7/30/2019 01023854

1/5

New Formula to Determine the Minimum Capacitance Required for Self-ExcitedInduction Generator

ALI M. ELTAMALY,PhDElectrical Engineering Department, Faculty of EngineeringElminia University, Elminia, Egypt

E-mail: eltamalv@,vahoo. omAbstract- Induction generator is the most common generatorin wind energy systems because of its simplicity, ruggedness,little maintenance, price and etc. The main drawbacks ininduction generator is its need of reactive power means tobuild up the terminal voltage. But this drawback is not anobstacle today where PWM inverters can accurately suppliesthe induction generator with its need from reactive power.The minimum terminal capacitor required for inductiongenerator to build up is the main concern. Most of previouswork uses numerical iterative method to determine thisminimum capacitor. But the numerical iteration takes long

time and divergence may be occurs. For this reason it cannotbe used online. A new simple formula for the minimum self-excited capacitor required for induction generator ispresented here. By using this formula there is no need foriteration and it can be used to obtain the minimum capacitorrequired online. Complete mathematical analysis for inductiongenerator to drive this new formula is presented. The resultfrom this new formula is typical as the results from iterativeprocesses.I. INTRODUCTION

Induction generator has a widely acceptance in usingwith wind energy conversion systems for many reasons.Induction generator is very simple, very rugged, reliable,cheap, lightweight, long lifetime (more than 50 years),produces high power per unite mass of materials andrequires very little maintenance. All above advantages arevery important especially in wind energy conversionsystems where the generator is in the top of he tower wherethe weight, maintenance and life time are very importantaspects. Induction generator can be used with stand alone aswell as grid connected wind energy conversion systems.Also, induction generator works with constant speedconstant frequency systems as well as variable speedconstant frequency systems. The main drawback ofinduction generator in wind energy conversion systemsapplications is its need for leading reactive power to buildup the terminal voltage and to generate electric power.Using terminal capacitor across generator terminals cangenerate this leading reactive power. The capacitance valueof the terminal capacitor is not constant but it is varyingwith many system parameters like shaft speed, load powerand its power factor. If the proper value of capacitance isselected, the generator will operate in self-excited mode.The capacitance of the excitation capacitor can be changedby many techniques like switching capacitor bank [ l], [2],thyristor controlled reactor [3] and thyristor controlled DCvoltage regulator [4]. In last decade many researches usesPWM technique to provide the desired excitation bycontrolling the modulation index and the delay angle of the

control waveform [ 5 ] . All previous techniques require anaccurate capacitance value for the terminal capacitor withchanging the system parameters. Many researches havebeen done to determine the minimum capacitor for selfexcited Induction generator [6,7,8,9,10]. Most of theseresearches use loop equations in the analysis of inductiongenerator equivalent circuiit [7], [8]. Most of theseresearches have much difkulty and it needs numericaliterative techniques to obtain the minimum capacitancerequired. Some of these researches require several minutesof computation by computer to obtain accurate value for theminimum capacitor required for this reason it is impossibleto uses these methods online [9].11. INDUCTIONENERATORQUIVALENTIRCUITThe structureof squirrel cage induction generator is sameas induction motor have alurrunum bar winding laid into theslots of the rotor core and short-circuited at both ends.Single-phase equivalent circuit of three-phase cagegenerator is similar to three-phase transformer equivalentcircuit with one winding is short-circuited, and the samecircuit models apply as shovm in Fig.1 (all reactances arerefereed to rated frequency,5 and the stator side). Theterminal capacitor shown in Fig.1 is to feed the inductiongenerator with the required reactive power.The circuit shown in Fig.1 can be used in steady stateoperation. But, in case of v,arying operating frequency ofthe generator, this circuit can be modified to be as'thecircuit shown in Fig.2 [ l l] . 'The elements of this circuit iscorresponding on the rated fiequency. In this circuit, themachine core losses have been ignored. In fact, forminimum capacitance requined, the machine must operateat threshold of saturation. Therefore, ignoring such losseswill result in no serious errors in estimating C,,, [6]. Thesuccessfully build up in sellf-excited induction generatoroccurs when X, have a value in saturation such as shown inFig.3 (i.e. X , should be less thanX,,)

Fig.1The equivalent circuit of one phase of three-phase inductiongenerator.

