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The multiple solution analysis of probabilistic load flow
Electric Power Systems Research, 13 (1987) 21 - 29
21
The Multiple Solution Analysis of Probabilistic Load Flow MARIAN SOBIERAJSKI Institute of Electric Power Engineering, I-8, Technical University of WrocIaw, Wyb. Wyspianskiego 27, 50-370WrocIaw (Poland) (Received February 28, 1987)
SUMMARY The paper presents a new method of load flow calculation in which the known bus powers are treated as random variables. The probabilistic load flow equations are obtained from the quadratic deterministic load flow equations and are quartic. The deterministic equations have 2 n i solutions for n buses, and therefore the probabilistic equations should have at least the same number of solutions. The mathematical model is presented and the detailed equations for a two-bus power system are analysed. It is shown that the solutions of probabilistic load flow may be inadmissable if the random variables of the bus powers violate the boundary solution curve. The probabilistic load flow equations for power systems of general size are also derived. Results are exemplified for the Klos-Kerner three-bus and the Ward-Hale six-bus power systems.
1. INTRODUCTION T h e c a l c u l a t i o n of probabilistic load flows is of a g r e a t i n t e r e s t since it enables us to avoid expensive s i m u l a t i o n c o m p u t a t i o n s which are n e c e s s a r y for the design of power systems under c o n d i t i o n s of r a n d o m bus power variations. In the first works c o n c e r n i n g the c a l c u l a t i o n of probabilistic load flows the probabilistic load flow e q u a t i o n s were based on the direct c u r r e n t model of electric p o w e r systems [ 1], in the l a t e r ones, on linearized load flow equations [2] and, in the most r e c e n t ones, on nonl i n e a r load flow e q u a t i o n s [3 - 5]. The l i t e r a t u r e lacks works discussing the solvability of probabilistic load flow e q u a t i o n s and t h e i r c o n n e c t i o n with d e t e r m i n i s t i c load 0378-7796/87/$3.50
flow equations. As is known, the d e t e r m i n i s t i c load flow e q u a t i o n s h a v e multiple solutions and m e t h o d s for finding the multiple solutions of q u a d r a t i c load flow e q u a t i o n s can be found in the l i t e r a t u r e [6 - 8]. The set of probabilistic load flow e q u a t i o n s w h i c h are o b t a i n e d from the d e t e r m i n i s t i c q u a d r a t i c load flow e q u a t i o n s must h a v e multiple solutions as well. In this p a p e r it will be shown t h a t the probabilistic load flow equations h a v e 2 n 1 solutions for n buses and some of the solutions m a y be inadmissable. The multiple solution of probabilistic load flow equations is c o n n e c t e d with the multiple s o l u t i o n of d e t e r m i n i s t i c load flow equations. W h e n bus power values violate a b o u n d a r y solution c u r v e t h e r e is no solution of the d e t e r m i n i s t i c load flow equations, and the solution of the probabilistic load flow e q u a t i o n s is inadmissable. Because of the dimension of the problem the detailed analysis of the multiple s o l u t i o n of deterministic and probabilistic load flow equations has been carried for a two-bus power system. The K l o s - K e r n e r three-bus and the W a r d - H a l e six-bus systems are used to illust r a t e the multiple solution of l a r g e r systems.
2. MULTIPLE SOLUTIONS OF DETERMINISTIC AND PROBABILISTIC LOAD FLOW EQUATIONS FOR A TWO-BUS POWER SYSTEM In this section, we consider a o n e - m a c h i n e infinite-bus system to i l l u s t r a t e the multiple solutions of load flow equations. It will be s h o w n t h a t t h e r e is a r e g i o n of o p e r a t i n g points for w h i c h a n analysis predicts two solutions, t h e r e is a b o u n d a r y c u r v e with one solution, and t h e r e is a region of no s o l u t i o n or inadmissable solutions. © Elsevier Sequoia/Printed in The Netherlands
22
2.1. Deterministic load flow equations F i g u r e s 1 a n d 2 p r e s e n t the s c h e m a for a twobus system. T h e c o m p l e x v o l t a g e at bus 1 is U~ + j U2 and the c o m p l e x v o l t a g e at the s l a c k bus h a s o n l y one real c o m p o n e n t e q u a l to unity. T h e buses are c o n n e c t e d t h r o u g h a t r a n s m i s s i o n line w i t h a d m i t t a n c e G + j B . W i t h o u t loss of g e n e r a l i t y , let G = 1 a n d B=-2. Bus 1 c a n be e i t h e r t y p e PQ or PU. T h e det e r m i n i s t i c load flow e q u a t i o n s h a v e the following forms.
solutions
~
Botmdary
Reot~g~arly
solution
in the r ~ g e
Area of inadmlssable
D~ Area of admissa~e solutions
1
slack bus distributed active power 0.1344P
~
and ~ . 6 ~ Violation of b o ~ d a ~ solution c~Ve
U2 2 -
U1+ 2U2
(3)
u~= u~+ u#
(4)
E q u a t i o n s (1) - (4) are q u a d r a t i c , a n d it is k n o w n from refs. 6 and 8 t h a t t h e y h a v e two solutions.
2.1.1. Multiple solution of P U bus load flow equations Q u a d r a t i c e q u a t i o n s m a y h a v e two solutions, one, or no solution, d e p e n d i n g on the v a l u e s of P and Q at bus 1. So we m a y s u p p o s e t h a t t h e r e is a b o u n d a r y s o l u t i o n c u r v e dividing the solution a r e a into two regions: (1) a n a r e a of real solutions, a n d (2) a n a r e a of no r e a l solutions.
and probablllstlc solutions
I
Rectangularly distributed II
slack bus
~ n w'w
/ / /
aCatcl;~.vand..... I
~.~
-I', Active
Fig. I. T h e b o u n d a r y equations.
P
1'
j~Bo"
'/
~
[ >
D = - 10U + 5
(5)
As a result, the e q u a t i o n of the following p a r a b o l o i d is obtained: 16P 2 - 16PQ + 4Q 2 - 20P - 40Q + D 2 - 25 = 0
(6)
PQ b u s
2.1.2 Multiple solution of P U bus load flow equations T h e b o u n d a r y s o l u t i o n c u r v e is t h e c u r v e of P , U 2 v a l u e s w h i c h h a v e o n l y one solution. To find the b o u n d a r y s o l u t i o n curve, the r e c t a n g u l a r c o m p o n e n t s [71 a n d [72 m u s t be e l i m i n a t e d f r o m eqns. (3) a n d (4) a n d t h e n i n t r o d u c e d into the f o r m u l a of the J a c o b i a n
'
undary solution curve
solution c u r v e of the
T h e b o u n d a r y s o l u t i o n c u r v e is the c u r v e c o m p o s e d of P,Q v a l u e s w h i c h h a v e o n l y one solution. To find the f o r m u l a of this curve, the r e c t a n g u l a r c o m p o n e n t s £71 a n d U2 m u s t be e l i m i n a t e d from eqns. (1) and (2), a n d t h e n i n t r o d u c e d into t h e f o r m u l a for the PQ bus load flow e q u a t i o n ' s J a c o b i a n
2
-bus
~=
3
E q u a t i o n (6) depicts a c u r v e d solid in the coord i n a t e s P, Q and D, c i r c u l a r w h e n seen f r o m the end, p a r a b o l i c w h e n seen f r o m the side. T h e b o u n d a r y s o l u t i o n c u r v e depicts the c u r v e of one s o l u t i o n w h i c h is the s o l u t i o n for the zero-value J a c o b i a n , D = 0 (see Fig. 1).
Area of deterministic
u
s~ ~ ~ :~ ~
2"
Fig. 2. The boundary solution curve of the PU bus load flow equations.
