Mathematical modeling of electrical machines using circle diagram

341 views

Published on

Published in: Technology, Business
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
341
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
16
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Mathematical modeling of electrical machines using circle diagram

  1. 1. International Journal of Electronics and Communication Engineering & Technology (IJECET), INTERNATIONAL JOURNAL OF ELECTRONICS AND ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME COMMUNICATION ENGINEERING & TECHNOLOGY (IJECET) ISSN 0976 – 6464(Print) ISSN 0976 – 6472(Online) Special Issue (November, 2013), pp. 173-181 © IAEME: www.iaeme.com/ijecet.asp Journal Impact Factor (2013): 5.8896 (Calculated by GISI) www.jifactor.com IJECET ©IAEME Mathematical Modeling of Electrical Machines using Circle Diagram R S Shekhawat Assistant Professor, Electrical Engineering Department, BKBIET, Pilani, Rajasthan, India rssraaz@gmail.com ABSTRACT: The circle diagram can be drawn for alternators, synchronous motors, transformers, induction motors. To perform no load and blocked rotor test on a machines and to determine the parameters of its equivalent circuits. The circle diagram helps to identify Max. Torques , Current , slip, power factor, Efficiency and electromagnetic values, which describe the machine’s operation mode at different slip values and gives them a better description when machine’s operating mode is changing. The circle diagram is also of great importance for studying the case, when the parameters of the asynchronous machines operating mode are assumed to be constant values. The circle diagram is the graphical representation of the performance of the electrical machine drawn in terms of the locus of the machine's input voltage and current. KEYWORDS: Circle diagram, electrical machines, mathematical modeling. I. INTRODUCTION In this paper, it will be shown that the performance characteristics of a motor are derivable from a circular locus. The data necessary to draw the circle diagram may be found from noload, blocked-rotor tests and resistance test, corresponding to the open-circuit and shortcircuit tests of a transformer. The stator and rotor Cu losses can be separated by drawing a torque line. The parameters of the motor, in the equivalent circuit, can be found from the above tests, as shown below. II. CIRCLE DIAGRAM FOR A SERIES CIRCUIT It will be shown that the end of the current vector for a series circuit with constant reactance and voltage, but with a variable resistance is a circle. With reference to Figure, it is clear that I= V V V X V = = × = Sinφ Z √R + X X √R + X X X Sinφ = √R + X International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 173
  2. 2. International Journal of Electronics and Communication Engineering & Technology (IJECET), ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME ∴I= V Sinφ X Fig. 1: (a) Equivalent circuit diagram (b) Impedance triangle. Tracing of polar curves: r=f(Φ ) I=V/X Sin Ф (r=f(θ); r = a Sin θ) Following procedure is employed for tracing these curves: A. Symmetry: The symmetry of this curve is decided by  If f (- θ) = f (θ) f(Ф)=V/X Sin Ф f(- Ф)= V/X Sin (-Ф) f(Ф) ≠ f(- Ф) The curve is not symmetrical about the initial line θ=0.  If f(π- θ)=f(θ) f(Ф)=V/X Sin Ф f(π - Ф)= V/X Sin (π -Ф)= V/X Sin Ф f(Ф) = f(π - Ф) The curve is symmetrical about the line θ= π/2.  If the equation of curve remains unchanged on replacing r by –r, then curve is symmetrical about the pole (r=0). I=V/X Sin Ф (r=a Sin θ) if on replacing of I by-I then curve is changed so curve is not symmetrical about the pole(r=0). B. Position of pole: If there exists values of θ (Say θ=α) for which f(α)=0 then curve passes through the pole and the line θ=α is tangent at pole. f(α)=V/X Sin α=0 Sin α=0; α=Sin-1(0)=0, π,2 π,3 π…… C. Asymptotes: If lim → = ∞, then there exists an asymptotes and is given by r sin(θ-α)=f’(θ). Where θ=α is the root of equation f(α)=0. Curve has no Asymptotes as r→∞ for any Ф. International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 174
  3. 3. International Journal of Electronics and Communication Engineering & Technology (IJECET), ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME D. Points on curve: Tabulate the value of Ф for various values of I laying between (0, 2π). Ф I 0˚ 0 30˚ V/2X 45˚ V/√2X 60˚ √3V/2X 90˚ V/X 120˚ √3V/2X 135˚ V/√2X 150˚ V/2X 180˚ 0 E. Region of existence of curve: Find the value of Ф for which I is imaginary, which will give the angular region in which no problem of curves lies. No part of the curve lies in this region. F. Direction of tangent: the slope of tangent is given by tan ∅ = = / ; Ф =θ. G. Loop: The curve containing the terms like sin nθ or cos nθ passes loop n or 2n in Number. n=1 in V/X sinФ loop is one. H. Region of extent: Find the limit of I and Φ   Find the greatest numerical value of I, so that the curve lies within the radius I. Find the value of Φ for which I is imaginary. This gives us an idea of region in to which the curve does not extent. I. Variation in r: Trace the variation of I for some values of Φ.   If dI/dΦ >0 then I increases as Φ increases. If dI/dΦ <0 then I decrease as Φ increases. Fig. 2: (a) Circle diagram with lagging current (b) Circle diagram with leading current. In other words, as resistance R is varied (which means, in fact, Φ is changed), the end of the current vector lies on a circle with diameter equal to V/X. For a lagging current, it is usual to orientate the circle of fig. (a) such that its diameter is horizontal and the voltage vector takes a vertical position, as shown in fig. (b). There is no difference between the two so far as the magnitude and phase relationships are concerned. III. CIRCLE DIAGRAM FOR THE APPROXIMATE EQUIVALENT CIRCUIT The approximate equivalent diagram is redrawn in Fig. It is clear that the circuit to the right of point’s ab is similar to a series circuit, having a constant voltage V1 and reactance X01 but International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 175
