CHAPTER 2 CAPACITANCE REQUIREMENTS OF SIX-PHASE SELF-EXCITED INDUCTION GENERATORS

Transcription

1 9 CHAPTER 2 CAPACITANCE REQUIREMENTS OF SIX-PHASE SELF-EXCITED INDUCTION GENERATORS 2.. INTRODUCTION Rapidly depleting rate of conventional energy sources, has led the scientists to explore the possibility of utilizing non-conventional energy sources. Wind energy, largely available in our atmosphere, can be harnessed to generate electric power using induction generator. Brush-less rotor construction, absence of separate source for excitation, ease of maintenance and reduced unit cost are some of the major advantages of induction generator that makes it suitable for power generation particularly in remote areas. Even though three-phase induction generators are used for this purpose, many research articles have been reported [9-4] on multiphase (more than three phase) induction machine due to their advantages like higher power rating and improved reliability. The generator scheme presented by Ojo et al. [33] is based on dual stator winding induction machine with displaced power and control three-phase winding. Basic et al. [34] and Levy [35] used the double stator machine with extended rotor common to both stators. In all the cases, the output is three phase. However, the studies of using number of phases higher than three in multi-phase ac machines in general and induction machines in particular are of great importance. The research in

2 this area is still in its infancy. Recently, some theoretical and experimental work on six-phase induction generator has been reported by Singh GK et al. [37-4]. Regarding the mathematical modeling of six-phase self-excited induction generators, the concept of three-phase and single-phase self-excited induction generator modeling [-4] can be utilized. However, they need separation of the real and imaginary components of the complex impedance or admittance of the Fig. 2.. Schematic diagram of six-phase self-excited induction generator. equivalent circuit to derive specific models which are tedious. Singaravelu et al. [5-6] and Velusami et al. [7-8] made an attempt to overcome the complication of three-phase and single-phase SEIG models by introducing the concept of graph theory which avoids the lengthy and tedious mathematical derivations of nonlinear equations. G.K. Singh et al. [38-4] followed the graph theory approach for the steady-state modeling of six-phase SEIG.

3 In this Chapter, a further simplified mathematical model of six-phase SEIG in matrix form is developed using nodal admittance method based on inspection. In the proposed model [43], the nodal admittance matrix can be formed directly from the equivalent circuit of six-phase SEIG by inspection rather than deriving it from the concept of graph theory. Moreover, this model is also flexible such that any equivalent circuit elements can be easily included or eliminated. Genetic algorithm (GA) is used to determine the capacitive VAr requirement of six-phase SEIG. Experiments are conducted on a prototype six-phase SEIG and the experimental results are found to be good agreement with the analytical results PROPOSED MATHEMATICAL MODELING A generalized mathematical model of a six-phase SEIG (Fig. 2.) using nodal admittance method based on inspection is developed from the equivalent circuit of the generator. The developed model results in a matrix form that proves convenient for computer solutions. Fig Steady-state equivalent circuit of six-phase self-excited induction generator.

4 2 The steady-state equivalent circuit of the six-phase SEIG with four nodes is shown in Fig. 2. branch admittances of the equivalent circuit are: Y = / [R r / (F - ) +j X r ]; Y 2 = / [j X M ]; Y 3 = / [j X lm ]; Y 4 = / [R S2 / F +j X S2 ]; Y 5 = / [R S / F + j X S ]; Y 6 = / [- j X C2 / F 2 ]; Y 7 = / [R L2 / F + j X L2 ]; Y 8 = / [- j X C / F 2 ]; Y 9 = / [R L / F + j X L ]. The matrix equation based on nodal admittance method for the equivalent circuit can be expressed as [Y] [V] = [I S ] (2.) where [V] is the node voltage matrix, [I S ] is the source current matrix, and [Y] is the nodal admittance matrix. The [Y] matrix can be formulated directly from the equivalent circuit (Fig.2.2) by inspection [47] as Y + Y 2 + Y 3 - Y 3 [Y] = - Y 3 Y 3 + Y 4 + Y 5 - Y 5 - Y 4 (2.2) - Y 5 Y 5 + Y 8 + Y 9 - Y 4 Y 4 + Y 6 +Y 7 where Y ii Y ij = - th node th node and j th node