P = £712+ U22 - U~ + 2U2
o
1
(2)
P U bus:
~.o o~
0
(1)
Q = 2Ux 2 + 2U22 - 2£71 - £72
=
and U-I
CuFve
PQ bus: P = £712+
2
D = -4U1-
2U2
As a r e s u l t the e q u a t i o n p a r a b o l o i d is obtained:
(7) of the following
load flow
p2 + U 4 _ 2P U 2 _ 5 U 2 + D2/4 = 0
(8)
23
F i g u r e 2 s h o w s the b o u n d a r y s o l u t i o n c u r v e w h e n D = 0. It is s e e n t h a t if the v a l u e s of P a n d U 2 are placed: (i) inside t h e p a r a b o l o i d , t h e n the l o a d equat i o n s h a v e t w o solutions; (ii) on t h e b o u n d a r y s o l u t i o n curve, t h e n the l o a d flow e q u a t i o n s h a v e one solution; (iii) outside t h e p a r a b o l o i d , t h e n the load flow e q u a t i o n s h a v e no solution. T h e s e facts m u s t h a v e some i m p a c t on the s o l u t i o n of p r o b a b i l i s t i c l o a d flow e q u a t i o n s .
t i o n on f u n c t i o n s (1) a n d (2), we o b t a i n Y~ = P = E(P) = E(U, 2 + U2s - U~ + 2/32) = -X1+2X2+X3+X4
(12)
]72 = - 2 X , - X~. + 2X3 + 2X4
(13)
T h e e q u a t i o n s for the s e c o n d - o r d e r m o m e n t are o b t a i n e d by m e a n s of the a v e r a g e v a l u e o p e r a t i o n on the products: Y3 = E(Ps) = E(U1 s + U2s - U , + 2U2) s = _ 2X~ 4 - 4)/12)/22- 2Xs a + 4X~ 3 - 8X~2X2
2.2. Probabilistic load flow equations P r o b a b i l i s t i c load flow e q u a t i o n s a r e the l o a d flow e q u a t i o n s w h e r e the bus p o w e r s a r e c h a r a c t e r i z e d by a v e r a g e v a l u e s a n d secondo r d e r m o m e n t s . I n o r d e r to e q u a t e t h e n u m b e r of u n k n o w n s a n d the n u m b e r of e q u a t i o n s , it m u s t be a s s u m e d t h a t the p r o b a b i l i t y distribut i o n of t h e r e c t a n g u l a r c o m p o n e n t s of t h e b u s v o l t a g e s is a m u l t i d i m e n s i o n a l n o r m a l distribut i o n [3, 5]. T h e n the third- a n d f o u r t h - o r d e r mom e n t s c a n be e x p r e s s e d by m e a n s of the average values and the second-order moments
+ 4X~X22 - 8X23 - 6 X I X 3 - 2 X I X 4 + 8XIX5 + 8)[2)[3 + 21X2X4 - 4X2X5 + 3)[3 s + 3X42 + 2X3X4 + 4)[52 + X3 + 4X4 - 4)/5
(14)
Y4 = - 8X, 4 - 16X~2Xs 2 - 8X2a + 16)/13 + 8XI2Xs + 16XIXs 2 + 8Xs 3 - 24)[1)[3 - 8XIX4 8X~X5 - 4X2X3 - 12XsX4 - 16)/2)(5
-
+ 12X3 s + 12X4 s + 8XaXa + 16Xs s + 4)/3 + X4 + 4X~
(15)
[9]: y~ = - 4 x ~ 4 - 8 x ~ x ~ s - 4 x s 4 + 8 x s 4 + 8 x /
m~j =/~ij + x i x j
(9) - 6x/xs
m i j k = rnijx--kk -{- m~,~ + m j k x i -- 2 X i X j X k
+ 8x~xs s - 6xs ~ - 12xix~
(10) - 4x, x4 + 6x~x~ + 3xsx~ + 9xsx4
m~j,~ = m~jmk~ + mi,mj~ + m~mj~ - 2 x~ xj x~ x~ -
(11) w h e r e t h e s e c o n d - o r d e r m o m e n t is rn~ = E(x~x~)
the s e c o n d - o r d e r c e n t r a l m o m e n t is p~ = E [ ( x / - x--~)(xj - ~)] t h e t h i r d - o r d e r m o m e n t is mij~ = E(xixjx~) t h e f o u r t h - o r d e r m o m e n t is mi~k~ = E(xix~xkxl)
a n d x/, xj, x~ a n d x~ a r e the a v e r a g e v a l u e s of t h e r a n d o m v a r i a b l e s xi, xj, x~ a n d x~. 2.2.1. Equations for the P Q bus L e t t h e bus p o w e r s be Y I = - P , Y s = Q , Y3 = mpp, Y4 = mQ__QQand Y5= mpQ, and the bus voltages be XI = UI, X2 = U2, )/3 = m11,X4 = m22 and X~ = m12. Then, performing the average value opera-
8 X s X ~ + 6Xs 2 + 6X4 2 + 4X3X4 + 8X5 2
+ 2Xa - 2X4 - 3X5
(16)
F r o m f o r m u l a e (12) - (16) it c a n be s e e n t h a t the P Q bus p r o b a b i l i s t i c load flow e q u a t i o n s are q u a r t i c . 2.2.2. Equations for the P U bus M a k i n g u s e of the s u b s t i t u t i o n for t h e bus powers: Y1 = P, Y2 = U 2, Y3 = m p p , Y4 = rnu2v2 a n d Y s = rnpu2; a n d for t h e bus voltages: XI=U1, X 2 = U 2 , X3=m11, X 4 = m 2 2 a n d )/5 = m12; a n d p e r f o r m i n g t h e a v e r a g e v a l u e ope r a t i o n on f u n c t i o n s (3) a n d (4) a n d on t h e i r products, we o b t a i n five q u a r t i c e q u a t i o n s : YI = - XI + 2X2 + X3 + X4
(17)
Y2 = X~ + X4
(18)
Y3 = - 2)(1 a - 4X~2X22 - 2X24 + 4X13 - 8)[12)[2 +4x~xs
- 8xs - 6xiz3
- 2x~z4 + 8x~x5
+ 8)[22'[3 + 21X2X4 - 4X2X5 + 3X32 + 3X4 s + 2X3Xa + 4)/52 + )/3 + 4X4 - 4Xs
(19)
24
Y4 = - 2X14 - 4X12X22 - 2)(24 + 3)(32
-4- 2X3X4 + 4)(52 + 3)(42
(20)
dom p o w e r s r e c t a n g u l a r l y d i s t r i b u t e d in t h e ranges 0.134 ~< P ~< 1.866
Y5 = -2X14 - 4X12X22 - 2X24 + 2X13
i.e.
+ 2)(1)(22 - 4X12X2 - 4X23 + 3)(32
0.134 ~< Q ~< 1.866
- 2X2X5 - X , X4 + 6X2X4 + 4X1X5 (21)
2.3. Solution of the probabilistic load flow equations T h e p r o b a b i l i s t i c load flow e q u a t i o n s (12) (16) a n d (17) - (21) c a n be put d o w n in t h e i r g e n e r a l form as
i.e.
