Modern Aspects of Power System Frequency Stability and Control
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110
Modern Modern Aspects of Power Power System Frequency Frequency Stabilit Stabilityy and Control Control
solution f (t ) which represents the required evolution of the power frequency with time.
6.4.6 6.4.6 The Lapl Laplace ace transfo transform rm of con consta stant nt terms terms Applying the definition of Eq. Eq. (6.70), (6.70), the Laplace transform ℒ of some constant c is ℒðcÞ 5
1 s
ð6:71Þ
where s is the new Laplace operator-transformed variable that was defined by Eq. by Eq. (6.70). (6.70).
6.4.7 6.4.7 The Lapl Laplace ace transfo transform rm of terms terms line linear ar in in f As a remi reminde nder, r, we reca recall ll that that f represe represents nts the unknown unknown functi function on f (t ), ) , the power system frequency , which we are trying to discover by our subsequent calcula calculatio tions. ns. Conseq Consequent uently, ly, we shall shall leave leave the Laplace Laplace transfo transform rm of c f , where c is some some cons consta tant nt,, as just just cUℒð f Þ for the time being, until all the required terms have been collected together. So, for now, we simply write
ð6:72Þ
ℒðcU f Þ 5 cUℒðf Þ
6.4.8 6.4.8 The Lapl Laplace ace transfo transform rm of the the unit unit step step functio function n u (t 2 t trip trip) unit step step functio function n (also The unit (also known known as “Heavi “Heaviside side unit unit functio function”— n”—see see for instance Ref. [5] [5],, p. 8, for more details) represents a function that has the value zero for t less than t trip trip and the value 1 for t greater than or equal to t trip trip. The Laplace transform of the unit step function c u(t 2 t trip trip) is
ℒ cUu t t trip trip
c
5
s
Uexp 2 sUt trip trip
ð6:73Þ
where t trip trip is the time of trip (either of generation or demand), and c is some constant.
6.4.9 6.4.9 The Lapl Laplace ace transfo transform rm of terms terms line linear ar in in t The Laplace transform of a term of the form c t , where c is some constant, and t represents represents time, is ℒðcUt Þ 5
c
ð6 74Þ
111
Analytical methods for solving the “swing equation” Chapter | 6
6.4.10 6.4.10 The Lapl Laplace ace tran transfor sform m of terms terms propor proportio tional nal to t 2 The Laplace transform of the “quadratic form” c t 2 is 2
ℒ cUt
where c is some constant.
5 2U
c
ð6:75Þ
s3
6.4.11 6.4.11 The Lapl Laplace ace tran transfor sform m of terms terms of highe higherr order order in t For the record the Laplace transform of terms in the nth power of t is given by the general formula: n
ℒðcUt
Þ 5 cUn!=sn11
ð6:76Þ
This formula applies in general where n is a nonnegative integer .
6.4.12 6.4.12 The Lapla Laplace ce transf transform orm of terms terms propor proportio tional nal to the the first first derivative of f with with respect to t The Laplace transform of terms proportional to the first derivative of f with respect to t is ℒ
cU
d f dt
5 cU
sUℒðf Þ f ð0Þ
ð6:77Þ
where c (as always) is some constant, and f (0) (0) is the value of f at t 5 0. All or most of the results earlier should be familiar to any student who has attended an introductory course in Laplace transforms, but for extra clarity we have listed the results here.
