Chapter 2 AUTOMATIC LOAD FREQUENCY CONTROL

Transcription

1 Chater 2 AUTOMATIC LOAD FEQUENCY CONTOL

2 . INTODUCTION Thi chater deal with the control mechanim needed to maintain the ytem frequency. The toic of maintaining the ytem frequency contant i commonly known a AUTOMATIC LOAD FEQUENCY CONTOL (ALFC). It ha got other nomenclature uch a Load Frequency Control, Power Frequency Control, eal Power Frequency Control and Automatic Generation Control. The baic role of ALFC i:. To maintain the deired megawatt outut ower of a generator matching with the changing load. 2. To ait in controlling the frequency of larger interconnection. 3. To kee the net interchange ower between ool member, at the redetermined value. The ALFC loo will maintain control only during mall and low change in load and frequency. It will not rovide adequate control during emergency ituation when large megawatt imbalance occur. We hall firt tudy ALFC a it alie to a ingle generator ulying ower to a local ervice area.

3 2. EAL POWE CONTOL MECHANISM OF A GENEATO The real ower control mechanim of a generator i hown in Fig.. The main art are: ) Seed changer 2) Seed governor 3) Hydraulic amlifier 4) Control valve. They are connected by linkage mechanim. Their incremental movement are in vertical direction. In reality thee movement are meaured in millimeter; but in our analyi we hall rather exre them a ower increment exreed in MW or.u. MW a the cae may be. The movement are aumed oitive in the direction of arrow. Correonding to raie command, linkage movement will be: A move downward; C move uward; D move uward; E move downward. Thi allow more team or water flow into the turbine reulting incremental increae in generator outut ower. When the eed dro, linkage oint B move uward and again generator outut ower will increae.

4 Fig. Functional diagram of real ower control mechanim of a generator

5 2. SPEED GOVENO The outut commend of eed governor i ΔP g which correond to movement Δx C. The eed governor ha two inut: ) Change in the reference ower etting, ΔP ref 2) Change in the eed of the generator, Δf, a meaured by Δx B. It i to be noted that a oitive ΔP ref will reult in oitive ΔP g A oitive Δf will reult in linkage oint B and C to come down cauing negative ΔP g. Thu ΔP g = ΔP ref - Δf () Here the contant ha dimenion hertz er MW and i referred a eed regulation of the governor. Taking Lalace tranform of eq. yield ΔP g () = ΔP ref () - Δf () (2)

6 ΔP g = ΔP ref - Δf () Here the contant ha dimenion hertz er MW and i referred a eed regulation of the governor. Taking Lalace tranform of eq. yield ΔP g () = ΔP ref () - Δf () (2) The block diagram correonding to the above equation i hown in Fig. 2. Δf () ΔP ref () + _ ΔP g () Fig. 2 Block diagram of eed governor

7 2.2 HYDAULIC VALVE ACTUATO The outut of the hydraulic actuator i ΔP v. Thi deend on the oition of main iton, which in turn deend on the quantity of oil flow in the iton. For a mall change Δx D in the ilot valve oition, we have ΔP v = k H Δx D dt (3) The contant k H deend on the orifice, cylinder geometrie and fluid reure. The inut to Δx D are ΔP g and ΔP v. It i to be noted that for a oitive ΔP g, the change Δx D i oitive. Further, for a oitive ΔP v, more fuel i admitted, eed increae, linkage oint B move downward cauing linkage oint C and D to move downward reulting the change Δx D a negative. Thu Δx D = ΔP g - ΔP v (4)

8 Lalace tranformation of the lat two equation are: k ΔP v () = H Δx D () Δx D () = ΔP g () - ΔP v () Eliminating Δx D and writing ΔP v () in term of ΔP g (), we get ΔP v () = T H ΔP g () (5) where T H i the hydraulic time contant given by T H = k H (6)

9 In term of the hydraulic valve actuator tranfer function G H (), eq. 5 can be written a G H () = ΔP ΔP () v = g () T H (7) Hydraulic time contant T H tyically aume value around 0. ec. The block diagram of the eed governor together with the hydraulic valve actuator i hown in Fig. 3. Δf () ΔP ref () + _ ΔP g () G H ΔP v () Fig. 3 Block diagram of eed governor together with hydraulic valve actuator

10 2.3 TUBINE GENEATO In normal teady tate, the turbine ower P T kee balance with the electromechanical air-ga ower P G reulting in zero acceleration and a contant eed and frequency. During tranient tate, let the change in turbine ower be ΔP T and the correonding change in generator ower be ΔP G. The accelerating ower in turbine generator unit = ΔP T - ΔP G Thu accelerating ower = ΔP T () - ΔP G () (8) If ΔP T - ΔP G i negative, it will decelerate.

11 The turbine ower increment ΔP T deend entirely uon the valve ower increment ΔP v and the characteritic of the turbine. Different tye of turbine will have different characteritic. Taking tranfer function with ingle time contant for the turbine, we can write ΔP T () = G T ΔP v () = T T ΔP v () (9) The generator ower increment ΔP G deend entirely uon the change ΔP D in the load P D being fed from the generator. The generator alway adjut it outut o a to meet the demand change ΔP D. Thee adjutment are eentially intantaneou, certainly in comarion with the low change in P T. We can therefore et ΔP G = ΔP D i.e. ΔP G () = ΔP D () (0)

12 In view of equation 8,9 and 0, Accelerating ower = ΔP T () - ΔP G () (8) ΔP T () = G T ΔP v () = T T ΔP v () (9) ΔP G () = ΔP D () (0) the block diagram develoed i udated a hown in Fig. 4. Thi correond to the linear model of rimary ALFC loo excluding the ower ytem reone. Δf () ΔP G () = ΔP D () ΔP ref () + _ ΔP g () ΔP v () ΔP T () G H G T + _ ΔP T () ΔP D () Fig. 4 Block diagram correonding to rimary loo of ALFC excluding ower ytem reone

13 Δf () ΔP D () = ΔP G () ΔP ref () + _ ΔP g () ΔP v () ΔP T () G H G T + _ ΔP T () ΔP D () 3. STATIC PEFOMANCE OF SPEED GOVENO The reent control loo hown in Fig. 4 i oen. We can neverthele obtain ome intereting information about the tatic erformance of the eed governor. The relationhi between the tatic ignal (ubcrit 0 ) i obtained by letting 0. A G H (0) = G T (0) = we obtain directly from Fig. 4 ΔP T 0 = ΔP ref 0 - Δf 0 () Note that at teady tate, ΔP T i equal to ΔP G. i.e. ΔP T 0 = ΔP G 0 We conider the following three cae.