Q = 1,
Xi+ 1 = Xi + Ji[Y - F(Xi)]
(23)
w h e r e J is the J a c o b i a n m a t r i x of eqn. (22) a n d i is the i t e r a t i o n n u m b e r . Set (22) h a s m a n y solutions and, d e p e n d i n g on the s t a r t i n g point, the i t e r a t i v e process converges to different solutions. The s o l u t i o n s of the P Q bus load flow equations are g i v e n in T a b l e 1. D e t e r m i n i s t i c solutions a r e r e l a t e d to the v a l u e s P = 1 and Q = 1. P r o b a b i l i s t i c s o l u t i o n s are r e l a t e d to the ran-
w h e r e a is the s t a n d a r d deviation. T h e s o l u t i o n s of the P Q bus p r o b a b i l i s t i c load flow, e q u a t i o n s a r e g i v e n in T a b l e 2 for
U1 U2 a1 a2 p Kind of solution
ap = 0
and
Q = -0.25,
aV = 0
t h a t is, for the v a l u e s w h i c h h a v e no deterministic s o l u t i o n s of the load flow equations. The r a n g e s of the r a n d o m c h a n g e of P,Q values are also s h o w n in Fig. 1. F r o m T a b l e s 1 a n d 2 and Fig. 1 it c a n be seen that: (i) the p r o b a b i l i s t i c e q u a t i o n s h a v e two solutions c o n n e c t e d w i t h two s o l u t i o n s of the det e r m i n i s t i c load flow e q u a t i o n s in the case w h e n the r a n d o m v a l u e s of the bus p o w e r s do not v i o l a t e the b o u n d a r y s o l u t i o n curve; (ii) the p r o b a b i l i s t i c e q u a t i o n s h a v e two ina d m i s s a b l e s o l u t i o n s despite t h e r e b e i n g no r e a l d e t e r m i n i s t i c s o l u t i o n in the case w h e n r a n d o m v a l u e s of the bus p o w e r s v i o l a t e the b o u n d a r y s o l u t i o n curve. The p r o b a b i l i s t i c s o l u t i o n s are i n a d m i s s a b l e since t h e y do n o t fulfil the r e q u i r e m e n t s of the definition of the c o v a r i a n c e m a t r i x b e c a u s e the s t a n d a r d d e v i a t i o n s h a v e i m a g i n a r y values. The s o l u t i o n s of the P U bus load flow equations are g i v e n in T a b l e 3 for r e c t a n g u l a r l y
1
The multiple solutions of PQ bus load flow equations for rectangularly a n d 0.134 ~< Q ~< 1.866 Variable
I~QQ = 0.25,
(22)
w h e r e Y is the v e c t o r of g i v e n quantities: the a v e r a g e v a l u e s a n d the s e c o n d - o r d e r m o m e n t s of the k n o w n variables; X is the v e c t o r of s e a r c h e d quantities: the a v e r a g e v a l u e s and the s e c o n d - o r d e r m o m e n t s of the u n k n o w n variables; and F is a n o n l i n e a r f u n c t i o n of f o u r t h order. T h e set of n o n l i n e a r e q u a t i o n s c a n be solved only i t e r a t i v e l y :
TABLE
aQ = 50%,
mQQ = Q 2 -{- ~IQQ = 1.25
P = -1, Y = F(X)
~pp = 0.25,
500//0,
=
mpp = p2 + ppp = 1.25
+ 3)(42 + 4X52 + 2X3X4 - 3)(1)(3
+ 2)(2)(3
ap
P = 1,
Deterministic
solns.
d i s t r i b u t e d b u s p o w e r s i n t h e r a n g e s 0 . 1 3 4 ~< P ~< 1.866
Probabilistic solns.
First
Second
First
Second
1.4 0.2 0 0 --
-- 0.4 0.2 0 0 --
1.3572 0.2 0.1486 0.2265 - 0.39 Admissable
--0.3572 0.2 0.1486 0.2265 0.39 Admissable
25 TABLE 2 The multiple solutions of P Q bus load flow equations for P = - 1, ap = 0 and Q = -0.25, aQ = 0 Variable
Deterministic solns. First
Second No No No No No
U2 ~i a2 p Kind of solution
Probabilistic solns.
solution solution solution solution solution
First
Second
0.9119 -0.35 j0.5861 0.0361 j0.11 Inadmissable
0.0881 -0.35 j0.5861 0.0361 -j0.11 Inadmissable
TABLE 3 The multiple solutions of P U bus load flow equations for U = 1 and rectangularly distributed P in the range 0.134 ~
U1 U2 a1 a2 p Kind of solution
Deterministic solns.
Probabilistic solns.
First
Second
First
Second
0.8944 0.4472 0 0 --
-0.8944 -0.4472 0 0 --
0.8724 0.4362 0.0946 0.1993 --I Admissable
-0.8724 -0.4362 0.0946 0.1993 -- I Admissable
TABLE 4 The solutions for P U bus load flow equations for U = 0.6 and rectangularly distributed P in the range 0.134 ~
UI U2 a1 a2 p Kind of solution
Deterministic solns.
Probabilistic solns.
First
Second
First
Second
0.3437 0.4918 0 0 --
- 0.5997 0.0202 0 0 --
0.2911 0.4655 0.2032 0.1315 - 1.30 Inadmissable
- 0.5471 0.0465 j0.0759 0.2572 - j0.09 Inadmissable
distributed active bus power in the range 0.134 ~ P ~< 1.866 a n d t h e c o n s t a n t b u s v o l t a g e magnitude U = 1. T h e s o l u t i o n s f o r t h e s a m e
solution curve by random values of P. As a result of this violation we obtain two inadmiss-
r a n g e o f P a n d U = 0.6 a r e g i v e n i n T a b l e 4. Figure 2 depicts both cases. It is seen that low-
first solution the correlation c o e f f i c i e n t exceeds minus one and in the case of the second s o l u t i o n t h e s t a n d a r d d e v i a t i o n o f [/1 h a s a n imaginary value.
e r i n g t h e b u s v o l t a g e m a g n i t u d e f r o m U -- 1 t o U = 0.6 e f f e c t s t h e v i o l a t i o n o f t h e b o u n d a r y
able probabilistic
solutions. In the case of the
26 3. PROBABILISTICLOAD FLOWEQUATIONS FOR POWER SYSTEMS OF GENERALSIZE Load flow equations can be written in their general form as a single equation:
(24)
Sk = ~ (bhlKkl + CklLkl) l=l
where if Sk=Ph t h e n bkl=Gkt and Ckl ~ Bkl (25) if sk =Qk
then
bkl = --Bkl
and
Ckl ~
Sk = Uk 2 then bkl
=
0
and
bhk=l
3.2. Equations for second-order moments of bus powers The equation for the second-order moment is obtained by means of the average value operation on the product SkSm, where k and m are the bus numbers. The average value of the product is
Vkl (26)
if
Hence, the equations for the average values of the bus powers are linear in the second-order moments of the rectangular components of the bus voltages.
SkSm = a [
i (bklKkl -4- ehlLkl)
and
Ckl = 0
× j ~: l (bmjKmj -4- cmjLmj ) ]
(27)
as well as g , , = V~lVtl + V,2V,~
Lkt =
-
= ~
(28)
+ chtbmjL~tKmj + CktC,njLhtLmj)
Ukl Ul2 "~- Uk2Ull
kl=k+k-1, ll=l+l-1,
(33)
where
k2=k+k, 12=l+l
~'~ (bklbmjKklKmj -at- bklCmjKklLmj
/=lj=l
(29)
Uk~ and Ua are the real components of the voltages at buses k and l; Uh2 and Um are the imaginary components of the voltages at buses k and l; Uk is the bus voltage magnitude at bus k; k and l are the successive numbers of the buses; and n is the number of buses. Application of the numbers of rectangular components of the bus voltages as in (29) makes it possible to remember them in the Tables in the following order: real component, imaginary component, real component, imaginary component, . . . . Form (24) is very convenient when deriving probabilistic load flow equations since the equations may be recorded in a shortened form without losing any of the minuteness of detail.
KktKmj = E [(Uhl Ua + UkeUm)(Um~Vii + Vm 2 Vj2)] = mklllmljl -4- mklllm2j2 ~- mk212rnlj 1 "~- mk212m2j2
(34)
Kk~Lmj = E[(Uh, Ua + Uh2U~2) × ( - Um~V~ + Um~V~l)] -- mklllmlj2 -~- mklllm2jl -- mk212mlj2 -4- mk2mm2jl
(35)
LkzKmj = E[( - Ukl Um + Uk2Un) × (Um~U~I + Um~Ujz)] = -- mkll2mljl -- mkll2m2j2 + mk2llmljl "4- mk2llm2j2
(36)
Lk~Lml = E[( - Ukl U~2 + Uk2Ua) × ( - UmIV~ + Vm~Vj~)l
3.1. Equations for average values of bus powers Performing the average value operation on function (22) we obtain Sk = ~ (bklKkl "4- ChlLkl ) l=l
(30)
where
Kkl = E(UklUtl + Uk2Um) = mkm + mk2m
(31)
Lkt = E( - Ukl Um+ Uh2Ua) = - m k t l 2 + mk2n (32)
mkll2mlj2 -- mkll2rn2j 1
- mk2tlmU2 + mk2nm2jl
(37)
As the rectangular components of the bus voltages have a normal probability distribution, the third and fourth-order moments can be expressed by the formulae (10) and (11) and the number of unknowns will be equal to the number of equations. The set of nonlinear equations (30) and (33) can be solved iteratively by using formula (23)
27 TABLE 5 Input data for the three-bus power system Bus to bus
R
X
B/2
Bus
P
Q
1 1 2
0.00538 0.0075 0.00269
0.0625 0.0662 0.0312
1.6 1.64 0.8
1 2 3
2 1.5 Slack bus
1
2 3 3
U
0"p
0"~
10% 10%
10% 10%
1
TABLE 6 The multiple solutions of the three-bus load flow equations with 0"p = 0"Q 10% =
Bus
Soln. No.
