Tare A Lab

EXPERIMENT 9

LOAD-FREQUENCY DYNAMICS OF SINGLE-AREA AND TWO-AREA POWER SYSTEMS 9.1 AIM To become familiar with the modelling and analysis of load-frequency and tie-line flow dynamics of a power system with load-frequency controller (LFC) under different control modes and to design improved controllers to obtain the best system response. 9.2 OBJECTIVES i. To study the time response (both steady state and transient) of area frequency deviation and transient power output change of regulating generator following a small load change in a single-area power system with the regulating generator under “free governor action”, for different operating conditions and different system parameters. ii. To study the time response (both steady state and transient) of area frequency deviation and turbine power output change of regulating generator following a small load change in a single- area power system provided with an integral frequency controller, to study the effect of changing the gain of the controller and to select the best gain for the controller to obtain the best response. iii. To analyse the time response of area frequency deviations and net interchange deviation following a small load change in one of the areas in an inter connected two-area power system under different control modes, to study the effect of changes in controller parameters on the response and to select the optimal set of parameters for the controller to obtain the best response under different operating conditions. 9.3 SOFTWARE REQUIRED ‘LOAD FREQUENCY CONTROL’ module of AU Powerlab or equivalent. 9.4 THEORETICAL BACKGROUND 9.4.1 Introduction Active power control is one of the important control actions to be performed during normal operation of the system to match the system generation with the continuously changing system load in order to maintain the constancy of system frequency to a fine tolerance level. This is one of the foremost requirements in providing quality power supply. A change in system load causes a change in the speed of all rotating masses (Turbine – generator rotor systems) of the system leading to change in system frequency. The speed change from synchronous speed initiates the governor control (Primary control) action resulting in all the participating generator – turbine units taking up the change in load, stabilizing the system frequency. Restoration of frequency to nominal value requires secondary control action which adjusts the load-reference set points of selected (regulating) generator – turbine units. The primary objectives of automatic generation control (AGC) are to regulate system frequency to the set nominal value and also to regulate the net interchange of each area to the scheduled value by adjusting the outputs of the regulating 9-1

units. This function is referred to as load – frequency control (LFC). The details of modelling and analysis of LFC are briefly presented in the following sections. 9.4.2. Load-Frequency Control In An Interconnected Power System An interconnected power system is divided into a number of “control areas” for the purpose of load- frequency control. When subjected to disturbances, say, a small load change, all generator – turbine units in a control area swing together with the other groups of generator – turbine units in other areas. Hence all the units in a control area are represented by a single unit of equivalent inertia and characterized by a single (area) frequency. Since the area network is “strong”(all the buses connected by adequate capacity lines), all the bus loads in a control area are assumed to act at a single load point and characterized by a single equivalent load parameter. The different control areas are connected by relatively “weak” tie-lines. A typical n-area power system is shown in Fig9.1.

Area 1 Other areas

Area i

PNIi

Area n Fig 9.1 Multi- Area Power System For successful operation of an interconnected power system the following operating principles are to be strictly followed by the participating areas: i. Under normal operating conditions each control area should strive to meet its own load from its own spinning generators plus the contracted (scheduled) “interchange” (import / export) between the neighboring areas. ii. During emergency conditions such as sudden loss of generating unit, area under emergency can draw energy as emergency support from the spinning reserves of all the neighboring areas immediately after it is subjected to the disturbances but should bring into the grid the required generation capacity from its “hot” and “cold” reserves to match the lost capacity and to enforce operating principle (i)

9-2

Satisfaction of principle (i) during normal operation requires a load-frequency controller for each area which not only drives the area frequency deviation to zero but also the “net interchange” of that area to zero under steady- state condition. “Net interchange” of area i, NIi is defined as the algebraic sum of the tie- line flows between area i and other connected areas (Fig 9.1) with tie-line flow out of area i taken as positive and is given by NIi = ™3 j ¼ .i

(9.1)

ij

ZKHUH .i is the set of all areas connected to area i 9.4.3. Area Load-Frequency Mechanism For the purpose of development of model for load-frequency dynamics let us consider a two – area power system connected by a tie-line (Fig 9.2)