6.4.13 6.4.13 Putting Putting it all togeth together er Let us now proceed to find our analytical solution f (t ) for Eq. (6.53), (6.53), using the individual results [Eqs. [Eqs. (6.71)(6.77) (6.77)]] that we have laid out earlier. Applyin Applying g the results results from from Eqs. Eqs. (6 (6.7 .71) 1) to (6 (6.7 .77) 7) inclusiv inclusivee to Eq. (6. (6.53) 53) produces the following initial result: c1 1 c2 Uℒðf Þ 1
c3
s
Uexp 2 sUt trip trip 1
c4 s2
1 2U
c5 s3
1 6U
c6 s4
1
? 5 ℒ f U
d f : dt
ð6 78Þ
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Modern Modern Aspects of Power Power System Frequency Frequency Stabilit Stabilityy and Control Control
6.4. 6.4.14 14 Deal Dealin ingg with with the f the f df /dt /dt term term We would now like to expand the Laplacian term on the right-hand side of Eq. (6.78), (6.78), and by doing so convert Eq. convert Eq. (6.78) to (6.78) to a form containing the variable s only throughout, rather than the current mixture of s and t . However, in its present form, converting the term ℒ f Ud f =dt to a function of s gives us a bit of a challenge. Although the Laplace transform of f is simply F (s), and the the Laplace transform of d f /dt is s F (s) 2 f (0) ( 0) [see [see Eq. (6.76)], (6.76)], the Laplace transform of the product f d f /dt requires requires some additional thought. thought. Essentially, what we need to do is to evaluate the Laplace transform of a product of two functions. Specifically, we need to find the Laplace transform ℒf f ðt ÞUgðt Þg Þg, where in our case f (t ) 5 t and g(t ) 5 d f /dt . Now the closest general formula to the Laplace transform of a product of Convolution theorem theorem (see Ref. [5] two functions is the Convolution [5],, Appendix A on p. 243 for more details). Could the convolution theorem help us in our case? Let’s see
6.4. 6.4.15 15 Tr Tryi ying ng to use th thee convolution theorem for for Laplace transforms on the f df /dt /dt term term The “convolution theorem” of Laplace transform theory can be stated in the following general way: First, let h(t ) be the convolution of f (t ) and g(t ) (see Ref. [5] [5] p. p. 243), that is t t
hðt Þ 5 ðf gÞðt Þ
ð
f ðuÞUgðt 2 uÞUdu
ð6:79Þ
0
With the definition given in Eq. in Eq. (6.78) the (6.78) the convolution theorem states that
ð6:80Þ
H ðsÞ 5 F ðsÞUGðsÞ
In which F ðsÞ 5 ℒf f ðt Þg Þg, GðsÞ 5 ℒfgðt Þg Þg, and H ðsÞ 5 ℒfhðt Þg Þg. Can we use the above result to help us find the Laplace transform of the product f (t ) g(t ) where f (t ) 5 t , and g(t ) 5 d f /dt ? Unfortunately, the convolution theorem cannot be applied in a straightforward manner to our case, because, in keeping with the definition of the convolution of f (t ) and g(t ) as given by Eq. by Eq. (6.78), (6.78), the subject of that integral is not the simple “ f (t ) g(t )” )” but instead “ f (u) g(t 2 u).” This presents us with a difficulty. In fact, Eqs. fact, Eqs. (6.78) and (6.79) are (6.79) are of much more practical use as an aid to the calcula calculatio tion n of inverse inverse Laplac Laplacee transfo transform rmss F (s) f (t ) than than to Lapl Laplac acee -
178
Modern Modern Aspects of Power Power System Frequency Frequency Stabilit Stabilityy and Control Control K3 5 TSTEP (2K_1 (F0 1 0.5 K2) 1 K_2)/(2 H F0) K4 5 TSTEP (2K_1 (F0 1 0.5 K3) 1 K_2)/(2 H F0)
% Calc Calcula ulate te the frequen frequency cy F(1) F(1) for the first first step: step: F(1)
5
F0 1 (K1 1 2 K2 1 2 K3 1 K4)/6
% Calcula Calculate te the frequen frequency cy for for three three time-steps: time-steps: for k 5 1,3 % Calcula Calculate te the coeffic coefficients ients for this this time step: step: K1 5 TSTEP (2K_1 F(k) 1 K_2)/(2 H F0) K2 5 TSTEP (2K_1 (F(k) 1 0.5 K1) 1 K_2)/(2 H F0) K3 5 TSTEP (2K_1 (F(k) 1 0.5 K2) 1 K_2)/(2 H F0) K4 5 TSTEP (2K_1 (F(k) 1 0.5 K3) 1 K_2)/(2 H F0) % Cal Calcula culate te the freque frequency ncy at the the next time-s time-step: tep: F(k 1 1)
5
F(k) 1 (K1 1 2 K2 1 2 K3 1 K4)/6
end % The val values ues f0, f1 f1, , f2 and f3 of the frequenc frequency y requi required red for sta starti rting ng the %
Milne-Hamming algorithm are now all known.
