ECE 422/522 Power System Operations & Planning/ Power Systems Analysis II 4 Active Power and Frequency Control Spring 2014 Instructor: Kai Sun 1
References Chapter 12 of Saadat s book Chapter 11.1 of Kundur s book (understand examples) Chapter 4 (Frequency Control) of the EPRI Tutorial 2
Background The frequency of a system depends on real power balance. Changes in real power affect mainly the system frequency, while reactive power is less sensitive to changes in frequency and is mainly dependent on changes in voltage magnitude. 3
Frequency Deviations Under normal conditions, the power system frequency in a large Interconnection (e.g. the EI) varies approximately 0.03Hz from the scheduled value When abnormal events, e.g. loss of a large generator unit, the frequency experiences larger deviations. 4
Control of Frequency As frequency is a common factor throughout the system, a change in real power demand at one point is reflected through the system by a change in frequency In an interconnected system with two or more independently controlled areas, in addition to control of frequency, the generation within each area has to be controlled so as to maintain scheduled power interchange. The control of generation and frequency is commonly referred to as Load Frequency Control (LFC), which involves Speed governing system with each generator Automatic Generation Control (AGC) for interconnected systems 5
Generator Control Loops For each generator, real power (or frequency) and reactive power (or voltage) outputs are controlled separately by LFC (Load Frequency Control) loop AVR (Automatic Voltage Regulator) loop. The LFC and AVR controllers are set for a particular steadystate operating condition to maintain frequency and voltage against small changes in load demand. Cross-coupling between the LFC and AVR loops is negligible because the excitation-system time constant is much smaller than the prime mover/governor time constants 6
Speed Governing System 7
Generator Model Initial values: P 0 = 0 T 0 P= r T P=P 0 + P, T=T 0 + T r = 0 + r P 0 + P=( 0 + r )(T 0 + T) 0 T 0 + 0 T + r T 0 so P= 0 T + r T 0 P m = 0 T m + r T m0 P e = 0 T e + r T e0 ( r T 0) =0 P m - P e = 0 ( T m - T e )+ r (T m0 -T e0 ) = T m - T e in per unit ( 0 =1) = T m -T e P m P e 8
Consider a frequency-dependent load model P e = P L +D r P L Frequency-insensitive load change D r Frequency-sensitive load change D Load damping constant, typically at 1~2, i.e. 1~2% change in load per 1% frequency change P m P e P m - P e =2Hs r P L D P m - P L -D r =2Hs r P m - P L =(2Hs+D) r =(Ms+D) r 9
Relationship between Load and Frequency D=2 10
Kundur s Example 11.1 A small system consists of 4 identical 500MVA generating units feeding a total load of 1,020MW. The inertia constant H of each unit is 5.0 on 500MVA base. The load varies by 1.5% for a 1% change in frequency. When there is a sudden drop in load by 20MW a. Determine the system block diagram with constants H and D expressed on 2,000MVA base b. Find the frequency deviation, assuming that there is no speedgoverning action 11
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=. /. 13
Governor Model See Bergen and Vittal s book for the model with time constants of key parts Classic Watt Centrifugal Governing System Speed changer Linkage mechanism Speed governor Hydraulic Amplifier 14
Governor Model P ref P v Without a governor, the generator speed drops when load increases The speed governor closes the loop for negative feedback control For stable operation, The governor reduces (rather than eliminate) the speed drop due to load increase. Usually, speed regulation R is 5-6% from zero to full load Governor output r /R is compared to the reference set power P ref P g = P ref - r /R Then, P g is transformed through the hydraulic amplifier to the steam valve/gate position command P v with time constant g r /R r (s) 15
Turbine Model P v P m The prime mover, i.e. the source of mechanical power, may be hydraulic turbines at water falls, steam turbines burning coal and nuclear fuel, or gas turbines The model for the turbine relates changes in mechanical power output P m to changes in gate or valve position P V T is in 0.2~2.0 seconds 16