14 ΔP T 0 = ΔP ref 0 - Δf 0 () Cae A The generator i ynchronized to a network of very large ize, o large in fact, that it frequency will be eentially indeendent of any change in the ower outut of thi individual generator ( infinite network). Since Δf 0 = 0, the above eq. become ΔP T 0 = ΔP ref 0 (2) Thu for a generator oerating at contant eed,(or frequency) there exit a direct roortionality between turbine ower and reference ower etting.

15 Thu for a generator oerating at contant eed, (or frequency) there exit a direct roortionality between turbine ower and reference ower etting. ΔP T 0 = ΔP ref 0 i.e. when the generator i oerating at contant frequency, if the eed changer etting i INCEASED,(DECEASED) turbine outut ower will increae (decreae) to that extent. Examle A 00-MW, 50-Hz generator i connected to infinite network. How would you increae it turbine ower by 5 MW? Solution It turbine ower can be increaed by 5 MW by imly giving a raie ignal of 5 MW to the eed changer motor.

16 Cae B Now we conider the network a finite. i.e. it frequency i variable. We do, however, kee the eed changer at contant etting. i.e. ΔP ref = 0. From eq. () ΔP T 0 = ΔP ref 0 - Δf 0 () we obtain ΔP T 0 = - Δf 0 (3). The above eq. how that for a contant eed changer etting, the tatic increae in turbine ower outut i directly roortional to the tatic frequency dro. 2. The above eq. (3) can be rewritten a Δf 0 = - ΔP T 0. Thi mean that the lot of f 0 with reect to P T 0 (or P G 0 ) will be a traight line with loe of. We remember that the unit for i hertz er MW. In ractice, both the frequency and the ower can be exreed in er unit.

17 Examle 2 Conider 00-MW 50-Hz generator in the reviou examle. It ha a regulation arameter of 4 %. By how much will the turbine ower increae if the frequency dro by 0. Hz with the eed changer etting unchanged. Solution egulation i 4%. 4% of 50 = 2 Hz. Thi mean that for frequency dro of 2 Hz the turbine ower will increae by 00 MW. 52 Hz 2 Thu = = 0.02 Hz er MW 00 It i given Δf 0 = - 0. Hz 50 Hz 0. Hz? 0 00 MW In eq. (), ΔP T 0 = ΔP ref 0 - Δf 0 etting ΔP ref a zero, ΔP T 0 = - Δf 0 i.e. ΔP T 0 = (- 0.) = 5 MW Thu the turbine ower will increae by 5 MW. We can alo get the reult uing ymmetrical triangle.

18 Examle 3 Conider again 00-MW, 50-Hz generator in the reviou examle. If the frequency dro by 0. Hz, but the turbine ower remain unchanged, by how much hould the eed changer etting be changed? Solution A ΔP T 0 = 0, from eq.(), ΔP T 0 = ΔP ref 0 - Δf 0 we have ΔPref 0 = Δf 0 Given that = 0.02 Hz er MW and Δf 0 = - 0. Hz ΔP ref 0 = 0.02 x (- 0.) = - 5 MW 52 Hz 50 Hz 0. Hz Therefore, eed changer etting mut be lowered by 5 MW? 0 00 MW

19 Cae C In general cae, change may occur in both the eed changer etting and frequency in which cae the relationhi ΔP T 0 = ΔP ref 0 - Δf 0 alie. For a given eed changer etting, ΔP ref 0 = 0 and hence Δf 0 = - ΔP T 0. In a frequency-generation ower grah, thi rereent a traight line with a loe = -. For a given frequency, Δf 0 = 0 and hence ΔP T 0 = ΔP ref 0. Thi mean that for a given frequency, generation ower can be increaed or decreaed by uitable raie or lower command. Thu the relationhi ΔP T 0 = ΔP ref 0 - Δf 0 rereent a family of loing line a deicted in Fig. 5, each line correonding to a ecific eed changer etting.

20 Percent of rated frequency Seed changer i et uch that at 00% rated frequency, outut i 00% rated outut ower Percent of rated outut Seed changer i et uch that at 00% rated frequency, outut i 50% rated outut ower. Fig. 5 Static frequency-ower reone of eed governor ( = 0.04.u.) The thick line how that correonding to 00% rated frequency, the outut ower i 00 % of rated outut. But for the new eed changer etting a hown by the dotted line, for the ame 00% rated frequency, outut ower i 50 % of rated outut. Hence the ower outut of the generator at a given frequency can be adjuted at will, by uitable eed changer etting. Such adjutment will be extreme imortance for imlementing the load diviion a decided by the otimal olicy.

21 Let the governor characteritic of two unit be 0 and 2 0. Let the oerating frequency be f r. Then load hared by unit and 2 are P 0 and P 0 2. If the economic diviion of load dictate the load haring a P and P 2, the governor characteritic hould be hifted to and 2 a hown in Fig. below. 0 f r P 0 P P 2 0 P 2 Shifting of governor characteritic

22 Examle Two ynchronou generator oerating in arallel uly a total load of 200 MW.The rating of the machine and 2 are 00 MW and 200 MW. Machine and 2 have governor droo characteritic of 4% and 3% reectively, from no load to full load. Aume that at full load, machine run at rated eed and the ytem frequency i 50 Hz. Calculate the load taken by each machine and the oerating frequency. Solution % of rated frequency % of rated frequency x 00 0 P G in MW x 200 P G in MW The figure how the characteritic of the machine.

23 % of rated frequency % of rated frequency x 00 0 P G in MW x 200 P G in MW Let x MW be the load taken by the machine. Then the load taken by the machine 2 i 200-x MW. Both hould oerate at ame eed and frequency. 4 3 Equating the common frequency: 04 x 03 (200 x ) On olving the above, x = MW and x = MW. Thu Load on machine = MW; Load on machine 2 =27.27 MW 4 Oerating frequency = 04 (72.73) = 0.09% = (0.09) 00 = Hz.