“
”
% % Progr Program ammin ming g the Milne-Hamming algorithm: “
”
% for k 5 3, 199 DFDT(k)
5
(2K_1 F(k) 1 K_2)/(2 H F0)
DFDT(k21) 5 (2K_1 F(k21) 1 K_2)/(2 H F0) DFDT(k22) 5 (2K_1 F(k22) 1 K_2)/(2 H F0) PF(k 1 1) 5 F(k23) 1 (4 TSTEP/3) (2 DFDT(k) DFDT(k21) 1
2 DFDT(k22))
PDFDT(k 1 1) 5 (2K_1 PF(K 1 1) 1 K_2)/(2 H F0) F(k 1 1) 5 (1/8) (9 F(k) F(k22)) 1
(3 TSTEP/8) (PDFDT(k 1 1) 1 2 DFDT(k) DFDT(k21))
end % The res result ults s for frequenc frequency y F can now be plotted plotted against against tim time e t.
7.11.7 7.11.7 Ins Instruc tructio tions ns for the use userr In each case, enter the input data in per unit (pu) when prompted and process the output data in such a way that a graph of “system frequency” against time may be plotted out.
7.11. 7.11.8 8 In the ne next xt se sect ctio ion n In the next section, we look at the “AdamsBashforthMoulton” method.
Numerical methods for solving the “swing equation” Chapter | 7
179
7.12 7.12 Th Thee “Ada “Adams msBashforthMoulton” multistep predictorcorrector method 7.12.1 7.12.1 A brief brief backgr background ound to the the “Adams “AdamsBashforthMoulton” method 1. Composition The “Adams “AdamsBashforthMoul Moulto ton” n” me metho thod d is a “m “mul ultis tiste tep p predi predicctorcorrector” method which is made up from an “AdamsBashforth” formula formula for the “predi “predicto ctor” r” and an “Adam “AdamssMoulton Moulton”” formul formulaa for the “corrector” (see Ref. [1] [1],, p. 181, for more details on this topic). 2. Possible combinations There are se seven each of of the “AdamsBashf ashfor orth th”” and “AdamsMoulto Moulton” n” formula formulas, s, but for solvin solving g the “swing “swing equatio equation,” n,” we shall be concentrating on the fourth-order formulas that have a four-step “AdamsBashforth” predictor stag stagee in combi combina natio tion n with with a three-step “AdamsMoulton” corrector stage.
7.12. 7.12.2 2 Th Thee de defi fini niti tion on of the fourth-order the fourth-order “AdamsBashforthMoulton” method The fourth-order “AdamsBashforthMoult oulton on”” method thod is defi define ned d as follows: The “pre “predi dict ctor or” ” for for the the “Ada “Adams ms Bashforth Moulton” fourth-order 1. The method The “pre “predic dicto tor” r” for for the the “Adam “AdamssBashforthMoulton” fourth-order method is defined as pyn11 5 yn 1
h
24
U
55U yn0 59U yn0 21 1 37U yn0 22 9U yn0 23
ð7:44Þ
where p denotes, as usual, a “predicted” value. See Ref. [1] [1],, p. 182, for further details. The “c “cor orre rect ctor or” ” for for the the “Ada “Adams ms Bashforth Moulton” fourth-order 2. The method The “cor “corre recto ctor” r” for for the the “Ada “Adams msBashforthMoulton” Moulton” fourth-order fourth-order method is defined as yn11 5 yn 1
h
24
0 0 0 0 U 9U pyn11 1 19U yn 5U yn21 1 yn22
where again denotes a “predicted” value.
ð7:45Þ
180
Modern Modern Aspects of Power Power System Frequency Frequency Stabilit Stabilityy and Control Control
7.12.3 The “truncation “truncation errors errors”” associa associated ted with the fourth-order fourth-order “AdamsBashforthMoulton” method 1. The
“truncation error” of the “predictor” for t he “Adams Bashforth Moulton” fourth-order method The “truncation error” associated with the “predictor” of the fourthorder “Adams “AdamsBashforthMoulton” method is given by T ð x ; hÞ 5
251Uh5 720
U y
½5
ðζ Þ
ð7:46Þ
fifth deri deriva vati tive ve and where the superscript [5] indicates the fifth ( xn23 , ζ , xn11). “truncation error” of the “corrector” for t he 2. The “Adams Bashforth Moulton” fourth-order method The “truncation error” associated with the “corrector” of the fourthorder “Adams “AdamsBashforthMoulton” method is given by T ðx; hÞ 5 2
19Uh5 720
U y
½5
ðζ Þ
fifth deri deriva vati tive ve where the superscript [5] indicates the fifth ( xn22 , ζ , xn11). See Ref. [1] [1],, p. 182, for further details on this topic.
