# Power System Operation & Control

Power System Operation & Control
Sun Chunshun
Changsha University of Science & Technology
13974819267, suncs65@csust.edu.cn

1

? Basic Concepts ? Parameters & Equivalent Circuits SteadyState ? Simple Power System Analysis ? Power Flow Analysis ? Active Power & Frequency Regulation ? Reactive Power & Voltage Regulation ? Faults (Symmetrical & Unsymmetrical)

Power System Analysis

? Short-Circuit of Synchronous Machines Transient ? Practical Calculation for f(3) ? Symmetrical Components and Sequence Networks ? Unsymmetrical Fault Calculations ? Basic Concepts Stability ? Steady-State or Small Signal Stability ? Transient Stability ? Dynamic Stability
2

? Power Flow Analysis(2) ? Economic DIspatch(2) Operation ? Unit Commitment(2) ? Optimal Power Flow(2)

Power System
º¬Interchange of power and energy

º¬AGC

? Control of Interconnected Systems(4) ? Voltage & Reactive Power Control(4) ? Advanced Topics(6)

3

Normal Secure State Economic operation Alert State Preventive control Emergency State Emergency control action (heroic measures) In Extremis State Emergency control action should be directed at avoiding total collapse. Restorative State Restorative control

All equality (E) and inequality (I) constraints are satisfied. The security level is below some threshold of adequacy. Inequality (I) constraints are violated.

States of Power System Operation

Both E and I constraints are violated. The violation of equality constraints implies that parts of system load are lost. This is a transitional state in which I constraints are met from the emergency control actions taken but the E constraints are yet to be satisfied.
4

Equality constraints (E) express balance between the generation and load demand. Inequality constraints (I) express limitations of the physical equipment.
5

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References
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7

9 Power System Control and Stability µçÁ¦ÏµÍ³¿ØÖÆÓëÎÈ¶¨ÐÔ (471p,1977,Iowa State University Press, P.M.Anderson, A.A.Fouad) 10 Power System Stability and Control µçÁ¦ÏµÍ³ÎÈ¶¨ÐÔÓë¿ØÖÆ (1196p,1994,McGraw-Hill Professional, Prabha Kundur) 11 Power System Stability and Control µçÁ¦ÏµÍ³ÎÈ¶¨ÐÔÓë¿ØÖÆ (353p,2006,CRC Press, Leonard L.Grigsby) 12 Power Systems in Emergencies£ºFrom Contingency Planning to Crisis Management ½ô¼±¿ØÖÆ(399p,2001,Wiley,U.G.Knight) 13 Real-Time Stability Assessment in Modern Power System Control Centers ÏÖ´úµçÁ¦ÏµÍ³¿ØÖÆÖÐÐÄÊµÊ±ÎÈ¶¨ÐÔÆÀ¹À(456p,2009,Wiley-IEEE Press,S.C.Savulescu) 14 Voltage Stability of Electric Power Systems (375p,1998,Springer,Thierry Van Cutsem, Costas Vournas) 15 Nonlinear Control Systems and Power System Dynamics ·ÇÏßÐÔ¿ØÖÆÏµÍ³ ÓëµçÁ¦ÏµÍ³¶¯Ì¬(201x2p,2001,Springer,Qiang Lu Â¬Ç¿, Yuanzhang Sun ËïÔª ÕÂ, Shengwei Mei Ã·ÉúÎ°)

8

16 Robust Control in Power Systems µçÁ¦ÏµÍ³Â³°ô¿ØÖÆ (207p,2005,Springer,Bikash Pal,Balarko Chaudhuri) 17 Robust Power System Frequency Control Â³°ôµçÁ¦ÏµÍ³ÆµÂÊ¿ØÖÆ (225p,2009,Springer,Hassan Bevrani) 18 Power Electronic Control in Electrical Systems µçÁ¦ÏµÍ³ÖÐµÄµçÁ¦µç×Ó¿ØÖÆ (451p,2002,Newnes,Enrique Acha,etc) 19 Inter-area Oscillations in Power Systems£ºA Nonlinear and Nonstationary Perspective µçÁ¦ÏµÍ³ÇøÓò¼äÕñµ´(278p,2009,Springer,Arturo Roman Messina) 20 HVDC and FACTS Controllers£ºApplications of Static Converters in Power System ¸ßÑ¹Ö±Á÷ÓëFACTS¿ØÖÆÆ÷(322p,2004,Kluwer,Vijay K.Sood) 21 Adaptive Voltage Control in Power Systems£ºmodeling,design and applications µçÁ¦ÏµÍ³×ÔÊÊÓ¦µçÑ¹¿ØÖÆ(170p,2007,Springer,Giuseppe Fusco,Mario Russo) 22 Optimal Economic Operation of Electric Power Systems(Mathematics in Science and Engineering,Vol.142) µçÁ¦ÏµÍ³×îÓÅ¾­¼ÃÔËÐÐ (298p,1979,Academic Press,M.E.El-Hawary,G.S.Christensen)