0-7803-7262-X/02/$10.002002 EEE. 106

7/30/2019 01023854

2/5

__.____I , _.____________________.~~Fig.2 Modified equivalent circuit of induction generator.

Where th e slipS s shown in equation (1)

-I

I Rr

Divided (1) by Ns p Then;a - bS=-a

I Im

( 5 )

(6)

P(X,,a) =(4X, 4 ) a 3+AJ, +A4)a2+(4X, %>a +(4X, $1 =0

+(B,X, +B,)aZ + ( B , X m + B * ) a + B ,= oQ ( X , ,u ) =( B I X ,+B2)a4+ B 3 X ,+B 4 ) a 3

The values of coefficientsA l to A8 and Bl to B9 regiven in [7].

Knowing the relationship betweenX , and (Via)(It canbe experimentally determined [7]) it is possible to computeC,, by using the following iterative procedure: -(i) Assume initial value of terminal capacitor C and solve( 5 ) and (6) for X , and a. The initial value of C should

be large enough to cause self-excitation of self-excitedinduction generator; X , has a value that lies in thesaturation region.

(ii) Gradually decrease the value of C in steps and computeX, corresponding to each value of C. A plot of X ,versus C is thus obtained.

(iii) C,, is obtained from such a plot as the intersection ofX , versus C curve and the lineX,=X,,, where X,, isthe maximum saturated reactance of the machine.

In this method the variation of X , with VJfis taken intoaccount.

The magnetizing reactance X , decreases with increasingsaturation as shown in Fig.4. The value of X , correspondsto an operating point tangent to the magnetizing curve isX,, that can be experimentally determined [7]. The valueof X , of the machine varies with operating conditions; theassumption of single value of X, in the analysis isacceptable [6] and [lo].

Fig.3 Th e saturation characteristicsfor induction generator.III. CALCULATING BY USINGLOOPANALYSISThe calculation of C, , by using loop analysis technique

has been presented in many researches [6], [7], [8] and [lo].This technique is listed in this section to explain itsdrawbacks.

The loop equation forIsof Fig.2 can be written as :-Where Z is the loop impedance seen by the current, IsI,Z=O (3 )

z=z, + ZLC + ZS (4)and can be obtained as in (4)Where

Z =(+]Il[$+ j X L ) And Z,=S+j X ,a

In steady state operation IS f 0 otherwise there is nogenerated voltage. Then; from (3); Z has to be equal to zero.By equating both the real and imaginary parts of (3) by zerowe get two nonlinear equations ( ( 5 ) and (6)) in function ofX, and a. Solving ( 5 ) and (6) together yields the values ofX, and a.

If we use the value of X , in. he calculation of C,, thenthe result is the minimum the minimum capacitancerequired (Cmifl) or successful build up in self-excitedinduction generator. According to this assumption (X , =constant) reference [6] modifies ( 5 ) and (6) to be as shownin (7) and (8) are function in a andXc .- la3+a2a2+ a3X , +a4>a , X,=o (7)-b1a4+ b 2 a 3+ ( b 3X , +b4)a2+(b,X , +b,)a-b, X , = O

Then by separation of Xc in (7) and (8) we can get thefollowing two equations: -

ala3 a2a2- 4aa3a-a5

b1a4- 2a3- 4a2- ,ab, a' - 5a- ,

X , =

x,=(9 )

The Coefficientsal to a5and bl to b7are positive realBy equating the right hand sides in (9) and (10) then weconstants given in [61.

have the following equations:-

107

7/30/2019 01023854

3/5

(11)ala3- 2a2- ,U - b1a4- ,a3- 4a2- 6aa3a -a , b 3 a 2 b,a-b,-(a,b3-a3 b,)a4 (a, b3+a , , -a3 b, -a 5 b,)a3

-(a3b, + a , b, -a4 b, - a 2 b , )a+ (a ,b, +a , b7)=0+ U, b5+a3b, - , b, - , b, - , b, )a Z (12)

From (12) the frequency can be calculated and thensubstitute this ftequency in (9) or (10) to calculateXc andC,;,,. Then [6] can eliminate the iteration process andnumerical solution in [7] and [8].IV.NEW FORMULAO CALCULATE C M ~ ,Y USINGNODALANALYSIS