ð7:47Þ and
7.12.4 7.12.4 The “stabi “stabilit lity” y” of the “Adams “AdamsBashforthMoulton” method 1. Introduction The The “Ada “Adams msBashforthMoulton” method is one one of the more stable methods (see Ref. [1] [1],, Fig. 4.19 on p. 187, for the “region of stability” in the complex hλ plane for the fourth-order “AdamsBashforth” plus “AdamsMoulton” combination). Comparative “stability” “stability” of the “Adams Bashforth Moulton” method 2. Comparative The “regio gion of stabi ability” of the “Adam damsBashforthMoulton” metho thod is about one-third tth he size of that of the fourth-order “RungeKutta” method (see Ref. [1] [1],, p. 120) and about twice that of the “MilneHamming” method (see Ref. [1] [1],, p. 179). trade-off between “accuracy” “accuracy” and “stability” “stability” 3. The trade-off Please note that increasing the “accuracy” of the “AdamsBashforthMoulton” method is of course achievable by reducing the “step size” h, but at the expense of decreasing its “region of sta-
187
Numerical methods for solving the “swing equation” Chapter | 7 % In this this cas case e the par paramet ameters ers a 0and a1are NOT both zero. % The dat data a for this case case are listed listed in Section Section 7.11.2 7.11.2 (d) (d) (ii) above. above. % % Ent Enter er the the dat data a in in per unit format: “
”
% PNR 5 input( Please enter the non-responsive generation in per unit ) ‘
’
PR 5 input( Please enter the responsive generation in per unit ) ‘
’
PL 5 input( Please enter the system load in per unit ) ‘
’
K1 5 input( Please enter the value of the constant k1 in per unit ) ‘
’
FN 5 input( Please enter the nominal frequency in per unit ) ‘
PTRIP
5
’
input( Please enter the generation tripped in per unit ) ‘
’
A_0 5 input( Please enter the value of the constant a0 in per unit ) ‘
’
A_1 5 input( Please enter the value of the constant a1 in per unit ) ‘
’
H 5 input( Please enter the value of the H-constant in seconds ) ‘
’
F0 5 input( Please enter the value of the initial frequency in per unit ) ‘
’
% % Calcula Calculate te the power power bala balance nce at nominal frequenc frequency y as a paramet parameter: er: % PBAL 5 PNR 1 PR PL PTRIP % % Star Start t off off the the routi routine ne using using the Classic Fourth-Order Runge-Kutta “
”
% alg algor orit ithm: hm: % % Calc Calcula ulate te the par paramet ameters ers for for the the initial condition : “
”
% Inpu Input t the size size of the time time step step for for the the Classic Fourth-Order Runge Kutta “
% meth method od: : % TSTEP
5
input( Please enter the value of the time step h in seconds ) ‘
’
% % Calc Calcula ulate te the valu value e of the the parame parameter ter k 1 in per unit : “
”
% K_1 5 1.0 PL % % Calc Calcula ulate te the the par paramet ameter er k 2 in per unit : “
”
% K_2 5 PBAL 1 K_1 FN 1 A_0 % % Calc Calcula ulate te the par paramet ameters ers for for the the initial condition of the Classic “
”
Fourth-Order % Rung Runge-K e-Kutta utta alg algorit orithm: hm: K1 5 TSTEP (2K_1 F0 1 K_2)/(2 H F0) K2 5 TSTEP (2K_1 (F0 1 0.5 K1) 1 K_2)/(2 H F0) K3 5 TSTEP (2K_1 (F0 1 0.5 K2) 1 K_2)/(2 H F0) K4 5 TSTEP (2K_1 (F0 1 0.5 K3) 1 K_2)/(2 H F0)
“
”
188
Modern Modern Aspects of Power Power System Frequency Frequency Stabilit Stabilityy and Control Control
% Calcula Calculate te the coeffic coefficients ients for this this time step: step: K1 5 TSTEP (2K_1 F(k) 1 K_2)/(2 H F0) K2 5 TSTEP (2K_1 (F(k) 1 0.5 K1) 1 K_2)/(2 H F0) K3 5 TSTEP (2K_1 (F(k) 1 0.5 K2) 1 K_2)/(2 H F0) K4 5 TSTEP (2K_1 (F(k) 1 0.5 K3) 1 K_2)/(2 H F0) % Calc Calcula ulate te the frequen frequency cy at the next next time-s time-step: tep: F(k 1 1)
5
F(k) 1 (K1 1 2 K2 1 2 K3 1 K4)/6
end % The The valu values es f 0, f1, f2 and f3 of the frequency required for starting the %
Adams-Bashforth-Moulton algorithm are now all known.