Deterministic solns.
Probabilistic solns.
Vl
Us
Vl
Us
0"1
0"2
P
1 2 3 4
0.9194 0.0553 0.3410 0.1589
-0.0913 --0.0771 --0.1313 -0.1262
0.9190 0.0557 0.3191 0.1708
-0.0913 -0.0771 -0.1307 -0.1268
0.0105 0.0100 0.0567 0.0548
0.0161 0.0161 0.0249 0.0272
0.4 -0.4 0.9 --0.9
1 2 3 4
0.9441 0.6430 0.0311 0.0345
-0.0595 -0.0669 - 0.0394 - 0.0431
0.9439 0.6430 0.0316 0.0343
-0.0595 -0.0688 - 0.0398 - 0.0426
0.0071 0.0077 0.0055 0.0071
0.0089 0.0105 0.0077 0.0089
0.3 0.1 - 0.2 - 0.2
1
1
0
1
0
0
0
m
2 3 4
1 1 1
0 0 0
1 1 1
0 0 0
0 0 0
0 0 0
m
a n d , a c c o r d i n g to t h e s t a r t i n g p o i n t , d i f f e r e n t s o l u t i o n s m a y be obtained.
4. APPLICATION OF GENERAL PROBABILISTIC LOAD FLOW EQUATIONS TO TYPICAL SMALL SIZE SYSTEMS The Klos-Kerner three-bus system was a n a l y s e d [8]. M u l t i p l e s o l u t i o n s of t h i s s y s t e m were obtained by the Tamura and Iwamoto m e t h o d d e s c r i b e d i n ref. 6. I n p u t d a t a of t h e s y s t e m a r e s h o w n i n T a b l e 5. T h e r e s u l t s o f t h e a n a l y s i s a r e s h o w n i n T a b l e 6. I t is s e e n t h a t t h e p r o b a b i l i s t i c l o a d flow e q u a t i o n s o f t h e three-bus system have four solutions connected w i t h t h e s o l u t i o n s of t h e d e t e r m i n i s t i c l o a d flow e q u a t i o n s . A l l t h e s o l u t i o n s a r e a d m i s s able b e c a u s e of the small i n p u t s t a n d a r d deviat i o n s e q u a l t o 10%.
m
4.2. S i x - b u s p o w e r s y s t e m The W a r d - H a l e six-bus system was analysed. I n p u t d a t a a n d m u l t i p l e d e t e r m i n i s t i c solut i o n s w e r e p r e s e n t e d i n ref. 7. T h e r e s u l t s o f t h e probabilistic analysis are shown in Tables 7 a n d 8. F r o m T a b l e 7 i t is s e e n t h a t t h e first a n d second solutions are very different in relation to the a v e r a g e v a l u e s a n d s t a n d a r d deviations. F r o m T a b l e 8 i t is s e e n t h a t i n c r e a s i n g t h e s t a n d a r d d e v i a t i o n s of the bus powers leads to loss of c o n v e r g e n c e of t h e i t e r a t i v e p r o c e s s , a s w a s t h e c a s e for t h e s e c o n d s o l u t i o n . T h e s e c o n d s o l u t i o n is s t e a d y s t a t e u n s t a b l e a n d m a y b e t h i s is t h e r e a s o n for t h e w o r s e conv e r g e n c e [7].
5. CONCLUSIONS In this paper the general nonlinear probab i l i s t i c e q u a t i o n s for l o a d flow c a l c u l a t i o n s
28 TABLE 7 Two solutions of the load flow e q u a t i o n s for the six-bus system with 0-p = 0-Q = 10% Bus
Deterministic solns.
Probabilistic solns.
V 1
V 2
V 1
V 2
0" I
0-2
1 2
1.0950 -0.2352
-1.047 -1.0746
1.0919 -0.2009
-0.1047 - 1.0729
0.0052 0.0897
0.0546 0.0168
2
1 2
0.8446 0.1830
-0.1965 -0.3556
0.8427 0.1826
-0.1966 -0.3482
0.0129 0.0485
0.0230 0.0236
0.82 -0.92
3
1 2
0.9495 0.3611
-0.1662 -0.2714
0.9478 0.3632
-0.1662 -0.2663
0.0113 0.0285
0.0186 0.0191
0.84 -0.92
4
1 2
0.8853 -0.0084
-0.2063 -0.4470
0.8828 0.0065
-0.2064 -0.4488
0.0135 0.0344
0.0304 0.0303
5
1 2
0.9227 0.1666
-0.2048 -0.3011
0.9208 0.1771
-0.2048 -0.3027
0.0133 0.0305
0.0223 0.0187
6
1 2
1.05 1.05
0 0
0 0
1
Soln. No.
0 0
1.05 1.05
0 0
-
-
1.00 1.00
0.67 0.85
0.88 -0.70 m m
TABLE 8 Two solutions of the load flow equations for the six-bus system with 0-p : Bus
Soln. No.
Deterministic solns.
Probabilistic solns.
vl
u2
Vl
0-Q
=
u2
20%
al
62
P
1
1 2
1.0950 - 0.2352
-0.1047 - 1.0746
1.0827 -0.1049 No iterative solution
0.0106
0.1096
1.00
2
1 2
0.8446 0.1830
-0.1965 -0.3556
0.8366 -0.1969 No iterative solution
0.0262
0.0460
0.82
3
1 2
0.9495 0.3611
-0.1662 -0.2714
0.9425 -0.1661 No iterative solution
0.0229
0.0373
0.84
4
1 2
0.8853 -0.0084
-0.2063 -0.4470
0.8752 -0.2067 No iterative s o l u t i o n
0.0273
0.0608
0.67
5
1 2
0.9227 0.1666
-0.2048 -0.3011
0.9149 -0.2048 No iterative solution
0.0269
0.0445
0.88
6
1 2
1.05 1.05
1.05 1.05
0 0
0 0
m
0 0
have been presented. It has been shown on a two-bus power system that the probabilistic load flow equations are quartic and have multiple solutions. Depending on the chosen starting point, different solutions can be obtained with the same input data. The probabilistic multiple solutions are connected with the deterministic multiple solutions. A lack of deterministic solutions leads to inadmissable probabilistic solutions.
0 0
E
F u r t h e r investigation should be made as to what kind of impact the multiple probabilistic solutions have on power system operations.
REFERENCES 1 B. Borkowska, Probabilistic load flow, I E E E Trans., P A S - 9 3 (1974) 752. 2 J. F. Dopazo, O. A. Klitin and A. M. Sasson, Stochastic load flows, I E E E Trans., P A S - 9 4 (1975) 299.
29 3 M. Sobierajski, Optimal stochastic load flows, Electr. Power Syst. Res., 2 (1979) 71. 4 M. Brucoli, F. Torelli and R. Napoli, Quadratic probabilistic load flow with linearly modelled dispatch, Electr. Power Energ. Syst., 3 (1985) 138. 5 M. Sobierajski, Probabilistic analysis of load flow solutions, Arch. Elektrotech. (Berlin), 69 (1986) 407. 6 Y. Tamura and S. Iwamoto, A method for finding multiple load flow solutions for general power systems, IEEE PES Winter Meeting, New York, 1980, Paper No. A 80043-0.
7 Y. Tamura, H. Mori and S. Iwamoto, Relationship between voltage instability and multiple load flow solutions in electrical power systems, IEEE Trans., PAS-102(1983) 1115. 8 A. Klos and A. Kerner, The non-uniqueness of load flow solutions, Proc. PSCC V, Dept. of Electr. and Electron. Eng., Queen Mary College, Univ. of London, 1975, 3.1/8. 9 K. S. Miller, Multidimensional Gaussian Distributions, Wiley, New York, 1964, p. 71.
