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Switching Transients on Induction Motors With Power Factor Correction Capacitors by Larry J. Burleson B.S., University of Iowa, 1972 A thesis submitted to the Faculty of the Graduate School of the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Master of Science Department of Electrical Engineering and Computer Science 1989
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This thesis for the Master of Science degree by Larry J. Burleson has been approved for the Department of Electrical Engineering and Computer Science by Marvin F. Anderson Date
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Burleson, Larry J. (M.S., Electrical Engineering) Title: Switching Transients on Induction Motors with Power Factor Correction Capacitors Thesis directed by Associate Professor Pankaj K. Sen For years it has been a standard practice in industry to utilize power factor correction capacitors on induction motors. A capacitor of appropriate size is applied at or near the terminals of the machine. These capacitors act as "VAR Generators" which neutralize the lagging component of the motor to varying degrees depending on the capacitor size. In applying such capacitors it has been widely accepted that one must be careful not to oversize these capacitors; otherwise, when the capacitor/motor combination is disconnected from the source, selfexcitation can occure and damaging overvoltages may result. This thesis discusses this problem using digital simulation techniques to quantify this situation as applied to a motor of current design. Theoretical results are also varified by the experiments performed on a small 1.5 HP, 230 volt, 3phase induction motor using a highspeed digital oscilloscope. The results obtained from the digital simulations and varified by experimential results
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indicates that very little overvoltaqe is encountered with correction up to 100 percent for small motors. The major effect of these capacitors (at no load) is that the terminal voltaqe is sustained for several seconds before finally decaying to zero. The form and content of this abstract are approved. I recommend its publication. Faculty m ber in charge of thesis
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To Ruth Magnuson gone but not forgotten
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ACKNOWLEDGEMENTS The author would like to acknowledge the indivduals or corporations who assisted in this thesis by donating time or equipment. Tim Gargiulo Wazee Electric Co. 2020 W. Barberry Pl. Denver, Colorado Mr. Gargiulo donated the 1.5 HP electric motor used in this study. Vito Raso US Motors Corporation 2600 S. Parker Rd. Aurora, Colorado Mr. Raso provided the motor equivalent circuit parameters for the same 1.5 HP motor. HewlettPackard Corporation HewlettPackard donated several 54501A Digitizing Oscilloscopes to the University of Colorado, one of which was used in the experimental portion of this thesis.
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CONTENTS CHAPTER 1. INTRODUCT,ON ................................... 1 1. 1 Approach ................................. 1 1.2 Methodology .............................. 4 2. LITERATURE REVIEW ............................. 5 3. COMPUTER MODEL DEVELOPEMENT ................... 7 3.1 Reference Frame Transformation ........... 7 3.2 Equation Developement ................... 10 3.2.1 Case 1No Capacitor ................ ; .. 11 3.2.2 Case 2Wi th Capacitor ................. 17 3.2.3 Rotor Motion .......................... 23 3.2.4 Saturation Effects .................... 28 4. COMPUTER HARDWARE AND SOFTWARE ................ 31 4.1 Software ....... ......................... 31 4.2 Hardware ................................ 34 5. DIGITAL SIMULATION AND EXPERIMENTAL RESULTS ...................................... 37 5.1 Digital Simulation Results .............. 38 5.2 Experimental Results .................... 39 6. ANALYSIS AND CONCLUSION ...................... 41 BIBLIOGRAPHY ........................................ 45
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viii APPENDIX A. EXCERPTS FROM BIBLIOGRAPHY REFERENCE [1] ..... 47 B. PASCAL PROGRAM USED TO IMPLEMENT CASE 1 NO CAPACITOR ................................ 51 C. PASCAL PROGRAM USED TO IMPLEMENT CASE 2 WITH CAPACITOR .............................. 56 D. PASCAL SUBROUTINE USED TO INCLUDE REFERENCE FRAME TRANSFORMATION AND SATURATION .......... 61 E. COMPUTER SIMULATED AND EXPERIMENTALLY OBSERVED WAVEFORMS FOR THE 1.5 HP MOTOR ...... 64 F. MISCELLANEOUS COMPUTER SIMULATED PLOTS FOR THE 1.5 HP MOTOR ......................... 73 G. COMPUTER SIMULATED PLOTS FOR A 3 AND 50 HP MOTOR ............................ 82 H. 1.5 HP MOTOR DATA ............................ 89
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FIGURES Figure 1.1 Motor circuit to be modeled with no capaci tor (easel) ........................... 2 1.2 Motor circuit to be modeled with a capaci tor (case2) ........................... 3 3.1 Relationship between the three phase quantities and the rotating two phase orthogonal axes quantities .................. 9 3.2 Single phase stator equivalent circuit with no capacitor .......................... 11 3.3 Single phase stator equivalent circuit with a capacitor ........................... 17 4.1 Flow chart of computer program "NOCAP" .... 35 5.1 1.5 HP motor steady state equivalent circuit .................................... 37 E.1 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, no capacitor, no load, and no saturation ................. 65 E.2 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 1 per unit capacitor, no load, and no saturation ...... 65 E.3 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 2 per unit capacitor, no load, and no saturation ...... 66 E.4 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 3 per unit capacitor, no load, and no saturation ...... 66 E.5 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, no capacitor., no load, and with saturation .................. 67
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E.6 E.7 E.8 E.9 E.l0 E.11 E.12 E.13 E.14 E.1S E.16 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 1 per unit capacitor, no load, and with saturation .... 67 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 2 per unit capacitor, no load, and with saturation .... 68 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 3 per unit capacitor, no load, and with saturatiop .... 68 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, no capacitor, full load, and with saturation ............. 69 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 1 per unit capacitor, full load, and with saturation ............................ 69 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 2 per unit capacitor, full load, and wi th saturation ............................ 70 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 3 per unit capacitor, full load, and wi th saturation ............................ 70 Actual wavefo,rm of motor terminal volts vs. time after switch opening: 1.5 HP, no capacitor, and no load .......... 71 Actual waveform of motor terminal volts vs. time after switch opening: 1.5 HP, 1 per unit capacitor, and no load ................................ 71 Actual waveform of motor terminal volts vs. time after switch opening: 1.5 HP, 2 per unit capacitor, and no load ................................ 72 Actual waveform of motor terminal volts vs. time after switch opening: 1.5 HP, 3 per unit capacitor, and no load ................................ 72 x
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F.l Computer simulated plot of the 1.5 HP motor speed/torque curve showing free no load, and with saturation ................................. 74 F.2 Computer simulated plot of the 1.5 HP motor speed/torque curve showing free full load, and with saturation ................................. 74 F.3 Plot of the 1.5 HP motor saturation F.4 F.5 F.6 F.7 F.8 F.9 ........Computer simulated plot of motor torque vs. time: 1.5 HP, no capacitor, no load, and with saturation ............... 75 Computer simulated plot of motor torque vs. time: 1.5 HP, no capacitor, full load, and with saturation ............. 76 Computer simulated plot of motor torque vs. time: 1.5 HP, 1 per unit capacitor, no load, and with saturation .............. _.76 Computer simulated plot of motor torque vs. time: 1.5 HP, 1 per unit capacitor, full load, and with saturation ............. 77 Computer simulated plot of motor speed vs. time: 1.5 HP, no capacitor, no load, and with saturation ............... 77 Computer simulated plot of motor speed vs. time: 1.5 HP, no capacitor, xi full load, and with saturation ............. 78 F.IO F.ll F.12 Computer simulated plot of motor line current vs. time: 1.5 HP, no capacitor, no load, and with saturation ............... 78 Computer simulated plot of motor line current vs. time: 1.5 HP, no capacitor, full load, and no saturation ............... 79 Computer simulated plot of motor line current vs. time: 1.5 HP, no capacitor, full load, and with saturation ............. 79
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.13 F.14 F.1S G.1 G.2 G.3 G.4 G.5 G.6 Computer simulated plot of motor line current vs. time: 1.5 HP, 1 per unit capacitor, no load, and with saturation .... 80 Computer simulated plot of motor line current vs. time: 1.5 HP, 1 per unit capacitor, full load, and with saturation ................................. 80 Computer simulated plot of motor line current vs. time: 1.5 HP, 1 per unit capacitor, full load, and. with no saturation .............................. 81 Computer simulated plot of motor terminal volts VS. time: 3 HP, 1 per unit capacitor, no load, and no saturation ...... 83 Computer simulated plot of motor terminal volts vs. time: 3 HP, 2 per unit capacitor, no load, and no saturation ...... 83 Computer simulated plot of motor terminal volts vs. time: 3 HP, 3 per unit capacitor, no load, and no saturation ...... 84 computer simulated plot of motor line current vs. time: 3 HP, no capacitor, full load, and with no saturation .......... 84 Computer simulated plot of the 3 HP motor speed/torque curve showing free acceleration; no load, and no saturation ................................. 85 Computer simulated plot of the 3 HP motor speed/torque curve showing free full load, and no saturation ................................. 85 G.7 Computer simulated plot of motor terminal volts vs. time: 50 HP, 1 per unit capacitor, no load, and no saturation ...... 86 G.8 Computer simulated plot of motor terminal volts vs. time: 50 HP, 2 per unit capacitor, no load, and no saturation ...... 86 G.9 computer simulated plot of motor terminal volts vs. time: 50 HP, 3 per unit no load, and no saturation ...... 87
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G.lO G.1l G.12 Computer simulated plot of motor line current vs. time: 50 HP, no capacitor, xiii full load, and with no saturation .......... 87 Computer simulated plot of the 50 HP motor speed/torque curve showing free acceleration; no load, and no saturation ................................. 88 Computer simulated plot of the 50 HP motor speed/torque curve showing free acceleration; full load, and no saturation ................................. 88
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CHAPTER 1 INTRODUCTION For years it has been a standard practice in industry to utilize power factor correction capacitors on induction motors. A capacitor of appropriate size is applied or near the terminals of the machine and is switched with the machine. These capacitors act as "Var Generators" which neutralize the lagging component of the motor current to varying degrees depending on the capacitor size. It has been widely accepted that one must be careful not tb over size these otherwise when the capacitor/motor combination is disconnected from the source, self excitation can occur and damaging overvoltages may result. This thesis will revisit this old problem using digital simulation techniques in an attempt to quantify the situation as applied to a motor of current design. 1.1 Approach. This thesis develops the induction machine model with and without a capacitor. The first model as shown in figure 1.1 is the machine without a capacitor and the second model is as shown in figure 1.2 with a
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__/'0 ........ ( ....... , c Figure 1.1 Motor circuit to be modeled with no capacitor
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( c ic c Figure 1.2 Motor circuit to be modeled with a capacitor w
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capacitor. The computer simulation developed enables machine variables to be plotted as a function of time (t). At t=O switch S1 closes, the motor is allowed to reach steady state operating conditions at which time the same switch is reopened. The model which will be developed will allow the effects of saturation to be observed and will consider both the no load and the full load case. 1.2 Methodology. In developing a model for this simulation the work of Krause [1] is used extensively. For reference, Krause's basic induction motor equations are reprinted in appendix A. As mentioned, figure 1.1 shows the induction machine circuit with no capacitor and figure 1.2 shows the machine circuit with the power factor capacitor installed. The impedance req & Leq have been included to take into account the power system impedance. The rotor circuit is identical for both models.