Load Frequency Control block Diagram (s) 17
Load Frequency Control block Diagram (s) How to choose the value of R for a stable speed governing system? 1 1 2 1 1 1/ For a step load change, i.e. = / lim (s) = / (final value theorem) If the load is supported by n generators 1 1 1 1 1 2 18
Saadat s Example 12.1 19
(s) The open-loop transfer function is 20
Review: Stability of a Linear System Characteristic equation: 3 2 s s s K + 7.08 + 10.56 + 0.8+ = 0 A necessary and sufficient condition for a linear system to be stable: Poles of the system transfer function (i.e. roots of the characteristic equation) are only in the lefthand portion of the s-plane (i.e. having negative real parts) 21
Review: Routh-Hurwitz Stability Criterion Characteristic equation a n s n +a n-1 s n-1 + +a 1 s+a 0 =0 (a n >0) Routh table: 3 2 s s s K + 7.08 + 10.56 + 0.8+ = 0 For i>2, x ij =(x i-2,j+1 x i-1,1 x i-2,1 x i-1,j+1 )/x i-1,1 where x ij is the element in the i-th row and j-th column s s s s 3 2 1 0 1 10.56 7.08 0.8+ K 73.965-K 7.08 0 0.8+ K 0 Routh-Hurwitz criterion: No. of roots of the equation with positive real parts = No. of changes in sign of the 1 st column of the Routh table Necessary and sufficient condition for a linear system to be stable: The 1 st column only has positive numbers 22 s 1 row>0 if K<73.965 s 0 row>0 since K>0 So R=1/K>1/73.965=0.0135
Review: Root-Locus Method -z i is the i -th zero and -p j is j -th pole The locus of roots of 1+KG(s)H(s) begins at KG(s)H(s) s poles and ends at its zeros as K=0 No. of separate loci = No. of poles; root loci must be symmetrical with respect to the real axis The root locus on the real axis always lies in a section of the real axis to the left of an odd number of poles and zeros Linear asymptotes of loci are centered at a point (x, 0) on the real axis with angle with respect to the real axis x=[ j=1~n (-p j ) - i=1~m (-z i ) ]/(n-m) = (2k+1)/(n-m) k=0, 1,, (n-m-1) 23 When s= j3.25, R min =1/K=0.0135 So R>0.0135
Closed-loop transfer function with R=0.05pu (>0.0135): D w() s (1+ 0.2)(1 0.5) () s + = T s = s -D P ( s) (10s+ 0.8)(1+ 0.2 s)(1+ 0.5 s) + 1/ 0.05 L 2 0.1s + 0.7s+ 1 3 2 s s s = + 7.08 + 10.56 + 20.8 Steady-state frequency deviation due to a step input: 1 1 D wss = lim sd w( s) =-D PL =- 0.2 =-0.0096 p.u. s 0 D + 1/ R 20.8 D f =- 0.0096 60 = 0.576 Hz Note: The frequency is not restored to 60Hz (there is an offset) 24
Using the MATLAB toolbox with Saadat s book chp12.ex1.m sim12ex1.mdl 60 Frequency deviation step response Without LFC (Open-loop) Freq., pu Hz 55 50 45 0 20 40 60 80 100 t, sec 25
Saadat s Example 12.2 Note: two generators use different MVA bases. Select 1000MVA as the common MVA base R base1 Dw Sbase 1Dw Sbase 1 Dw Sbase 1 Dw = = = = DP DP S DP/ S S DP base1 base2 base2 base2 base2 R S = R base1 base1 base2 Sbase2 R 1000 1000 = (0.06) = 0.1 pu R = (0.04) = 0.08 pu 600 500 1 2 90 D P L = = 0.09 pu 1000 26
(a) D=0 -DP L -0.09 D wss = = =-0.004 pu 1 1 + 10+ 12.5 R R 1 2 D f =- 0.004 60 =-0.24 Hz f = f0 +D f = 60-0.24 = 59.76 Hz Dw -0.004 D P1 =- =- = 0.04 pu = 40 MW R 0.1 1 Dw -0.004 D P2 =- =- = 0.05 pu = 50 MW R 0.08 2 (b) D=1.5(900+90)/1000=1.485 (frequency dependent) -DPL -0.09 D wss = = =-0.00375 pu 1 1 + + D 10+ 12.5+ 1.485 R R 1 2 D f =- 0.00375 60 =-0.225 Hz f = f0 +D f = 60-0.225 = 59.775 Hz Dw -0.00375 D P1 =- =- = 0.0375 pu=37.5mw R 0.1 1 Dw -0.00375 D P2 =- =- = 0.0469 pu=46.9mw R 0.08 2 Unit 1 supplies 540MW and unit 2 supplies 450MW at the new operating frequency of 59.76Hz. Unit supplies 537.5MW and unit 2 supplies 446.9MW at the new operating frequency of 59.775Hz. The total change in generation is 84.4MW, i.e. 5.6MW less than 90MW load change, because of the change in load due to frequency drop. Dw D =- 0.00375 1.485 =-0.005572 pu = -5.6MW 27
Dw D P1 =- R D P = D P 1 Dw D P2 =- R 2 R R 1 2 2 1 D=1.485 D=0 Adjusting R 1 and R 2 may change the generation dispatch between Units 1 and 2 for economic concerns 28