5 3 Since [Y] is symmetric, Y ji = Y ij. If there is no branch between any two nodes, then the corresponding element in the matrix is zero. Since, the equivalent circuit does not contain any current sources, [I S ] = [] and hence Eq. (2.) is reduced as [Y] [V] = (2.3)! " # Eq. (2.3), [Y] should be a singular matrix i.e., determinant of [Y] =. It implies that both the real and the imaginary components of det [Y] should be independently zero. Therefore to obtain required parameter which results det [Y] =, genetic algorithm based approach is implemented which is discussed in the next section. 2.3 APPLICATION OF GENETIC ALGORITHM FOR THE STEADY- STATE PERFORMANCE ANALYSIS Genetic algorithm (GA) is a stochastic global search method that mimics natural biological evolution. The GA operates on a population of potential solutions [48] by applying the principle of survival of the fittest to converge to an optimal solution. An application of GA method to obtain det[y] =, which provides solution for unknown quantities, is illustrated in Fig. 2.3, which indicates the handling of population size and generating new set of chromosomes at the end of every iteration until convergence is reached. Thus satisfying an operating point, the objective function whose value is to be minimized is given by Eq. (2.4). g (F,X M or X C ) = abs{real(det[y])} + abs{imag(det[y])} (2.4)

6 4 In many optimization problems to obtain initial estimates suitably, certain trials may be required. However, in the present problem of the SEIG, it is easy to START read all required parameters for matrix Y read population size (PS) and ranges for F and X M or X C generation (GEN) = generate the initial population population count (POP) = evaluate objective function g(f, X M ) or g(f, X C ) whether g(f, X M) or g(f, X C ) reached desired tolerance NO POP = POP + YES compute performance and print results STOP YES is POP < PS NO GEN = GEN + generate new set of chromosomes using Genetic Algorithm process Fig Flow chart for minimization of the objective function using genetic algorithm (GA). give the range for the unknown variables F and X M or X C because in well-designed ( self-excited induction generators, it is known that the slip {(F $ % & ' is small and

7 5 operation of the machine is only in the saturated region of the magnetization characteristics. So, the ranges for F can be given as.8 to.999 times the value of % and for X M as 25% to % of critical magnetizing reactance X MO. Similarly for X C, the same range 25% to % of C MAX can be used, where C MAX is the maximum capacitance required under any conditions. Thus, starting from such initial estimates, the final value of F and X M or X C is obtained through GA. The air gap voltage V g can be determined from the magnetization characteristics corresponding to X M, as described in section 2.5. Once the air gap voltage V g is calculated, the equivalent circuit can be completely solved to determine the steady-state performance of SEIG WINDING DESIGN OF SIX-PHASE INDUCTION MACHINE The six stator phases are divided into two three-phase winding sets (set-: abc and set-2: xyz). The a, b and c phases of winding set- are displaced by 2 o from each other. Similarly, x, y and z phases of winding set-2 are displaced by 2 o from Fig Vector diagram of stator winding set- (abc) and winding set-2 (xyz).

8 6 each other. The phase angle between winding set- and winding set-2 is 6 o as shown in Fig The Fig. 2.4 shows that the x, y and z phases of winding set-2 are displaced by 8 o with a, b and c phases of winding set- respectively. The winding diagram of each phases of winding set- and winding set-2 are shown in Fig. 2.5 and Fig. 2.6 respectively. The complete winding diagram of six-phase SEIG is shown in Fig Number of slots (N s ) = 36 Number of poles (P) = 4 Pole pitch = Number of slots / pole = 36/4 = 9 Slot pitch angle = 36 o /N s = 36 o /36 = o (Mechanical) = o x P/2 = o x 4/2 = 2 o (Electrical) Winding set- (abc): Starting slot of a-phase Starting slot of b-phase Starting slot of c-phase = st slot = (2 o /2 o ) + = 7 th slot = (24 o /2 o ) + = 3 th slot Winding set-2 (xyz): Starting slot of x-phase Starting slot of y-phase Starting slot of z-phase = (8 o /2 o ) + = th slot = (3 o /2 o ) + = 6 th slot = (6 o /2 o ) + = 4 th slot