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CHAPTER 2 LITERATURE REVIEW At the beginning of this project, a search was done to see what pertinent studies, information, books, theses, dissertations or articles were available. Going back 30 years, the only work that was found dealing directly with capacitors applied in parallel with the motor was the work of Walsh and deMello [4]. Their modeling used an analog computer to simulate switching motor terminal voltage after switch opening and at the torque reclosing transients produced when the switch was reclosed on a decaying field. Their results predicted that capacitors did not create significant overvoltages when applied to induction motorscontrary to what was generally thought at the time. Another paper relevant to this study is the work by Ortmeyer and Calabrese [5]. This paper uses digital simulation to model an induction motor with both shunt and series capacitive compensation. Their study was done to determine the effect of the capacitors in both configurations on the effects of motor starting.
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However, while Ortmeyer and Calabrese used methods similar to those employed herein, their field of inquiry is substantially different from that of this study. Various other works have assisted and furthered the development of the state model of the induction machine, many of which are listed in the bibliography for future reference purposes only.
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CHAPTER 3 COMPUTER MODEL DEVELOPMENT 3.1 Reference Frame Transformation. The concept of reference frame transformation is adopted to simplify the model developement. This reduces time varying quantities such as winding inductances, which are dependent on rotor position, to constants in this new reference frame. The general equation of transformation for the three phase stator circuits used throughout this work is as followsz andz Ks wherez f qd08 [fqs fds fos] = fabos a [fas fbs fes] 21r cos6 COS(6) 3 2 2n sin6 sin(6) 3 3 1 1 2 2 2n cos(6) 3 2n sin(B) 3 1 2
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fsa,fba, and fea refer to phase a,b, and c stator quantities respectively. 8 The transformation which returns variables from the rotating reference frame to the stationary reference frame is: cose sine 1 Figure 3.1 shows the relationship between the three phase quantities and the two phase orthogonal axes quantities plus the noncoupled owinding for both the stator and rotor.
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9 +q STATOR ROTOR 3.1 Relationship between the three phase quantities and the rotating two phase orthogonal axes quantities. The equation of transformation for the rotor circuits are similar to the stator circuits except the angle 8 is replaced with the angle 8 where 8 = 88 and 8 is the angular displacement of the rotor with respect to the stationary reference frame. faber" Kr"fqdor
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cos8 cos 2 Si+sin sin8 3 2 2 2 cos8 sin8 1 3.2 Equation Development. It is important to write the equations that describe the motor circuits in state variable form so that the system of first order differential equations can be solved using standard programs. 10
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11 There are four systems of equations that must be solved. These are: Case 1a: no capacitorbefore 51 opens Case 1b: no capacitorafter 51 opens Case 2az with capacitorbefore 51 opens Case 2b: with capacitorafter 51 opens These cases will be developed one at a time as follows. 3.2.1 Case 1No Capacitor. The situation that exists without a capacitor and before the switch opens is shown (for one phase) in figure 3.2. v Ls Figure 3.2 5ingle phase stator equivalent circuit with no capacitor. It should be noted that in this case the stator current and the line current are the same. The rotor is, of course, coupled to this circuit through the stator winding represented by La. The system of equations may be written:
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dIaa V lareq LeqVa dt Va = V lareq LeqpIa (3.21) (3.22) (where p is equal to the derivative operator with respect to time) In terms of all three phases and to chanqe reference frames the followinq matrix equations are neededs and Makinq the appropriate substitutions into equation (3.22) and assuminq identical windinqs qives: premultiplyinq by Ka qivesl Vqdoa VqdO reqlqdoa Leqplqdoa (3.23) 12
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Using the definition of Ks it can be shown that: Substituting this into (3.23) results in : req [:1 : I Vos Vo los 0 0 0 los [ II Leqp Ids (3.24) los or in expanded form: Vqs ,. V q reqlqs LeqWlds Leqplqa Vds Vd reqlds Leqwlqs Leqplda Vos = Vo reqlos Leqplos rearranging: 1 plqs .. Leq 1 plds .. [Vd Vds reqlda Leqwlqa] Leq (3.25) (3.26) (3.27) (3.28) (3.29) 13 ploa .. Vos reqloa] Leq (3.210)
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14 w+ .. .. (3.211) d W + d. [ d) (3. 2 12 ) = W+.S .. ] w+.r (3.214) W+dr (3.215)
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(3.216) 1 pIqs Leq 1 pIds = [Vd Vds reqlds Leq(,)eIqs] Leq pIos = [Vo Vos reqIos] Leq (3.28) (3.29) (3.210) 15 At the instant the switch opens, the source is removed (i.e. current is set equal to zero). This effectively removes equations (3.28), (3.29), and the model then becomes equations (3.211) through (3.216). The last values of the dependant variables before the switch opens are then passed to the new model as initial conditions and the solution continues. Case lb. (3.211) (3.212)
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+ 3 2 13 1 = w+.r (3.2141 (3.2151 w+or (3.2161
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17 3.2.2 Case 2With Capacitor. The situation that exists with a capacitor before the switch opens is shown in figure 3.3. v c Figure 3.3 Single phase stator equivalent circuit with a capacitor. With the addition of another branch another equation is necessary to compensate for the additional unknown. The following 'equation can be written to bring the effect of the additional branch (e.g. the capacitor) into the model given in figure 3.3: integrating this once gives: dVs C... I Is dt or in terms of three phase variables: (3.217) (3.218)
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CpVabcs = labc labcs (3.219) Making the substitution to change reference frames gives: taking the indicated derivative on the left side: lqdos) or: expanding this gives: : I (3.221) o 0 0 Vos Vos lo los or in expanded forml CpVos = lo los (3.222) (3.223) (3.224) 18
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19 rearranging: 1 1 Iq Iqs (3.225) c c 1 IdS) pVds = WeVqs C(Id 1 1 = weVqs Id Ids (3.226) C C 1 los) pVos = e[Io 1 1 = Io los (3.227) c c Equations (12.21), (12.22), and (12.23) from appendix A are rewritten below: 1 Iqs ::0 [I/IqsI/Imq) XIs (3.228) 1 Ids = ( I/Id 8 I/Imd) X18 (3.229) 1 I08 I/Ios (3.230) XIS If equations (3.228), (3.229), and (3.230) are substituted into equations (3.225), (3.226) and (3.227) respectively, results in the elimination of the stator currents as dependant variables (3.231)
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20 pVds = f.)eVqs Id ["'dS tPmd) (3.232) 1 S 1 1 pVos = I o tPos (3.233) e ex 1 S The above equations (3.231), (3.232) and (3.233) combined with equations (3.28), (3.29), (3.210), (3.211), (3.212), (3.213), (3.214), (3.215), and (3.216) can be used to describe case 2. It should again be noted that equation (1.2) has been modified due to the fact that the line current is no longer the same as the stator current with the additional branch. CASE 2a: (3.213)
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1 pI= Leq pIo = Vos reqIo] Leq 1 1 pVos Io !JIos (3.214) (3.215) (3.216) (3.28) (3.29) (3.210) (3.231) (3.232) (3.233) As in case 1, to simulate the switch opening the current I is effectivly set to zero. The model is changed by a similar deletion of equations (3.28), (3.29), and (3.210). This then leaves equations 21
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w+ .. ..  ] (3.211) Pt w+ .. .. (3.212) = w+o. [::}o.] w+.r (3.214) w+.r ) [w::wr (3.215) w+o r 22
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23 (3.234) (3.235) pVos CXls (3.236) The above equations form the basis for the computer solution to this problem. These are nine (case 1a), six (case 1b),twelve (case 2a) or nine (case first order, linear differential equations which are in state variable form. These equations were solved iteratively over the solution interval by a state variable differential equation solver as published by Hultquist [3]. This method uses a Pascal written, RungeKutta 4th/5th order differential equation solver called "RKF45". 3.2.3 Rotor Motion. In order to incorporate the effect of the rotor motion into the above equations, an expression for is needed which is in terms of flux linkages. This is necessary because the angular velocity of the rotor as it appears in the above equations is not a constant. It is a function of torque which can be itself expressed directly in terms of flux linkages. To derive this relationship first begin with the "swing equation" as presented by Fouad [8] :