21
The Multiple Solution Analysis of Probabilistic Load Flow MARIAN SOBIERAJSKI Institute of Electric Power Engineering, I-8, Technical University of WrocIaw, Wyb. Wyspianskiego 27, 50-370WrocIaw (Poland) (Received February 28, 1987)
SUMMARY The paper presents a new method of load flow calculation in which the known bus powers are treated as random variables. The probabilistic load flow equations are obtained from the quadratic deterministic load flow equations and are quartic. The deterministic equations have 2 n i solutions for n buses, and therefore the probabilistic equations should have at least the same number of solutions. The mathematical model is presented and the detailed equations for a two-bus power system are analysed. It is shown that the solutions of probabilistic load flow may be inadmissable if the random variables of the bus powers violate the boundary solution curve. The probabilistic load flow equations for power systems of general size are also derived. Results are exemplified for the Klos-Kerner three-bus and the Ward-Hale six-bus power systems.
1. INTRODUCTION T h e c a l c u l a t i o n of probabilistic load flows is of a g r e a t i n t e r e s t since it enables us to avoid expensive s i m u l a t i o n c o m p u t a t i o n s which are n e c e s s a r y for the design of power systems under c o n d i t i o n s of r a n d o m bus power variations. In the first works c o n c e r n i n g the c a l c u l a t i o n of probabilistic load flows the probabilistic load flow e q u a t i o n s were based on the direct c u r r e n t model of electric p o w e r systems [ 1], in the l a t e r ones, on linearized load flow equations [2] and, in the most r e c e n t ones, on nonl i n e a r load flow e q u a t i o n s [3 - 5]. The l i t e r a t u r e lacks works discussing the solvability of probabilistic load flow e q u a t i o n s and t h e i r c o n n e c t i o n with d e t e r m i n i s t i c load 0378-7796/87/$3.50
flow equations. As is known, the d e t e r m i n i s t i c load flow e q u a t i o n s h a v e multiple solutions and m e t h o d s for finding the multiple solutions of q u a d r a t i c load flow e q u a t i o n s can be found in the l i t e r a t u r e [6 - 8]. The set of probabilistic load flow e q u a t i o n s w h i c h are o b t a i n e d from the d e t e r m i n i s t i c q u a d r a t i c load flow e q u a t i o n s must h a v e multiple solutions as well. In this p a p e r it will be shown t h a t the probabilistic load flow equations h a v e 2 n 1 solutions for n buses and some of the solutions m a y be inadmissable. The multiple solution of probabilistic load flow equations is c o n n e c t e d with the multiple s o l u t i o n of d e t e r m i n i s t i c load flow equations. W h e n bus power values violate a b o u n d a r y solution c u r v e t h e r e is no solution of the d e t e r m i n i s t i c load flow equations, and the solution of the probabilistic load flow e q u a t i o n s is inadmissable. Because of the dimension of the problem the detailed analysis of the multiple s o l u t i o n of deterministic and probabilistic load flow equations has been carried for a two-bus power system. The K l o s - K e r n e r three-bus and the W a r d - H a l e six-bus systems are used to illust r a t e the multiple solution of l a r g e r systems.
2. MULTIPLE SOLUTIONS OF DETERMINISTIC AND PROBABILISTIC LOAD FLOW EQUATIONS FOR A TWO-BUS POWER SYSTEM In this section, we consider a o n e - m a c h i n e infinite-bus system to i l l u s t r a t e the multiple solutions of load flow equations. It will be s h o w n t h a t t h e r e is a r e g i o n of o p e r a t i n g points for w h i c h a n analysis predicts two solutions, t h e r e is a b o u n d a r y c u r v e with one solution, and t h e r e is a region of no s o l u t i o n or inadmissable solutions. © Elsevier Sequoia/Printed in The Netherlands
22
2.1. Deterministic load flow equations F i g u r e s 1 a n d 2 p r e s e n t the s c h e m a for a twobus system. T h e c o m p l e x v o l t a g e at bus 1 is U~ + j U2 and the c o m p l e x v o l t a g e at the s l a c k bus h a s o n l y one real c o m p o n e n t e q u a l to unity. T h e buses are c o n n e c t e d t h r o u g h a t r a n s m i s s i o n line w i t h a d m i t t a n c e G + j B . W i t h o u t loss of g e n e r a l i t y , let G = 1 a n d B=-2. Bus 1 c a n be e i t h e r t y p e PQ or PU. T h e det e r m i n i s t i c load flow e q u a t i o n s h a v e the following forms.
solutions
~
Botmdary
Reot~g~arly
solution
in the r ~ g e
Area of inadmlssable
D~ Area of admissa~e solutions
1
slack bus distributed active power 0.1344P
~
and ~ . 6 ~ Violation of b o ~ d a ~ solution c~Ve
U2 2 -
U1+ 2U2
(3)
u~= u~+ u#
(4)
E q u a t i o n s (1) - (4) are q u a d r a t i c , a n d it is k n o w n from refs. 6 and 8 t h a t t h e y h a v e two solutions.
2.1.1. Multiple solution of P U bus load flow equations Q u a d r a t i c e q u a t i o n s m a y h a v e two solutions, one, or no solution, d e p e n d i n g on the v a l u e s of P and Q at bus 1. So we m a y s u p p o s e t h a t t h e r e is a b o u n d a r y s o l u t i o n c u r v e dividing the solution a r e a into two regions: (1) a n a r e a of real solutions, a n d (2) a n a r e a of no r e a l solutions.
and probablllstlc solutions
I
Rectangularly distributed II
slack bus
~ n w'w
/ / /
aCatcl;~.vand..... I
~.~
-I', Active
Fig. I. T h e b o u n d a r y equations.
P
1'
j~Bo"
'/
~
[ >
D = - 10U + 5
(5)
As a result, the e q u a t i o n of the following p a r a b o l o i d is obtained: 16P 2 - 16PQ + 4Q 2 - 20P - 40Q + D 2 - 25 = 0
(6)
PQ b u s
2.1.2 Multiple solution of P U bus load flow equations T h e b o u n d a r y s o l u t i o n c u r v e is t h e c u r v e of P , U 2 v a l u e s w h i c h h a v e o n l y one solution. To find the b o u n d a r y s o l u t i o n curve, the r e c t a n g u l a r c o m p o n e n t s [71 a n d [72 m u s t be e l i m i n a t e d f r o m eqns. (3) a n d (4) a n d t h e n i n t r o d u c e d into the f o r m u l a of the J a c o b i a n
'
undary solution curve
solution c u r v e of the
T h e b o u n d a r y s o l u t i o n c u r v e is the c u r v e c o m p o s e d of P,Q v a l u e s w h i c h h a v e o n l y one solution. To find the f o r m u l a of this curve, the r e c t a n g u l a r c o m p o n e n t s £71 a n d U2 m u s t be e l i m i n a t e d from eqns. (1) and (2), a n d t h e n i n t r o d u c e d into t h e f o r m u l a for the PQ bus load flow e q u a t i o n ' s J a c o b i a n
2
-bus
~=
3
E q u a t i o n (6) depicts a c u r v e d solid in the coord i n a t e s P, Q and D, c i r c u l a r w h e n seen f r o m the end, p a r a b o l i c w h e n seen f r o m the side. T h e b o u n d a r y s o l u t i o n c u r v e depicts the c u r v e of one s o l u t i o n w h i c h is the s o l u t i o n for the zero-value J a c o b i a n , D = 0 (see Fig. 1).
Area of deterministic
u
s~ ~ ~ :~ ~
2"
Fig. 2. The boundary solution curve of the PU bus load flow equations.