H1 , D1 , R1 ,

P12

P21

TG1, TT1

H2 , D2 , R2 , TG2, TT2

Fig 9.2 A Two- Area Power System Let us first model the load-frequency mechanism of area i ; i=1,2... To simplify the model let us assume that area i has got only one “regulating” generator – turbine unit (Reference1 gives the details of model where many regulating units are present) )RUDVPDOOFKDQJHLQFUHDVH LQDUHDORDG û3Di MW, the speed of the rotating units (or the equivalent unit) changes (reduces) due to change (depletion) of kinetic energy stored; leading to FKDQJHIDOO LQDUHDIUHTXHQF\ ûIi,Hz. The generator senses the change (reduction) in speed and changes (increases) the steam / water valve / gate opening leading to change (increase) in turbine RXWSXW û3Ti , MW. During the transient period following the load change, the power balance equation of area i at time instant t may be written as û3Ti (t) –û3Di (t) = d/dt (Ei) + DiûIiW  û3ij (t)

MW ;

i = 1, 2

where d/dt (Ei) = rate at which the kinetic energy Ei of the equivalent unit in area i changes DiûIi

= change in area load consumption in MW due to frequency-sensitive “old” load

Di = ˜3'i / ˜Ii = load-damping constant in MW / Hz which gives the change in load due to change in frequency. û3ij = change in tie-line flow Pij from area i to area j measured at area i. 9-3

(9.2)

Since the kinetic energy is proportional to the square of the speed or frequency and noting that the predisturbance values of Ei and fi are E0i and f0i, we can write Ei = E0i (fi / f0)2 = E0i ((f0 ûIi) / f0)2 = Ei ûIi / f0)2 Ei §( 0i ûfi / f0)

dEi / dt = (2E0i / f0 GGW ûIi)

(9.3)

Substituting equation (9.3) in equation (9.2), dividing equation (9.2) by rated area capacity Pri and dropping the argument t, we get û3Ti –û3Di = (2Hi / f0 GGW ûIi) + DiûIi û3ij

p.u. MW

7KH û3¶VDUHQRZLQSHUXQLWDQG'LQSHUXQLW0:SHUKHUW]+ constant” of area i and is defi ned as Hi û( 0i / Pri

i

(9.4)

is called the per unit “inertia

MWs / MW (or sec)

(9.5)

Laplace transformation of equation (9.4) yields û3Ti (s) –û3Di (s) = (2 Hi / f0 V ûIi(s) + DiûIiV  û3ij (s) p.u. MW ; i=1,2

(9.6)

Tie – line flow model In the two-area system with the resistance of tie line neglected û321 = -û312 MW

(9.7)

+HQFHLWLVHQRXJKZHGHYHORSH[SUHVVLRQIRU

û312 alone.

The active power flow in line i-j (with negligible resistance) is given by P12 = (V1 V2 / x12 VLQ /1 –/2) = Pmax12VLQ /12

(9.8)

where V1, V2 = Voltage magnitude of buses 1 and 2 x12 = reactance of line 1-2 /12 OLQHSKDVHDQJOH 

/1 –/2

Pmax12 = tie line capacity $VVXPLQJWKDWWKHSUHGLVWXUEDQFHOLQHSKDVHDQJOHLV the disturbance is given by

/012, the tie- line power derivation due to

û312 = (˜312 / ˜/12  û/12 = P0s12û/12  /012

9-4

MW

(9.9)

where P0s12 = Pmax12FRV /012 = “synchronizing coefficient” of the line The freqXHQF\GHYLDWLRQ ûILVUHODWHGWRWKHUHIHUHQFHDQJOH

û/E\

ûI  Œ GGW /0 û/   Œ GGW û/ +] or



t û/  œ ûIGWUDG



t t û312  Œ3 s12 ( œ ûI1 dt - œ ûI2 dt )

(9.10)

Expressing tie-OLQHSRZHUGHYLDWLRQLQWHUPVRI 0

ûIHTXDWLRQ EHFRPHV

MW

On Laplace transformation  û312V   Œ3 s12V  ûI1 (s) –ûI2(s))

(9.11)

Equation (9.11) is in MW or p.u. MW depending on whether the parameter P0s12 is in MW or p.u. MW (to a base of area 1 capacity) 9.4.4