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d"'r P [ ] = 2J Ta (3.237) where Ta is the accelerating torque, which equals the difference between the electromagnetic torque and the load torque (TeTl). Equation (3.237) can then be writtenl (3.238) after evaluating the integrals: (3.239) (3.240) where is the solution step size of the program and "'1 is the previous (accumulated) value of "'r. In this particular problem, the mechanical load torque Tl was assumed to be constant throughout the motor speed range in order to simplify the solution. This assumption introduced a slight aberration which can be seen in the speed/torque curve in figure F.2. Since the load torque is not zero at zero speed, a slight negative acceleration takes place until the electromagnetic torque (Te) becomes significant. This situation could be corrected by appropriate load
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25 modeling. An equation for Tl could be introduced in the program which ties the load torque to rotor speed; such as Tl = knx where x would vary from 0 to 2 depending on the load to be modeled. A fan or pump, say, would be in the range of x=1.5 to 2. The electromagnetic torque equation is taken from appendix A: (3.241) In order to solve this equation expressions for the rotor currents are obviously needed. These can be obtained from the following five equations from appendix A [equations (12.24, 5, 6, 7, & 8)]. 1 Iqr = ["'q r "'mq] Xl r (3.242) 1 Idr = [Wdr"'md] Xl r (3.243) 1 lor ["'or] Xl r (3.244) and: Wmq = Xm [Iqs+lq r] (3.245) "'md = Xm [Ids+ldr] (3.246) By utilizing equations (3.242, 43, & 44), the currents can be eliminated from equations (3.245 & 46). If the
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results of this are then substituted back into equations (3.242, 43, 44) the result [ 1 ][ Iqr = + Xir XIs Xir [1][ Idr = + Xir XIS Xlr lor = Xl r where: XSq = Xad = Xm X 1 a r] is: (3.247) (3.248) (3.249) The order of solution on each iteration of the program is as follows: first the new value of the state variables are found by the state variable solver previously mentioned (RKF45). For case 1, this returns the current values of Iqs, Ida, & los. For case 2, this returns these values are returned by RKF45 the rotor currents are then solved for using equations (3.247), (3.248) & (3.249) above. These are then used to solve for the electromagnetic torque giverr by equation (3.241). This value is substituted into (3.240) to determine the
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27 current value of rotor angular velocity (wr).Angular velocity can then be converted to RPM by the equation: (3.250) It should be noted that in case one (equations 3.28,9, 10, 11, 12, 13, 14, 15, 16), the way the equations were written, the machine terminal voltage is not included as a state variable. Since this is obviously changing with time it must be solved on each iteration to be reincluded in equations (3.211), (3.212), & (3.213) for the next solution interval. The equations which were used to accomplish this are again taken from appendix A [equations (4.522, 23, & 24) ] : 1 VqS rsIqs ,pds PIPqs (3.251) Wb Wb 1 Vds rsIds ,pqs PIPdS (3.252) Wb Wb 1 Vos = rsIos PIPos (3.253) Wb In order to use these equations, the published program RKF45 was modified to return the current values of PIPqS_, PIPds, & PIPos (RKF45a). Also note that in case 2 (equations 3.28, 9, 10, 11, 12, 13, 14, 15, 16, 31, 32, 33), that the stator current is not included as a state
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variable. In order to observe this current, as the solution advances, describing equations are needed. Using a development similar to the one used to obtain equations (3.247), (3.248), & (3.249) these are given as: r] Iqs Xls Xls Xlr (3.254) ] Ids Xls Xls Xl r (3.255) los Xl s (3.256) 3.2.4 Saturation Effects. In order that a realistic model of the induction machine is developed, the effects of saturation must be included in the simulation. These effects are included by noting the last term in equations (4.528), (4.529), (4.531), (4.532) (appendix A). e.g: These terms may be taken as the magnetizing flux and are denoted by and respectively as
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shown by equations (3.245) and (3.246). These are the same terms which appear in equations (3.211), (3.212), (3.214), and (12.215) of case 1; and equations (3.211), (3.212), (3.214), (3.215), (3.231), and (3.232) of case 2. The magnitizing flux equation which was actually used is given by the last terms in equations (3.247) and (3.248) which is equation (3.245) and (3.246) with the currents eliminated. x [ !/IqS ] !/Imq = Xl s Xl (3.257) [ ] !/Imd = Xad Xl s Xl (3.258) For the induction machine used in this study an open circuit test was done to determine the saturation curve. This curve was then adjusted to compensate for the stator losses (rs and Xls) and established empirically as a plot of !/I vs. !/I. This curve was then modeled by an exponential equation using the methods outlined in [6]. A plot of this curve in the first quadrant only is shown in figure F.3. The resulting equation was implemented with the Pascal functions "Fmq" and "Fmd" located in subroutine "TRANS" as listed in appendix D. After each computer iteration a new
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30 "calculated" value of the magnitizing flux is given by equations (3.257) and (3.258), the "actual" value used in the next iteration is then returned by the modeled saturation curve. It should be pointed out that these functions ("Fmq" and "Fmd") were written so a saturated or an unsaturated value is returned as directed by the main program. This is to allow comparision between the effects of a saturated vs. a theoretical machine in which there is no saturation. It should be mentioned that in this simulation the effects of core loss and no load rotational losses have been ignored.
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CHAPTER 4 COMPUTER HARDWARE AND SOFTWARE 4.1 Software. The Pascal programs listed in appendices B, C and D are used for the digital simulation in this study. Since these programs are similar, only one (program NOCAP) will be described here. %IHCLUDE 'TRAMS.PAS'; The subroutine 'TRANS.PAS', as previously mentioned contains procedures Ks, Ks_l, K r & Kr_1 which are used to transform actual stator and rotor quantities (abc) to and from the rotating reference frame (qdo). Also contained here are functions Fmq & Fmd which are used by the differential equations to account for saturation in both the direct and quadrature axes. PROCEDURE GETDATA; This procedure is used to initialize the variables and to set program control functions. The solution interval is set by variables "STARTTIME" and
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32 "FINISHTIME". Step size is set by "DELTAt" and the time of switch opening is set by "OPENSWITCH". "ABSERR" and "RELERR" are the absolute and relative error respectively used by the "RKF4S" program. "SAMPLERATE" is used to determine how often the solution is sampled and stored by "PROCEDURE RETAINDATA" for later plotting. This procedure is used to store the sampled dependent and independent variable to a "flat file" for later access and plotting. These values in order are: accumulated time stator current rotor current rotor revolutions per minute electromagnetic torque and motor terminal voltage s ) Procedure "DIFFEQ" contains the differential equations required for the solution of the problem. These are in first order state variable form. This procedure is called by the equation solver "RKF4S" as necessary to find the next value of the dependent variables over the selected step size. Notice the "IF" statement that determines the set of equations to be used. This is dependent on the time in relation to when
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the switch opens. This is the previously mentioned Kutta differential equation solver for a system of first order differential equations. For further description of this program, reference [3] should be consulted. 33 This program is userinteractive to the extent that you are prompted for: motor load torque; a file name in which you would like the data points stored; and whether to include the effects of saturation. For the time interval before the switch opens, the motor supply voltage (V) is taken as three equal sinusoids for the three phases, each displaced 120 degrees from each other. These are: and Where V mag is the single phase magnitude, which in this case is 132.8 volts (230 volts line to line). The moment of inertia (J) which is used in the swing equation to calculate rotor speed is selected between two values. If the motor is unloaded, then the value of 0.005 Newton meters is used. This is equal to the moment of inertia of the rotor for this particular machine. If the program is run with a load (any load)
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the value of is 0.01. This assumes the moment of inertia of the load is the same as that of the rotor and that J is the sum of these two values. The rest of the program is relatively straightforward, using the equations as presented in the previous sections. 4.2 Hardware. 34 These programs were run on a VAX 8700, using the resident Pascal compiler. To obtain the plots necessary for the results, a VAXbased MATLAB [9] program was used. This was done because Pascal, on this particular machine, does not support graphics. To get the desired plots, a small routine written in the Matlab environment would call up the data file created by PROCEDURE RETAINDATA and plot the results as desired. Any of the stored parameters may be accesed in this fasion and plotted.