9 Fig Winding diagram of Phase-A, B, and C of winding set-. 7

10 Fig Winding diagram of Phase-X, Y, and Z of winding set-2. 8

11 Fig 2.7. Complete winding diagram of six-phase SEIG: winding set- (set abc) and winding set-2 (set xyz). 9

12 / #. + / - ) # EXPERIMENTAL SETUP AND MACHINE PARAMETERS The machine used for the purpose of conducting experiments (Fig. 2.8) is a six-phase, 4-pole, delta connected, 2.2kW, 23V, 5A, 5Hz induction generator which consists of two identical three-phase stator winding sets namely winding set abc (set-) and winding set xyz (set-2). The base values of the six-phase SEIG are V base = rated phase voltage = 23V I base ) ' * + Z base = V base / I base = ) / P base = V base x I base =.664 kw N base = 5 rpm f base = 5 Hz,, - The parameters of the equivalent circuit of the test machine, obtained from the results of the standard tests are: R S = R S2 =. R r 2 # + / X S = X S2 = X r + X lm # -, + /. Fig View of the laboratory experimental setup.

13 4 2 Magnetization characteristics of the machine play a vital role in the analysis of six-phase SEIG. The magnetization curve (Fig. 2.9) of the six-phase SEIG can be determined experimentally by conducting synchronous speed test. The machine is made to run at rated synchronous speed corresponding to rated frequency, and the magnetizing reactance for different input voltages is measured. The variation of V g /F (in p.u) with X M (in p.u) is non-linear due to magnetic saturation and approximated as two straight lines using piecewise linearization technique as given by Eq. (2.5) and (2.6). V g /F =.2 $.22 X M, X M <.82 (2.5) V g /F = 2.66 $.97 X M, X 3 M.82 (2.6).2 V g /F (p.u) V g /F = X M V g /F = X M - Expt.points X M (p.u) Fig Variation of air gap voltage with magnetizing reactance CAPACITANCE REQUIREMENTS TO MAINTAIN DESIRED TERMINAL VOLTAGE It is observed that the six-phase SEIG is able to self-excite if proper value of shunt capacitor connected to either one of the three-phase winding set or both the

14 winding sets. Moreover, the capacitance requirements vary depends upon the variation in speed and the required constant terminal voltage. Selection of excitation capacitance and effect of speed on terminal voltage are presented in this section. Also, the capacitance and VAr requirements to maintain constant terminal voltage for different constant speeds are computed for the following configurations. Selection of excitation capacitance. Effect of speed on terminal voltage. Capacitance and VAr requirements when both winding sets are excited and both are loaded equally. both winding sets are excited and any one of the winding set is loaded. any one of the winding set is excited and both winding sets are loaded equally. any one of the winding set is excited and loaded. All the above configurations are analyzed using the generalized mathematical model of six-phase SEIG given by Eq. (2.3) and genetic algorithm based approach as discussed in section 2.3. For obtaining the capacitive VAr requirement to maintain the required terminal voltage under varying load conditions, det [Y] is solved and the unknown variables X C and F are computed using the genetic algorithm method. After obtaining X C and F, the equivalent circuit (Fig. 2.2) is completely solved to determine the capacitive VAr requirement. The analysis is performed under three different speeds (below rated speed, rated speed, above rated speed). Moreover, for all the

15 above cases, the same mathematical model can be used with modifications in few terms to consider the different capacitance and loading configurations Selection of Excitation Capacitance a).4.2 Speed = p.u, No load V T = V T2 Terminal voltage (p.u) Expt Capacitance (6 F) b).2 Speed = p.u, No load V T Terminal voltage (p.u) Expt Capacitance (µf) Fig. 2.. No load terminal voltage when capacitance connected to: (a) both winding sets, (b) only set abc.

16 24 The value of shunt excitation capacitance (C) must be selected to obtain the desired no load voltage at the given speed. For this purpose, the capacitive reactance (X C ) and the frequency (F) are chosen as unknown quantities for a particular speed. The variation of no load terminal voltage (V T ) with shunt excitation capacitance bank connected across both the winding sets abc and xyz is shown in Fig.2.(a). It is observed that no load terminal voltage increases with the increase in capacitor value. From Fig.2.(a), value of excitation capacitance that corresponds to no load terminal voltage of pu is 339 F. The variation of no load terminal voltage (V T ) with shunt excitation capacitance bank connected across only one winding set abc and set xyz is kept open is shown in Fig.2.(b). From this, excitation capacitance value corresponding to no load terminal voltage of pu is found to be 6 µf Effect of Speed on Terminal Voltage For fixed value of capacitance, speed is varied under no load condition and it is found that the terminal voltage decreases with decrease in speed. Variation of no load terminal voltage with prime mover speed for various values of excitation capacitance C and C 2 when capacitor bank is connected across both winding sets abc and xyz is shown in Fig. 2.(a). Fig. 2.(b) shows the no load terminal voltage with prime mover speed for different excitation capacitance when winding set abc is excited and other set xyz is kept open. In both the cases, it is observed that for a given no load terminal voltage, value of excitation capacitance must be increased with the decrease in speed.