24 Examle 4 Two generator are ulying ower to a ytem. Their rating are 50 and 500 MW reectively. The frequency i 50 Hz and each generator i half-loaded. The ytem load increae by 0 MW and a a reult the frequency dro to 49.5 Hz. What mut the individual.u. regulation be if the two generator hould increae their turbine ower in roortional to their rating? Solution Generator rating: 50 MW 500 MW ; Initial loading: 25 MW 250 MW Change in load = 0 MW; Change in frequency = Hz equired change in turbine ower: 0 MW 00 MW (Proortional to rating) Since Δf 0 = - ΔP T 0, egulation = - (change in frequency) / (change in ower) Smaller unit: = = 0.05 Hz er MW; = / 50 0 / 50 = 0.05.u. Hz /.u.mw Larger unit: = - = Hz er MW; = / / 500 = 0.05.u. Hz /.u.mw The above reult teache u that generator working in arallel hould have ame regulation (exreed in.u. baed on their own rating) in order to hare the load change in roortional to their rating.

25 3. CLOSING THE ALFC LOOP We oberved earlier that the loo in Fig. 4 i oen. We now roceed now to cloe it by finding a mathematical link between ΔP T and Δf. A our generator i ulying ower to a conglomeration of load in it ervice area, it i neceary in our following analyi to make reaonable aumtion about the lumed area behavior. We make thee aumtion:. The ytem i originally running in it normal tate with comlete ower balance, that i, P 0 G = P 0 D + loe. The frequency i at normal value f 0. All rotating equiment rereent a total kinetic energy of W 0 kin MW ec. 2. By connecting additional load, load demand increae by ΔP D which we hall refer to a new load. (If load demand i decreaed new load i negative). The generation immediately increae by ΔP G to match the new load, that i ΔP G = ΔP D.

26 3. It will take ome time for the control valve in the eed governing ytem to act and increae the turbine ower. Until the next teady tate i reached, the increae in turbine ower will not be equal to ΔP G. Thu there will be ower imbalance in the area that equal ΔP T - ΔP G i.e. ΔP T ΔP D. A a reult, the eed and frequency change. Thi change will be aumed uniform throughout the area. The above aid ower imbalance get aborbed in two way. ) By the change in the total kinetic.e. 2) By the change in the load, due to change in frequency. Since the.e. i roortional to the quare of the eed, the area.e. i W kin = W 0 f kin ( 0 ) 2 MW ec. (4) f The old load i a function of voltage magnitude and frequency. Frequency deendency of load can be written a D = P D f MW / Hz (5)

27 Thu d ΔP T ΔP D = (Wkin ) + D Δf (6) dt Noting that f = f 0 + Δf W kin = W 0 0 f f kin ( 0 f d 2 W (Wkin ) = 0 dt f 0 kin ) 2 = W 0 2 f f kin [ ( 0 ) 2 ] W 0 f kin ( ) (7) f f f d ( Δf ) dt Subtituting the above in eq. (6) 2 W ΔP T ΔP D = 0 f 0 kin d ( Δf ) + D Δf MW (8) dt By dividing thi equation by the generator rating P r and by introducing erunit inertia contant H = W P 0 kin r MW ec / MW (or ec) (9) it take on the form 2 H ΔP T ΔP D = 0 f d ( Δf ) + D Δf u MW (20) dt

28 2 H ΔP T ΔP D = 0 f d ( Δf ) + D Δf u MW (20) dt The ΔP are now meaured in er unit (on bae P r ) and D in u MW er Hz. Tyical H value lie in the range 2 8 ec. Lalace tranformation of the above equation yield 2 H ΔP T () ΔP D () = 0 f Δf () + D Δf () (2) 2 H = [ 0 f + D ] Δf () i.e. Δf () = 2H f 0 D [ ΔP T () ΔP D () ] Δf () = G () [ ΔP T () ΔP D () ] (22)

29 Δf () = G () [ ΔP T () ΔP D () ] (22) where G () = 2H f 0 D = /D 2H 0 f D = T (23) = D (24) T = 2 H f 0 D (25) Equation (22) rereent the miing link in the control loo of Fig. 4. By adding thi, block diagram of the rimary ALFC loo i obtained a hown in Fig. 6. ΔP ref () + _ Δf () ΔP D () _ ΔP g () ΔP v () ΔP T () ΔP T () - ΔP D () G H G T + T Power ytem Fig. 6 Block diagram correonding to rimary loo of ALFC

30 Examle 5 Primary ALFC loo arameter for a control area are: Total rated caacity P r = 2000 MW Normal oerating load P 0 D = 000 MW Inertia contant H = 5.0 ec. egulation = 2.0 Hz /.u. MW Aume that the load-frequency deendency i linear, meaning that the load would increae one ercent for one ercent frequency increae. Obtain the ower ytem tranfer function. Solution D = f P 0 D = % of 000 % of = = 20 MW / Hz = = 0.0.u. MW / Hz P = = 00 Hz /.u. MW; T = D 2 H f D 0 = 2 x 5 50 x 0.0 = 20 ec. Thu G () = T = 00 20

31 4. PIMAY ALFC LOOP UNCONTOLLED CASE The rimary ALFC loo in Fig. 6 ha two inut ΔP ref and ΔP D and one outut Δf. Δf () ΔP D () ΔP ref () = 0 + _ ΔP g () ΔP v () _ ΔP T () G H G T + T For uncontrolled cae, (i.e. for contant reference inut) ΔP ref = 0 and the block diagram hown in Fig. 6 can be imlified a hown. - ΔP D () Δf () From thi imlified diagram, we can write + - G P Δf () = - G G H G T G ΔP D () (27) GH G T Fig. 6 a educed block diagram

32 4. STATIC FEQUENCY DOP DUE TO STEP LOAD CHANGE For a te load change of contant magnitude ΔP D = M, we have ΔP D () = M Uing the final value theorem, we readily obtain from eq. (27), Δf () = - G G H G T G lim Δf 0 = [ Δf () ] = - 0 ΔP D () (27) M = - the tatic frequency dro a M D Hz (28) We introduce here the o-called Area Frequency eone Characteritic (AFC) β, defined a β = D +.u. MW / Hz (29) Then the tatic frequency dro i given by Δf 0 = - M Hz (30) β