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Iset initial values (embedded in program) operator enter: Tl, file name to recieve data, include saturation ? Itime .. Atl Is time openswitch ? initial values .. previous final values Solve diff. equations for case 1(a) (equationsl 3.28,9,10,11,12,13,14,15,16) solve diff. equations case 1(b) (equations: 3.211,12,13,14,15,16) = 0 IIabcs = Ksl'Iqdosl J, Figure 4.1 Flow chart of computer program "NOCAP", 35
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Solve for Vqs,Vds,Vos (equations: 3.251,52,53) IVabcs = solve for Iqr,Idr,Ior (equationsl 3.247,48,49) lIabcr Ksl'Iqdor! solve for torque (Te) (equation: 3.241) store for later ploting: solve for Wr time,Iabcs,Iabcr,RPM,Te,Vs (equation 3.240) ,,, solve for RPM (equationl 3.250) !time time At 1 NO/iS time finishtime Figure 4.1 (cont.) Flow chart of program "NOCAP". 36
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CHAPTER 5 DIGITAL SIMULATION AND EXPERIMENTAL RESULTS The equivalent circuit parameters which were used in the computer solution are shown in figure 5.1. J"\. 42 Figure 5.1 circuit. 1.5 HP motor steady state equivalent These parameters were obtained from the open circuit test previously mentioned and information supplied by manufacturer. The capacitor size which was selected for the computer runs was a multiple of a one per unit capacitor. It was assumed that a one per unit capacitor would be (in ohms) equal to the magnitizing reactance
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38 Xm Assuming the effects of the leakage reactances to be negligable, it would be approximately this value of capacitor that would be needed to neutralize the lagging component of motor current and raise the power factor to unity. This capacitance was determined by the equation: 1 Xm = or C = For the motor in this study, C is approximately equal to 65#f. Applying a capacitor of this size on the motor used in this study would approximately equal to 1.296 KVAR's of correction [KVAR 5.1 Digital Simulation Results. The motor terminal voltage plots for the 1.5 HP motor are shown in appendix E. Appendix F contains plots of the 1.5 HP motor speed/torque curves showing acceleration from zero speed to full speed for the full and no load case. The saturation curve used in this simulation is shown along with torque vs. time curves for the no capacitor case and the 1 per unit capacitor. Also shown are plots of motor speed vs. time (no load and full load) and plots of motor line current vs. time. For the plots depicting the motor no load case, the switch was opened at 0.6 seconds and 0.8 seconds
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for full load. Also included for the above plots are the similar runs comparing the saturated vs. the unsaturated machine. 39 In addition to the 1.5 HP motor which is is of primary concern here, motor parameters were also obtained [1] for a 3 and a 50 HP motor. For purposes of comparison these values were plugged into the programs and the results were also plotted. These plots are shown in appendix G. The 3 HP motor used was a 230 volt three phase machine, while the 50 HP was connected for 460 volts three phase. For both motors the following digitally simulated plots were done: terminal volts vs. time for 1, 2, & 3 per unit capacitors (unloaded case only), motor line current vs. time for the full load case, and a speed vs. torque plot for both the full load and no load case. All of these plots were done with saturation negelected since saturation information was unavailable. 5.2 Experimental Results. In an at temp to varify the accuracy of the digital simulation, the circuit as shown in figure 1.2 was duplicated in the lab. An "event" recording type oscilloscope (HewlettPackard 54501A) was connected to the motor terminals. After allowing the motor to reach steady state the switch was opened and the resulting
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waveform recorded. This was done for capacitor values of 0, 1, 2, and 3 per unit. These waveforms are shown in appendix E (figures E.13, E.14, E.15, & E.16). The motor was operated at no load condition for this experiment. 40 Comparison of the digitally simulated plots (Figure E.5, E.G, E.7, & E.8) with the actual motor waveforms taken in the lab (Figure E.13, E.14, E.15, & E.16) indicates that the computer model developed herein is reasonably accurate in reproducing the behavior of the actual machine. When comparing these plots note that the scales are different.
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CHAPTER 6 ANALYSIS AND CONCLUSION From the digitally simulated plots of motor terminal volts vs. time, with saturation neglected, one can see that the voltage does indeed rise after the switch is particularly for the 2 & 3 per unit capacitors A peak rise takes place with a 3 per unit capacitor (computer plots were done for larger values to varify this). This effect is especially dramatic for the 3 and 50HP motors shown in appendix G. By comparing the digitally simulated plots of terminal volts for the saturated (Figure E.l, E.2, E.3, & E.4) and the unsaturated case (Figure E.5, E.6, E.?, & E.8) it is observed that the effects of saturation tend to limit or clip the post switching voltage transients. The results of these plots which have incorporated saturation are in close agreement with the results obtained by Walsh and deMello [4] using an analog computer. Their results predicted that for capacitor values of 1, 2 & 3 per unit, peak voltages of approximately 0.95, 1.2, and 1.3 per unit would be seen at the motor terminals.
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Comparison of the no load (figure E.5, E.6, E.7, & E.8) vs. the full load case (figure E,9, E.10, E.ll, & E.12), indicates that the no load scenario is clearly the worst case situation. These plots show that motor load serves to dampen any post switching transients making capacitive effects negligable. By direct comparison of the plots of motor terminal volts for the unsaturated (figure E.1, E.2, E.3, & E.4) vs. the saturated case (figure E.5, E.6, E.7, & E.8), the clipping effect due to saturation can clearly be seen. This indicates that if not for the effects of saturation, over voltages would occur at the machine terminals. One should note, however, that even these "over voltages" which occur for the actual machine are in fact not as large as the voltage seen by the motor on startup due to the large inrush currents. The comparison of the post switching voltage transient (neglecting saturation) of the 3 motors (1.5, 3, & 50 HP) indicates that similar, albeit proportionally larger over voltages occure for the 3 and 50 HP machines as well. If saturation curves are also proportionally similar, one would expect to see a similar clipping of motor terminal voltage. It should be mentioned that the fact saturation may limit the voltage rise for a 2 and 3 per unit capacitor is somewhat irrelevant from the standpoint of
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43 power factor correction. One would rarely over correct a motor to this degree as this would increase the apparent power to the motor on the leading side of the power triangle, thereby defeating the whole purpose of the correction. The indication provided by the computer plots and experimental results is that no problem is encountered with correction up to 100% and very little, if any, above that. The major nhazardous" effect of these capacitors (no load case only) is that the terminal voltage is sustained for several seconds before finally decaying to zero. However, when a load is this too appears not be a problem. From the results presented in this thesis one can speculate that while the problem explored here may have been significant in the machines constructed 40 or 50 years ag?, it is probably no longer of concern in the modern machine design. The reason for this could possibly be because modern machines are physically hence constructed with less steel and are therefore operated into the saturation regio,n. The steel saturation characteristics thereby providing a built in voltage limiting effect. Due to time and constraints this study was concluded with modeling and testing of the 1.5 HP motor. The logical extension of this study would be to
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study the effects of transient torques developed when, after disconnecting the motor from the source, it was reconnected while the terminal voltage was decaying, becoming more and more out of phase with the source. This would be a straightforward study using the model developed here. 44 The model developed here would also allow easy observation of other motor parameters such as rotor and stator current, speed, developed torque, as well as terminal voltage.
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BIBLIOGRAPHY [1] P. C. Krause, Analysis of Electric Machinery. McGrawHill Inc., New York, N.Y.,1986. [2] R. H. Park, "TwoReaction Theory of Synchronous MachinesGeneralized Method of AnalysisPart I," AlEE Trans., Vol. 48, July 1929, pp. 716727. [3] P. F. Hultquist, Numerical Methods for Engineers and Computer Scientists, Benjamin/Cummings Publishing Inc.,Menlo Park, California, 1988. [4] G. W. Walsh and F. P. deMello,"Reclosing Transients in Induction Motors with Terminal Capacitors", IEEE Trans. on Power Appratus and Systems. Vol.74, Feb. 1961, pp. 12061213. [5] T. H. Ortmeyer, nEffects of Reactive Compensation on Induction Motor Dynamic Performance", IEEE Trans. on Power Apparatus and Systems. Vol 99, May/June 1980, pp. 841846. [6] W. K. Macfadyan, "Representation of Magnetisation Curves By Exponential Series", Proc. lEE, vol. 120, No.8, Aug. 1973, pp. 902904. [7] P. M. Derusso, State Variables for Engineers, John Wiley & Sons, Inc., New York, N.Y., 1965. [8] P. H. Anderson & A. A. Fouad, Power System Control and Stability, Iowa State University Press, Ames, Iowa, 1977. [9] MATLAB Program, Copyright The Mathworks, Inc. 19841988, Version 3.34, March 1988.