  4. 4. International Journal of Electronics and Communication Engineering & Technology (IJECET), ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME variable resistance (corresponding to different values of slip s). Hence, the end of current vector for I2′ will lie on a circle with a diameter of V/X01. In Fig, I2′ is the rotor current referred to stator, I0 is no-load current (or exciting current) and I1 is the total stator current and is the vector sum of the first two. When I2′ is lagging and Φ2 = 90˚, then the position of vector for I2′ will be along OC i.e. at right angles to the voltage vector OE. For any other value of Φ2, point A will move along the circle shown dotted. The exciting current I0 is drawn lagging V by an angle Φ0. If conductance G0 and susceptance B0 of the exciting circuit are assumed constant, then I0 and Φ0 are also constant. The end of current vector for I1 is also seen to lie on another circle which is displaced from the dotted circle by an amount I0. Its diameter is still V/X01 and is parallel to the horizontal axis OC. Hence, we find that if an induction motor is tested at various loads, the locus of the end of the vector for the current (drawn by it) is a circle. Fig. 3: Approximate equivalent circuit. IV. CONSTRUCTION OF CIRCLE DIAGRAM Conduct No load test and blocked rotor test on the induction motor and find out the per phase values of no load current I0, short circuit current ISC and the corresponding phase angles Φ0 and ΦSC. Also find short circuit current ISN corresponding to normal supply voltage. With this data, the circle diagram can be drawn as follows. 1. With suitable scale, draw vector OA with length corresponding to I0 at an angle Φ0 from the vertical axis. Draw a horizontal line AB. 2. Draw OS equal to ISN at an angle ΦSC and join AS. 3. Draw the perpendicular bisector to AS to meet the horizontal line AB at C. 4. With C as centre, draw a portion of circle passing through A and S. This forms the circle diagram which is the locus of the input current. 5. From point S, draw a vertical line SL to meet the line AB. 6. Divide SL at point K so that SK: KL = rotor resistance: stator resistance. International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 176
  5. 5. International Journal of Electronics and Communication Engineering & Technology (IJECET), ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME Location of K  Squirrel-cage Rotor. Stator resistance/phase i.e. R1 is found from stator-resistance test. Now, the short circuit motor input Ws is approximately equal to motor Cu losses (neglecting iron losses). =  =3 = −3 −3 3 Wound Rotor. In this case, rotor and stator resistances per phase R2 and R1 can be easily computed. For any values of stator and rotor currents I1 and I2 respectively, we can write = = = × 1 = ⁄ = = = ℎ = ℎ Value of K may be found from short-circuit test itself by using two ammeters, both in stator and rotor circuits. Let us assume that the motor is running and taking a current OP Then, the perpendicular GD represents fixed losses, GF is stator Cu loss, PF is the rotor input, EF is rotor Cu loss, PE is rotor output and PD is the total motor input. From our knowledge of the relations between the above-given various quantities, we can write: √3.VL.PD=Motor input. √3.VL.GD=Fixed losses. √3.VL.GF=Stator Cu losses. √3.VL.EF=Rotor Cu losses. √3.VL.ED=Total losses. √3.VL.PE=Mechanical output. √3.VL.PF=Rotor input α Torque. PE Mechanical output = = Ef icency PD Motor input = = = =1− = = = ( ) ℎ . International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 177
  6. 6. International Journal of Electronics and Communication Engineering & Technology (IJECET), ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME 7. For a given operating point P, draw a vertical line PEFGD as shown. Then PE = output power, EF = rotor copper loss, FG = stator copper loss, GD = constant loss (iron loss + mechanical loss) 8. Maximum values  Maximum output: It occurs at point U where the tangent is parallel to the output line AS. Point U may be located by drawing a line CU from point C such that it is perpendicular to the output line AS maximum output is represented by the vertical UM.  Maximum torque or rotor input: It occurs at point V where the tangent is parallel to the output line AK. Point V may be located by drawing a line CV from point C such that it is perpendicular to the output line AK maximum torque is represented by the vertical VN. Maximum torques is also known as stalling or pulls out torque.  Maximum input power: It occurs at the highest point of the circle i.e. at point W where the tangent to the circle is horizontal. It is proportional to WT. As the point W is beyond the point of maximum torque the motor is unstable here. Fig. 4: Circle diagram of equivalent circuit International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 178