33 Examle 6 Find the tatic frequency dro for 2000 MW ytem in the reviou examle following load increae of % of ytem rating. Solution Load increae M = % of 2000 MW = 20 MW = 0.0.u. MW A in reviou examle, D = 0.0.u. MW / Hz; = 2 Hz /.u. MW β = D + = = 0.5.u. MW / Hz 2 Therefore Δf 0 = = Hz; 0.5 or frequency dro = % of normal frequency

34 Examle 7 What would the frequency dro in the reviou examle if the eed-governor loo were non-exitent or oen? Solution Δf () ΔP D () ΔP ref () + _ ΔP g () ΔP v () _ ΔP T () G H G T + T ALFC loo reduce to - ΔP D () T Δf () Δf () = - T ΔP D () For a udden load increae of M, ΔP D () = M Then Δf 0 = lim [ Δf () ] = - M = - 0 M D Hz; Thu β = D = 0.0.u. MW / Hz Therefore Δf 0 = = -.0 Hz; or frequency dro = 2 % of normal frequency 0.0

35 5. DYNAMIC ESPONSE OF PIMAY ALFC LOOP - UNCONTOLLED CASE We know that Δf () = - G G H G T G ΔP D () (27) Finding the dynamic reone, for a te load, i quite traight forward. Eq. (27) uon invere Lalace tranformation yield an exreion for Δf (t). However, a G H, G T and G contain at leat one time contant each, the denominator will be a third order olynomial reulting in unwieldy algebra.

36 We can imly the analyi coniderably by making the reaonable aumtion that the action of eed governor lu the turbine generator i intantaneou comared with the ret of the ower ytem. The latter, a demontrated in Examle 5 ha a time contant of 20 ec, and ince the other two time contant are of the order of ec, we will erform an aroximate analyi by etting T H = T T = 0. From eq. (27), Δf () = - G G H G T G ΔP D () (27) we get Δf () - T T M = - ( T ) M (3)

37 Δf () - T T M = - ( T ) M (3) Dividing numerator and denominator by T we get Δf() = - T M ( T ) (32) Uing the fact A ( α ) = A [ - α α ] and noting A α M T T M equation (32) become Δf() = - M [ - T ] (33)

38 Δf() = - M [ - T ] (33) Making ue of reviou numerical value: M = 0.0.u. MW; = 2.0 Hz /.u. MW; = 00 Hz /.u. MW; T = 20 ec. M 2 = 02 = 0.096; T = 02 = Δf() = [ ] The aroximate time reone i urely exonential and i given by Δf(t) = ( e t ) Hz (34)

39 Alternatively, the above reult can be obtained from the reduced block diagram hown in Fig. 6 a. Then - ΔP D () + - G P Δf () Δf () GH G T 0.5 Fig. 6 a educed block diagram Thu Δf() = (20 5) 0.05 ( 2.55) 0.05 = [ ] 0.096[ ] Δf(t) = ( e t ) Hz (34)

40 Fig. 7 how thi reone. For comarion, the reone with the incluion of the time contant T H and T T i alo hown. Fig. 7 eone of rimary loo of ALFC It i to be oberved that the rimary loo of ALFC doe not give the deired objective of maintaining the frequency contant. We need to do omething more to bring the frequency error to zero. Before dicuing the neceary control which can make the frequency error to zero, we hall hed ome light on to the hyical mechanim in the rimary loo of ALFC.

41 5. PHYSICAL INTEPETATION OF ESULTS When the load i uddenly increaed by % (20 MW), where did thi ower come from? Certainly it mut have come from omewhere a the load increae of 20 MW ha been met with intantaneouly. In the milliecond following the cloure of the witch, the frequency ha not changed a meaureable amount, eed governor would not have acted and hence turbine ower would not have increaed. In thoe firt intant the total additional load demand of 20 MW i obtained from the tored kinetic energy, which therefore will decreae at an initial rate of 20 MW. eleae of E will reult in eed and frequency reduction. A een in eq.(34), Δf(t) = ( e t ) Hz (34) initially frequency reduce at the rate of x 2.55 = 0.05 Hz / ec. The frequency reduction caue the team valve to oen and reult in increaed turbine ower. Further, the old load decreae at the rate of D = 20 MW / Hz.

42 In concluion, the contribution to the load increae of 20 MW i made u of three comonent:. ate of decreae of kinetic energy from the rotating ytem 2. Increaed turbine ower 3. eleaed old cutomer load Initially the comonent 2 and 3 are zero. Finally, the frequency and hence the E ettle at a lower value and the comonent become zero. In between, comonent kee decreaing and comonent 2 and 3 kee increaing. Let u comute comonent 2 and 3 at teady tate condition. We know that with 4% regulation = 4%of = 2000 = 0.00 Hz / MW. Further Δf 0 = Hz. Increaed generation ΔP G = ΔP T = - Δf 0 = = 9.6 MW 0.00 Value of D (Examle 5) = 20 MW / Hz eleaed old cutomer load = D x Δf 0 = 20 x = MW Thee two comonent add u to 20 MW. Note that the larget contribution i from new generation.

43 6. POPOTIONAL PLUS INTEGAL CONTOL ( Secondary ALFC loo) It i een from the reviou dicuion that with the eed governing ytem intalled in each area, for a given eed changer etting, there i coniderable frequency dro for increaed ytem load. In the examle een, the frequency dro i Hz for 20 MW. Then the teady tate dro in frequency from no load to full load ( 2000 MW ) will be.96 Hz. Sytem frequency ecification i rather tringent and therefore, o much change in frequency cannot be tolerated. In fact, it i exected that the teady tate frequency change mut be zero. In order to maintain the frequency at the cheduled value, the eed changer etting mut be adjuted automatically by monitoring the frequency change.