P = £712+ U22 - U~ + 2U2
o
1
(2)
P U bus:
~.o o~
0
(1)
Q = 2Ux 2 + 2U22 - 2£71 - £72
=
and U-I
CuFve
PQ bus: P = £712+
2
D = -4U1-
2U2
As a r e s u l t the e q u a t i o n p a r a b o l o i d is obtained:
(7) of the following
load flow
p2 + U 4 _ 2P U 2 _ 5 U 2 + D2/4 = 0
(8)
23
F i g u r e 2 s h o w s the b o u n d a r y s o l u t i o n c u r v e w h e n D = 0. It is s e e n t h a t if the v a l u e s of P a n d U 2 are placed: (i) inside t h e p a r a b o l o i d , t h e n the l o a d equat i o n s h a v e t w o solutions; (ii) on t h e b o u n d a r y s o l u t i o n curve, t h e n the l o a d flow e q u a t i o n s h a v e one solution; (iii) outside t h e p a r a b o l o i d , t h e n the load flow e q u a t i o n s h a v e no solution. T h e s e facts m u s t h a v e some i m p a c t on the s o l u t i o n of p r o b a b i l i s t i c l o a d flow e q u a t i o n s .
t i o n on f u n c t i o n s (1) a n d (2), we o b t a i n Y~ = P = E(P) = E(U, 2 + U2s - U~ + 2/32) = -X1+2X2+X3+X4
(12)
]72 = - 2 X , - X~. + 2X3 + 2X4
(13)
T h e e q u a t i o n s for the s e c o n d - o r d e r m o m e n t are o b t a i n e d by m e a n s of the a v e r a g e v a l u e o p e r a t i o n on the products: Y3 = E(Ps) = E(U1 s + U2s - U , + 2U2) s = _ 2X~ 4 - 4)/12)/22- 2Xs a + 4X~ 3 - 8X~2X2
2.2. Probabilistic load flow equations P r o b a b i l i s t i c load flow e q u a t i o n s a r e the l o a d flow e q u a t i o n s w h e r e the bus p o w e r s a r e c h a r a c t e r i z e d by a v e r a g e v a l u e s a n d secondo r d e r m o m e n t s . I n o r d e r to e q u a t e t h e n u m b e r of u n k n o w n s a n d the n u m b e r of e q u a t i o n s , it m u s t be a s s u m e d t h a t the p r o b a b i l i t y distribut i o n of t h e r e c t a n g u l a r c o m p o n e n t s of t h e b u s v o l t a g e s is a m u l t i d i m e n s i o n a l n o r m a l distribut i o n [3, 5]. T h e n the third- a n d f o u r t h - o r d e r mom e n t s c a n be e x p r e s s e d by m e a n s of the average values and the second-order moments
+ 4X~X22 - 8X23 - 6 X I X 3 - 2 X I X 4 + 8XIX5 + 8)[2)[3 + 21X2X4 - 4X2X5 + 3)[3 s + 3X42 + 2X3X4 + 4)[52 + X3 + 4X4 - 4)/5
(14)
Y4 = - 8X, 4 - 16X~2Xs 2 - 8X2a + 16)/13 + 8XI2Xs + 16XIXs 2 + 8Xs 3 - 24)[1)[3 - 8XIX4 8X~X5 - 4X2X3 - 12XsX4 - 16)/2)(5
-
+ 12X3 s + 12X4 s + 8XaXa + 16Xs s + 4)/3 + X4 + 4X~
(15)
[9]: y~ = - 4 x ~ 4 - 8 x ~ x ~ s - 4 x s 4 + 8 x s 4 + 8 x /
m~j =/~ij + x i x j
(9) - 6x/xs
m i j k = rnijx--kk -{- m~,~ + m j k x i -- 2 X i X j X k
+ 8x~xs s - 6xs ~ - 12xix~
(10) - 4x, x4 + 6x~x~ + 3xsx~ + 9xsx4
m~j,~ = m~jmk~ + mi,mj~ + m~mj~ - 2 x~ xj x~ x~ -
(11) w h e r e t h e s e c o n d - o r d e r m o m e n t is rn~ = E(x~x~)
the s e c o n d - o r d e r c e n t r a l m o m e n t is p~ = E [ ( x / - x--~)(xj - ~)] t h e t h i r d - o r d e r m o m e n t is mij~ = E(xixjx~) t h e f o u r t h - o r d e r m o m e n t is mi~k~ = E(xix~xkxl)
a n d x/, xj, x~ a n d x~ a r e the a v e r a g e v a l u e s of t h e r a n d o m v a r i a b l e s xi, xj, x~ a n d x~. 2.2.1. Equations for the P Q bus L e t t h e bus p o w e r s be Y I = - P , Y s = Q , Y3 = mpp, Y4 = mQ__QQand Y5= mpQ, and the bus voltages be XI = UI, X2 = U2, )/3 = m11,X4 = m22 and X~ = m12. Then, performing the average value opera-
8 X s X ~ + 6Xs 2 + 6X4 2 + 4X3X4 + 8X5 2
+ 2Xa - 2X4 - 3X5
(16)
F r o m f o r m u l a e (12) - (16) it c a n be s e e n t h a t the P Q bus p r o b a b i l i s t i c load flow e q u a t i o n s are q u a r t i c . 2.2.2. Equations for the P U bus M a k i n g u s e of the s u b s t i t u t i o n for t h e bus powers: Y1 = P, Y2 = U 2, Y3 = m p p , Y4 = rnu2v2 a n d Y s = rnpu2; a n d for t h e bus voltages: XI=U1, X 2 = U 2 , X3=m11, X 4 = m 2 2 a n d )/5 = m12; a n d p e r f o r m i n g t h e a v e r a g e v a l u e ope r a t i o n on f u n c t i o n s (3) a n d (4) a n d on t h e i r products, we o b t a i n five q u a r t i c e q u a t i o n s : YI = - XI + 2X2 + X3 + X4
(17)
Y2 = X~ + X4
(18)
Y3 = - 2)(1 a - 4X~2X22 - 2X24 + 4X13 - 8)[12)[2 +4x~xs
- 8xs - 6xiz3
- 2x~z4 + 8x~x5
+ 8)[22'[3 + 21X2X4 - 4X2X5 + 3X32 + 3X4 s + 2X3Xa + 4)/52 + )/3 + 4X4 - 4Xs
(19)
24
Y4 = - 2X14 - 4X12X22 - 2)(24 + 3)(32
-4- 2X3X4 + 4)(52 + 3)(42
(20)
dom p o w e r s r e c t a n g u l a r l y d i s t r i b u t e d in t h e ranges 0.134 ~< P ~< 1.866
Y5 = -2X14 - 4X12X22 - 2X24 + 2X13
i.e.
+ 2)(1)(22 - 4X12X2 - 4X23 + 3)(32
0.134 ~< Q ~< 1.866
- 2X2X5 - X , X4 + 6X2X4 + 4X1X5 (21)
2.3. Solution of the probabilistic load flow equations T h e p r o b a b i l i s t i c load flow e q u a t i o n s (12) (16) a n d (17) - (21) c a n be put d o w n in t h e i r g e n e r a l form as
i.e.
