Electrical Power Systems

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Bibliography Bibliography 465 1. O.I. Elgard, “Electric Energy Systems Theory: An Introduction”, McGraw-Hill, New York, 1982. 2. T. Gonen, “Electric Power Distribution System Engineering”, McGraw-Hill, New York, 1986. 3. T. Gonen, “Electric Power Transmission System Engineering: Analysis and Design”, John Wiley & Sons, New York, 1988. 4. P. Kundur, “Power System Stability and Control”, McGraw-Hill, New York, 1994. 5. C.L. Wadhwa, “Electrical Power Systems”, New Age International, New Delhi, 2000. 6. J.D. Glover and M.S. Sarma, “Power System Analysis and Design”, Brooks/Cole, Pacific Grove, U.S.A., 2002. 7. W.D. Stevenson, Jr., “Elements of Power System Analysis”, McGraw-Hill, New York, 1982. 8. B.M. Weedy and B.J. Cory, “Electric Power Systems”, John Wiley & Sons, New York, 2002. 9. General Electric Company, “Electric Utility Systems and Practices”, Wiley, New York, 1983. 10. C.A. Gross, “Power System Analysis”, Wiley, New York, 1979. 11. M.V. Deshpande, “Electrical Power System Design”, Tata McGraw-Hill, New Delhi, 1984. 12. R.L. Sullivan, “Power System Planning”, McGraw-Hill, New York, 1977. 13. A.T. Starr, “Generation, Transmission and Utilization of Electric Power”, Pitman, 1962. 14. H. Cotton and H. Barber, “The Transmission and Distribution of Electrical Energy”, Hodder and Stoughton, 1970. 15. I.J. Nagrath and D.P. Kothari, “Power System Engineering”, Tata McGraw-Hill, New Delhi, 1994. 16. A.R. Bergen and V. Vittal, “Power System Analysis”, Pearson Education, Singapore, 2001. 17. M.A. Pai, “Computer Techniques in Power System Analysis”, Tata McGraw-Hill, New Delhi, 1979. 18. A.J. Wood and B.. Wollenberg, “Power Generation Operation and Control”, John Wiley & Sons, New York, 1996. 19. A.S. Prabla, “Electric Power Distribution”, Tata McGraw-Hill, New Delhi, 2004. 20. S.C. Tripathy, “Electric Energy Utilization and Conservation”, Tata McGraw-Hill, New Delhi, 1991. 21. J. Arrillaga and C.P. Arnold, “Computer Analysis of Power System”, John Wiley & Sons, New York, 1990. 22. J.A. Harrison, “The Essence of Electric Power Systems”, Prentice-Hall of India, New Delhi, 1997. 23. Y. Wallach, “Calculations and Programs for Power System Networks”, Prentice-Hall, New Jersey, 1986. 24. Z.A. Yamayee, “Electromechanical Energy Devices and Power Systems”, John Wiley & Sons, New York, 1994. 25. R.D. Shultz and R.A. Smith, “Introduction to Electric Power Engineering”, Harper & Row Publishers, New York, 1985.
466 Electrical Power Systems 26. H.M. Rustebakke, “Electric Utility Systems and Practices”, John Wiley & Sons, New York, 1983. 27. M.E. El-Hawary, “Electrical Power Systems Design and Analysis”, Reston Publishing Company, Virginia, 1983. 28. A.E. Guile and W. Paterson, “Electrical Power Systems”, Oliver & Boyd, Edinburgh, 1969. 29. H. Sadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. 30. J. Kumar, K.H. Ng and G. Sheble, “AGC Simulator for Price Based Operation Part I and Part II”, IEEE Transactions on Power Systems, Vol. 12, May 1997, pp. 527-538. 31. B.H. Bakken and O.S. Grande, “Automatic Generation Control in Deregulated Power System”, IEEE Transactions on Power Systems, Vol. 14, October 1999, pp. 1495-1503. 32. R.D. Christie and A. Bose, “Load requency Control Issues in Power System Operation After Deregulation”, IEEE Transactions on Power Systems, Vol. 11, August 1996, pp. 1191-1200. 33. V. Donde, M.A. Pai and I.A. Hiskens, “Simulation and Optimization in An AGC System After Deregulation”, IEEE Transactions on Power Systems, Vol. 16, August 2001, pp. 481- 489.
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To My Wife Shanta Son Debojyoti and
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Preface During the last fifty years
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Contents Preface vii 1. Structure o
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Contents xi 9. Symmetrical Componen
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Structure of Power Systems and ew O
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1.2 REASONS OR INTERCONNECTION Stru
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Case-1: If P 1 = P 2 = P 3 = ... =
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Load factor, L = Maximum load = 3 M
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\ LD = 1 MW (c) rom eqn.(1.13), coi
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Case-2: Very short lasting peak. He
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1.9 DISADVANTAGES O LOW POWER ACTOR
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Resistance and Inductance of Transm
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2.4.1 Internal Inductance igure 2.1
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Therefore, we can write L 1 = L 11
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L a = 0.4605 n log R S| T| Resistan
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L = 0.4605 L Resistance and Inducta
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=(2Dd 1 · 2Dd 1) 1/4 = (2 Dd 1) 1/
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Solution. The distances from strand
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Solution: Using eqn. (2.14), Hx = I
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= 0.4605 log 6611 . 04 . Resistance
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Since from symmetry D b1= D b2, l 1
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\ D m = D D D D Resistance and Indu
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dg 1 2 \ Dsa = 0. 7788 ´ 0. 015 ´
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Capacitance of Transmission Lines 5
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V Ki = 1 pÎ N å q m 2 o m = 1 3.3
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Similarly for the 2nd transposition
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C an = log R S| T| 00242 . 1 1 1 1
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\ V12 = q ln pÎo \ C 12 = q \ C 12
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I chg = jw C anV LN \ |I chg| = w C
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Line length is 200 km. \ Can (total
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Now applying eqn. (3.7), we have C
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= 0.008993 m/km. Capacitance of Tra
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Synchronous Machine: Steady State a
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\ V = 138 Using eqn. (4.18), . 3 =
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om ig. 4.6, Synchronous Machine: St
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\ i asy max = from which Synchronou
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Postfault voltage Synchronous Machi
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or the reference phase a, E a = (Z
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5.3 THE PER-UNIT (pu) SYSTEM Power
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Power System Components and Per Uni
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Z L (pu) = Z L (ohm) (. 08+ j 03 .)