The proposed technique uses nodal analysis instead ofloop analysis to obtain just one formula for the minimumcapacitance required for induction generator operation atdifferent load and speed conditions. In this technique, theoperating frequency can be obtained directly from equatingthe real part of admittance with zero where the real partdoes not function in &, then use the imaginary part tocalculate the value of XC. The new proposed method isexplained in the following:-

Applying the nodal analysis at the terminal voltage V, ofthe circuit shown in Fig.3 we get the following equation:-- Y , = o,U

Where Y, =K n Y , + Y , (all these admittances areshown in Fig.2)

Then, Real of Y, =0 (14)And Imaginaryof =0 (15)After some algebraic operations we can get the

following:1 From real part we get the following equation:-c, a4+C3a3+C2a2+C p+CO=0 (16)

The Coefficients C,,, =O, 1, 2, 3, 4 are shown inAppendix 1. The coefficients of this equation do notcontain Xi.The frequency can be obtained directly bysolving (16) to get the operating frequency. There are fourroots; the positive real roots only have the physicalmeaning. If there is no any positive real root, then there isno self-excitation.2- From the imaginary part we can drive a simple formula

for the minimum value of terminal capacitor as shownin (17).

The coefficients M I,MI,h43 and M4 of (17) are shown inAppendix I .In this method we used the real part of Y,=O to determinethe frequency due to the resultant equation does notcontains X c and substituting this frequency in imaginarypart to calculate C,;,, in a simple form as shown in (17).

This new formula can be used on line to calculate theminimum capacitor required for induction generator tobuild up. This new formula does not require any numericalanalysis iteration.

v. APPLICATIONAND RESULTSTo validate the above formula (17) we can use the same

machine in [6] and [8]. The data are: X , =3.23 pu, R,N= 1800 revlmin, 6=60Hz,R,:= pu, XL=2 pu, b =1 pu.

Applying these data to (1 6) and solve it for the frequencywe get only two positive real roots which are (~)~=0.5191and (~)~=0.9937.

Applying these frequencies to (17) we get thecorresponding capacitors C1=:200pF and C2=42.95 pF.

Then, the minimum capacitor is Cmi,,=42.95 F. The sameresults are obtained in the two methods loop and nodalanalysis. Also it is the same results as shown in the tworeferences [6] and [8]. The consistence in the results from(17) and results of (3) can prove the accuracy of the newformula (17). The variation of the minimum capacitancerequired with rotational speed in the same generator [6] and[8] at RL=l pu, XL=Opu is shlown in Fig.4. It is clear that therequired capacitance is inversely proportional with thespeed.

Fig.5 shows the variation of the minimum capacitancerequired with rotational speed in the same generator for atno load. It is clear also the steady state operation ofinduction generator at no loads requires a terminal capacitorinversely proportional with the shaft speed.

Fig.6 shows the variation of reactive power required forthe induction generator and the output power with -rotational speed at RL=lpu and unity power factor. It isclear fiom Fig.6 that the active and reactive power isincreasing with speed.

~ 0 . 0 7PU,R,=0.088 1 PU, X, =X , =O.1813 PU,2b343.3

Fig.7 shows the variation of reactive power required forthe induction generator and the output power withrotational speed at different values of load resistance(RL=0.7,1and 2pu) and unity power factor. It is clear fromFig.7 that the active and reactive power is increasing withspeed.

The variation of the stator frequency with rotationalspeed is shown in Fig.8.

108

7/30/2019 01023854

4/5

C , F I: Active powbr10080

6040200 50 100 150 200

%speedFig.4 variation of c,, wth rotational speed at RL=lpu and unity power

factor.

80

604020

I I50 100 150 200

%SpeedFig.5 Variation of c wth rotational speed at no load.

50 100 150%speed

200Fig.6 Variation of reactive power and output powerwthrotational speedat RL=lpu and unity power factor.

3 I I: Active pcjwer

50 100 150%Speed

Fig.7 Variation of reactive power and output powerwth rotational speedat different values of load resistance (R~=0.7,1 nd 2pu) and unity powerfactor.

109

7/30/2019 01023854

5/5

1.8 41.4I No oad /AA: L = l p u0

N * W O O C -~ a x p ~ E ~ ~ ~ ~ ~ 2Fig.8 Stator frequencywth rotational speed is for different loads.