17 25 C.2 = C 2 = 45 µf V T = V a) T2 C No Load = C 2 = 33 µf C = C 2 = 25 µf Terminal voltage (p.u) : ; < ; = - Expt Prime mover speed (p.u).2 b) V T No Load C = 7 µf C = 6 µf C = 5 µf Terminal voltage (p.u) : ; < ; = - Expt Prime mover speed (p.u) Fig. 2.. No load terminal voltage when capacitance connected to: (a) both winding sets, (b) only set abc Capacitance and VAr Requirements when Both Winding Sets are Excited and Both are Loaded Equally In this configuration, both the winding sets abc and xyz of six-phase SEIG are connected with shunt capacitances. The six-phase SEIG is subjected to equal resistive loading on both winding sets abc and xyz. The analysis is carried out to find the range

18 26 of values of capacitance and VAr requirement to maintain the rated terminal voltage of p.u for three different constant speeds (.95,.,.5 p.u). The variations of shunt capacitance and VAr with output power to maintain the terminal voltage constant at p.u. at different constant speeds are shown in a) VAr (p.u) & Capacitance (p.u) V T =V T2 =. p.u... VAr -- Capacitance Speed=.95 p.u Speed=.95 p.u Speed=. p.u Speed=.5 p.u Speed=. p.u Speed=.5 p.u x, >, + - Expt Output power (p.u) b).6.4 V T =V T2 =. p.u I S = I S2 Speed=.95 p.u Speed=. p.u Stator current (p.u) Speed=.5 p.u.2 x, >, + - Expt Output power (p.u) Fig Variation of: (a) VAr and capacitance, and (b) stator current with output power (Both the winding sets abc and xyz are excited and subjected to equal loading).

19 27 Fig.2.2(a). The capacitance and capacitive VAr requirements increase as the speed decreases and as the output power increases. From Fig.2.2(a), it is observed that a voltage regulator should be capable of providing a range of VAr from 2. to about 3.55 p.u. to maintain the terminal voltage at p.u. under varying speeds from.95 to.5 p.u with maximum output power of 3.35 p.u. The required shunt capacitance ranges from.6 to.3 p.u. The variations of stator current with output power at constant terminal voltage are shown in Fig. 2.2(b). From this, it is observed that over a range of speed from.95 to.5 p.u, the stator currents are well below the rated value Capacitance and VAr Requirements when Both Winding Sets are Excited and Any One of the Winding Set is Loaded For this configuration, both the winding sets abc and xyz are excited by shunt capacitance, but any one winding set (set abc) is subjected to resistive load. The shunt capacitance and VAr requirement to maintain rated terminal voltage under three different speeds are computed. Fig. 2.3(a) shows the variation of capacitance and VAr with output power for varying speeds from.95 to.5 p.u. From Fig. 2.3(a), it is observed that the range of VAr and capacitance varies from 2. to about 3.8 p.u and.6 to.6 p.u respectively with maximum output power of 3.2 p.u. Fig. 2.3(b) shows the variations of stator current with output power at constant terminal voltage. It is observed that the stator currents are well below the rated value.

20 28 a) VAr (p.u) & Capacitance (p.u) V T =. p.u... VAr -- Capacitance Speed=.95 p.u Speed=. p.u Speed=.5 p.u Speed=.95 p.u Speed=. p.u Speed=.5 p.u x, >, + - Expt Output power (p.u) b) Stator current- (p.u) V T =. p.u... I S -- I S2 Speed=.95 p.u Speed=. p.u Speed=.5 p.u Speed=.95 p.u Speed=. p.u Speed=.5 p.u x, >, + - Expt Output power (p.u) Stator current-2 (p.u) Fig Variation of: (a) VAr and capacitance, and (b) stator current with output power (Both the winding sets abc and xyz are excited and only winding set abc is loaded) Capacitance and VAr Requirements when Any One of the Winding Set is Excited and Both Winding Sets are Loaded Equally For this type of configuration, only one set (set abc) is excited by shunt capacitance and the six-phase SEIG is subjected to equal resistive loading on both winding sets abc and xyz.

