVVBVV ++=
)sin()0sin()sin(
06661.0
322332222
2122112
2
δδδδ −+°+−=
+=
BVVBVBVV
P
)sin(*10**05.10)sin(*10*05.1 3232 δδδ −++= V - ①
· ∑=
=+3
133333 sin
jjjjLG BVVPP δ
333333323223313113 sinsinsin δδδ BVVBVVBVV ++=
0)sin()sin(
8653.20
233223133113
3
+−+−=−=
δδδδ BVVBVVP
)sin(*10*05.1*)sin(*10* 23333 δδδ −+= VV - ②
• ∑=
−=+3
133333 cos
jjjJLG BVVQQ δ
( )333333323223313113 coscoscos δδδ BVVBVVBVV ++−=
1*)88.19(*)cos(*10*05.1*cos10*1*
2244.102
323333
3
−+−+−=
−=
VVV
Q
δδδ - ③ Three unknown variables ( 332 ,, δδ V ) and Three equations
→ ( 332 ,, δδ V )=(-3.0023°, 0.9502, -9.9924°) from numerical tech.
°−∠°−∠
°∠=
9924.99502.00023.305.1
01
BUSV
The other variables
• ?1 =GP
1987.2)994.9sin(*10*9502.0*1)0023.3sin(*10*05.1*10
sinsinsin
sin0
13133112122111111
3
11111
=°+°+=
++=
=+ ∑=
δδδ
δ
BVVBVVBVV
BVVPj
jjjG
• ?1 =GQ
∑=
−=+3
11111 cos0
jjjjG BVVQ δ
[ ]1365.0
coscos 1313311212211111
=++−= δδ BVVBVVBVV
• ?2 =GQ
∑=
−=3
12222 cos
jjjjG BVVQ δ
6395.1−=
Transmission loss = 321 LGG PPP ++
= 8653.26661.01987.2 −+
= 0005.0
→ nearly zero (Q transmission line has no resistance )
BUSBUSBUS VYI =
<Procedure of power flow study>
1) Prepare BusY
2) Determine slack bus and Set )( SlackV = 1 °∠0
3) Set LiLiLi QPS += of Load buses
4) Set GiGi VP , of Generators ( i ≠ 1)
5) Apply G-S or N-R technique to find BusV
→ Assume initial unknown voltage variable
for numerical technique (usually °∠= 01iV )
6) Find the other circuital variables with BusV and BusY
Discussion)
1. 수학적으로 정확한 식인가?
2. 답이 나오지 않거나 현실적으로 적절하지 않은 결과 (부하모선 전압, 1 번 발전기
출력)가 나오면?
< Power flow solution with Newton-Raphson >
1++= lg nnn
# of voltage variable : gl nn +2
T
nn VVVX ],,,,,,,[ng) (nl
32
ng) (nl
32 444 3444 21444 3444 21++
⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅∆ δδδ
)(2 gl nn + → 식 유도 후, 나중에 불필요한 변수 제외!
)()sincos(1
XPBGVVP in
jijijijijjii ∆+= ∑
=δδ
∑=
−=n
jijijijji YVV
1)cos( θδ
)()cossin(1
XQBGVVQ in
jijijijijjii ∆−= ∑
=δδ
∑=
−=n
jijijijji YVV
1)sin( θδ
ni ,,3,2 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=
=> )(2 gl nn + variables and )(2 gl nn + equations
Given nonlinear equations
0)(
)()(
)(
)()(
)()(
0)(
0)(0)(
0)(0)(
2
3
2
22
33
22
=∆
∆∆
∆
∆
∆∆
∆∆
→
=−
=−=−
=−=−
XFXQXP
XQ
XQXP
XPXP
QXQ
QXQPXP
PXPPXP
n
nformvector
nn
nn
M
M
M
M
where,
×−×−∆×−∆
1)1(21)1(
)()(
1)1()(
nn
XFXQ
nXP
update rule
)()(
)()(1
11
kkk
kkkk
XFXJX
CXFXJXX−
−+
−=
−−=
where,
=
∂∂
∂∂
−−∂∂
∂∂
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=∂
∂=
2221
1211
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
|
|
|
)()(
)()(
)()(
)()(
)()(
)()(
)()(
)()(
)()(
JJJJ
VQQ
VPP
VXQ
VXQ
VXQ
VXQ
XQXQ
XQXQV
XPV
XP
VXP
VXP
XPXP
XPXP
XXFXJ
n
nn
n
n
nn
n
n
nn
n
n
nn
n
δ
δ
δδ
δδ
δδ
δδ
L
M
L
L
M
L
L
M
L
L
M
L
.)(,)(,)(,)( 12211211
VXQJXQJ
VXPJXPJ
∂∂
=∂
∂=
∂∂
=∂
∂=
δδ
Let pqJ designate the pq elements of J to avoid confusion ij in )(),( XQXP .