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[10] S. Ertem, Y. Baqhzouz, "Simulation of Induction Machinery For Power System Studies", IEEE Transations on Energy Conversion, Vol. 4, No.1, March 1989, pp. 8894. [11] A. Keyhani, H. Tsai, "IGSPICE Simulation of Induction Machines with Saturable Inductances", IEEE Transactions on Energy Conversion. Vol. 4, No.1, March 1989, pp. 118125. 46
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APPENDIX A EXCERPTS FROM BIBLIOGRAPHY REFERENCE [1] The basic voltage equations describing an induction machine with no capacitor have been derived in reference [1] and are given below. These equations have been transformed to the syncronously rotating reference frame. 1 Vqs rsIqa !/Ids p!/lqa (4.522) Wb Wb 1 Vds raIds !/IqS P!/lds (4.523) Wb Wb 1 Vos = rsIos (4.524) Wb 1 Vqr rsIqr !/Idr p!/lqr (4.525) Wb 1 Vdr rsIqs !/Iqr p!/ldr (4.526) Wb 1 Vor = rsIor p!/lor (4.527) Wb Wb is the base angular frequency used to calculate the
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48 inductive reatances, We is the angular frequency of the syncronously rotating reference frame and Wr an9ular frequency of the rotor. The flux linkages have been expressed as flux linkages per second which allows the mutual inductance term (M) to be replaced with the magnitizing reactance (4.528) (4.529) !/los = Xlslos (4.530) (4.531) (4.532) !/lor = XlsIor (4.533) If these equations are solved for currents the following equations result: los = Pos Xl s (12.21) (12.22) (12.23)
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49 1 Iqr (IPqrIPmq ] (12.24) Xl r 1 Idr (IPdrIPmct] (12.25) Xl r 1 lor .por (12.26) Xl r where: IPmq (12.27) (12.28) If the above equations for currents are substituted into equations (4.522) to (4.527), the equations in terms of flux linkages result. (12.29) PIPds = W+dS (WdS WMd] [::]WQS] (12.210) PIPos = Wb Vos [::}os] (12.211)
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50 w+.r .. ) [W::W}dr] (12.2121 W+dr [::r] [W::W}.r] (12.2131 Wb[vor (12.2141 s '" r] = 1 Xl =
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APPENDIX B PASCAL PROGRAM USED TO IMPLEMENT CASE 1 NO CAPACITOR The following is the Pascal program which was used to analyize the motor performance without a capacitor (case 1). PROGRAM CONST Wb .. Rs,..!. 63 Rr=1.125 P=4 ReqO.006 TYPE VECTORaARRAY[l .. NEQN] OF MULTIVECTOR=ARRAY[DIRECTION] OF double; VAR RPM,U26,ABSERR,RELERR,OLDABSERR,OLDRELERR, Xrr,Xss,Xaq,Xad,STARTTIME,OPENSWITCH,TIME,TIN, TOUT,Wr,Te,DELTAt,FINISHTIME, DISPLAYRATE,COUNT1,FLAG,OLDFLAG,
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LETTER: CHAR; Y,YP,YO,YPZEROzVECTOR; DATAFILE:PACKED ARRAY[l .. 35] OF CHAR; %INCLUDE 'TRANS.PAS' PROCEDURE GETDATA; BEGIN DELTAt:=O.OOOl; STARTTIME:ClO.O; SAMPLERATEz=lO; OPENSWITCHz=O.S; FINISHTIMEz=2.0; TIHE:=DELTAt; ABSERR:=l.OE8; RELERR:=l.OES; Fs[q):=O.O; Fr[q)z=O.O; Is[q):=O.O; Vr[q):=O.O; Vs[q]:ClO.O; Wrz=O.O; END; Fs[d)z=O.O; Fr[d):=O.O; Is[d)z"O.O; Vr[d)z=O.O; Vs[d):O.O; PROCEDURE RETAIN_DATA; BEGIN Fs[o)zO.O; Fr[o]:=O.O; Is[o)zO.O; Vr[o):=O.O; Vs[o)z=O.O; WRITELN(F,TIMEzO:6,' ',Is[a):O:l,' ',Ir[a):O:l, ',RPMzO:l,' ',Te:O:l,' ',Vs[a]:O:l); END; PROCEDURE VAR Y,YP:VECTOR); BEGIN IF TIME OPENS WITCH THEN BEGIN 52 Fmq(Y[l],Y[4],SATURATION))(We/Wb)*Y[2)); YP[2].Wb*(Vs[d](Rs/Xls)*(Y[2) Fmd(Y[2],Y[S),SATURATION)+(We/Wb)*Y[l)); YP[3]z=Wb*(Vs[o](Rs/Xls)*Y[3]); Fmq(Y[l],Y[4],SATURATIONWeWr)/Wb)*Y[S]); YP[S]:=Wb*(Vr[d](Rr/Xlr)*(Y[5] Fmd(Y[2],Y[5],SATURATION)+WeWr)/Wb)*Y[4]); YP[6]:=Wb*(Vr[o)(Rr/Xlr)*Y[6]); YP[7]:=l/Leq*(V[q]Vs[q]Req*Y[7]Leq*We*Y[8]); YP[S]:al/Leq*(V[d]Vs[d]Req*Y[8]+Leq*We*Y[7]); YP[9]:=l/Leq*(V[o]Vs[o]Req*Y[9]); END; IF TIME OPENSWITCH THEN BEGIN
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YP[l]:=Wb*(Vs[q](Rs/Xls) *(Y[1]Fmq(Y[1],Y[4],SATURATION)) ( We Wb ) Y [ 2 ] ) YP[2]&=Wb*(Vs[d](Rs/Xls) YP[4]:=Wb*(Vr[q](Rr/Xlr) *(Y[4]Fmq(Y[l],Y[4],SATURATION)) ( (WeWr) IWb) *y [ 5 ] ) YP[S]:=Wb*(Vr[d](Rr/Xlr) *(Y[S]Fmd(Y[2],Y[S],SATURATION)) %INCLUDE 'RKF4Sa.PAS' PROCEDURE BEGIN Xss:. Xrr:=Xlr+Xm; WRITE('ENTER LOAD TORQUE IN NH WRITE('ENTER FILE TO RECEIVE DATA POINTS WRITE('INCLUDE THE EFFECTS OF SATURATION?'); WRITE(' (y for yesany key for no) READLN(LETTER}; WRITELN; IF LETTER IN ['y','Y'] THEN ELSE SATURATION :=FALSE; IF Tl 0.1 THEN J:aO.01 ELSE J:0.005; OPEN(F,DATAFILE); REWRITE ( F) ; GETDATA; FINDU26; COUNT1:=0; WRITELN('RUNNING, DO NOT ..
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REPEAT IF TIME < OPENS WITCH THEN BEGIN V[a]:=(Vmaq*SQRT(2*COS(We*TIME); V[b]s=(Vmaq*SQRT(2))*COS(We*TIME2*PI/3); V[c):=(Vmaq*SQRT(2))*COS(We*TIME+2*PI/3); Ks (We, "TIME, V) ; END; YO [ 1 ] : =F s [ YO[2]s=Fs[d]; YO[3]:=Fs[o); YO[4]:=Fr[q]; YO[5]s=Fr[d); YO [ 6 ) : =F r [ 0 ) ; IF TIME < OPENSWITCH THEN BEGIN YO[7]:=Is[q]; YO(8):=Is[d); YO[9]:=Is[o]; END; IF TIME >= OPENS WITCH THEN BEGIN YO[7]:=O.O; YO(8):=O.O; YO(9):=O.O; END; FLAG: =1; RKF45(NEQN,YPZERO,YO,Y,STARTTIME, TIME,RELERR,ABSERR,FLAG); IF FLAG IN [3,4,5,6,7] THEN BEGIN 54 WRITELN('TROUBLE IN RIVER CITY, FLAG= ',FLAG); HALT; END; Fs[q]s=Y[l]; Fs[d) :=Y[2]; Fs[o]:=Y[3); Fr[q]:Y[4]; Fr[d]s=Y[5); Fr[o]:=Y[6]; Is[q] &=Y[7]; Is[d]:=Y[8); Is[o]:=Y[9]; KS_l(We,TIME,Is); Vs[q):=Rs*Is[q]+(We/Wb)*Fs[d]+(l/Wb)*(YPZERO[l]); Vs[d]:=Rs*Is[d](We/Wb)*Fs[q]+(1/Wb)*(YPZERO[2]); Vs[o]:=Rs*Is[o]+(1/Wb)*(YPZERO[3]); Ks_l(We,TIME,Vs);
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55 Ir[d):=(l/Xlr)*(Fr[d)Xad*(Fs[d)/Xls+Fr[d)/Xlr; Ir[o)I=(l/Xlr)*Fr[o); Kr_l(We,Wr,TIME,Ir); Te:=(3/2)*(P/2)*(1/Wb)*(Fr[q)*Ir[d)Fr[d)*Ir[q); Wr:=(P/(2*J*(TeTl)*DELTAt+Wr; RPM:=( (Wr/(2*PI) IF COUNT1=SAMPLERATE THEN RETAIN_DATA; COUNT1:=O; END; COUNT1:=COUNT1+l; TIME:=TIME+DELTAt; UNTIL TIME> FINISHTIME; CLOSE (F) ; END; BEGIN MAIN END.