Composite Frequency Response Characteristic (FRC) When analyzing LFCs for a multi-generator system, we may assume the coherent response of all generators to changes in system load represent them by an equivalent generator. M eq =2H eq = sum of the inertia constants of all generators = 29
Frequency response characteristic (FRC) or Frequency bias factor =D+1/R eq = P L / f (Unit: MW/0.1 Hz) FRC can be developed for any section of a power system. It relates the MW response of the system (or section of the system) to a change in frequency. FRC depends on: The governor droop settings of all on-line units in the system. The condition of the power system when the frequency deviation occurs. The condition of the power system includes current generator output levels, transmission line outages, voltage levels, etc. The frequency response of the connected load in the system. 30
FRCs of Different Interconnections 31
Limitations of Governor Frequency Control Governors do not recover frequency back to the scheduled value (60Hz) due to the required % droop characteristic. Governor control does not adequately consider the cost of power production so control with governors alone is usually not the most economical alternative. Governor control is intended as a primary means of frequency control. As such governor control is course and not suited to fine adjustment of the interconnected system frequency Other limitations (see Sec. 4.3 in EPRI Tutorial) Spinning Reserve is not considered Governors have dead-bands (not functioning in 60 0.03~0.04Hz) Depends on the type of Unit (Hydro: very responsive; Combustion turbine: may or may not be responsive; Steam: varies depending on the type) Governors may be blocked: a generator operator can intentionally prevent the unit from responding to a frequency disturbance 32
Automatic Generation Control (AGC) Adding supplementary control on load reference set-points of selected generators Controlling prime-mover power to match load variations As system load is continually changing, it is necessary to change the output of generators automatically Primary objective: LFC, i.e. regulating frequency to the specified nominal value, e.g. 60Hz, and maintaining the interchange power between control areas at the scheduled values by adjusting the output of selected generators Secondary objective: Generation dispatch, i.e. distributing the required change in generation among generators to minimize operation costs. AGC is bypassed during large disturbances and emergencies, and other emergency controls are applied. 33
AGC for an Isolated Power System An integral controller is added with gain K I Dw() s s(1 + tgs)(1 + tts) = -D P () s s(2 Hs+ D)(1 + t s)(1 + t s) + K + s/ R L g T I Applied to the system in Example 12.1 (Example 12.3) with K I =7 34
LFC for a Two-Area System Generators in each area is coherent, i.e. closely coupled internally Two areas are represented by two equivalent generators (modeled by a voltage source behind an equivalent reactance) interconnected by a lossless tie line P = E E sin d 1 2 12 12 XT X X X X T 12 1 2 1 tie 2 dp D P» D d = PD d = P( Dd -Dd ) 12 12 12 s 12 s 1 2 dd12 d 12 0 P s dp dd 12 1 2 = = cosd 12 d 12 0 E X E T d 12 0 P 12 P 12,max Slope=P s P s is the synchronizing power coefficient P 12,0 35 12,0 12
LFC with only the Primary Loop Consider a load change P L1 in area 1. Both areas have the same steady-state frequency deviation D w=d w1=dw2 DP -DP -D P =DwD m1 12 L1 1 D P +DP - 0 =DwD m2 12 2 D P12 =DwD2 -DP m 2 The change in mechanical power is determined by the governor speed characteristics -Dw D P m 1 = D P m 2 = R Solve and P 12 1 1 2 1 R2 -Dw R D P P w = -D 1 1 = -D ( + D ) + ( + D ) b + b R L1 L1 2 1 2 1 -( + D ) DP b D P = = (-D ) =0 2 L1 R2 2 12 PL 1 1 1 ( + D 1 2 1) + ( + D2) b + b R1 R2 36
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AGC with Frequency Bias Tie-Line Control The objective is to restore generation-load balance in each area A simple control strategy: Keep frequency approximately at the nominal value (60Hz) Maintain the tie-line flow at about schedule Each area should absorb its own load changes Area Control Error (ACE): supplementary control signal added to the primary LFC through an integral controller ACE n å = D P + BDw i ij i j= 1 B i : frequency bias factor (may or may not equal i ) Any combination of ACEs containing P ij and will result in steady-state restoration of the tie line flow and frequency deviation (the integral control action reduces each ACE i to 0) What composition of ACE signals should be selected is more important from dynamic performance considerations. 38