9

23 Optimization of Power System Operation µçÁ¦ÏµÍ³ÔËÐÐÓÅ»¯ (623p,2009,IEEE-Wiley,Jizhong Zhu) 24 Market Operations in Electric Power Systems£º Forecasting,Scheduling,and Risk Management µçÁ¦ÏµÍ³ÊÐ³¡ÔËÓª£ºÔ¤²â¡¢µ÷ ¶ÈÓë·çÏÕ¹ÜÀí(549p,2002,Wiley,Mohammad Shahidehpour,etc) 25 Modern Heuristic Optimization Techniques£ºTheory and Applications to Power Systems(616p,2008,Wiley-IEEE, Kwang Y.Lee, Mohamed A.ElSharkawi) 26 Reliability Evaluation of Power Systems_2nd Edition µçÁ¦ÏµÍ³¿É¿¿ÐÔÆÀ¹À (534p,1996,Plenum Press, Roy Billinton, Ronald N.Allan) 27 New Computational Methods in Power System Reliability µçÁ¦ÏµÍ³¿É¿¿ÐÔ ÐÂ¼ÆËã·½·¨(418p,2008,Springer,David Elmakias) 28 Risk Assessment of Power Systems£ºModels, Methods, and Applications µçÁ¦ÏµÍ³·çÏÕÆÀ¹À£ºÄ£ÐÍ¡¢·½·¨ÓëÓ¦ÓÃ(347p,2005,Wiley-IEEE,Wenyuan LiÀî ÎÄãä) 29 Power Distribution System Reliability£ºPractical Methods and Applications ÅäµçÏµÍ³¿É¿¿ÐÔ£ºÊµÓÃ·½·¨ÓëÓ¦ÓÃ(556p,2009,Wiley-IEEE,Ali A.Chowdhury,Don O.Koval)
10

30 Emerging Techniques in Power System Analysis(209p,2010,¸ßµÈ½ÌÓý³ö °æÉç+Springer, Zhaoyang Dong¶­³¯Ñô,Pei Zhang) 31 Power System State Estimation£ºTheory and Implementation µçÁ¦ÏµÍ³ ×´Ì¬¹À¼Æ£ºÀíÂÛÓëÊµÏÖ(336p,2004,Marcel Dekker,Ali Abur, Antonio Gomez Exposito) 32 Embedded Generation (IET Power and Energy,31)(293p,2008,IET,Nick Jenkins,etc)

11

Chap.1 Power Flow Analysis
1.1 Network Equations
1.1.1 Nodal voltage equations based on a nodal admittance matrix

& & I = YU
& & ? I1 ? ? Y11 Y12 L Y1n ? ?U1 ? ?? & ? ?& ? ? I 2 ? ?Y21 Y2 2 L Y2 n ? ?U 2 ? ? = ?? M ? ?M? ? L ? ? ? ?? ? & ? ?Yn1 Yn 2 L Yn n ? ?U ? & ? ?In ? ? ?? n?
& Ii Yi i = & & U i U j = 0 ( j = 1, 2 ,L, n , j ¡Ù i ) ¡ª self-admittance
& Ij Y ji = & & U i U j = 0 ( j = 1, 2 ,L, n , j ¡Ù i )

& & ? I1 ? ? Y11 Y12 Y13 ? ?U1 ? ?? & ? ?& ? ? I 2 ? = ?Y21 Y2 2 Y23 ? ?U 2 ? ? ?? & ? ? 0 ? ?Y ? ? ? 31 Y32 Y33 ? ?U 3 ?

? Properties of a nodal admittance matrix

symmetric sparse
12

1.1.2 Nodal voltage equations based on a nodal impedance matrix

& & I = YU

& & Y ?1 I = U

& & ZI =U

& & ? Z11 Z12 L Z1n ? ? I1 ? ?U1 ? ?? & ? ? & ? ? ? Z 21 Z 2 2 L Z 2 n ? ? I 2 ? = ?U 2 ? ? ?? M ? ? M ? L ? ?? ? ? ? & & Z n1 Z n 2 L Z n n ? ? I n ? ?U n ? ? ? ?? ? ? ?
& Ui Zi i = & & I i I j = 0 ( j = 1, 2 ,L, n , j ¡Ù i )

¡ª self-impedance (input impedance)

& Uj Z ji = & & I i I j = 0 ( j = 1, 2 ,L, n , j ¡Ù i ) ¡ª mutual impedance

? Z11 Z12 ? ? Z 21 Z 2 2 ?Z ? 31 Z 32

& Z13 ? ? I1 ? ? ?& ? Z 23 ? ? I 2 ? Z 33 ? ? 0 ? ?? ?

& ?U1 ? ?& ? = ?U 2 ? & ?U ? ? 3?

(transfer impedance)

? Properties of a nodal impedance matrix

symmetric full

13

1.2 Nodal Power Equations
~ S G 1 = PG 1 + j Q G 1 ? ? ? ~ S G 2 = PG 2 + j Q G 2 ? ?

~ S L1 = PL1 + j Q L1 ? ? ? ~ S L 2 = PL 2 + j QL 2 ? ?

~ S 1 = P1 + j Q1 = ( PG 1 ? PL1 ) + j ( Q G 1 ? Q L1 ) ~ S 2 = P2 + j Q 2 = ( PG 2 ? P L 2 ) + j ( Q G 2 ? Q L 2 )

? ? ? ? ?