i) q
ppq
XPJ
δ∂
∂=
)(11
∑=
+∂
∂=≠
n
jpjpjpjpjjp
qBGVVqp
1)sincos()( δδ
δ
)sincos(
)(
pqpq
qp
pqpqqpq
BGVV δδδ
δδ
+∂
∂=
−=
)1(cos)1(sin −⋅+−⋅−= pqpqpqpqqp BGVV δδ
)cossin( pqpqpqpqqp BGVV δδ −=
∑=
+∂
∂=
∂
∂==
n
jpjpjpjpjjp
pp
p BGVVXP
qp1
)sincos()(
)( δδδδ
++
++
+
∂∂
=
)sincos(
)sincos(
)sincos( 11111
pnpnpnpnnp
pppppppppp
ppppp
p
BGVV
BGVV
BGVV
δδ
δδ
δδ
δO
O
+−+
+
+−
=
)cossin(
0
)cossin( 11111
pnpnpnpnnp
ppppp
BGVV
BGVV
δδ
δδ
O
O
∑≠=
+−=n
pjjpjpjpjpjjp BGVV
,1)cossin( δδ
formXpQ
n
pjjpjpjpjpjjp BGVV
같은와)(,1
)cossin(∑≠=
−−= δδ
)10(
])cossin()cossin([)10(
,1
⋅−⋅+
−+−−=
⋅−⋅=≠=
∑
pppppp
ppBppGpVpV
ppppppppppn
pjjpjpjpjpjjp
BGVV
BGVVBGVV444444 3444444 21
δδδδ
ppp
n
jpjpjpjpjjp BVBGVV 2
1)cossin( −−−= ∑
=
δδ
pppp BVXQ 2)( −−=
ⅱ) q
ppq V
XPJ
∂∂
=)(12
∑=
+∂
∂=≠
n
jpjpjpjpjjp
qBGVV
Vqp
1)sincos()( δδ
)sincos( pqpqpqpqqpq
BGVVV
δδ +∂∂
=
)sincos( pqpqpqpqp BGV δδ +=
∑=
+∂
∂=
∂
∂==
n
jpjpjpjpjjp
pp
p BGVVVV
xPqp
1)sincos(
)()( δδ
∑≠=
+=n
pjjpjpjpjpjj BGV
;1)sincos( δδ
)sincos( ppppppppppp
BGVVV
δδ +∂
∂+
)(2)sincos(;1
pppn
pjjpjpjpjpjj GVBGV ++= ∑
≠=δδ
ppp
completeto
n
pjjppppjpjpjpjj GVGVBGV +
∑
+⋅++= ∑≠= 444444444 3444444444 21;1
)01()sincos( δδ
∑=
++=n
jppppjpjpjpjj GVBGV
1)sincos( δδ
ppppjpjpjpjjpp
GVBGVVV
++= ∑ )sincos(1 δδ
pppp
p GVV
XP+=
)(
iii) q
ppq
XQJ
δ∂
∂=
)(21
∑=
−∂∂
=≠n
jpjpjpjpjjp
q
BGVVqp1
)cossin()( δδδ
)cossin( pqpqpqpqqpq
BGVV δδδ
−∂∂
=
)1(*sin)1(*cos −+−= pqpqpqpqqp BGVV δδ
)sincos( pqpqpqpqqp BGVV δδ +−=
cf) ,)()(
),&( 2112
q
pq
q
p
VXP
VXQ
JJofdiagonaloffqp∂
∂−=
∂
∂≠
δ
∑=
−∂
∂==
n
jpjpjpjpjjp
pBGVVqp
1)cossin()( δδ
δ
∑≠=
+=n
pjjpjpjpjpjjp BGVV
;1)sincos( δδ
444444 3444444 210
)cossin(
=
+∂
∂+ pppppppppp
pBGVV δδ
δ
∑≠=
−+⋅++=n
pjjppppppppjpjpjpjjp GVGVVBGVV
;1
2)01()sincos( δδ
∑=
−+=n
jppppjpjpjpjjp GVBGVV
1
2)sincos( δδ
pppp GVxP 2)( −=
iv) q
ppq V
XQJ
∂
∂=
)(22
)( qp ≠ ∑=
−∂
∂=
n
jpjpjpjpjjp
gBGVV
V 1)cossin( δδ
)cossin( pqpqpqpqqpg
BGVVV
δδ −∂
∂=
)cossin( pqpqpqpqp BGV δδ −=
)( qp = ∑=
−∂
∂=
n
jpjpjpjpjjp
pBGVV
V 1)cossin( δδ
44 344 21ppBpV
pppp
n
pjjpjpjpjpjj BV
VBGV
2
2
,1)0()cossin(
−=
≠=−
∂∂
+−= ∑ δδ
ppppppn
pjjpjpjpjpjj BVBVBGV −−+−= ∑
≠=)0()cossin(
,1δδ
pppn
jpjpjpjpjj BVBGV −−= ∑
=1)cossin( δδ
pppp
p BVV
XQ−=
)(
cf) Jacobian with polar admittance
ijijijijij YjBGY θ∠=+=
==
pqpqpq
pqpqpq
YBYG
θθ
sincos
→ (6.55)~(6.62) P.234
q
ppq
XPJ
δ∂
∂=
)(11
)( qp ≠ )cossin( pqpqpqpqqp BGVV δδ −=
)cossinsincos( pqpqpqpqpqpqqp YYVV δθδθ −=
)cossinsin(cos pqpqpqpqpqqp YVV δθδθ −=
)sin( pqpqpqqp YVV θδ −= (6.56)
)( qp = pppn
jpjpjpjpjjp BVBGVV 2
1)cossin( −−−= ∑
=δδ
∑=
−−−=n
jppppppjpjpjpjpjjp YVYVV
1