“
”
% % Sett Setting ing up the time vect vector: or: % for k 5 1, 200 T(k) 5 TSTEP k end % % Progr Program ammin ming g the Adams-Bashforth-Moulton algorithm “
”
% for k 5 3, 199 DFDT(k)
5
(2K_1 PF(k) 1 A1 T(k) 1 K_2)/(2 H F0)
DFDT(k 21) 5 (2K_1 PF(k21) 1 A1 T(k21) 1 K_2)/(2 H F0) DFDT(k 22) 5 (2K_1 PF(k22) 1 A1 T(k22) 1 K_2)/(2 H F0) DFDT(k 23) 5 (2K_1 PF(k23) 1 A1 T(k23) 1 K_2)/(2 H F0) % %
The Predictor : “
”
% PF(k 1 1) 5 F(k) 1 (TSTEP/24) (55 DFDT(k) 1
2
59 DFDT(k21)
37 DFDT(k22) 9 DFDT(k23))
% PDFDT(k 1 1) 5 (2K_1 PF(k 1 1) 1 A1 T(k 1 1) 1 K_2)/(2 H F0) % %
The Corrector : “
”
% F(k 1 1) 5 F(k) 1 (TSTEP/24) (9 PDFDT(k 1 1) 1 19 DFDT(k) 5 DFDT(k21) 1 DFDT(k22))
2
end % % The res result ults s for frequenc frequency y F can now be plotted plotted out against against time time T. %
7.13.7 7.13.7 Ins Instruc tructio tions ns for the use userr
Numerical methods for solving the “swing equation” Chapter | 7
189
7.14 Advantag Advantages es and and disadva disadvantage ntagess of the numer numerical ical methods methods 7.14. 7.14.1 1 Adva Advanta ntage gess Compar Compared ed with the analyti analytical cal me metho thods, ds, the numeri numerical cal method methodss cope cope more more easily with step changes in the power balance on the system being modeled. The new power balance can be simply introduced into the equation at the next next time time step step,, with withou outt havi having ng to be conc concer erne ned d abou aboutt havi having ng to sati satisf sfy y boundary conditions.
7.14.2 7.14.2 Disadv Disadvant antage agess The disadvantages of the numerical methods are (1) that the solutions generated are not exact and their accuracy depends upon the step size chosen; (2) there is no guarantee of numerical stability, although the risk of this can be minimized by choosing a method with a sufficiently large region of stability.
7.15 The next next chapt chapter: er: the the “contr “control ol diagra diagram” m” metho method d In the next chapter, we investigate a method that can be used to solve the “swing equation” based on a “control diagram” approach.
References Books [1]
L. La Lapi pidu dus, s, J. J.H. H. Se Sein infe feld ld,, Nu Nume meri rica call So Solu luti tion on of Or Ordi dina nary ry Di Diff ffer eren enti tial al Eq Equa uati tion ons, s, Academic Press, 1971 (Library of Congress Catalog Card Number: 73-127689).
[2]
J.D. Lambert, Lambert, Nume Numerical rical Methods Methods for Ordinary Differenti Differential al Syst Systems: ems: The Init Initial ial Value Problem, Wiley, 1991. ISBN: 0471929905.
[3]
R. Bronson, Bronson, (Schau (Schaum’s m’s Outline Outline Series) Theory Theory and Problems Problems of Differential Differential Equat Equations, ions, second seco nd ed. ed.,, McG McGraw raw-H -Hill ill Inc Inc., ., New Yor York, k, St. Lou Louis, is, San Fra Franci ncisco sco,, Auc Auckla kland, nd, Bog Bogot ota´ , Caracas, Caraca s, Lisb Lisbon, on, London, Madrid, Mexico City, Milan Milan,, Mont Montreal, real, New Delhi Delhi,, San Juan, Singapore, Sydney, Tokyo & Toronto, 1994. ISBN: 0-07-008019-4.
[4]
K.K. Mishra, A Handb Handbook ook on Numer Numerical ical Technique Technique Lab: MATL MATLAB AB Based Exper Experimen iments, ts, second ed., I. K. International Publishing House, 2007. ISBN: 978-8189-866433. 2015.
Paper [5]
R.M.. War R.M Warten ten,, Aut Autom omati aticc ste step-s p-size ize con contro troll for Run Rungege-Kut Kutta ta Int Integr egrati ation, on, IBM J. Res Res.. Develop. Devel op. 2 (1963 (1963)) 340.