& & & I 1 = I G1 ? I L 1 ? ? & & ? I 2 = I G2 ?I L 2 ? ?

14

1.2.1 Nodal power equations
? ~ & S1 = U 1 I1 2 2 = y s U1 sin¦Á s + y m U 1U 2 sin[ (¦Ä 1 ? ¦Ä 2 ) ? ¦Á m ] + j { y s U1 cos¦Á s ? y m U 1U 2 cos[ (¦Ä 1 ? ¦Ä 2 ) ? ¦Á m ]} = ( PG 1 ? PL1 ) + j (Q G 1 ? Q L1 )

? P ? P = y U 2 sin¦Á + y U U sin[ (¦Ä ? ¦Ä ) ? ¦Á ] L1 s 1 s m 1 2 1 2 m ? G1 ? 2 ? Q G 1 ? Q L1 = y s U1 cos¦Á s ? y m U 1U 2 cos[ (¦Ä 1 ? ¦Ä 2 ) ? ¦Á m ] ?

Number of variables: 4£¨6£© Number of equations: 2

Pi ( PG i ? PL i )

¡¢Qi ( QG

i

? Q Li )

¡¢ U i ¡¢¦Ä i

1.2.2 Classification of node types
PQ nodes: P¡¢Q are specified as known parameters PV nodes: P¡¢V are specified as known parameters Slack node: V=constant, ¦Ä = 00

15

1.2.3 Constrains for power flow calculation
Vi?mi n ¡Ü Vi ¡Ü Vi?m a x (PQ nodes)

? PG i?mi n ¡Ü PG?i ¡Ü PG i?m a x ( Slack node ) ? ? ? QG i?mi n ¡Ü QG?i ¡Ü QG i?m a x ( PV nodes¡¢Slack node ) ?

¦Äi ?¦Ä j < ¦Äi ?¦Ä j

ma x

( Some transmission lines )

16

1.3 Jacobi method & Gauss-Seidel method
1.3.1 Jacobi method
f 1 ( x1 , x 2 , L, x n ) = 0 ? ? f 2 ( x 1 , x 2 ,L , x n ) = 0 ? ? LL ? f n ( x1 , x 2 , L, x n ) = 0 ? ?

x1 = g 1 ( x 1 , x 2 , L , x n ) ? ? x 2 = g 2 ( x 1 , x 2 , L, x n ) ? ? LL ? x n = g n ( x 1 , x 2 , L, x n ) ? ?

( ( ( ( x11) = g 1 ( x1 0) , x 20) ,L, x n0) ) ? ? (1) ( 0) (0) ( 0) x 2 = g 2 ( x1 , x 2 ,L, x n ) ? ? LL ? ? ( ( ( ( x n1) = g n ( x1 0) , x 20) ,L, x n0) ) ?

¡­¡­

( ( ( ( x1 k +1) = g 1 ( x1 k ) , x 2k ) ,L, x nk ) ) ? ? ( k +1) (k ) (k ) (k ) x 2 = g 2 ( x1 , x 2 ,L, x n ) ? ? LL ? ? ( ( ( ( x nk +1) = g n ( x1 k ) , x 2k ) ,L, x nk ) ) ?

? Convergence condition

x i( k +1) ? x i( k ) < ¦Å 1 ( i = 1, 2 ,L, n )
( ( ( or ma x{ f i ( x1 k +1) , x 2k +1) ,L, x nk +1) ) } < ¦Å 2

17

1.3.2 Gauss-Seidel method ? Basic concept of Gauss-Seidel method
( ( ( ( x1 k +1) = g 1 ( x1 k ) , x 2k ) ,L, x nk ) ) ( x 2k +1)

( x nk +1)

? ? ( ( ( = g 2 ( x1 k +1) , x 2k ) ,L, x nk ) ) ? ? LL ? ( k +1) ( k +1) ( k +1) (k ) ? = g n ( x1 , x 2 ,L, x n?1 , x n ) ?

? Power flow calculation based on Gauss-Seidel method
~ & S i = U i ¡ÆY i j U j
j =1 n ? ?

( i = 1, 2 , L , n )

1 & Ui = Y ii

?? ? n ? Si ? & ? ? ? ¡Æ Y i jU j ? ( i = 1, 2 ,L, n ) ?U ? j =1 ? i j ¡Ùi ? ? ?

? ? n ? Pi ? j Q i i ?1 ? 1 & & & U i( k +1) = ? ¡Æ Y i jU (j k +1 ) ? ¡Æ Yi jU (j k ) ? ( i = 1, 2 ,L, n ) ? Y ii ? ? ( k ) ? j =1 j =i +1 ? U ? ? ?

ma x { U i( k +1) ? U i( k ) } < ¦Å

18

1.4 Newton-Raphson method
1.4.1 Basic concept of Newton method

f ( x) = 0
x ( 0)
? x ( 0) ( = x ? x ( 0) )

f ( x ) = f ( x(0) ) +

f ' ( x ( 0 ) ) ( 0 ) f '' ( x ( 0) ) f n ( x(0) ) ?x + ( ?x ( 0 ) ) n + L = 0 (?x ( 0) ) 2 + L + 1! 2! n!