VI CONCLUSIONSIn this paper a new formula for the m i n i capacitancerequired for self-excited induction generator is presented.This new formula is simple and it does not need numericaliteration. For this reason this new formula helps todetermine the minimum capacitance required for selfexcited induction generator on line. The new formula givestypical results as the results obtained fiom iterativetechnique without any iterationor divergence problem.LISTOF SYMBOLSActual or (generated) fiequency,P*N&20.

Actual or (generated) rational speed, 120 fJP.Rated fiequency of induction generator ,P*N,420.Rated speed of nduction generator, 120f/P.Synchronous speed corresponding to actualfiequency.Synchronous speed corresponding to ratedfiequencyPu iequencyfdfrPu speedNJNsr.Slip of induction generator.Load resistance.Load inductance.Inductance of terminal capacitor.Stator resistance.Stator inductance.Magnetizing inductance.Rotor resistance.Rotor inductance.Terminal voltage.Maximum value of magnetizing inductance.Magnetizing current.Loop impedance.Output real power.Reactive power required for induction generator.Capacitance of the terminal capacitor.Minimum capacitance of the terminal capacitor.

REFERENCES[11 R. M Hilloowala, A. M. Sharaf Modeling, simulation and analysis ofvariable speed constant frequ.ency wind energy conversion schemeusing self excited induction generator, 1991. Proceedings, Twenty-Third Southeastern Symposium on System Theory, Page(s): 33 -38.[2] E. Muljadi and J. Sallan, M. Sam and C. P. Butterfield Investigationof Self-Excited Induction Generators for Wind TurbineApplications, LPS conference, IEEE, 3-7 October 1999, Phoenix,Arizona USA.[3] A. A. Shaltout and M. A. Abdel-Halim,Solid-state control of a winddriven self-excited induction generator nternational Journal onElectric machines and po wer .systems,vol. 23, 1995, pp. 571-582.[4] N. Ammasaigounden, M. Subbiah Chopper-controlled wind-drivenself-excited induction generators IEEE Transactions on Aerospaceand Electronic Systems, 1989 Volume: 252, Page@): 268 -276.[5] S. Wekhande V. Agarw Wind Driven Self-Excited InductionGenerator wth Simple De-Coupled Excitation Control USconference, IEEE, 3-7 October 1999, Phoenix, Arizona USA.[6] A. K. Al. Jabri and A. L. Alolah Capacitor requirement for isolatedself-excited induction generator,ZEE proceedings, Vol. 137, pt. B,No. 3, May 1990[7] N. H. Mal& and S. E. Haque Steady state analysis and perfomanceof an isolated self excited induction generator IEEE Trans.,

[8]N.H. Malik and A. A. Mazi Capacitance requirements for isolatedexcited induction generators, IEEE Trans., 1987, EC-2 (I), pp. 62-69.[9] M. Orabi design of wind energy conversion system, MSc. ElectricalEngineering Department, Faculty of Engineering, ElminiaUniversity, Elminia, Egvp t,2000.[lo] S. S. Murthy and N. H. Malilc and A. K. Tandon Analysis of selfexcited induction generators,ZEEproceedings, Vol. 129, pt. C, No.6, November 1982.

1986,EC-1, (3). pp. 134-139.

[113 Say, M. G. Alternating Current Machines,book, pitman, 1976.Appendix1The coefficients of equation (:16)c4=X ; R , ( L ~, - L , ) + X ; R , L: +R, L ; ,C , = X ;R , v( L , - L , L , ) - 2 v ( X ~ R , L : + R L L ~ ) ,c,=R;(R, r: -RJ, R ~ L , ~Y;R, (R;+g v z )

+ 2 *RLR, R,(L, L,-4 ) RLL(L& +R:L, +R :c ,C , =RZR, v(L ,- , L3 )- 2v R, R,L: ( R, +R, ) and,CO=(R: +L, v2) R, R, (R , t.R , )The coefficientsof equation (17)M , = R , R , - f ~ f - v ) L , YM 2 =R r f L3 +R,(f 9M 3 = R i +X: f andM4 = R , *M 2 L2 f- )A4,Where,L1= x, X , +X, ) +x, ,L2 =X , +X , and,L 3 = X , +x,

110