Modelling of Governor and Turbine

Governor with speed – droop characteristics Governor is provided with a speed- droop characteristics so as to obtain stable load deviation between units operating in parallel. The ideal steady- state speed versus load characteristics of the generating unit is shown in Fig 9.3 fNL Slope = -R

Frequency or speed (p.u) fFL

0

1.0 Power output (or) valve / gate position (p.u)

Fig 9.3 Steady-State Speed-Load Characteristics of a Governor with Speed Droop 9-5

The negative slope of the curve, R, is referred to as “ Percent speed regulation or droop” and is expressed as Percent R = Percent speed or frequency change Percent power output change

x 100

=((fNL – fFL) / f0 ) x 100 where fNL = steady-state frequency at no load fFL = steady- state frequency at full load f0 = nominal or rated frequency For example a 5% droop means that a 5% frequency deviation causes 100% change in valve position (û;v RUSRZHURXWSXW û3 Control of generating unit power output The output of a generating unit at a given system frequency can be varied only by changing its “load reference point” which is integrated with the speed governing mechanism. The block diagram of a governor with the governor droop R, the time constant of hydraulic amplifier TG and the load reference set point is shown in Fig 9.4

ûI

1/R

GG(s)

-

1/(1+sTG)

û;v

+ Load reference

Fig 9.4 Governor with Speed Load Reference Set Point The adjustment of load reference set point is accomplished by operating the “speed changer motor”. This in effect moves the speed droop characteristics up and down. Turbine model For the purpose of load-frequency dynamics the turbine may be modelled by an approximate model with a single time constant TT as given by equation û3T(s) = GTV  û;v (s) = (1/ (1+sTT  û;v (s)

9-6

(9.12)

9.4.5

Modelling and Analysis of Single-Area Load-Frequency Control

The block diagram for single-area load-frequency control is assembled by combining equation  DIWHUGURSSLQJ û3tie), equation (9.12) and Fig 9.4. The block diagram is given in Fig 9.5. Secondary LFC loop Primary LFC loop

û3D(s)

1/R

-KI /s Controller

û3ref(s) +

™

û;v(s)

GG

û3T(s) + GT ™

Kp 1+sTp

ûI (s)

Power System Fig 9.5 Block Diagram for Single-Area Load – Frequency Control In the above diagram, all powers are in per unit to area rated capacity and the frequency deviation is in hertz. Kp = 1/D

Hz / p.u.MW

Tp = (2H/f0 D) sec The load damping constant D is normally expressed in percent and typical values of D are 1 to 2 percent. A value of D = 1.5 means that 1.0 percent change in frequency would cause a 1.5 percent change in load. The dashed portion of the diagram marked as the secondary loop represents the integral controller whose gain is KI. This controller actuates the load reference point until the frequency deviation becomes zero. Steady-state analysis with governor control Let the disturbance be a step increase in load, M p.u. MW. With only governor control (integral controller deactivated) the frequency deviation will not be made zero. The steady-state frequency deviation ûIs can be determined by applying final value theorem in s-domain ûIs OLPV s 0

ûIV

(9.13)

9-7

From the block diagram Fig 9.5 ûIV  

Gp(s) 1+Gp(s) (1/R) GG(s).GT(s)

[-û3D(s)]

)XUWKHU û3D(s) = (M/s)

(9.15)

Substituting equations (9.14) and (9.15) in equation (9.13) we obtain ûIs = -0 +]

ZKHUH  $UHD)UHTXHQF\5HVSRQVHFRHIILFLHQW = D+(1/R)

(AFRC)

Hz / p.u. MW

M is in p.u. MW Example 9.1 The data for a single-area power system is given below Rated area capacity = Pr = 2000MW Nominal operating load = P0D = 1000MW f0 = 50 Hz; D=1 %; R = 3%; H = 5 sec Load increase = M= 20 MW Compute steady-state frequency deviation. Solution $)5& 

 '5

D = (1/100) * 1000

= 20 MW/Hz

(1/100) * 50 D in p.u. MW / Hz = (20/2000) = 0.01 (1/R) = 2000

= (3/100) * 50

(9.14)