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APPENDIX C PASCAL PROGRAM USED TO IMPLEMENT CASE 2 WITH A CAPACITOR The following is the Pascal program which was used to analyze the motor performance with a capacitor (case 2). PROGRAM CONST HP= 1. 5 We=377, Wb=377; Rs=1. 63, Xls=1.6375, Rr=1.12S, Xm=42.0, PI=3.1415926S, P=4 Vmag"132.S, NEQN12, Req .. O.006, Leq=O.OOOlS TYPE VECTORARRAY[l .. NEQN] OF double; DIRECTION=(a,b,c,q,d,o); MULTIVECTOR=ARRAY[DIRECTION] OF double, VAR Fs,Fr,V,Vs,Vr,I,Is,Ir:MULTIVECTOR" RP,M, U26, ABSERR, RELERR, OLDABSERR, OLDRELERR, Xrr,CC,Xss,Xaq,Xad,STARTTIME,OPENSWITCH,TIME, TIN,TOUT,Tl,J,Wr,Te,DELTAt,FINISHTIMEIDOUBLE, SATURATION,INIT:BOOLEAN, COUNTl,FLAG,OLDFLAG,EXITFLAG,SAMPLERATE:INTEGER; F:TEXT, LETTER:CHAR; Y,YP,YO,YPZERO:VECTOR; ARRAY[1 .. 35] OF CHAR;
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%INCLUDE 'TRANS. PAS' PROCEDURE GETDATA; BEGIN DELTAt:=O.OOOl; SAMPLERATE:=10; OPENSWITCH:=0.8; FINISHTIME:=2.0; TIME:=DELTAt; ABSERR,=1.0E8; RELERR:=1.0E8; Fs[q], .. O.O; Fr[q):=O.O; Is[q]:=O.O; Vr[q):=O.O; Vs[q]:=O.O; I[q):=O.O; Wr:=O.O; END; Fs[d],=O.O; Fr[d]:=O.O; Is[d),O.O; Vr[d]:=O.O; Vs[d]:"O.O; I[d]:O.O; PROCEDURE RETAIN_DATA; BEGIN Fs[o]I!=O.O; Fr[o):=O.O; Is[o]:=O.O; Vr [ 0 ) .. 0.0 ; Vs[o]:=O.O; I[o):=O.O; WRITELN(F,TIME:0:6,' ',Is[a]:Oll,' ',Ir[a]:Oll, ,RPM:O: 1,' ,Te:O: 1,' ,Vs[a) :0: 1, ,I[a] :0: 1); END; PROCEDURE DIFFEQ(T:double; VAR Y,YP:VECTOR); BEGIN IF TIME < OPENSWITCH THEN BEGIN *(Y[1]Fmq(Y[l],Y[4],SATURATION (We/Wb)*Y[2]); YP[2],Wb*(Y[ll](Rs/Xls) *(Y[2]Fmd(Y[2],Y[S],SATURATION +(We/Wb)*Y[l]); YP[3]:"Wb*(Y[12](Rs/Xls)*Y[3]); YP[4]:=Wb*(Vr[q](Rr/Xlr) *(Y[4]Fmq(Y[l],Y[4],SATURATION ((WeWr)/Wb)*Y[S]); YP[S]:=Wb*(Vr[d](Rr/Xlr) +((WeWr)/Wb)*Y(4]); YP[6]:=Wb*(Vr(o](Rr/Xlr)*Y[6]); YP[7]:=(1/Leq)*(V[q]Y[10]Req*Y[7]Leq*We*Y[8]);
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YP(9]1=(1/Leq)*(V(o]Y[12)Req*Y[9]); YP[10):=We*Y[11]+(1/CC)*Y[7](1/(CC*Xls*(Y[1] Fmq(Y[1],Y[4],SATURATION; YP[11):a+We*Y[10]+(1/CC)*Y[8](1/(CC*Xls*(Y[2] Fmd(Y[2],Y[5],SATURATION, END; IF TIME OPENSWITCH THEN BEGIN YP[l]:=Wb*(Y[lO](Rs/Xls) (Y [1] Fmq (Y [1] Y [4] SATURATION) ) (We/Wb)*Y[2]), YP[2]:=Wb*(Y[ll](Rs/Xls) *(Y[2]Fmd(Y[2],Y[S],SATURATION +(We/Wb)*Y[l]); YP[3]:=Wb*(Y[12](Rs/Xls)*Y[3]); YP[4]:=Wb*(Vr[q](Rr/Xlr) *(Y[4]Fmq(Y[1],Y[4],SATURATION ((WeWr)/Wb)*Y[S]), YP[5]:=Wb*(Vr[d](Rr/Xlr) *(Y(5]Fmd(Y(2],Y[S],SATURATION +((WeWr)/Wb)*Y[4]), YP[6]:=Wb*(Vr[o](Rr/Xlr)*Y[6]), YP[7]:=0.O, YP[9]1=0.O, YP[lO]:=We*Y[ll](l/(CC*Xls *(Y[1]Fmq(Y[l],Y[4],SATURATION, YP[ll]:=+We*Y[lO](l/(CC*Xls YP[12]:=(1/(CC*Xls*Y[3], END; END, %INCLUDE 'RKF45.PAS' PROCEDURE MAIN, BEGIN Xss:=Xls+Xm, Xaq:=l/(l/Xm+l/Xls+l/Xlr), Xad:Xaq, WRITELN, 58 ... ) ) ; READLN(CC), WRITELN, WRITE('ENTER LOAD TORQUE IN NM READLN(Tl);WRITELN; WRITE('ENTER FILE TO RECEIVE DATA POINTS '); READLN(DATAFILE)i WRITELN; OPEN(F,DATAFILE),
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WRITE('INCLUDE SATURATION? WRITE('(y for yesany other for no) READLN(LETTER}; IF LETTER IN ['Y','y'] THEN SATURATION:=TRUE ELSE IF T1 > 0.1 THEN J:=O.Ol ELSE WRITELN; WRITELN('RUNNING, DO NOT DISTURB ... REPEAT IF TIME OPENSWITCH THEN BEGIN V[a]:=(Vmag*SQRT(2*COS(We*TIME); V[b]:=(Vmag*SQRT(2*COS(We*TIME2*PI/3); V[c]:=(Vmag*SQRT(2*COS(We*TIME+2*PI/3); Ks(We,TIME,V) END; YO [ 1 ) : =F s [ YO[2).""Fs[d); YO [ 3 ) : =F s [ 0 ) YO [4): aFr[q]; YO [ 5 ) : =F d ) ; YO [ 6 ] : =F 0 ] IF TIME OPENS WITCH THEN BEGIN YO(7]:=I[q]; END; IF TIME >=OPENSWITCH THEN BEGIN YO[8]:::::I0.O; YO[9]:=0.0; END; YO[10]:=Vs[q); YO [ 11 ) : =Vs [d] ; YO[12]:aVs[o]; FLAG:"l ; RKF45(NEQN,YO,Y,STARTTIME,TIME,RELERR, IF FLAG IN [3,4,5,6,7] THEN BEGIN 59 WRITELN('TROUBLE IN RIVER CITY, FLAG= ',FLAG); END;
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Fs[d] :=Y[2] Fs[o] :=Y[3]: Fr[q]:=Y[4]: Fr[d] :=Y[S]: Fr[o]:=Y[6]; I[q]:=Y[7]; I [ d ] : =Y [ 8] : 0 ] : =Y [ 9 ] Vs[d]:=Y[ll]; Vs[o]p'Y[12; 60 Is[q]:=(l/Xls)*(Fs[q]Xaq*(Fs[q]/Xls+Fr[q]/Xlr)); Is[d]:=(l/Xls)*(Fs[d]Xad*(Fs[d]/Xls+Fr[d]/Xlr)); Is[o]:=(l/Xls)*(Fs[o]); KS_l(We,TIME,Is); Ir[q]:=(l/Xlr)*(Fr[q]Xaq*(Fs[q]/Xls+Fr[q]/Xlr)); Ir[d]:=(l/Xlr)*(Fr[d]Xad*(Fs[d]/Xls+Fr[d]/Xlr)); Kr_l(We,Wr,TIME,Ir); KS_l(We,TIME,I); Te:=(3/2)*(P/2)*(1/Wb)*(Fr[q]*Ir[d]Fr[d]*Ir[q]); Ks_l(We,TIME,Vs); Wr:=(P/(2*J))*(TeTl)*DELTAt+Wr; RPM: = ( (W r ( 2 PI) ) 60 ) ( P ) ; IF COUNT1=SAMPLERATE THEN BEGIN RETAIN_DATA; COUNT1:=O; END; COUNT1:=COUNT1+l; TIMEI=TIME+DELTAt: UNTIL TIME> FINISHTIME; CLOSE (F) ; END; BEGIN MAIN END.
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APPENDIX D PASCAL SUBROUTINE USED TO INCLUDE REFERENCE FRAME TRANSFORMATION AND SATURATION The Following is a Pascal subroutine TRAMS contains several procedures and functions. Procedure Ks and Kr are used to transform stator and rotor variables to the rotating reference frame implementing the equation of transformation: fqdos = Ksfabes and Procedure Ks_1 and Kr_1 are used to transform stator rotor variables from the rotating reference frame to the stationary reference frame implementing the inverse transformationz fabes = (Ks ] fqdOS and faber := (Kr) fqdor Function Fmq & Fmd are used by the main program to incorporate the effects of satuaration into the solution.