2 sin)cossinsin(cos θδθδθ
pppppn
jpjpjpjjp YVYVV θθδ sin)sin( 2
1−−−= ∑
=
)sin()sin(1
ppppppppn
jpjpjpjjp YVVYVV δθδθ −−−= ∑
=
∑≠=
−=n
pjjpjpjpjjp YVV
,1)sin( δθ (6.55)
Ex1) A Two Bus System,
+
-
222 δ∠= VV111 δ∠= VV +
-
1I 2IjX
Sending Power
*111 IVS ⋅= ,
jXVV
I 211
−=
*2211
1
∠−∠=
jXVV
Vδδ
jX
VVVjXVV
V−
−∠−=
−
−∠−−∠=
)( 21212
122111
δδδδ
−−−−=
XVjVVVV
j)sin()cos( 21212121
21 δδδδ
−−+
−=
XVVV
jX
VV )cos()sin( 21212
12121 δδδδ
111221
211221 cossin
jQPXVVV
jX
VV+=
−+=
δδ
Similary,
222112
222112*
222cossin
jQPXVVV
jX
VVIVS +=
−+=⋅=
δδ
XVV
XVVP 12211221
11
1 cossin δδδδ
=
∂∂
=∂∂
XVV
XVVP 12211221
22
1 cossin δδδδ
−=
∂∂
=∂∂
XV
XVV
VVP 1221221
11
1 sinsin δδ=
∂∂
=∂∂
Ex2) Result of ex1) from Jacobian.
−
−=
Xj
Xj
Xj
Xj
YBUS 11
11
XB
XB
XB
XB 1,1,1,1
22211211 −===−=
)1(,)()( 2
1
1 =−−=∂
∂pBVXQ
XPppppδ
112
12
111111 )cossin( BVBGVV
jjjjjj −−−= ∑
=δδ
112
12
1111 cos BVBVV
jjjj −= ∑
=δ
112
11211221
1111111 coscos BVBVVBVV
X
−+===
δδ321
X
VV 121 cosδ=
)2,1(),cossin(2
1 ==−=∂∂
qpBGVVP
pqpqpqpqqp δδδ
121221 cosδBVV−=
X
VV 1221 cosδ−=
)1(,)()(
1
1 =+=∂
∂ pGVV
XPV
XPppp
p
p
∑=
++=2
11111111 )sincos(
jjjjjj GVBGV δδ
∑=
=2
111 sin
jjjj BV δ
1212211111 sinsin δδ BVBV +=
XV 122 sinδ
=
< Decoupled power flow >
Jacobian matrix can be approximated simply.
≅
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=∂
∂=
22
11
2221
1211
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
00
)()(
)()(
)()(
)()(
)()(
)()(
)()(
)()(
)()(
JJ
JJJJ
VXQ
VXQ
VXQ
VXQ
XQXQ
XQXQV
XPV
XP
VXP
VXP
XPXP
XPXP
XXFXJ
n
nn
n
n
nn
n
n
nn
n
n
nn
n
L
M
L
L
M
L
L
M
L
L
M
L
δδ
δδ
δδ
δδ
i) Transmission line is mostly reactive. )0(,0 ≠≅ ijij BG ii) Voltage angle difference is small
)0(cos0sin10 ≠≅→°<< ijijij δδδ
0)sincos()(12 ≅+=
∂
∂= pqpqpqpqp
q
ppq BGV
VxP
J δδ
0)sincos(1)(≅++=
∂
∂= ∑ ppppjpjpjpjjp
pp
p GVBGVVVV
xPδδ
q
ppq
XQJ
δ∂
∂=
)(21 0)sincos( ≅+−= pqpqpqpqqp BGVV δδ
0)sincos()(
1
2 ≅−+=∂
∂= ∑
=
n
jppppjpjpjpjjp
p
p GVBGVVXQ
δδδ
Update rule can be decoupled.
)()(
)()(1
11
kkk
kkkk
XFXJX
CXFXJXX−
−+
−=
−−=
where, ],,,,,,,[ng) (nl
32
ng) (nl
32 444 3444 21444 3444 21++
⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= nn VVVX δδδ
-> T
nX ],,,[ 32 δδδδ ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅= , T
nV VVVX ],,,[ 32 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=
)()()( ,,111
,1, kkkk XFXJXX δδδδδ−
+ −=
)()()( ,,122
,1, kVVkVkVkV XFXJXX −+ −=
-> The decoupled power flow require more iteration, but less time per iteration.
Analysis problem of Power system can be decoupled also !
P depends on δ dominantly.
Q depends on V dominantly.