44 For thi uroe, INTEGAL CONTOLLE i included. In the integral controller the frequency error i firt amlified and then integrated. Further, a negative olarity i alo included o that a NEGATIVE frequency deviation will give rie to AISE command. The ignal fed into the integrator i referred a Area Controlled Error (ACE). For thi cae ACE = ΔP ref I Δ f dt Taking Lalace tranformation ΔP ref ( ) I ΔF( ) Δ f. Thu The gain contant I control the rate of integration and thu the eed of reone of the loo. (35) (36)

45 ΔP ref ( ) I ΔF( ) (36) For thi ignal Δf () i fed to an integrator whoe outut control the eed changer oition reulting in the block diagram configuration hown in Fig. 8. Δf () Δf () ΔP D () Δf () - I ΔP ref () + _ ΔP g () G HT ΔP T () + _ G Fig. 8 Block diagram correonding to comlete ALFC A long a an error remain, the integrator outut will increae, cauing the eed changer to move. When the frequency error ha been reduced to zero, the integrator outut ceae and the eed changer oition attain a contant value. Integral controller will give rie to ZEO STEADY STATE FEQUENCY EO following a te load change becaue of the reaon tated above.

46 eferring to the block diagram of ingle control area with integral controller hown in Fig. 8, inut to G HT i - Δ f ( ) I - Δf () i.e. - [ I + ] Δf (). Uing thi, the block diagram in Fig. 8 can be reduced a hown in Fig. 9 and 0. Δf () Fig. 9 educed block diagram - ( I ) ΔP T () _ ΔP D () G HT + G - ΔP D () + _ G Δf () Fig. 0 educed block diagram ( I ) GHT

47 + - ΔP D () _ G Δf () Fig. 0 educed block diagram ( I ) GHT G Therefore Δf () = - ΔPD () I GHT G GHT G (37) The above equation i much more general. For a given ΔP D (), Δ f (t ) can be obtained by taking Lalace invere tranform. By etting I a zero, we get the exreion for Δf () uncontrolled cae, a een by eq. (27). Δf () = - G G H G T G ΔP D () (27) correonding to

48 Δf () = - I G HT G G G HT G ΔP D () (37) 6. STATIC FEQUENCY DOP FOLLOWING A STEP LOAD CHANGE M Let the te load change be ΔP D, which i equal to M. Then ΔP D () = Uing final value theorem, lim Δf 0 = [ Δf () ] (38) 0 = - I G HT G G M G HT G = - I 0 0 M = 0 (39) Thu tatic frequency dro due to te load change become zero, which i a deired feature we were looking. Thi i made oible becaue of the integral controller that ha been introduced.

49 6.2 DYNAMIC ANALYSIS Δf () = - I G HT G G G HT G ΔP D () (37) Let u aume time contant T H and T T a zero a we did in Section 5. Then G H T =. For a te load change of ΔP D = M, from eq. (37) we get Δf () = - I T T T M (40) Multilying the numerator and denominator by ( + T ) we get Δf () = - ( T ) M I (4)

50 Δf () = - I ) T ( M (4) Dividing the numerator and denominator by T the above equation become I 2 T T M T Δ f ( ) I 2 T T M T (42)

51 M Δf() = (42) T 2 I T T We obtain the time reone Δ f (t ) uon Lalace invere tranformation of thi exreion in the right hand ide of above equation. Since reone deend uon the ole of the denominator olynomial, let u examine it. 2 I 0 (43) T T A econd order ytem of tye a 2 + b + c = 0 will be table if the coefficient a, b and c are greater than zero. Since thi condition i met with, the ytem under conideration i STABLE. For a econd order equation 2 + b + c = 0, the root are b b 2,2 c 2 4 The nature of the root deend on b 2 4 c

52 0 T T I 2 For critical cae c 4 b 2 = 0 i.e. b 2 = 4 c Now, both the root are real, equal and negative. For thi critical cae 2 ) T ( = I T 4 Thu I crit 2 ) ( T 4 (44) Suer critical cae When b 2 < 4 c the root are comlex conjugate and the olution i exonentially damed inuoidal. For thi cae the integral gain contant I i obtained from 2 ) T ( I T 4 i.e. I I crit (45) When I I crit the ytem i aid to be uer-critical. Due to daming reent, the final olution will tend to zero. However, the olution will be ocillatory tye.

53 Sub-critical cae When b 2 > 4 c the root are real and negative. For thi cae, the integral gain contant I i given by ( T ) 2 4 I T i.e. I I crit (46) Thi cae i referred a ub-critical integral control. In thi cae the olution contain term of the tye e αt and e α2t and it i non-ocillatory. However, finally the olution will tend to zero. Thu the integral controller ytem i STABLE and ISOCHONOUS i.e. following a te load change, the frequency error alway return to zero. The dynamic reone for different value of I are hown in Fig..

54 In ractical ytem T H and T T will not be zero; but will have mall value. When T H = 80 m ec; T T = 0.3 ec and T = 20 ec., dynamic reone for different integral gain contant I will be a hown in Fig. 2.

55 7. TWO-AEA SYSTEM A control area i characterized by the ame frequency throughout. Thi tantamount to aying that the area network i rigid or trong. In the inglearea cae we could thu rereent the frequency deviation by the ingle variable Δf. In the reent cae we aume each area individually trong. Having interconnected them with a weak tie-line therefore lead u to the aumtion that the frequency deviation in the two area can be rereented by two variable Δf and Δf 2 reectively. 7. MODELING THE TIE-LINE and BLOC DIAGAM FO TWO-AEA SYSTEM In normal oeration the ower flow in the tie-line connecting the area and 2 i given by 0 0 V P 0 V = in(δ δ2 ) (47) X where δ 0 and δ 0 2 are the angle of end voltage V and V 2 reectively. The order of the ubcrit indicate that the tie-line ower i defined in direction to 2.

56 nowing dy/dx = Δy/Δx, for mall deviation in angle δ and δ 2 the tie-line ower change by an amount 0 0 V V2 0 0 ΔP 2 co(δ δ2 )(Δδ Δδ 2 ) (48) X We now define the ynchronizing coefficient of a line a 0 0 V T 0 V2 0 0 = co(δ δ2 ) MW / rad. (49) X Then the tie-line ower deviation i ΔP 2 = T 0 (Δδ Δδ 2 ) MW (50) We like to have ΔP 2 in term of frequency deviation Δf and Δf 2. We know f = ω 2π 2π dδ ; dt i.e. dδ 2Π f dt Thu δ 2π t 0 f dt (5) and hence t Δδ = 2 π Δf dt (52) Exreing tie-line ower in term of Δf and Δf 2 we get t ΔP 2 = 2 π T 0 ( t Δf dt - Δf 2 dt ) (53)

57 t ΔP 2 = 2 π T 0 ( t Δf dt - Δf 2 dt ) (53) Taking Lalace tranformation of the above eq. we get ΔP 2 () = 2 π T 0 [ Δf () Δf 2 () ] (54) ereenting thi equation in term of block diagram ymbol yield the diagram in Fig. 3. Δf () + ΔP 2 () 2 π T 0 _ Δf 2 () Fig. 3 ereentation of Tie-lone ower flow Tie-line ower ΔP 2 hall be treated a load in area. Similar to ower balance eq. Δf () = G () [ ΔP T () ΔP D () ] (22) we can write Δf () = G () [ ΔP T () ΔP D () ΔP 2 ()] (54 a) Block diagram for two-area uncontrolled ytem i hown in Fig. 4.