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These are as follows: PROCEDURE Ks(We,T:double; VAR F:MULTIVECTOR); BEGIN F[q):=2/3*(F[a]*COS(We*T)+F(b]*COS(We*T2*PI/3) +F[c)*COS(We*T+2*PI/3)); F[d)I=2/3*(F[a)*SIN(We*T)+F[b)*SIN(We*T2*PI/3) +F[c)*SIN(We*T+2*PI/3)); F[O)la1/3*(F[a)+F[b)+F[c)); END; PROCEDURE KS_l(We,T:double; VAR FIMULTIVECTOR); BEGIN F[a):=F[q)*COS(We*T)+F[d)*SIN(We*T)+F[o); F[b):=F[q]*COS(We*T2*PI/3)+F[d) *SIN(We*T2*PI/3)+F[o); F[c]:=F[q]*COS(We*T+2*PI/3)+F[d) *SIN(We*T+2*PI/3)+F[o); END; PROCEDURE Kr(We,Wr,T:double; VAR F:MULTIVECTOR); BEGIN F[q):=2/3*(F[a)*COSWeWr)*T)+F[b) *COSWeWr)*T2*PI/3) +F[c)*COS( (WeWr)*T+2*PI/3)); F[d)I=2/3*(F[a)*SINWeWr)*T)+F[b) *SINWeWr)*T2*PI/3) +F[c)*SINWeWr)*T+2*PI/3)); 0 ) : = END; PROCEDURE Kr_l(We,Wr,Tldouble; VAR FIMULTIVECTOR); BEGIN F[a]:aF[q]*COSWeWr)*T)+F[d)*SINWeWr)*T)+F[o]; F[b]:sF[ql*COSWeWr)*T2*PI/3)+F[d] *SINWeWr)*T2*PI/3)+F[o); F[c]I=F[q)*COSWeWr)*T+2*PI/3)+F[d) *SINWeWr)*T+2*PI/3)+F[o); END; FUNCTION Fmq(Fqs,Fqr:double; yes:boolean)ldouble; VAR Kl,K2,K3,K4;H,OFFSET,SLOPE:DOUBLE; BEGIN K11=129.7997; K2:=O.00614; 62
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IF YES THEN BEGIN END IF (H > OFFSET) AND (H < OFFSET) THEN IF H >= OFFSET THEN Fmq:=OFFSET*SLOPE+Kl*ilEXP(K2*(HOFFSET) IF H <= OFFSET THEN Fmq:= 1.0*(OFFSET*SLOPE+Kl *(1EXP(K2*(ABS(H)OFFSET) ELSE FUNCTION VAR BEGIN IF. YES THEN BEGIN END IF (H > OFFSET) AND (H < OFFSET) THEN IF H OFFSET THEN Fmd:=OFFSET*SLOPE+Kl*(1EXP(K2*(HOFFSET) +K3*(1EXP(K4*(HOFFSET); IF H <= OFFSET THEN Fmd:=1.0*(OFFSET*SLOPE+Kl *(1EXP(K2*(ABS(H)OFFSET) ELSE Fmd:=H; END;
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APPENDIX E COMPUTER SIMULATED AND EXPERIMENTALLY OBSERVED VOLTAGE WAVEFORMS FOR THE 1.5 HP MOTOR
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300 250 0 200 t 0 150 t e 100 50 a 0 0 50 t s 100 150 200 no capacitor no load no saturation ___ _______ ______ __ ____ 4 _______ o o ,0 0 0 t ___ ____ 0 0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8 seconds 65 300 20' 0 0 0 e a 0 9 300 1.5 hp C "" 1 p.u. o no load no saturation j o 0 j 0 o 0 ______ ___ __ J ___ __ __ ____ J _______ ______ 0 0 0.6 0.8 1 1.2 1.4 1.6 1.8 2 o 0.2 0.4 seconds
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400 300 0 t 0 200 t e a 1 v 0 1..5 hp c p.u. no load no saturati.on __ 4 __ _______ ________ _______ 4 _______ ______ 66 1 100 s 200 ____ ____ __ Figure volts and no o 0.2 0.4 0.6 O.B 1. 1..2 1..4 1..6 seconds B.3 simulated plot of motor 2 per unit capacitor, Computer vs. time: 1.5 HP, saturation. 1..5 hp C 3 p.u. no load no saturation 2 terminal no load, 0 0 ___ 4 ______ _____ ___ ______ 4 ______________ e a 0 s 300 Figure volts and no o 0.2 0.4 0.6 O.B 1. 1.2 1..4 1..6 seconds vs. saturation. simulated plot of motor 3 per unit capacitor, E.4 Computer time: 1.5 HP, loB terminal no load,
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300 250 0 200 t 0 150 t e 100 50 a 1 0 0 50 1 t s 100 150 200 1.5 hp no capacitor no load with saturation j _______ HKIHIHtKIKt _____ _______ _______ ________ _______ 4 __ ____ ______ ____ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 seconds 2 Figure .5 volts vs. time: saturation. Computer 1. 5 HE>, simulated plot no capacitor, of motor terminal no load, and with m 0 0 t e m a v 0 1 s 300 200 100 0 100 200 300 1.5 hp c 1 p.u. no load with saturation t .. t 0.6 0.8 1.2 1.4 1.6 1.8 o 0.2 0.4 seconds Figure .6 volts vs. and with Computer time: 1. 5 HE>, saturation. simulated 1 per plot of motor unit capacitor, terminal no load,
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68 1.5 hp c 2 p.u. no 1.0ad. with saturaton 400 m 300 0 t. 0 200 __ _______ 4 _________________ _____ _______ ___ t e 100 m a 0 1 0 1 100 t s 200 I t I 1 300 0 0.4 0 6 0.8 1 1.2 2 second.s Figure E.l Computer simulat.d plot of motor terminal volts vs. time: 1.5 HP, 2 per unit capacitor, no load, and with saturation. 1.S hp c 3 p.u. no load. with saturation 300 200 0 t 0 100 e 0 a 1 100 v 0 1 200 t s 300 ___ _______ ' o 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 second.s Figure E.B Computer simulated plot of motor terminal volts vs. time: 1.5 HP, 3 per unit capacitor, no load, and with saturation.
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300 250 0 200 t 0 150 t e 100 50 a 0 0 1 50 s 100 150 200 0 1.5 hp no capacitor full load with saturation ,r,__ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 seconds .. ,_ .. _0.8 0.9 1 Figure E.9 Computer simulated plot of motor terminal volts vs. time: 1.5 HP, no capaCitor, full load, and with saturation. m o o e a 1 s 1.5 c 1 p.u. full load 200 100 o 100 200 300 III ............. i" ............ :.............. 1.. .. ... .. .. 11____ ____ ____ ____ o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 seconds 69 Figure E.I0 Computer simulated plot of mQtor terminal volts vs. time: 1.5 HP, 1 per unit capacitor, full load, and with saturation.