? x ( 0) = ?
( 0)

f ( x (0) ) f ¡ä( x ( 0 ) )

x £½x
x
( k +1)

(1)

( 0)

+ ?x
?

=x

( 0)

f ( x (0) ) ? f ¡ä( x (0) )

£½x

(k )

f (x (k ) ) f ¡ä( x ( k ) )

? Convergence condition

f ( x (k ) ) < ¦Å 1
or ? x (k ) < ¦Å 2

19

1.4.2 Newton method for simultaneous nonlinear equations
? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?

f1( x1 , x 2 ,L, x n ) = 0 f 2 ( x 1 , x 2 , L, x n ) = 0 LL f n ( x 1 , x 2 ,L , x n ) = 0
( f1( x10) , x (0) ,L, x (0) ) ? n 2 ? ( 0) ( 0 ) ( 0) ? f 2 ( x 1 , x 2 , L, x n ) ? ? LL ? ( f n ( x10) , x (0) ,L, x(0) ) ? n ? 2 ?
F X ( k ) = ?J ( k )¦¤X ( k ) X ( k +1) = X ( k ) +¦¤X ( k )

( )

? ? f1 ? ? ? x1 0 ?? f2 ? = ? ? ? x1 0 ? ? ??f n ? 0 ? ? x1

? f1 ?x 2 0 ?f2 ?x 2 0 LL ?fn ?x 2 0

L L

? f1 ?x n ?f2 ?x n ?fn ?x n

L

? ? ( 0? ?? x10) ? ? ? ? ? ( 0) ? ? ?x 0? ? 2 ? ? ? ?? M ? ? ? ( 0) ? ? ?? x n ? ? ? 0? ?

? Convergence condition
( ( ( ma x f i x1 k ) , x 2k ) ,L, x nk )

{ (

) }< ¦Å

1

( i = 1, 2 ,L, n )

or ma x ? x i( k ) < ¦Å 2 ( i = 1, 2 ,L, n )
20

{

}

1.4.3 Power flow calculation based on Newton-Raphson method
n ? ? ~ &i ¡ÆY i j U j S i = Pi + jQi = U j =1

( i = 1, 2 ,L, n )
? Polar coordinates form
n ? Pi = U i ¡ÆU j (G i j cos¦Ä i j + Bi j si n¦Ä i j ) ? j =1 ? ? n ?Q = U ¡ÆU (G si n¦Ä ? B cos¦Ä ) i j ij ij ij ij ? i j =1 ?

? Rectangular coordinates form
n n ? Pi = ei ¡Æ (Gi j e j ? Bi j f j ) + f i ¡Æ (Gi j f j + Bi j e j ) ? j =1 j =1 ? ? n n ? Q = f ¡Æ (G e ? B f ) ? e ¡Æ (G f + B e ) i ij j ij j i ij j ij j ? i j =1 j =1 ?

? Nodal power error equations (polar coordinates form)
??Pi = Pi ? Vi ¡ÆV j (Gi j cos ¦Ä i j + Bi j sin ¦Ä i j ) = 0 j¦Å i ? ? ??Qi = Qi ? Vi ¡ÆV j (Gi j sin ¦Ä i j ? Bi j cos ¦Ä i j ) = 0 j¦Å i ?

21

? Process of solving equations
( ( U 1( 0) ,U 20) ,L,U m0) 0) ¦Ä 1( 0 ) , ¦Ä 2( 0) ,L, ¦Ä n(?1
(0) ? N ( 0 ) ? ?? ¦Ä ? ?1 ? ? L(0) ? ?U D ?U (0) ? ? ?

?? P (0) ? ? H ( 0) ? ? = ? ? (0) (0) ?? Q ? ?K ? ?

U i(1) = U i( 0) + ?U i( 0)

(i = 1,2,L, m)

¦Ä i(1) = ¦Ä i( 0) + ?¦Ä i( 0) (i = 1,2,L, n ? 1)

m a x ? Pi ( k ) , ?Qi( k ) < ¦Å
H = (Hi j )
( n ?1)¡Á( n ?1)

{

}

? ? ? Pi =? ? ?¦Ä j ?
? ? ? ?

? ? ? ?

( n ?1) ¡Á ( n ?1)

N = ( Ni j )

( n ?1)¡Ám

? ? ?Q i K = ( K i j ) m¡Á( n ?1) = ? ? ?¦Ä j ?

m¡Á( n ?1)

? ? ? Pi ? ? = ?U j ? ?U j ? ? ?
? ? ? ?

( n ?1) ¡Á m

L = (L i j )

m¡Ám

? ? ?Q i = ?U j ? ?U j ?

m¡Ám

? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

Hi j =

? ? Pi ?¦Ä j

= ?U i U j ( G i j s in¦Ä i j ? B i j c os¦Ä i j ) = ?U i U j ( G i j c os¦Ä i j + B i j s in¦Ä i j )

Ni j = U j Ki j =

? ? Pi ?U j

? ?Q i ?¦Ä j

= U i U j ( G i j c os¦Ä i j + B i j s in¦Ä i j ) = ?U i U j ( G i j s in¦Ä i j ? B i j c os¦Ä i j )

Lij = U j

? ? Pi ?U j

? ? ? ? ? ? ? ? ? ? ? ? ?