58 Δf () = G () [ ΔP T () ΔP D () ΔP 2 ()] (54 a) Δf () ΔP D () ΔP ref () _ + ΔP g () G HT ΔP T () + _ ΔP 2 () _ G 2πT 0 + _ Δf () - ΔP ref 2 () + _ ΔP g2 () G HT2 ΔP T2 () _ + _ ΔP 2 () G 2 Δf 2 () 2 Δf 2 () ΔP D2 () Fig. 4 Block diagram rereentation of two area ytem

59 Similarly we need to add ΔP 2 in area 2. Defining the tie-line ower in direction 2 to a ΔP 2. ΔP 2 = - ΔP 2 (55) For thi reaon, tranfer function of - i introduced in the block diagram. Further, we remember that the ower in the ingle-area diagram were exreed in er unit of area rating. The arameter, D and H, likewie were baed on the ame bae ower. When two or more area of different rating, are involved, we mut refer all ower and arameter to the one choen bae ower.

60 7.2 STATIC ESPONSE OF UNCONTOLLED TWO-AEA SYSTEM We hall firt invetigate the tatic reone of the two-area ytem with fixed eed changer oition; i.e. ΔP ref = ΔP ref 2 = 0 We aume that the load in each area are uddenly increaed by the contant incremental te ΔP D = M and ΔP D2 = M 2. At teady tate, a een from Fig. 5, Δf 0 = Δf 2 0 = Δf 0. We hall reently limit our analyi to finding the tatic change in frequency and tie-line ower denoted by Δf 0 and ΔP 2 0 reectively. Δf 0 ΔP D 0 _ ΔP g 0 _ Δf 0 G HT ΔP T 0 + ΔP 2 0 _ D Δf 0 G + Note that ΔP T 0 = - Δf0 and ΔP T2 0 = - Δf0 (56) 2 nowing the outut of G a Δf 0 inut to G = D Δf 0. Similarly, inut to G 2 = D 2 Δf 0.

61 Conider the umming oint with ΔP T 0, ΔP D 0, and ΔP 2 0. Δf 0 ΔP D 0 Δf 0 _ ΔP g 0 _ G HT ΔP T 0 + ΔP 2 0 _ D Δf 0 G + - Δf0 - M - ΔP 2 0 = D Δf 0 and - Δf0 M 2 + ΔP 2 0 = D 2 Δf 0 i.e. 2 D Δf 0 + Δf0 + ΔP 2 0 = M ; i.e. β Δf 0 + ΔP 2 0 = M (57) D 2 Δf 0 + Δf0 - ΔP 2 0 = M 2 ; i.e. β 2 Δf 0 - ΔP 2 0 = M 2 (58) 2 where area frequency reone characteritic (AFC) of each area are defined a β = D + β 2 = D 2 + (59) 2

62 β Δf 0 + ΔP 2 0 = M (57) β 2 Δf 0 - ΔP 2 0 = M 2 (58) Solving the above equation (57) and (58) for Δf 0 and ΔP 2 0 we get Δf 0 = - M β M β 2 2 (60) ΔP 2 0 = - ΔP 2 0 = β M β 2 β β 2 2 M (6) Eqn. (60) and (6) become imle if we aume identical area arameter; i.e. = 2 = ; D = D 2 = D; Then β = β 2 = β

63 We then get Δf 0 = - M M2 2β (62) ΔP 2 0 = - ΔP 2 0 = M2 M 2 (63) For examle, if a te load change occur only in area 2, we get Δf 0 = - M 2 Hz (64) 2β ΔP 2 0 = - ΔP 2 0 = M 2.u. MW (65) 2 The above two equation tell u, in a nuthell, the advantage of ool oeration:. The frequency dro will be only half that would be exerienced if the area were oerating alone % of the added load in area 2 will be ulied by area via. the tie-line.

64 Examle 8 A 2000 MW control area i interconnected with a 0000 MW area 2. The 2000 MW area ha the ytem arameter a = 2.0 Hz /.u. MW; D = 0.0.u. MW / Hz Area 2 ha the ame arameter, but on a bae of 0000 MW. A 20 MW load increae take lace in area. Find tatic frequency dro and tie-line ower change. Solution Let u take 2000 MW a bae ower. β = = 0.5.u. MW / Hz 2 β 2 = 0.5.u. MW / Hz on a bae of 0000 MW = 0.5 x 0000 = 2.55.u. MW / Hz on a bae of 2000 MW 2000

65 20 M = = 0.0.u. MW 2000 Δf 0 = = Hz ΔP 2 0 = - ΔP 2 0 = x = u. MW = MW Note that the frequency dro i only one ixth of that exerienced by area oerating alone (0.096 Hz comare Examle 6). Alo note that thi frequency uort i accomlihed by an added delivery of 6.67 MW from the larger area.

66 7.3 DYNAMIC ESPONSE OF UNCONTOLLED TWO-AEA SYSTEM With the very imle turbine model that we have ued, the two area ytem in Fig. 4 i of eighth order for a te load change. It would therefore meaningle, to attemt a direct analytic aroach for finding the dynamic reone of the ytem. Tyical imulation reult for a two-area ytem are hown in Fig. 5. Fig. 5 Simulation reult of a two-area ytem

67 Fig. 5 Simulation reult of a two-area ytem It can be een that the olution for Δf (t) and Δf 2 (t) are ocillatory. However, becaue of ytem daming, finally they ettle at a teady tate value. Similarly, the olution of ΔP 2 (t) ha ocillation at beginning and finally ettle at a teady tate value.