21 29 The variation of capacitance and VAr with output power to maintain terminal voltage at rated value of p.u is shown in Fig. 2.4(a). From this, it is observed that the VAr and capacitance are varying from 3.8 to about 5.5 p.u and. to 2. p.u a) 6 V T =V T2 =. p.u Speed=.95 p.u Speed=. p.u VAr (p.u) & Capacitance (p.u) VAr -- Capacitance Speed=.95 p.u Speed=.5 p.u Speed=. p.u Speed=.5 p.u x, >, + - Expt Output power (p.u) 2 b) Stator current- (p.u) V T =V T2 =. p.u Speed=.95 p.u Speed=.95 p.u Speed=. p.u Speed=. p.u Speed=.5 p.u Speed=.5 p.u... I S -- I S2 x, >, + - Expt Output power (p.u) Stator current-2 (p.u) Fig Variation of: (a) VAr and capacitance, and (b) stator current with output power (Only winding set abc is excited and both the winding sets abc and xyz are subjected to equal loading).

22 3 respectively under varying speeds from.95 to.5 p.u with maximum output power of 2.7 p.u. Fig. 2.4(b) shows the variations of stator current with output power and the stator currents are found to be within rated value Capacitance and VAr Requirements when Any One of the Winding Set is a) VAr (p.u) & Capacitance (p.u) V T =. p.u... VAr -- Capacitance Speed=.95 p.u Speed=. p.u Speed=.95 p.u Speed=. p.u Speed=.5 p.u Speed=.5 p.u x, >, + - Expt Output power (p.u) b) 2 I S2 = Speed=.95 p.u Speed=. p.u Stator current- (p.u).5.5 Speed=.5 p.u x, >, + - Expt Output power (p.u) Fig Variation of: (a) VAr and capacitance, and (b) stator current with output power (Only winding set abc is excited and subjected to load with winding set xyz is kept open).

23 3 Excited and Loaded Here, winding set abc is alone excited and the same winding set is subjected to resistive loading with winding set xyz is kept open. The variations of shunt capacitance and VAr with output power are shown in Fig 2.5(a). From this, it is found that VAr varies from 3.8 to about 5.3 p.u. to maintain the rated terminal voltage with maximum output power of.6 p.u. Also it is observed that the range of capacitance is. to 2 p.u. Fig. 2.5(b) shows the variations of stator current with Table 2.: Comparative evaluation of variation of capacitance requirement and VAr requirement of six-phase SEIG from no load to full load under different modes of operation. Shunt Capacitance (C sh ) Loading Configuration Speed (p.u) Capacitance Requirement (p.u) (µf) VAr Requirement (p.u) Max. output power (p.u) Connected to both winding sets abc and xyz Connected to winding set abc only Both winding sets abc and xyz are subjected to equal loading Only winding set abc is subjected to load Both winding sets abc and xyz are subjected to equal loading Only winding set abc is subjected to load with winding set xyz is kept open????????????????????????

24 32 output power and it is observed that over a range of speed from.95 to.5 p.u, the stator currents are below the rated value. From the above discussion, it is observed that the terminal voltage of sixphase SEIG depends on change in load, speed and shunt capacitance. The analytical and experimental study shows that the capacitance requirement varies from.6 to.6 p.u (24 to 64µF) when both winding sets are excited and. to 2 p.u (44 to 8µF) when one winding set is excited to maintain the rated terminal voltage of p.u with speed varying from.95 to.5 p.u. The results of all the configurations are summarized in Table CONCLUSION Generalized mathematical model using nodal admittance method based on inspection and genetic algorithm based computation are proposed to determine the necessary capacitance and VAr requirement to maintain constant terminal voltage for different constant speeds. From the obtained range of capacitance values, the shunt capacitance is selected such that to get rated value of terminal voltage at no load for the steady-state performance of six-phase SEIG which is discussed in the Chapter 3.