  7. 7. International Journal of Electronics and Communication Engineering & Technology (IJECET), ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME Efficiency line  The output line AS is extended backwards to meet the X-axis at O′.  From any convenient point on the extended output line, draw a horizontal line QT so as to meet the vertical from O′. Divide the line QT into 100 equal parts.  To find the efficiency corresponding to any operating point P, draw a line from O′ to the efficiency line through P to meet the efficiency line at T1. Now QT1 is the efficiency. Fig. 5: Circle diagram for efficiency. Slip Line  Draw line QR parallel to the torque line, meeting the vertical through A at R. Divide RQ into 100 equal parts.  To find the slip corresponding to any operating point P, draw a line from A to the slip line through P to meet the slip line at R1. Now RR1 is the slip.  To find the slip corresponding to point at maximum power point, draw a line from A to Pm to meet the slip line at SPmax. Now RSPmax is the slip.  To find the slip corresponding to point at maximum torque point, draw a line from A to Tm to meet the slip line at STmax. Now RSTmax is the slip. International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 179
  8. 8. International Journal of Electronics and Communication Engineering & Technology (IJECET), ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME Fig. 6: Circle diagram of slip. Power Factor Curve  Draw a quadrant of a circle with O as centre and any convenient radius. Divide OCm into 100 equal parts.  To find power factor corresponding to P, extend the line OP to meet the power factor curve at C′. Draw a horizontal line C′C1 to meet the vertical axis at C1. Now OC1 represents power factor. Fig. 7: Circle diagram of power factor International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 180
  9. 9. International Journal of Electronics and Communication Engineering & Technology (IJECET), ISSN 0976 – 6464(Print), ISSN 0976 – 6472(Online), Special Issue (November, 2013), © IAEME V. CONCLUSION Electrical machines circle diagram is a graphical representation of its equivalent circuit. This means that whatever information can be obtained from the equivalent circuit, the same can also be obtained from the circle diagram. The advantages of a circle diagram are its simplicity and quick estimation of the machines operating characteristics. Though a computer can rapidly provide the performance characteristics of a machine from its equivalent circuit, yet the graphical solution through the use of circle diagram is even now popular in several organizations. It is because a circle diagram gives the results which are sufficient accurate for practical purpose, despite the fact that an approximate equivalent circuit is used in a circle diagram and provides a panoramic view of how operating characteristics are affected by changes in the machines parameters, voltage, frequency etc. REFERENCES [1] Behrend, B.A. (1921). The Induction Motor and Other Alternating Current Motors, their Theory and Principles of Design. McGraw-Hill. p. ix. Retrieved 4 January, 2013. [2] Heyland, A. (1894). "A Graphical Method for the Prediction of Power Transformers and Polyphase Motors)". ETZ 15. pp. 561–564. Retrieved 4 January 2013. [3]Terman, Frederick Emmons; Freedman, Cecil Louis; Lenzen, Theodore Louis ; Rogers, Kenneth Alfred (Jan. 1930). The General Circle Diagram of Electrical Machinery 49. American Institute of Electrical Engineers, Transactions of the. [4] S. K. Bhattacharya. Electrical Machines (2008 edition). Tata McGraw-Hill Education. p. 359. ISBN 0-07-066921-X. [5] Heyland, Alexander; Translated by G.H. Rowe and R.E. Hellmund (1906). A Graphical Treatment of the Induction Motor. McGraw Publishing Company. Retrieved 10 January 2013. [6] Phase to Phase BV (2006). "The Asynchronous Motor Model". pp. 5–6. Retrieved 10 January 2013. [7] Alger, Philip L. et al. (1949). "'Induction Machines' subsec. of sec. 7 - Alternating-Current Generators and Motors". In A.E. Knowlton (Ed.). Standard Handbook for Electrical Engineers (8th ed.). McGraw-Hill. pp. 710–711. [8] Fernandez, Francis M. "Construction of Circle Diagram". College of Engineering Trivandrum. Retrieved 10 January 2013. [9] http://www.naeeesfutmx.com/books/Theraja/part-2/ch-35.pdf BIOGRAPHY R S Shekhawat is Assistant professor, Department of Electrical Engineering at B K Birla Institute of Engineering & Technology; Pilani (Rajasthan) He had an experience of more than 6 year in teaching to undergraduate and diploma classes in the areas of electrical machines, transmission networks, utilization of energy, switchgear and protection, high voltage engineering, advance network analysis, economic operation of power system computer method of power system and machines etc. he had contribute papers in national and international journals. He was born in Pilani, Rajasthan, India. He received the B. E. degree in Electrical Engineering from University of Rajathan, Jaipur (Rajsthan), India in 2007. He received the M. Tech. degree in Power System (Electrical Engineering) from Singhania University, Jhunjhunu (Rajsthan), India in 2012. He is pursuing his P. hd. in Electrical Engineering. His current research interests focus on nonconventional energy sources. International Conference on Communication Systems (ICCS-2013) B K Birla Institute of Engineering & Technology (BKBIET), Pilani, India October 18-20, 2013 Page 181

×