4000

MW / Hz

3

(1/R) in p.u MW / Hz = (4000/2000) / 3 = 0.6666   SX0:+]

9-8

M = (20/2000) = 0.01 p.u.MW/Hz 

ûIs = -0     

- 0.01478 Hz

Steady-state analysis with integral control By reducing the full block diagram Fig 9.5 and by applying final value theorem in s-domain, one can show that the steady-state frequency deviation is made zero. Transient analysis The block diagram Fig 9.5 can be used to derive the state variable model with the following four states: x1  ûIs = frequency deviation

x2  û3T = Turbine power deviation

x3  û;v = Steam valve/water gate position x4  û3ref = Load-reference setting

Transient response for step change in load can be obtained by numerically integrating the four state equations through Runge – Kutta fourth order method or any other method. AU Powerlab or any available software can be used for this purpose. Example 9.2 For the system given in Example 9.1, assuming that the time constants TG and TT are small compared to TP, determine the critical value for controller gain KIcrit so that the response is critically damped. Solution The block diagram in Fig 9.5 is reduced by setting GG(s) and GT(s) to unity taking the reference input as – û3DV DQGRXWSXWDV ûIV :HJHWDEORFNGLDJUDPZLWKQHJDWLYHIHHGEDFNZLWK forward path transfer function G(s) and feedback path transfer function H(s) given by G(s) = Kp / (1+sTp) H(s) = (1/R + KI/s) The characteristic equation of the system can be obtained by setting 1+G(s)H(s) = 0 and is given by s2 + s(R+Kp)/(RTp) + Kp KI / Tp =0 (9.17) The system in equation (9.17) will have critical damping if (R+Kp)2 / (4R2 Tp2) = KpKIcrit / Tp which reduces to KIcrit = (1/4TpKp ) (1+Kp / R)2

9-9

Using the data given in Example 9.1 D = 0.01 p.u. MW / Hz ; R =1.5 Hz / p.u.MW Kp = 1/D = 100 Tp = (2H) / (f0D) = (2x5)/(50x0.01) = 20 KIcrit = (1+100/1.5)2 / (4x20x100) = 0.5723 Comments It is preferable to use “sub critical” gain settings, K I
Modelling and Analysis of Two-Area Load-Frequency Control

The block diagram for two-area load-frequency control is assembled by combining equations (9.6), (9.7) (9.11) and Fig 9.4 and is given in Fig 9.6

+ +

û3D1(s)

1/R1

B1 ™

-KI1/s ACE1

û3ref1 +

™

GG1

û3T1(s) + GT1 ™ -

Kp1 ûI 1(s)

1+sTp1

û312(s) Œ3

1/s

+

s1

™

-

-1

-1 +

+

0

û321(s)

™

-KI2 /s ACE2

B2

û3ref2 +

™

GG2 -

û3T2(s) + GT2 ™ -

Kp2 1+sTp2

û3D2(s)

1/R2

Fig 9.6 Block Diagram for Two-Area Load Frequency Control

9-10

ûI 2(s)

In the diagram Fig 9.6 the following points are to be noted: i. While in single-area diagram the powers and parameters R,D and H are expressed in per unit of area rating, in two-area diagram since their ratings may be different, we must refer all powers and parameters to the common chosen base power. ii. The dashed portion of the diagram gives one integral controller for each area. Since the objective of the controller is to maintain the frequency and tie-line power at scheduled values, the input signal to the controller is the Area Control Error (ACE) which is given by ACE1  û312 + B1ûI1

(9.18)

ACE2  û321 + B2ûI2

(9.19)

where B1 and B2 are frequency-bias parameters, which should be selected optimally to achieve better dynamic response. Steady-state analysis with governor control Let the disturbance be step load increases M1 and M2 p.u. MW in areas 1 and 2 respectively. With integral controllers deactivated the steady- VWDWH IUHTXHQF\ GHYLDWLRQ ûIs and tie-line GHYLDWLRQ û312s can be determined from the static loop gains by letting s tends to 0 in Fig 9.6 û3T1s = -ûIs / R1 û3T2s = -ûIs / R2

(9.20)