68 7.4 TIE-LINE BIAS CONTOL FO TWO-AEA SYSTEM The eritent tatic frequency error i intolerable. Alo, a eritent tatic error in tie-line ower flow would mean that one area would have to uort the other on a teady tate bai. To circumvent thi, ome form of reet integral control mut be added to the two-area ytem. The control trategy of tie-line bia control i baed uon the rincile that all oerating ool member mut contribute their hare to frequency control in addition to taking care of their own net interchange. Thi mean that for two-area ytem, at teady tate, both Δf 0 and ΔP 2 0 mut be zero. To achieve thee objective, the Area Control Error (ACE) for each area conit of a linear combination of frequency and tie-line error. Thu ACE = ΔP 2 + B Δf (66) ACE 2 = ΔP 2 + B 2 Δf 2 (67)

69 The eed changer command will thu be of the form ΔP ref = - I ( ΔP 2 + B Δf ) dt (68) ΔP ref 2 = - I2 ( ΔP 2 + B 2 Δf 2 ) dt (69) The contant I and I 2 are integrator gain and the contant B and B 2 are the frequency bia arameter. The minu ign mut be included to enure that, if there i oitive frequency deviation or tie-line ower deviation, then each area hould decreae it generation 7.5 STATIC SYSTEM ESPONSE WITH TIE-LINE BIAS CONTOL The choen trategy will eliminate the teady-tate frequency and tie-line deviation for the following reaon. Following a te load change in either area, a new tatic equilibrium can be achieved only after the eed-changer command have reached contant value.

70 But thi evidently require that both the integrant in eqn (68) and (69) be zero; i.e. ΔP B Δf 0 = 0 (70) and ΔP B 2 Δf 0 = 0 i.e. - ΔP B 2 Δf 0 = 0 (7) The above two condition can be met with only if Δf 0 = 0 and ΔP 2 0 = - ΔP 2 0 = 0 (72) Note that thi reult i indeendent of the value of B and B 2. In fact, one of the bia arameter can be zero, and we till have a guarantee that eq. (72) i atified. Having checked ue of the integral controller, let u ee how they can be included in the block diagram. Lalace tranform of Equation (68) and (69) give ΔP ref () = I [ΔP 2 () + B Δf ()]; ΔP ref 2 () = I2 [ΔP 2 () + B 2 Δf 2 ()]

71 ΔP ref () = I [ΔP 2 () + B Δf ()]; ΔP ref 2 () = I2 [ΔP 2 () + B 2 Δf 2 ()] B Δf () Δf () ΔP D () ΔP 2 () + + I _ + ΔP ref () ΔP g () G HT ΔP 2 () ΔP T () + ΔP 2 () G 2πT 0 + _ Δf () - - ΔP 2 () + + I 2 ΔP ref 2 () + _ ΔP g2 () G HT2 ΔP T2 () _ + _ ΔP 2 () G 2 Δf 2 () B 2 Δf 2 () 2 Δf 2 () ΔP D2 () Fig. 6 Block diagram rereentation of two area ytem with tie-line bia control

72 Quetion on Automatic Load Frequency Control. What i the baic role of ALFC? 2. Name the main art in the real ower control mechanim of a generator. 3. Draw the functional diagram of real ower control mechanim of a generator. Exlain how a raie command to the eed changer will reult in increaed generator outut ower. Dicu how eed dro will reult in increaed generator outut ower. 4. Draw the ortion of real ower control mechanim of a generator that correond to the eed changer and the eed governor and develo it block diagram. 5. Develo the block diagram correonding to eed changer, eed governor and the hydraulic valve actuator 6. Conider an uncontrolled eed governing ytem of a generator. Obtain the relationhi between frequency v outut ower at teady tate condition.

73 7. Conider the eed governing ytem of 50 Hz, 40 MW generator, having 4% eed regulation. Find the increae in turbine ower, if the frequency dro by 0.5 Hz with the eed changer etting unchanged. 8. Conider the eed governing ytem of 50 Hz, 00 MW generator, having 6% eed regulation. Find the increae in frequency, if the turbine ower i decreaed by 4 MW with the eed changer etting unchanged. 9. Two 50 Hz generator of rating 00 MW and 300 MW are oerating in arallel. Each of them ha a eed regulation of 5%. They uly a total load of 320 MW. Aume that the eed changer are et to give rated frequency at 00% rated outut ower. Determine the outut ower of each generator and the oerating frequency.

74 0. The following two ynchronou generator are oerating in arallel: Generator 50 MW 6 % eed regulation Generator 2 50 MW 3 % eed regulation i) Determine the load taken by each generator for a total load of 80 MW when the eed changer are et to give rated eed at 00 % rated outut. ii) The eed changer of generator i o adjuted that 80 MW load i equally hared. Find the outut of generator for rated eed it frequency at no load and the change to be made in the eed changer. iii) The eed changer of generator 2 i o adjuted that 80 MW load i equally hared. Determine the outut of generator 2 for rated eed, it frequency at no load and the change to be made in the eed changer.. Draw the block diagram rereentation of uncontrolled ingle area ower ytem. Develo the exreion for tatic frequency dro correonding to te load increae. Alo ketch the dynamic reone i) neglecting the time contant of eed governor and turbine generator ii) including the time contant of eed governor and turbine generator.

75 2. The data ertaining to an uncontrolled ingle area ower ytem are a follow: Total rated caacity Nominal oerating load Nominal frequency Inertia contant = 2500 MW = 500 MW = 50 Hz = 4 ec. Governor dro = 4 % Aume that the load frequency characteritic of the ytem i linear. For a decreae of 20 MW load, determine a) Steady tate frequency deviation and frequency. b) Change in generation ( in MW ) and increae in original load ( in MW ) under teady tate condition. c) Find the Tranfer Function of the ower ytem. d) Alo obtain the dynamic reone neglecting time contant of eed governor and turbine generator and ketch it.

76 3. Following data ertain to uncontrolled ingle area ower ytem. Sytem rating = 200 MW; Load = 00 MW; egulation = 4%; Sytem frequency = 50 Hz. Load increae = 0 MW; % frequency increae caue 0.8% load increae. Inertia contant = 4 ec. Find teady tate frequency deviation and the tranfer function of ower ytem. 4. What i the neceity of roortional lu integral controller? 5. Draw the comlete block diagram rereentation of ALFC of ingle area ytem and decribe the role of different comonent. 6. For a two area ytem, develo the exreion for change in tie-line ower flow in term of change in frequency of both the area and rereent it in the form a block diagram. 7. What are advantage of interconnected two area ytem over ingle area ytem? 8. Draw the block diagram rereentation of uncontrolled two area ytem and briefly exlain.