Q = 1,
Xi+ 1 = Xi + Ji[Y - F(Xi)]
(23)
w h e r e J is the J a c o b i a n m a t r i x of eqn. (22) a n d i is the i t e r a t i o n n u m b e r . Set (22) h a s m a n y solutions and, d e p e n d i n g on the s t a r t i n g point, the i t e r a t i v e process converges to different solutions. The s o l u t i o n s of the P Q bus load flow equations are g i v e n in T a b l e 1. D e t e r m i n i s t i c solutions a r e r e l a t e d to the v a l u e s P = 1 and Q = 1. P r o b a b i l i s t i c s o l u t i o n s are r e l a t e d to the ran-
w h e r e a is the s t a n d a r d deviation. T h e s o l u t i o n s of the P Q bus p r o b a b i l i s t i c load flow, e q u a t i o n s a r e g i v e n in T a b l e 2 for
U1 U2 a1 a2 p Kind of solution
ap = 0
and
Q = -0.25,
aV = 0
t h a t is, for the v a l u e s w h i c h h a v e no deterministic s o l u t i o n s of the load flow equations. The r a n g e s of the r a n d o m c h a n g e of P,Q values are also s h o w n in Fig. 1. F r o m T a b l e s 1 a n d 2 and Fig. 1 it c a n be seen that: (i) the p r o b a b i l i s t i c e q u a t i o n s h a v e two solutions c o n n e c t e d w i t h two s o l u t i o n s of the det e r m i n i s t i c load flow e q u a t i o n s in the case w h e n the r a n d o m v a l u e s of the bus p o w e r s do not v i o l a t e the b o u n d a r y s o l u t i o n curve; (ii) the p r o b a b i l i s t i c e q u a t i o n s h a v e two ina d m i s s a b l e s o l u t i o n s despite t h e r e b e i n g no r e a l d e t e r m i n i s t i c s o l u t i o n in the case w h e n r a n d o m v a l u e s of the bus p o w e r s v i o l a t e the b o u n d a r y s o l u t i o n curve. The p r o b a b i l i s t i c s o l u t i o n s are i n a d m i s s a b l e since t h e y do n o t fulfil the r e q u i r e m e n t s of the definition of the c o v a r i a n c e m a t r i x b e c a u s e the s t a n d a r d d e v i a t i o n s h a v e i m a g i n a r y values. The s o l u t i o n s of the P U bus load flow equations are g i v e n in T a b l e 3 for r e c t a n g u l a r l y
1
The multiple solutions of PQ bus load flow equations for rectangularly a n d 0.134 ~< Q ~< 1.866 Variable
I~QQ = 0.25,
(22)
w h e r e Y is the v e c t o r of g i v e n quantities: the a v e r a g e v a l u e s a n d the s e c o n d - o r d e r m o m e n t s of the k n o w n variables; X is the v e c t o r of s e a r c h e d quantities: the a v e r a g e v a l u e s and the s e c o n d - o r d e r m o m e n t s of the u n k n o w n variables; and F is a n o n l i n e a r f u n c t i o n of f o u r t h order. T h e set of n o n l i n e a r e q u a t i o n s c a n be solved only i t e r a t i v e l y :
TABLE
aQ = 50%,
mQQ = Q 2 -{- ~IQQ = 1.25
P = -1, Y = F(X)
~pp = 0.25,
500//0,
=
mpp = p2 + ppp = 1.25
+ 3)(42 + 4X52 + 2X3X4 - 3)(1)(3
+ 2)(2)(3
ap
P = 1,
Deterministic
solns.
d i s t r i b u t e d b u s p o w e r s i n t h e r a n g e s 0 . 1 3 4 ~< P ~< 1.866
Probabilistic solns.
First
Second
First
Second
1.4 0.2 0 0 --
-- 0.4 0.2 0 0 --
1.3572 0.2 0.1486 0.2265 - 0.39 Admissable
--0.3572 0.2 0.1486 0.2265 0.39 Admissable
25 TABLE 2 The multiple solutions of P Q bus load flow equations for P = - 1, ap = 0 and Q = -0.25, aQ = 0 Variable
Deterministic solns. First
Second No No No No No
U2 ~i a2 p Kind of solution
Probabilistic solns.
solution solution solution solution solution
First
Second
0.9119 -0.35 j0.5861 0.0361 j0.11 Inadmissable
0.0881 -0.35 j0.5861 0.0361 -j0.11 Inadmissable
TABLE 3 The multiple solutions of P U bus load flow equations for U = 1 and rectangularly distributed P in the range 0.134 ~
U1 U2 a1 a2 p Kind of solution
Deterministic solns.
Probabilistic solns.
First
Second
First
Second
0.8944 0.4472 0 0 --
-0.8944 -0.4472 0 0 --
0.8724 0.4362 0.0946 0.1993 --I Admissable
-0.8724 -0.4362 0.0946 0.1993 -- I Admissable
TABLE 4 The solutions for P U bus load flow equations for U = 0.6 and rectangularly distributed P in the range 0.134 ~
UI U2 a1 a2 p Kind of solution
Deterministic solns.
Probabilistic solns.
First
Second
First
Second
0.3437 0.4918 0 0 --
- 0.5997 0.0202 0 0 --
0.2911 0.4655 0.2032 0.1315 - 1.30 Inadmissable
- 0.5471 0.0465 j0.0759 0.2572 - j0.09 Inadmissable
distributed active bus power in the range 0.134 ~ P ~< 1.866 a n d t h e c o n s t a n t b u s v o l t a g e magnitude U = 1. T h e s o l u t i o n s f o r t h e s a m e
solution curve by random values of P. As a result of this violation we obtain two inadmiss-
r a n g e o f P a n d U = 0.6 a r e g i v e n i n T a b l e 4. Figure 2 depicts both cases. It is seen that low-
first solution the correlation c o e f f i c i e n t exceeds minus one and in the case of the second s o l u t i o n t h e s t a n d a r d d e v i a t i o n o f [/1 h a s a n imaginary value.
e r i n g t h e b u s v o l t a g e m a g n i t u d e f r o m U -- 1 t o U = 0.6 e f f e c t s t h e v i o l a t i o n o f t h e b o u n d a r y
able probabilistic
solutions. In the case of the
26 3. PROBABILISTICLOAD FLOWEQUATIONS FOR POWER SYSTEMS OF GENERALSIZE Load flow equations can be written in their general form as a single equation:
(24)
Sk = ~ (bhlKkl + CklLkl) l=l
where if Sk=Ph t h e n bkl=Gkt and Ckl ~ Bkl (25) if sk =Qk
then
bkl = --Bkl
and
Ckl ~
Sk = Uk 2 then bkl
=
0
and
bhk=l
3.2. Equations for second-order moments of bus powers The equation for the second-order moment is obtained by means of the average value operation on the product SkSm, where k and m are the bus numbers. The average value of the product is
Vkl (26)
if
Hence, the equations for the average values of the bus powers are linear in the second-order moments of the rectangular components of the bus voltages.
SkSm = a [
i (bklKkl -4- ehlLkl)
and
Ckl = 0
× j ~: l (bmjKmj -4- cmjLmj ) ]
(27)
as well as g , , = V~lVtl + V,2V,~
Lkt =
-
= ~
(28)
+ chtbmjL~tKmj + CktC,njLhtLmj)
Ukl Ul2 "~- Uk2Ull
kl=k+k-1, ll=l+l-1,
(33)
where
k2=k+k, 12=l+l
~'~ (bklbmjKklKmj -at- bklCmjKklLmj
/=lj=l
(29)
Uk~ and Ua are the real components of the voltages at buses k and l; Uh2 and Um are the imaginary components of the voltages at buses k and l; Uk is the bus voltage magnitude at bus k; k and l are the successive numbers of the buses; and n is the number of buses. Application of the numbers of rectangular components of the bus voltages as in (29) makes it possible to remember them in the Tables in the following order: real component, imaginary component, real component, imaginary component, . . . . Form (24) is very convenient when deriving probabilistic load flow equations since the equations may be recorded in a shortened form without losing any of the minuteness of detail.
KktKmj = E [(Uhl Ua + UkeUm)(Um~Vii + Vm 2 Vj2)] = mklllmljl -4- mklllm2j2 ~- mk212rnlj 1 "~- mk212m2j2
(34)
Kk~Lmj = E[(Uh, Ua + Uh2U~2) × ( - Um~V~ + Um~V~l)] -- mklllmlj2 -~- mklllm2jl -- mk212mlj2 -4- mk2mm2jl
(35)
LkzKmj = E[( - Ukl Um + Uk2Un) × (Um~U~I + Um~Ujz)] = -- mkll2mljl -- mkll2m2j2 + mk2llmljl "4- mk2llm2j2
(36)
Lk~Lml = E[( - Ukl U~2 + Uk2Ua) × ( - UmIV~ + Vm~Vj~)l
3.1. Equations for average values of bus powers Performing the average value operation on function (22) we obtain Sk = ~ (bklKkl "4- ChlLkl ) l=l
(30)
where
Kkl = E(UklUtl + Uk2Um) = mkm + mk2m
(31)
Lkt = E( - Ukl Um+ Uh2Ua) = - m k t l 2 + mk2n (32)
mkll2mlj2 -- mkll2rn2j 1
- mk2tlmU2 + mk2nm2jl
(37)
As the rectangular components of the bus voltages have a normal probability distribution, the third and fourth-order moments can be expressed by the formulae (10) and (11) and the number of unknowns will be equal to the number of equations. The set of nonlinear equations (30) and (33) can be solved iteratively by using formula (23)
27 TABLE 5 Input data for the three-bus power system Bus to bus
R
X
B/2
Bus
P
Q
1 1 2
0.00538 0.0075 0.00269
0.0625 0.0662 0.0312
1.6 1.64 0.8
1 2 3
2 1.5 Slack bus
1
2 3 3
U
0"p
0"~
10% 10%
10% 10%
1
TABLE 6 The multiple solutions of the three-bus load flow equations with 0"p = 0"Q 10% =
Bus
Soln. No.