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0 t 0 t e i a 0 t s 400 300 200 100 0 100 200 1.S hp c 2 p.u. ful.l. loac1 with saturation t ... ____ __ __ ____ __ o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 seconds 70 Figure volts load, E.11 Computer simulated plot of time: 1.5 HP, 2 per with saturation. motor terminal m 0 t 0 t e m a 0 t s vs. and 300 200 100 0 100 200 unit capacitor, full 1.S hp C ,. 3 p.u. full. l.oac1 with saturation _____ ______ ________ L ____ .. __ _______ ______ .. ... .. .. _._.,.. 300 t : _____ _____ _____ ____ _____ _L _____ o 0.1 0.3 Figure volts load, 0.4 0.5 0.6 0.7 0.8 0.9 1 seconds B.12 Computer simulated plot of motor terminal vs. and time: 1.5 HP, 3 per with saturation. unit capaCitor, full
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500.000 ms 0.00000 s 11)0 ms/div 500.000 0.00000 s 100 Rlsidl ... 500.000 rus 71
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P P e d U I) O()U , : 0.00000 s 100 m s/di'l 500.00 0 ms 72 1 : 1 : 0 50:)( 000 s I ms/div 5 0 0 .000 !Os
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APPENDIX F MISCELLANEOUS COMPUTER SIMULATED PLOTS FOR THE 1.5 HP MOTOR
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40 35 e e 30 c t 25 0 a 20 n e 15 t i c 10 t 0 5 u e 0 5 10 0 1.5 hp no capacitor no load with saturation _______ __ ______ ________ _______ 4 _______ ______ __ ________ _______ _______ J ______ 200 400 600 800 1000 1200 1400 1600 1800 RPM 2000 e 1 e c 0 a e t c 0 e 45 40 35 30 25 20 15 10 5 1.5 hp no capacitor full load with saturation .. .. ro 200 o 200 400 600 800 1000 1200 1400 1600 1800 RPM
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_m 1.5 hp saturation curve 250 200 150 100 50 .. _ _ ..... _ .. ............ .. 1_______________ .. _____ 50 100 150 200 250 300 350 400 450 Fmq FiqureF.3 Plot of the 1.5 HP motor saturation curve. e 1 e c o a e t c o e 1.5 hp no capacitor no load with saturation 35 .. .. .. 30 .. .. .... ............ .. .. .. .... .. .. .. ... .. ... .. .. 25 .... .... .... .. .. .. .. 20 15 r .. .. .. .. .. .. :.. .. .. r .... ...... .... .... .... 10 : r ls .. .. .. .... .... + .. .. ...... .. .. _ .... .... .. ................ .... .. .. .... ...... ...... _ .... o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 seconds r.4 Computer simulated 1.5 HP, no capacitor, Figure time: vs. saturation. plot of motor no load, and torque with
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45 40 e 1 e 35 c t r 30 0 a 25 n e 20 t i c 15 t 0 10 r u 5 e 0 5 0 1.5 hp no capacitor full load with saturation __ _____ ___ .. ___ A' ,rT,,r,,___ ... .. .. .. .. L _______ J4 ______ J ___ ___ L ____ .. __ WL 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 seconds Figure F.S Computer simulated vs. time: 1.5 HP, no capacitor, saturation. plot full of motor torque load, and with e 1 e c t o a e c t o e 1.5 hp c p.u. no load with saturation 35 ,.. 30 ______ ___ 25 20 _. _____ _________ .. ________ .. __ .. __ .. .. J __________ ___ .. _____ J ________ 15 10 .. .. :.. .... 5 .. .... r.. ,.. .. rr,o ... .. '" 1;: 1J 5 r.. .. .. .. .. .. .. o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 seconds Figure vs. time: simulated plot of per unit capacitor, F.6 Computer 1.5 HP, saturation. 1 motor torque no load, and with 76
PAGE 90
45 40 e 1 e 35 c t 30 0 a 25 e 20 t c 15 t 0 10 e 5 0 5 0 1.5 hp c 1 p.u. full load with saturation rt _____ .. _____ 6 _______ ________ _______ __ ____ 4 _______ ________ _______ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 seconds 1.5 hp no capacitor no load with saturation 1800 o t o 1600 ..1400 s P l200 e e d 1000 p m 800 400 200 ________ __ o 0.05 0.1 0.15 0.2 0.25 seconds 77
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o o s p e e 1.5 hp no capacitor full load with saturation 1600 1400 1200 1000 900 600 400 200 o t o 0.05 0.1 0.15 0.2 0.25 seoonds 1 c e 1.5 hp no oapacitor no load with saturatl.on 100 50 0 t 50 o 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 seconds
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1 i n e c u r r e n t a p e r e s no saturation 1.5 hp no capacitor full load 100 50 0 SO o 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 seconds 79 1 e c e 1.5 hp no capacitor full load with saturation 100 50 0 50 o 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 seconds
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1 e c 1.5 hp c p.u. no load with saturation 100 _________ _________ 4 __________ _________ _________ ________ 50 0 e 50 ................... : .................... .................. .................. ................ .. 100 __ __________ _________ p _________ t o 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 seconds Figure F.13 Computer simulated plot of motor line current VS. time: 1.S HP, 1 per unit capacitor, no load, and with saturation. 1 e c e 1.5 hp c p.u. full load with saturation 100 50 100 .. .. .. ...... .. .. .. .. .. ...... .. ...... .... .. __ .. .. __ .. _____ 1... .. .. .. .... ____ __ ____ .. _1 ________ o 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 seconds Figure F.14 current VS. full load, Computer simulated plot of motor time: 1.S HP, 1 per and with unit capacitor, line 80
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e c e 1.5 hp c 1 p.u. full load no saturation 100 so o so 100 to ...................................... t i"" .............. '"':.................. i"'" ............ .. ...... 1... ___ .. .. ___ .... ____ 1 ___ ___ ..... o 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 seconds Figure F.1S vs. load, and Computer simulated plot of motor time: 1.5 HP, 1 per with no saturation. unit capacitor, line full 81
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APPENDIX G COMPUTER SIMULATED PLOTS FOR A 3 AND 50 HP MOTOR
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400 300 m 0 t 0 200 e 100 m 0 a 100 0 s 300 400 0 3 hp C100e6 (1 p.u.) o o o no load no saturation o o 0 o 0 o o 0 1.6 seconds 83 Figure volts and no G.1 Computer vs. timer 3 HP, saturati"on. simulated plot of motor 1 per unit capacitor, no terminal load, 0 0 a o s 3 hp C200e6 (2 p.u.) no load no saturation o 0 600 o 0 0 o 0 0 o o 0 400 o 0 0 o 0 0 400 ... o o o 0 600 ..... ... ..... ... :_ ....... .o 0 o o 0 o o o 0 o o o o ... o 0 o 0 o 0 o 0.2 0.4 0.6 0.8 1.2 1.4 1.6 seconds Figure volts and no G.2 Computer vs. time: 3 HP, saturation. plot of motor unit capacitor, no simulated 2 per terminal load,
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0 t 0 t e a 1. o 1. s 84 3 hp C300e6 (3 p.u.) no 1.oacl no saturation 600 ________ 4 __________ _________ __ ________ _________ _________ 4 ________ 400 r,,r,400 600 .. .... _ .. :ro 0.2 0.4 0.6 0.8 1 seconcls Figure volts and no G.3 Computer vs. time: 3 HP, saturation. simulated plot of motor 3 per unit capacitor, terminal no load, 200 150 1. 100 e c 50 e 0 50 100 3 hp no capacitor fu1.1. 1.oacl no saturation ______ ___ .. ______ _________ _________ __________ _________ L _________ ________ 1_: .. ... Figure current with no o 0.1 0.2 0.3 0.4 0.5 G.4 simulated plot no capacitor, Computer vs. time: 3 HP, saturation. 0.6 0.7 0.8 of motor full line load, and
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e 1 e c t o a e c o q e 3 hp no capacitor no load no saturation 100 80 60 40 20 o o o o o .. o o o o o o o o o 0 0 o 0 o 0 0 o o o o o o o o o .. o 0 0 o 0 : 0 o 0 I 1.1 I o 0 o o o 0 o 0 0 o o I 1 I I o 0 85 o 200 400 600 800 1000 1200 1400 1600 1800 RPM e e c t o m a e t c o q e no capacitor 3 hp full load no saturation 140 120 100 80 60 40 20 o 20 o o o o o 0 o o o 0 0 0 .. 0 0 0 0 0 0 0 0 0 0 ,0 ,o 0 0 o 0 0 o 0 0 o 0 0 o 0 0 o 0 0 o 0 0 o 0 0 o 0 o 0 o 0 o o 0 o 0 0 o 0 o 0 ... ... .200 o 200 400 600 800 1000 1200 1400 1600 1800 RPM
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86 50 C200e6 (1 p.u.) no load m 0 t 0 t e m n a 200 0 t s 400 2.5 3 o 0.5 1.5 secondB Pigure G.7 Computer simulated plot of. motor terminal volts vs. 50 8P, 1 per unit capacitor, no load, and with no saturation. SO hp C400e6 (2 p.u.) no load no saturation 2000 m o 1500 ...... ,. ...... o 1000 ............ ,. .. e m a 500 1000 s 1500 2000 3 o 0.5 1.5 2.5 seconds Figure G.B Computer simulated plot of motor terminal volts vs. time; 50 8P, 2 per unit capacitor, no load, and with no saturation.
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50 hp C600e6 (3 p.u.) no load no saturation o 0.5 1.5 2 2.5 3 seconds 87 Figure G.9 Computer volts vs. timer 50 HP, and with no saturation. simulated plot 3 per unit of motor capacitor, terminal no load, 50 hp no capacitor t'ull load no saturation e c e 1000 500 o 500 Figure current with no o 0.2 0.4 0.6 0.8 1 seconds G.I0 Computer vs. times 50 HP, saturation. simulated plot no capacitor, 1.2 1.4 1.6 line and of motor full load,
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e e c o a e t c o e SO hp no capacitor no no saturation 1500 1000 500 o 500 t o 200 400 600 800 1000 1200 1400 1600 1800 2000 RPM Figure G.ll Computer simulated motor speed/torque curve showing no load, and no saturation. plot free of the 50 HP e e c 0 m a e c 0 e 50 hp no capacitor no saturation 1500 1000 500 0 500 ____ e. ............. ......... 1800 200 o 200 400 600 800 1000 1200 1400 1600 RPM Figure G.12 Computer simulated plot of the 50 HP motor speed/torque curve showing free acceleration; full load, and with no saturation. 88
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APPENDIX H The following is a list of the motor data for the motor used in this paper. US MOTORS UNIMOUNT 125 1.5 HP volts, wye conected Full load amps @ service factor 40 degree ambient, continuous duty Totally enclosed fan cooled (TEFC) Code letter K 1.25 service factor Nema design B Model F036 (JS87 P63CSD2911) Rotor moment of inertia = 0.OS09 Ibft2 Equivalent circuit parameters: stator resistance (rs) s 6.52 Q/phase rotor resistance (rr) = 4.5 Q/phase stator leakage reactance (XIs) = 6.55 Q/phase rotor leakage reactance (Xlr) = 6.55Q/phase magnitizing reactance (X 42 Q/phase (determined from open circuit test)