By adding the powers at the summing junction we get - (1/R1  ûIs – M1 = D1ûIs û312s

(9.21)

- (1/R2  ûIs – M2 = D2ûIs -û312s

Solving equation (9.21) fRU ûIsDQG û312s we get ûIs = - (M1 + M2  1 2) Hz

(9.22)

û312s = -û321s  1 M2 –2 M1  1 2)

p.u MW

(9.23)

Example 9.3 A two-area power system has two identical areas with parameters and operating conditions same as that given in Example 9.1. A load increase M1 0:RFFXUVLQDUHD'HWHUPLQHWKH ûIs DQG û312s. Solution D1 = D2 =0.01

p.u .MW / Hz

R1 = R2 = 1.5 Hz / p.u.MW 9-11

1  2 = (D+1/R) = 0.01+0.6666 = 0.6766 p.u. MW / Hz ûIs = - M1   - 0.01 / (2*0.6766) = -0.00739 Hz

û312s = -û321s = -2 M1   -M1 / 2 = -0.005 p.u. MW Comments

Comparing the results with that of ExamSOHZHQRWHWKDW ûIs has reduced by 50% because of WKHVXSSRUWUHFHLYHGIURPWKHDUHD7KHYDOXHRI û312s reveals that support received from area 2 is 50% of the load change in area 1. Steady-state analysis with integral control With integral controller adopting an error signal of ACEi  û3ij + BiûIi ;

i =1,2

One can show [2] that the steady-state frequency deviations and tie-line power deviation will be brought to zero irrespective of the value chosen for frequency bias factors B1 and B2. Transient analysis The block diagram Fig 9.6 can be used to derive the state variable model for two-area loadfrequency control using integral controller. The nine-state model comprises four states per area already introduced for single-area system and the tie- lineSRZHUGHYLDWLRQ û312. The transient response can be simulated using Runge-Kutta fourth order method or any other method. AU Powerlab or any available software can be used for this purpose. Different control modes In addition to “Free governor a ction” (only governer control) that can be obtained by deactivating the integral controller; the following alternative control modes may be simulated. i. Flat frequency control ii. Flat tie-line control iii. Frequency bias tie- line control In the flat-frequency mode, the ACE comprises area frequency deviation only. Hence only steady- state frequency deviation only. Hence only steady-state frequency deviation is brought to zero and the tie-line deviation remains non-zero. In flat tie-line mode the ACE comprises tie-line deviation only. Hence only steady-state tie-line deviation is brought to zero but not the frequency deviation. In the practical case of frequency bias tie-line control, the ACE comprises both the area frequency deviation and tie-line power deviation. Hence both the deviations are brought to zero under steady-state . Proper selection of the frequency bias factors B1 and B2 is important from the dynamic performance considerations. The performance of this mode of integral control is 9-12

examined [1] for diffeUHQWVHWWLQJVRIDUHDIUHTXHQF\ELDVIDFWRUVDQGWKHVHOHFWLRQRI%  AFRC is recommended as a logical choice.

WKH

)XUWKHUWKHVHOHFWLRQRI%IDFWRUVVLJQLILFDQWO\KLJKHUWKDQWKH$)5& PDNHVWKHFRQWURODFWLRQ unstable. High values of gain KI also will make the control unstable. 9.5 EXERCISES 9.5.1. It is proposed to simulate using the software available the load-frequency dynamics of a single-area power system whose data are given below: Rated capacity of the area Normal operating load Nominal frequency Inertia constant of the area Speed regulation (governor droop) of all regulating generators Governor time constant Turbine time constant

= = = =

2000 MW 1000 MW 50 Hz 5.0 s

= 4 percent = 0.08 s = 0.3 s

Assume linear load–frequency characteristics which means the connected system load increases by one percent if the system frequency increases by one percent. The area has a governor control but not a load-frequency controller. The area is subjected to a load increase of 20 MW. (a) Simulate the load-frequency dynamics of this area using available software and check the following: (i) Steady – state frequency deviation ∆fs in Hz. Compare it with the hand-calculated value using “Area Frequency Response Coefficient” (AFRC). (ii) Plot the time response of frequency deviation ∆f in Hz and change in turbine power ∆PT in p.u MW upto 20 sec. What is value of the peak overshoot in ∆f? (b) Repeat the simulation with the following changes in operating condition, plot the time response of ∆f and compare the steady-state error and peak overshoot. (i) Speed regulation (ii) Normal operating load