H ii =

? ? Pi ?¦Ä i

= ?U i2 B ii + Q i = ?U i2G ii ? Pi

Nii = U i Kii =

? ? Pi ?U i

? ?Q i ?¦Ä i

= U i2G ii ? Pi = U i2 B ii ? Q i

L ii = U i

? ? Pi ?U i

22

1.5 P-Q Decoupled method
?? P ? ?H ?? Q ? = ? ? K ? ? ? N ? ? ?¦Ä ? ? ? L ? ?U ?1?U ? ?? D ?

? Primary simplification ? P = ? H ?¦Ä ?? P ? ? H 0 ? ? ?¦Ä ? ?? Q ? = ? ? 0 L? ?U ?1?U ? ? Q = ? L U ?1?U D ? ? ? ?? D ? ? ? ? Secondary simplification ¡ú Fast decoupled method ? H i j ¡Ö U iU j Bi j ( i , j = 1, 2 ,L, n ? 1) ? ? ? L i j ¡Ö U i U j Bi j ( i , j = 1, 2 ,L, m ) ?
U1B12U 2 L U1B1, n ?1U n ?1 ? ? U1B11U1 ? ? U 2 B22U 2 L U 2 B2, n ?1U n ?1 ? ? U 2 B21U1 H= ? ? LL ? ? ?U n ?1Bn ?1,1U1 U n ?1Bn ?1,2U 2 L U n ?1Bn ?1,n ?1U n ?1 ? ? ? B12 L B1,n ?1 ? ?U1 0 ? ? B11 0 ? ?U1 ?? ? ?? B ? B22 L B2,n ?1 ? ? U2 U2 ? ? 21 ? =? ?? ? ?? ? LL O O ?? ? ?? ? U n ?1 ? ? Bn ?1,1 Bn ?1,2 L Bn ?1,n ?1 ? ? 0 U n ?1 ? ?0 ? ?? ? ?? = U D1 B¡äU D1

? U1B11U1 U1B12U 2 L U1B1mU m ? ? ? U B U U 2 B22U 2 L U 2 B2mU m ? L = ? 2 21 1 ? ? LL ? ? ?U m Bm1U1 U m Bm 2U 2 L U m BmmU m ? ? ? 0 ? ? B11 B12 L B1m ? ?U1 0 ? ?U1 ? ? ?? ?? U2 U2 ? ? ? ? B21 B22 L B2m ? ? = ?? ?? ? ? O LL O ?? ? ? ?? U m ? ? Bm1 Bm 2 L Bmm ? ? 0 Um ? ?0 ?? ? ? ?? = U D 2 B¡ä¡äU D 2

23

Chap.2 Economic Dispatch
2.1 Characteristics of Power Generation Units
2.1.1 Characteristics of Steam Units

Boiler-turbine-generator unit

Input-output curve of a steam turbine generator Heat=f(P), or Fuel cost=f(P)

24

Incremental heat (cost) rate characteristic H/ P =f(P), or F/ P =f(P) Unit (net) heat rate characteristic of a steam turbine generator unit H/P =f(P)

Approximate representations of the incremental heat rate curve H/ P =f(P)

25

2.1.2 Variations in Steam Unit Characteristics

Characteristics of a steam turbine generator with four steam valves

26

2.1.3 Cogeneration Plants

steam electricity

Industrial process¡¢district heating ¡¢

Fuel input required for steam demand and electrical output for a single extraction steam turbine generator
27

2.1.4 Hydroelectric Units

Hydroelectric unit input-output curve

Incremental water rate curve for hydroelectric plant

28

Input-output curves for hydroelectric plant with a variable head

Input-output characteristics for a pumped storage hydroplant with a fixed, net hydraulic head
29

2.2 Economic Dispatch of Thermal Units
2.2.1 The economic dispatch problem
Objective function

F = F + F +L+ F = ¡ÆF (P) T 1 2 N i i
i= 1

N

Constrain function

¦Õ = 0 = P ? ¡ÆP i load
i= 1

N

P,min ¡Ü P ¡Ü P,max i i i

Lagrange function

L = F + ¦Ë¦Õ T
?L dF (P) = i i ?¦Ë = 0 ?P dP i i
dF i =¦Ë dP i
equal incremental cost criterion

Optimal condition

or

30

2.2.2 Thermal system dispatching with network losses consideration
Objective function

F = F + F +L+ F = ¡ÆF (P) T 1 2 N i i
i= 1

N

Constrain function

¦Õ = 0 = P + P ? ¡ÆP load loss i
1 i=

N

P,min ¡Ü P ¡Ü P,max i i i
coordination equations

Lagrange function

L = F + ¦Ë¦Õ T
?L dF (P) ?P i i = + ¦Ë loss ?¦Ë ?P dP ?P i i i

Optimal condition

or

? ?P ? dF (P) i i ?¦Ë?1? loss ? = 0 ? dP ?P ? i ? i ? dF(P) 1 i i ? =¦Ë dP 1? ?P loss i ?P i =