Deterministic solns.
Probabilistic solns.
Vl
Us
Vl
Us
0"1
0"2
P
1 2 3 4
0.9194 0.0553 0.3410 0.1589
-0.0913 --0.0771 --0.1313 -0.1262
0.9190 0.0557 0.3191 0.1708
-0.0913 -0.0771 -0.1307 -0.1268
0.0105 0.0100 0.0567 0.0548
0.0161 0.0161 0.0249 0.0272
0.4 -0.4 0.9 --0.9
1 2 3 4
0.9441 0.6430 0.0311 0.0345
-0.0595 -0.0669 - 0.0394 - 0.0431
0.9439 0.6430 0.0316 0.0343
-0.0595 -0.0688 - 0.0398 - 0.0426
0.0071 0.0077 0.0055 0.0071
0.0089 0.0105 0.0077 0.0089
0.3 0.1 - 0.2 - 0.2
1
1
0
1
0
0
0
m
2 3 4
1 1 1
0 0 0
1 1 1
0 0 0
0 0 0
0 0 0
m
a n d , a c c o r d i n g to t h e s t a r t i n g p o i n t , d i f f e r e n t s o l u t i o n s m a y be obtained.
4. APPLICATION OF GENERAL PROBABILISTIC LOAD FLOW EQUATIONS TO TYPICAL SMALL SIZE SYSTEMS The Klos-Kerner three-bus system was a n a l y s e d [8]. M u l t i p l e s o l u t i o n s of t h i s s y s t e m were obtained by the Tamura and Iwamoto m e t h o d d e s c r i b e d i n ref. 6. I n p u t d a t a of t h e s y s t e m a r e s h o w n i n T a b l e 5. T h e r e s u l t s o f t h e a n a l y s i s a r e s h o w n i n T a b l e 6. I t is s e e n t h a t t h e p r o b a b i l i s t i c l o a d flow e q u a t i o n s o f t h e three-bus system have four solutions connected w i t h t h e s o l u t i o n s of t h e d e t e r m i n i s t i c l o a d flow e q u a t i o n s . A l l t h e s o l u t i o n s a r e a d m i s s able b e c a u s e of the small i n p u t s t a n d a r d deviat i o n s e q u a l t o 10%.
m
4.2. S i x - b u s p o w e r s y s t e m The W a r d - H a l e six-bus system was analysed. I n p u t d a t a a n d m u l t i p l e d e t e r m i n i s t i c solut i o n s w e r e p r e s e n t e d i n ref. 7. T h e r e s u l t s o f t h e probabilistic analysis are shown in Tables 7 a n d 8. F r o m T a b l e 7 i t is s e e n t h a t t h e first a n d second solutions are very different in relation to the a v e r a g e v a l u e s a n d s t a n d a r d deviations. F r o m T a b l e 8 i t is s e e n t h a t i n c r e a s i n g t h e s t a n d a r d d e v i a t i o n s of the bus powers leads to loss of c o n v e r g e n c e of t h e i t e r a t i v e p r o c e s s , a s w a s t h e c a s e for t h e s e c o n d s o l u t i o n . T h e s e c o n d s o l u t i o n is s t e a d y s t a t e u n s t a b l e a n d m a y b e t h i s is t h e r e a s o n for t h e w o r s e conv e r g e n c e [7].
5. CONCLUSIONS In this paper the general nonlinear probab i l i s t i c e q u a t i o n s for l o a d flow c a l c u l a t i o n s
28 TABLE 7 Two solutions of the load flow e q u a t i o n s for the six-bus system with 0-p = 0-Q = 10% Bus
Deterministic solns.
Probabilistic solns.
V 1
V 2
V 1
V 2
0" I
0-2
1 2
1.0950 -0.2352
-1.047 -1.0746
1.0919 -0.2009
-0.1047 - 1.0729
0.0052 0.0897
0.0546 0.0168
2
1 2
0.8446 0.1830
-0.1965 -0.3556
0.8427 0.1826
-0.1966 -0.3482
0.0129 0.0485
0.0230 0.0236
0.82 -0.92
3
1 2
0.9495 0.3611
-0.1662 -0.2714
0.9478 0.3632
-0.1662 -0.2663
0.0113 0.0285
0.0186 0.0191
0.84 -0.92
4
1 2
0.8853 -0.0084
-0.2063 -0.4470
0.8828 0.0065
-0.2064 -0.4488
0.0135 0.0344
0.0304 0.0303
5
1 2
0.9227 0.1666
-0.2048 -0.3011
0.9208 0.1771
-0.2048 -0.3027
0.0133 0.0305
0.0223 0.0187
6
1 2
1.05 1.05
0 0
0 0
1
Soln. No.
0 0
1.05 1.05
0 0
-
-
1.00 1.00
0.67 0.85
0.88 -0.70 m m
TABLE 8 Two solutions of the load flow equations for the six-bus system with 0-p : Bus
Soln. No.
Deterministic solns.
Probabilistic solns.
vl
u2
Vl
0-Q
=
u2
20%
al
62
P
1
1 2
1.0950 - 0.2352
-0.1047 - 1.0746
1.0827 -0.1049 No iterative solution
0.0106
0.1096
1.00
2
1 2
0.8446 0.1830
-0.1965 -0.3556
0.8366 -0.1969 No iterative solution
0.0262
0.0460
0.82
3
1 2
0.9495 0.3611
-0.1662 -0.2714
0.9425 -0.1661 No iterative solution
0.0229
0.0373
0.84
4
1 2
0.8853 -0.0084
-0.2063 -0.4470
0.8752 -0.2067 No iterative s o l u t i o n
0.0273
0.0608
0.67
5
1 2
0.9227 0.1666
-0.2048 -0.3011
0.9149 -0.2048 No iterative solution
0.0269
0.0445
0.88
6
1 2
1.05 1.05
1.05 1.05
0 0
0 0
m
0 0
have been presented. It has been shown on a two-bus power system that the probabilistic load flow equations are quartic and have multiple solutions. Depending on the chosen starting point, different solutions can be obtained with the same input data. The probabilistic multiple solutions are connected with the deterministic multiple solutions. A lack of deterministic solutions leads to inadmissable probabilistic solutions.
0 0
E
F u r t h e r investigation should be made as to what kind of impact the multiple probabilistic solutions have on power system operations.
REFERENCES 1 B. Borkowska, Probabilistic load flow, I E E E Trans., P A S - 9 3 (1974) 752. 2 J. F. Dopazo, O. A. Klitin and A. M. Sasson, Stochastic load flows, I E E E Trans., P A S - 9 4 (1975) 299.
29 3 M. Sobierajski, Optimal stochastic load flows, Electr. Power Syst. Res., 2 (1979) 71. 4 M. Brucoli, F. Torelli and R. Napoli, Quadratic probabilistic load flow with linearly modelled dispatch, Electr. Power Energ. Syst., 3 (1985) 138. 5 M. Sobierajski, Probabilistic analysis of load flow solutions, Arch. Elektrotech. (Berlin), 69 (1986) 407. 6 Y. Tamura and S. Iwamoto, A method for finding multiple load flow solutions for general power systems, IEEE PES Winter Meeting, New York, 1980, Paper No. A 80043-0.
7 Y. Tamura, H. Mori and S. Iwamoto, Relationship between voltage instability and multiple load flow solutions in electrical power systems, IEEE Trans., PAS-102(1983) 1115. 8 A. Klos and A. Kerner, The non-uniqueness of load flow solutions, Proc. PSCC V, Dept. of Electr. and Electron. Eng., Queen Mary College, Univ. of London, 1975, 3.1/8. 9 K. S. Miller, Multidimensional Gaussian Distributions, Wiley, New York, 1964, p. 71.