= 3 percent = 1500 MW

9.5.2. Assume that the single-area power system given in exercise 9.5.1 is provided with a load frequency controller (an integral controller) whose gain KI can be tuned. (a) Carryout the simulation for the same disturbance of load change of 20 MW for different values of KI, obtain the time response ∆f for each case, critically compare these responses and comment on their suitability for practical application. 9-13

(Hint: For choosing different values of KI, first set the governor and turbine time constants to zero and determine analytically the value of integral gain KI,cr to have critical damping on the response ∆f (t). Choose the range of KI to include KI,cr as 0 ≤ KI ≤ ( KI,cr + 1.0 ) ) (b) From the investigations made in (a) above, choose the best value of KI which gives an “optimal” response ∆f (t) with regard to peak overshoot, settling time, steady-state error and Mean Sum- Squared-Error (MSSE). 9.5.3

It is proposed to simulate the load frequency dynamics of a two-area power system. Both the areas are identical and has the system parameters given in exercise 9.5.1. Assume that the tie-line has a capacity of Pmax 1-2 = 200 MW and is operating at a power angle of ( δ01 - δ20 ) = 300. Assume that both the areas do not have load –frequency controller. Area 2 is subjected to a load increase of 20 MW. (a) Simulate the load-frequency dynamics of this system using available software and check the following: (i). Steady-State frequency deviation ∆fs in Hz and tie-line flow deviation, ∆P12,S in p.u. MW. Compare them with hand-calculated values using AFRC’s (ii) Compare result ∆fs with that obtained in single area simulation in exercise 9.5.1 (a), and comment on the support received from area 1 and the advantages of interconnecting with neighbouring areas . (iii) Plot the time responses, ∆f1(t), ∆f2 (t), ∆PT1(t) , ∆PT2(t) and ∆P12(t). Comment on the peak overshoot of ∆f1, and ∆f2.

9.5.4. Assume that the two areas of the system given in exercise 9.5.3. are provided with identical integral controllers whose gain KI and frequency bias B can be varied. (a) Carryout the simulation for the same disturbance in exercise 9.5.3 and obtain time response ∆f1 (t), ∆f2(t) and ∆P12 (t) for the following different control actions and assuming KI = 1.0, B = 60 MW / 0.1 Hz: (i). Flat Tie- line control (ii) Flat frequency control (iii) Tie-line bias control. Comment on the responses and their suitability for practical application. (b) Consider only the “Tie -line bias Control”. Demonstrate that selection of too large a value for B and/or KI will render the control loop unstable . (c) Choose the frequency bias factors; B to be equal to the AFRC, β and determine the best value of KI giving optimal response for ∆f1 (t), ∆f2 (t) and ∆P12 (t) with regard to peak over shoot, settling time, steady-state error and SSE. (d) Repeat exercise (c) above by choosing B = 0.5 β and obtain the best KI. 9-14

(e) Out of the responses obtained under (c) and (d) above, choose the best response and hence the best KI & B for the system. 9.5.5. Consider a two-area power system with unequal areas. A 2-GW control area (1) is interconnected with a 10- GW area (2). The 2-GW area has the system parameters given in exercise 9.5.1. The 10 GW area has the following data: Nominal operating load = 5000 MW ; H = 5 sec ; D = 1.0 % ; R = 4% ; TG = 0.08 sec ; TT = 0.3 sec. Design an integral load frequency controller for the system by considering a disturbance of 20 MW load increase in area 1. Determine the best controller parameters KI1, KI2, B1 and B2 after checking the time responses of ∆f1 (t), ∆f2 (t) and ∆ P12 (t) for different values of the controller parameters. 9.6. REFERENCES [1] Prabha Kundur, “Power System Stability and Control” , McGraw -Hill, Inc; New York, 1994. [2] O.I. Elgerd, “Electric Energy Systems Theory: An Introduction” , Tata McGra w Hill Publishing Co. Ltd, New Delhi, 2003.

9-15