"penalty factor" of bus i

31

2.2.3 Transmission system effects

Ploss ¡ú Pdemand ¡ü
cause transmission lines overloaded power flow equations generation scheduling equations ignore the constrains on flows loss formulae OPF constrains on power flow through the network elements

include the complete transmission system model with no transmission effects considered:

dF i =¦Ë dP i dF(P) 1 i i ? =¦Ë including the effects of incremental losses: ?P dP 1? loss i 1 ?P i pf j pfi = pfi ? Fi¡ä = pf j ? F j¡ä Fi¡ä = ? F j¡ä ?Ploss 1? pfi ?Pi
32

2.2.4 The ¦Ë-iteration method

¿ªÊ¼ ¸ø¶¨µü´ú³õÊ¼Öµ K=0

¦Ë ( 0)

¦Ë ( k +1) = ¦Ë ( k ) ? ?¦Ë
k = k +1

¦Ë ( k +1) = ¦Ë ( k ) + ?¦Ë
k = k +1
Y

ÇóÓë

¦Ë (k )

¶ÔÓ¦µÄ

Pi (k )
Y

K´ïµ½¹æ¶¨µÄ´ÎÊýÂð ´ïµ½¹æ¶¨µÄ´ÎÊýÂð? ´ïµ½¹æ¶¨µÄ´ÎÊýÂð N

¡Æ Pi ( k ) < Pload + ¦Å
i =1

N

?

N Y

¡Æ Pi ( k ) > Pload + ¦Å
i =1

N

?

N ½áÊø

33

2.2.5 Gradient methods of economic dispatch a scalar to guarantee that ? ?f ? the process converges ? Gradient search ? ?x ? ? 1? x (1) = x (0) ? ?f ? ¦Á ?f = ? M ? f ( x) ? ?f ? ? ? ? ?xn ? direction of maximum ascent ? ? direction of maximum descent ? Economic dispatch by gradient search
Lagrange function

L = F + ¦Ë¦Õ = ¡ÆF (P) + ¦Ë(P + ¡ÆP) T i i load i
i= 1 i= 1

N

N

? ?L ? ? d ? F1 ( P ) ? ¦Ë ? 1 ? ?P ? ? dP ? 1 ? ? 1 ? ? ?L ? ? d ? F2 ( P2 ) ? ¦Ë ? ? ?P ? ? dP ? 2? ? 2 ? ?L = ? M ? = ? M ? ? ? ? ? ?L ? ? d ? FN ( PN ) ? ¦Ë ? ? ?PN ? ? dPN ? ? ? ? ? N ?L ? ? ? ? Pload ? ¡Æ Pi ?? ¦Ë ? ? ? i =1 ? ? ? ?

x ( k +1) = x ( k ) - (?L)¦Á

?P ? 1 ?P ? ? 2? x=? M ? ? ? ? PN ? ?¦Ë ? ? ?

34

2.2.6 Newton method
g( x) = 0
g ( x + ?x ) = g ( x ) + [g¡ä( x )]?x = 0
?x = -[g¡ä( x )] g ( x )
?1

? g1 ( x1, x2 ,L, xn ) ? ? g ( x , x ,L, x ) ? n ? g( x) = ? 2 1 2 ? ? M ? ? g n ( x1, x2 ,L, xn )? ?

? Lagrange function

L = F + ¦Ë¦Õ = ¡ÆF (P) + ¦Ë(P T i i load
i= 1

N

? ?g1 ? ?x ? 1 ? ?g 2 g¡ä( x ) = ? ?x1 ? ? ? ? ?g n N ? + P) ? ?x1

¡Æi
i= 1

?g1 ?g1 ? L ?x2 ?xn ? ? ?g 2 ?g 2 ? L ?x2 ?xn ? ? M ? ? ?g n ?g n ? L ?x2 ?xn ? ?

? Aim of economic dispatch ?L =

?L =0 ?x

?d2L d2L d2L d2L ? L ? 2 ? dP dPdP dPdP dPd¦Ë ? 1 2 1 n 1 1 ? ? d2L d2L d2L d2L ? L ? ? 2 ?dPdP dP dPdP dPd¦Ë? 2 1 2 n 2 2 ? ? ? ? ? M ??x ?L? = ? ? ? ? 2 2 2 2 ?dL dL dL dL ? L 2 ? ? dP P dPdP dP dPd¦Ë ? n n ? n1 n 2 ? d2L d2L d2L d2L? L ? ? d¦ËdP d¦ËdP d¦ËdP d¦Ë2 ? ? 1 2 n ? ?