Comparing different B i s in ACE signals Consider a sudden load increase in Area 1: B i = i =D+1/R i b -DPL ACE =D P + bd w= (-D P ) + b =-DP 2 1 1 12 1 L1 1 L1 b1+ b2 b1+ b2 b -DP ACE =-D P + b D w=- (-D P ) + b = 0 2 L1 2 12 2 L1 2 b1+ b2 b1+ b2 Load change is taken care of locally B 1 =k 1, B 2 =k 2 b -D PL kb + b ACE =D P + kbd w= (-D P ) + kb =-DP 2 1 1 2 1 12 1 L1 1 L1 b1+ b2 b1+ b2 b1+ b2 PL ( k 1) ACE P k b -D - b =-D + bd w=- (-D P ) + kb =-DP 2 1 2 2 12 2 L1 2 L1 b1+ b2 b1+ b2 b1+ b2 Coefficient of ( 1 = 2 =20) k=2 k=1 k=1/2 1.5 1 0.75 0.5 0-0.5 What does k 1 mean? (k>1: the generator is more active in dynamics) 39
1 ~0 B i = i =D+1/R i 2 ~0 P ref1 P ref2 =0 P m1 >0 P m2 ~0 P 12 ~0 B i =2 i 1 2 In practice, only selected units participate in AGC, i.e. receiving supplementary control signals (ACE) P m1 P m2 40 P 12
NERC Balancing Authority A Balancing Authority (BA) is a part of an interconnected power system that is responsible for meeting its own load. Each BA operates an AGC system to balance its generation resources to its load requirements. The generation resources may be internal or purchased from other BAs and transferred over tie-lines between BAs. Similarly, load requirements may include internal customer load, losses, or scheduled sales to other BAs. 41
The control center is the headquarters of the BA, where the AGC computer system is typically located. All the data collected by the AGC system is processed in the control center. Based on the gathered data, the AGC signals are transmitted from the control center to the various generators currently involved in supplementary control to tell the generators what generation levels to hold (adjust the generator set-points). It is not necessary for the AGC system to regulate the output of all the generators in a BA. Most BAs have policies which require that as many units as needed are under control and able to respond to the BA s continual load changes. Those units that receive and respond to AGC signals are called regulating units. The number vary from a few for a small BA to 40~50 for the largest BA 42
NERC Balancing Authorities The EI is composed of approximately 90 BAs, which range in load size from over 130GW peaks to BAs that serve no load but simply use their generation for meeting interchange responsibilities. The WI (WECC) is composed of approximately 30 BAs with a distribution similar to the Eastern Interconnection. The ERCOT and Hydro Quebec are each operated as single BAs. 43
AGC for more than two areas By means of ACEs, the frequency bias tie-line control scheme schedules the net import/export for each area, i.e. the algebraic sum of power flows on all the tie lines from that area to the others 44
Influences from reserves Sufficient or insufficient spinning reserve Normal conditions: each area has sufficient generation reserve to carry out its supplementary control (AGC) obligations to eliminate the ACE Abnormal conditions: one or more areas cannot fully eliminate the ACE due to insufficient generation reserve; thus, there will be changes in frequency and tie-line flows (under both supplementary control and primary control) Operating reserve resources Spinning reserve: unloaded generating capacity (P ref,max -P ref ), interruptible load (controlled automatically) Non-spinning reserve: not currently connected to the system but can be available within a specific time period, e.g. 15 minutes. Examples are such as combustion turbines while cold standby and some interruptible load. Each BA shall carry enough operating reserves. 45
Kundur s Example 11.3 Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz (losing some spinning reserve) 46