Some important practical applications Chapter | 9
221
run by submitting prices, with “bids” and “offers” to have their outputs increased or decreased. b. Two “merit orders” However, since we also need to provide response, in practice there are two “merit orders” in operation at the same time: one for “fully loaded” generators and the other for “partly loaded” generators. part-loading c. The degree of part-loading In theory theory a gene genera rato torr ma may y be down downloa loade ded d to its its mini minimu mum m stab stabili ility ty limit, for example, 30% of full output, the precise figure depending on the the ma make ke and and type type of ma mach chine ine.. In prac practi tice ce,, most most gene genera rator torss used used for for respo respons nsee prov provis ision ion will will be downl downloa oaded ded to some some much much high higher er loadi loading ng,, typically 90% of its rated output. simple le “gen “gener erat ator or sche schedul dulin ing” g” proc proced edur ure e for esti estimat matin ing g frequ frequen ency cy 6. A simp response requirements In practice the selection of generation to meet demand is a kind of multiple optimization process involving many parameters, but to demonstrate strate the main main princi principles ples of “inert “inertia” ia” and “downlo “downloadi ading ng for respons response” e” and how they affect the subsequent behavior of system frequency, in the next next exer exerci cise se we shal shalll be ma maki king ng an init initia iall esti estima mate ted d mix mix of “ful “fully ly loaded” and “partly loaded” machines to meet the demand and estimated respo respons nsee requ requir irem ement entss and and then then refi refini ning ng this this esti estima mate te to find find the the best best ratio that just satisfies the statutory frequency requirements for the size of the loss involved. assumption about the nature of generator generator response 7. An assumption In the the next next exer exerci cise se,, we shal shalll be assu assumi ming ng that that all all the the gene genera rator torss available for responsive duty are traditional synchronous machines possessing a “ramped response profile.” “exact solution” of the “swing equation” equation” 8. Using the “exact To help us complete the next exercise ( MATLAB example simulation #10: calculating the “response requirements” for a “low frequency excursion”), we shall be using the “exact solution” of the “swing equation” as given at the beginning of this chapter: d f dt
5
t TRIP ΣPGENðNRÞ 1 ΣPGEN n ð R Þ 2 ΣPLOAD n 2 k 1 Uðf f n Þ 2 PTRIP Uuðt TRIP Þ 1 a0 1 a1 t 2U H U f 0
ð9:22Þ k 1 5 k UΣPGEN n ð RÞ 1 αUΣPLOAD n
ð9:23Þ
in the form given in Eq. in Eq. (6.27) of Chapter Chapter 6, 6, Analytical methods for solving the “swing equation”: f ð Þ
A
B
C
ðD
Þ
ð9 24Þ
222
Modern Modern Aspects of Power Power System Frequency Frequency Stabilit Stabilityy and Control Control
A1 5 f n 1
a0 1 ΣPGENðNRÞ 1 ΣPGEN nðRÞ 2 ΣPLOAD n
2 2U H U f 0 U
k 1
a1
ð9:25Þ
k 12 B1 5
C 1 5 f 0 f n 2
a1
a0 1 ΣPGENðNRÞ 1 ΣPGEN nðRÞ 2 ΣPLOAD n
1 2U H U f 0 U
k 1
a1
ð9:26Þ
k 1
ð9:27Þ
k 12
D1 5
2 k 1
2U H U f 0
ð9:28Þ
where k 1 5 ΣPLOAD n UαU 1 ΣPGEN nðRÞ Uk
ð9:29Þ
in which the constant “α” is defined implicitly by the equation ΣPLOAD 5 ΣPLOAD n ½1 1 αUðf f n Þ
ð9:30Þ
and the constant “k ” is defined by the equation ∆PRESP 5 2 PRESP Uk ∆ f
ð9:31Þ
where Dð%Þ 5
100 k ðpuÞ
ð9:32Þ
is the “generator governor droop” in “%” and ∆ f f 2 f n. Some typical values are α 5 2% per Hz and D 5 4%.
9.5 MATLAB MATLAB example example simula simulation tion #10: calculatin calculatingg the the “respons “responsee requirements” for a “low frequency excursion” 9.5. 9.5.1 1 In Intro troduc ducti tion on The objective of this exercise is to calculate how much response must be held in total on system generators to cover for the loss of one of the generators. This involves finding out how much responsive generation needs to be held by de-loading some machines to less than their rated MW outputs.