35

2.2.7 Economic dispatch with piecewise linear cost functions

(a) start with all of them at Pmin (b) then begin to raise the output of the unit with the lowest incremental cost segment (c) If this unit hits the right-hand end of a segment, or if it hits Pmax, we then find the unit with the next lowest incremental cost segment and raise its output (d) Eventually, we will reach a point where a unit's output is being raised and the total of all unit outputs equals the total load, or load plus losses
36

2.2.8 Economic dispatch using dynamic programming
? nonconvex input-output curves multiple values of MW output for any given value of incremental cost

? cannot use an equal incremental cost methodology

? dynamic programming

= an allocation problem

Ftotal = ¡Æ
N

Tmax

t =1 i =1

F ( Pit ) ¡Æ
( t = 1,2,L, tmax )

N

generate a set of outputs, at discrete points, for an entire set of load values

t ¡Æ Pit = Pload i =1

Pit +1 = Pi t + ?Pi
rate limit

? ?Pi max ¡Ü ?Pi ¡Ü ?Pi max

37

2.2.9 Base point & participation factors ? base point: a given schedule Load changes (by a reasonably small amount) how much each generating unit needs to be moved (i.e., "participate" in the load change)

? new schedule

?¦Ëi = ?¦Ë ? Fi¡ä¡ä(i0 )?Pi

?P = ?¦Ë¡ä¡ä , ?P2 = ?¦Ë¡ä¡ä ,L, ?PN = ?¦Ë¡ä 1 F F F¡ä
1 2 N

?PD = Pload + Ploss ?1? = ?P + ?P2 + L + ?PN = ?¦Ë ¡Æ ? ? 1 ? ? i =1 ? Fi¡ä¡ä?
N
1 1 ¡Æ (F ¡ä¡ä) N i =1
i

?Pi = ?PD

Fi¡ä¡ä

assume that both and F¡ä exist ¡ä
i

F¡ä i

38

2.2.10 Economic dispatch versus unit commitment Problem ED N units already connected to the system a given demand to be served Purpose find the optimum operating policy for these N units N units available Problem a forecast of the demand to be served
may be extended over some period of time ( 24 h or a week)

UC

Definition: Given that there are a number of subsets of the complete set of N generating units that would satisfy the expected demand, which of these subsets should be used in order to provide the minimum operating cost? more complex than ED more difficult to solve mathematically (integer variables)

39

Chap.3 Unit Commitment
3.1 Introduction
? Load variation: Hourly, daily, seasonally

? Unit commitment: commit enough units and leave them on line ? It is quite expensive to run too many generating units.

40

? Example 5-1: Unit combination
Unit 1:

P,min =150M , P,max = 600M W 1 W 1
H1 = 510 + 7.2P + 0.00142P2(M / h) Btu 1 1 Fuel cost1 =1.1 / M R Btu

Unit 2:

P2,min = 100MW, P ,max = 400MW 1
H 2 = 310 + 7.85 P2 + 0.00194 P22 ( MBtu / h )

Fuel cost 2 = 1.0R / MBtu
Unit 3:

P,min = 50M , P,max = 200M W 3 W 3
H3 = 78 + 7.97P + 0.00482P2(M / h) Btu 3 3 Fuel cost 3 =1.2R/ M Btu

41

? Example 5-2: Unit commitment schedule using shut-down rule
Unit 1 Unit 2 Unit 3 ¡Ì

PD > 1000MW
600MW < PD < 1000MW PD < 600MW

¡Ì ¡Ì ¡Ì

¡Ì ¡Ì

? other constraints ? ? other phenomena ?

42

3.1.1 Constraints in Unit Commitment
¡Ý %10 Pmax , or Pmax ,loaded
be allocated between fast- and slow-responding units

suncs1

? Spinning reserve

be spread around the power system quick-start diesel or gas-turbine units "scheduled reserves" or "off-line" reserves hydro-units pumped-storage hydro-units
time to come up to full capacity

Minimum up time

? Thermal Unit Constraints

Minimum down time Crew constraints

Hydro-Constraints

? Other Constraints

Must Run Fuel Constraints
43

»ÃµÆÆ¬ 43 suncs1 Each individual power system, power pool, reliability council, and so forth, may impose different rules on the scheduling of units, depending on the generation makeup, load-curve characteristics, and such.
suncs65, 2011-11-27

3.2 Unit Commitment Solution Methods

? Assumptions

have N units to commit & dispatch M load levels & operating limits on the N units: any one unit can supply the individual loads, and any combination of units can also supply the loads

? The total number of combinations enumeration (brute force)
2 N N C1 + C N + L + C N -1 + C N = 2 N ? 1 N

for each period (hour)

M =24 h

(2

N

?1

)

M

for the total period of M N

(2

N

?1

)

24

Priority-list schemes

5 10 20

6.2 ¡Á 1035

? Solution methods

Dynamic programming (DP) Lagrange relation (LR)

1.73 ¡Á 1072 3.12 ¡Á 10144
Too big

40

44

3.2.1 Priority-list Methods
? Example 5-3: Construct a priority list for the units of Example 5-1
the full-load average production cost Unit 1 2 3 Full Load Average Production Cost ( R / MWh ) 9.79 9.48 11.188

priority-list based on the average production cost Unit 1 2 3

commitment scheme (ignoring min up/down time, start-up costs, etc.) Combination 2+1+3 2+1 2 Min MW 300 250 100 Max MW 1200 1000 400