Notes on AGC In an interconnect system, all generators with governors may respond to a generation/load change due to either f/r 0 or P ref 0 For load increase or generation loss, only generators with spinning reserves may increase their outputs up to their maximum output limits (by either governors or AGC) (see EPRI tutorial Sec. 4.4.2: Spinning reserves consist of unloaded generating capacity that is synchronized to the power system. A governor cannot increase generation in a unit unless that unit is carrying spinning reserves. An AGC system cannot increase a unit s MW output unless that unit is carrying spinning reserves. ) For load decrease, all generators may reduce their outputs as long as higher than their minimum output limits 47
Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz ACE i = B i D f +DP ij =0 with sufficient reserve or 0, otherwise P ref1 P ref2 DPmi -D PLi = DiD f +DPij DP Gi Without supplementary control (AGC): å å å - D PLi, = ( 1/ Ri+ Di) Df i i i = (1/ R+ D) Df 48
Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz Loss of 1,000MW load 1000 Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz Online generators with active governor control 49
Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz Loss of 1,000MW load 1000 322.56 Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz 50
Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz Loss of 1,000MW load 1000 1000 Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz 51
Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz Loss of 500MW generation carrying part of spinning reserve 1000 1000? Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz (losing some spinning reserve) 833.33-500=333.33MW 52
Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz Loss of 2,000MW generation, not carrying spinning reserve 1000 1,937.50 Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz Only held for the area with sufficient spinning reserve 53
Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz X Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz 54
Spinning reserve: 1,000 of 4,000MW B 1 =250MW/0.1Hz X Spinning reserve: 1,000 of 10,000MW B 2 =500MW/0.1Hz 55
Frequency response following the loss of a generator 56
Impact of Abnormal Frequency Deviations Prolonged operation at frequencies above or below 60Hz can damage power system equipment. Turbine blades of steam turbine generators can be exposed to only a certain amount of off-frequency operation over their entire lifetime. Steam turbine generators often have under- and over-frequency relays installed to trip the unit if operated at off-frequencies for a period A typical steam turbine can be operated, under load, for 10 minutes over the lifetime at 58Hz before damage is likely to occur to the turbine blades 57
Frequency Decay Due to Generation Deficiency Severe system disturbances can result in cascading outages and isolation of areas to form electrical islands. If such an islanded area does not have sufficient generation (and spinning reserve), it will experience a frequency decline, which is largely determined by frequency sensitive characteristics of loads. 58
Underfrequency Load Shedding In many situations, the frequency decline may lead to tripping of steam turbine generators by underfrequency protective relays, thus aggravating the situation further Underfrequency Load Shedding (UFLS) is a protection program that automatically trips selected customer loads once frequency falls below a specific value. The intent of UFLS is not to recover the frequency to 60 Hz but rather to arrest or stop the frequency decline. Once UFLS has operated, manual intervention by the system operators is likely required to restore the system frequency to a healthy state. 59
A typical UFLS setting for a North American utility may include three steps conducted by under-frequency relays, e.g., shedding 10% of the load at 59.3 HZ shedding 10% additional load at 59.0 HZ, and shedding 10% more at 58.7Hz 60
UFLS and Automatic Load Restoration in the Western Interconnection Maximum delay Minimum waiting time 61
Homework Problems 12.3 and 12.5~12.10 in Saadat s book (3 rd ed., Page 619), due by April 1st (Tue) in class 62