77 9. Develo the exreion for tatic frequency deviation and tatic tie line ower deviation for two area ytem ubjected to udden load change. From thee exreion, jutify the advantage of interconnected ytem. 20. Two uncontrolled interconnected ower ytem, A and B, each ha a eed regulation of 5 Hz /.u. MW and tiffne D of.0.u. MW / Hz (on reective caacity bae). The caacity of ytem A i 500 MW and of B i 000 MW. The two ytem are interconnected through a tie line and are initially at 50 Hz. If there i a 00 MW load change in ytem A, calculate the change in teady tate value of frequency and tie line ower. 2. What do you undertand by tie-line bia control? 22. Draw the block diagram rereentation of uncontrolled two-area ytem. Exlain how thi could be modified to include the tie-line bia control. 23. Draw the block diagram rereentation of two area ytem with tie-line bia control.

78 24. Two uncontrolled area and 2 are connected by a tie-line. Sytem arameter are: Area ated caacity 5000 MW = 2.5 Hz /.u. MW D = 0.02.u. MW / Hz Area 2 ated caacity 2000 MW = 2 Hz /.u. MW D = 0.05.u. MW / Hz Taking 5000 MW a the bae, find the teady tate frequency change and the change in tie-line ower flow from area to 2 when i) 20 MW load increae take lace in area. ii) 20 MW load increae take lace in area 2. iii) 0 MW load increae take lace in area and 2. iv) 20 MW load deceae take lace in area.

79 ANSWES 7. 3 MW Hz MW; 240 MW; 50.5 Hz MW; MW 45 MW; 52.7 Hz 60 MW; 5.8 Hz Hz; Hz MW; MW ; ( e t ) Hz Hz; Hz; MW; MW 24. i) Hz MW ii) Hz iii) Hz iv) Hz 3.25 MW 3.25 MW MW

80 SOLUTION. Mention four requirement of a ower ytem. i) It mut uly ower, ractically everywhere the cutomer demand. ii) It mut be able to uly the ever changing load demand at all time. iii) The ower ulied hould be of good quality. iv) The ower ulied hould be economical. v) It mut atify neceary afety requirement. 2. Over voltage: Increaed motor eed; vibration and mechanical damage; Inulation failure Under voltage: Decreaed motor eed; heating due to increaed current

81 3. Mention three baic role of Automatic Load Frequency Control. i) To maintain deired outut ower of a generator. ii) To maintain the frequency contant. iii) To maintain deired tie-line ower. 4. What do you undertand by Unit Commitment Problem? Generator unit are to be witched on or off to match with the varying load. The chedule of witching on and witching off the variou generator unit over a eriod, ay one day, uch that total cot of oeration over the eriod i minimum ubjected to certain contraint.

82 5. Draw the chematic diagram of eed changer and eed governor and mark the linkage and linkage oint movement.

83 6. The eed governing ytem of a 40 MW, 50 Hz generator, ha 3.6% eed regulation. Find change in turbine ower, if the frequency increae by 0. Hz f 5.8? 0. ΔP 0 40 T ; ΔPT 0 =.6 MW P T Decreae in P T0 =.6 MW O 2.5 ΔP T 0 = - Δf 0 = = Hz / MW; Δf 0 = 0. Hz ΔP T 0 = = -.6 MW (decreae in turbine ower)

84 7. Draw the comlete block diagram of uncontrolled ingle area ALFC loo. Δf () ΔP D () ΔP ref () + _ ΔP g () ΔP v () _ ΔP T () G H G T + T

85 P T 50 P T 0 x x 50 Let load on Gen. be x MW; Load on Gen. 2 = 240 x; Equating common fre x 5.5 (240 x ) ; On olving P G = 5. MW ; P G2 = MW 2.5 Oerating frequency = 5.5 X 5.= 50.6 Hz. 200 Correonding to P G2 = 75 MW, frequency f.5 = 5.5 X 75 = Hz 50 When f = Hz, with original characteritic, P G i obtained from P G = 50.75; Thu P G = 40 MW. However, to atify the total load 200 of 240 MW, P G mut be 65 MW. Thi i achieved by giving AISE command of 25 MW.

86 % of 50 Hz = 0.0 x 50 = 0.5 Hz; 0.8% of 900 MW = x 900 = 7.2 MW = 7.2 / 400 = u. MW D = = u. MW / Hz; = = 0.5 D = 97.8 Hz /.u. MW T = 2H f 0 D = 2 x 4 50 x = 5.55 ec.; Thu G () = T =

87 ΔP D () ΔP ref () + _ ΔP g () ΔP v () _ ΔP T () G H G T + T - ΔP D () Δf () Thi can be reduced a Thu Δf () = - G G H G T G ΔP D () + - G P GH G T Let ΔP D = M ; Then ΔP D () = M and hence Δf () = - G G H G T G M Uing the final value theorem, the tatic frequency dro i lim Δf 0 = [ Δf () ] = - 0 M = - M D Hz

88 Fig. how the configuration of EMS / SCADA ytem for a tyical ower ytem. Planning and analyi function Local Area Network SCADA hot comuter Energy Management Sytem (EMS) Function CONTOL CENTE COPOATE OFFICES Communication Network SCADA Front end comuter TU TU IED IED TU Subtation Monitored Device Fig. EMS / SCADA ytem configuration

89 In general cae, change may occur in both the eed changer etting and frequency in which cae the relationhi ΔP T 0 = ΔP ref 0 - Δf 0 alie. For a given eed changer etting, ΔP ref 0 = 0 and hence Δf 0 = - ΔP T 0. In a frequencygeneration ower grah, thi rereent a traight line with a loe = -. For a given frequency, Δf 0 = 0 and hence ΔP T 0 = ΔP ref 0. Thi mean that for a given frequency, generation ower can be increaed or decreaed by uitable raie or lower command. Thu the relationhi ΔP T 0 = ΔP ref 0 - Δf 0 rereent a family of line with loe -, each line correonding to a ecific eed changer etting.