R / MWh Min MW
9.48 9.79 11.188 100 150 50

Max MW 400 600 200

45

? Priority-list Schemes ? At each hour when load is dropping, determine whether dropping the next unit on the priority list will leave sufficient generation to supply the load plus spinning-reserve requirements. If not, continue operating as is; if yes, go on to the next step. ? Determine the number of hours, H, before the unit will be needed again. That is, assuming that the load is dropping and will then go back up some hours later. ? If H is less than the minimum shut-down time for the unit, keep commitment as is and go to last step; if not, go to next step. ? Calculate two costs. The first is the sum of the hourly production costs for the next H hours with the unit up. Then recalculate the same sum for the unit down and add in the start-up cost for either cooling the unit or banking it, whichever is less expensive. If there is sufficient savings from shutting down the unit, it should be shut down, otherwise keep it on. ?Repeat this entire procedure for the next unit on the priority list. If it is also dropped, go to the next and so forth. Power Generation,Operation,and Control_2nd Edition ·¢µç¡¢ÔËÐÐÓë¿ØÖÆ(592p, 1996, Wiley, Allen J.Wood,Bruce F.Wollenberg), page141

46

? Priority-list Schemes (another description) Step (1): Compute the minimum average production cost of all units, and order the units from the smallest value of ¦Ì min . Form the priority list. Step (2): If the load is increasing during that hour, determine how many units can be started up according to the minimum downtime of the unit. Then select the top units for turning on from the priority list according to the amount of load increasing. Step (3): If the load is dropping during that hour, determine how many units can be stopped according to the minimum up time of the unit. Then select the last units for stopping from the priority list according to the amount of load dropping. Step (4): Repeat the process for the next hour.

Optimization of Power System Operation µçÁ¦ÏµÍ³ÔËÐÐÓÅ»¯(623p,2009,IEEE-Wiley, Jizhong Zhu), page254

47

? Priority-list Schemes (the third description) ? Check at the end of every hour of operation. if the load demand has fallen. If the demand has decreased check if the last unit in the priority list is dropped, the load demand can be met, satisfying the spinning reserve requirement. Status quo is maintained if the demand cannot be met. ? If it is possible to drop the unit in step I, then determine the number of hours '"h" before the unit is required again for service. If this "h" is less than the shut down and start up times for the unit, it has to be left in service without removal. ? Then, calculate the cost of floating the unit within the system without supplying any generation and the cost of shut down and start up processes and if there is sufficient savings from shutting it down, and starting it again for service it can be removed. ? The process is to be repeated for the next unit on the priority list and continued. Operation and Control in Power Systems(422p,2008,BS Publications, P.S.R.Murty) page175

48

3.2.2 Dynamic-Programming Solution
©~suitable only to multistage decision processes ©~a decision taken at one stage does not influence the preceding events in the sequence but affect only what follows later ? Example: The least cost route E 1 A
the cost of transport along the route

1 1 4 H 1

F 6 G 3 D
the goal

2 B 3 C 2

allowed Traffic direction

49

E 1
0 CDG = 3 = CDG

1 1 4 H 1

F 6 G 3 D

D¡úG

A

0 0 H¡úG via D CHG = CHD + CDG =1+ 3 = 4 2
0 0 C¡úG via D CCG = CCD + CDG = 2 + 3 = 5

B 3

C

2

F¡úG E¡úG B¡úG A¡úG

0 0 CFG = m CFG, CFH + CHG = m [6, 1+ 4] = 5 in in
0 CEG 0 0 + CFG , CEC + CCG = min [1 + 5, 4 + 5] = 6 EFHDG EF

[ = min [C
[

]

]

0 0 CBG = CBC + CCG = 3+ 5 = 8 BCDG
0 0 0 CAG = min CAB + CBG , CAE + CEG = min [2 + 8, 1 + 6] = 7 AEFHDG

]

Conclusion: the minimum cost to move from one location P to another location R via Q is given by
0 CPR = CPQ + CQR

50

? Unit Commitment by Dynamic Programming the number of units available ©~data required the cost characteristics of these units the load cycle ©~total load
N ?1 i =1

¡Æ Pi
i =1

N

¡Æ Pi

?N ? ©~total operating cost f N ? ¡Æ P ? i ? i =1 ?

? N?1 ? the cost of operating (N-1) units F ?1? ¡ÆP ? to supply a reduced demand N i ? i=1 ?

fN (P ) the cost of operating the Nth unit at a N power output of PN

N ?1 ?N ? ? ? ? N ?1 ? f N ? ¡Æ Pi ? = f N ? PN + ¡Æ Pi ? = f N ( PN ) + FN ?1 ? ¡Æ Pi ? ? i =1 ? ? i =1 ? ? i =1 ?

? ? N ?? ? ?N ? ? N ?1 ?? FN ? ¡Æ Pi ? = min ? f N ? ¡Æ Pi ?? = min ? f N ( PN ) + FN ?1? ¡Æ Pi ?? ? i =1 ? ? i =1 ?? ? ? i =1 ?? ?
51

? Process of solving UC problems by DP

F1 ( P ) = f1 ( P ) 1 1
F2 ( P + P2 ) = min [ f 2 ( P2 ) + F1 ( P )] 1 1 F3 ( P + P2 + P3 ) = min[ f 3 ( P3 ) + F2 ( P + P2 )] 1 1

M
? ?N ? ? N ?1 ?? FN ? ¡Æ Pi ? = min ? f N ( PN ) + FN ?1 ? ¡Æ Pi ?? ? i =1 ? ? i =1 ?? ?

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