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\ VL (pu) = 0. 2033 - 76. 68° pu P
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Base impedance for the load is ( )
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Now we can write, \ |I| cos f = P |
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This is a simple series circuit. Th
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Eqns (6.15) and (6.17) can be writt
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Characteristics and Performance of
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om eqn. (6.6), Characteristics and
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as: \ |V S| = 145.13 kV. \ Sending
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Sending end power factor angle = 8.
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\ I R = V R = 208 Characteristics a
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6.6 VOLTAGE WAVES Characteristics a
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6.9 ERRANTI EECT Characteristics an
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Load low Analysis 7.1 INTRODUCTION
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Applying KCL to the independent nod
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y 10 = y 20 = y 30 = y13 ¢ y12 ¢
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\ V i = 1 Y ii L NM P - jQ i i * Vi
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n 2 i ii ii ik i k ik k i \ Pi - jQ
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Load low Analysis 157 ig. 7.6: P-re
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Load low Analysis 159 are permitted
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1 1 y13 = y31 = = = ( 10 - j30) Z (
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(1) \ V3 = 10011 . - 2. 06° After
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Load low Analysis 165 Using eqn. (7
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Load low Analysis 167 \ Q 2 = -|V 2
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|V 1| = 1.05,|V 2| = 1.0,|V 3| (1)
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\ L NM e j ( 0) ( 0) ( 0) 1 1 1 2 n
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( p) p The terms DPi and DQi Load l
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Load low Analysis 175 P2 = -35.77
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(1) P2( cal ) = -2.62 (1) P3( cal )
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Load low Analysis 179 Use deoupled
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0 Dd2 \ Dd3 0 ( ) ( ) NM \ 0 Dd2 L
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Assuming q ik - d i + d k » d ik,
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Table 7.4: Line impedances BUS code
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Symmetrical ault 187 is drawing 10
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\ X TH = ig. 8.4: Thevenin equivale
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ig. 8.5: Circuit diagram of Example
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= | Vo| | Vo| | Z| ´ × (MVA) Base
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\ 1 - V 1f = j0.14 × (-j4.165) \ V
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Base current I B = Per unit reactan
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Symmetrical ault 199 Example 8.9: T
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Solution: Let Base MVA = 12 Base Vo
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Solution: Let Base MVA = 100 Base V
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Therefore, current to be interrupte
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Circuit model under fault condition
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Symmetrical ault 209 Example 8.16:
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8.4 SHORT CIRCUIT ANALYSIS OR LARGE
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om equations. (8.9) and (8.10), we
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Using equation (8.14), Z BUS = L NM
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Symmetrical ault 217 Solution: Inje
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Hence, new Z = BUS 8.6.3 Type-3 Mod
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or L V 1 V M NM 2 V 0 n O QP = L ol
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Solution: (a) Using equation (8.11)
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Symmetrical ault 225 8.7 ig. 8.31 s
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has the following properties R S| T
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T stands for transpose. Using eqn.
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Using eqn.( 2.46) as given in Chapt
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Also I n = I a + I b + I c Using eq
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Symmetrical Components 235 ig. 9.4:
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Symmetrical Components 237 (b) - co
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Solution: I a + I b + I c = 0, I a
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Also note that I ab1 = I a1 3 30°
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Base voltage of transmission line 1
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Symmetrical Components 245 Example
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Symmetrical Components 247 Example
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9.5 Draw the zero-sequence network
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Unbalanced ault Analysis 251 Assumi
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The symmetrical components of the f
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Unbalanced ault Analysis 255 The bo
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ig. 10.9: Two conductors open. Unba
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Z 1 = j0.691 0.23 ´ 0. 691+ 0. 23
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Unbalanced ault Analysis 261 Exampl
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\ 3 ´ 1.0 ´ j0.4 j0.3 ´ j0.4 + j
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Denominator of eqns. (i) and (ii) Z
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Hence, rom eqn. (i), V b = V c Unba
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ig. 10.18 shows the sequence networ
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\ V b = E a = 3752.8 0° 1763 . - 3
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Unbalanced ault Analysis 273 ig. 10
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Unbalanced ault Analysis 275 10.4 A
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where J = moment of inertia of roto
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Power System Stability 279 Pi - Pe
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Power System Stability 281 (a) ind
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On a per-phase basis, power at the
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x eq = 0.25 + 0.15 + 02 . ´ 02 . 0
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Power System Stability 287 system.
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Power System Stability 289 or P i (
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\ d1 Pi (dc - d0 ) = z Pmax dc sin
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dcr z 1 d0 cr zbPi-Bgdd d dm = C -
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We know, K1 = Pmax In ig. 11.14, P
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Solution: Prefault operation x A =
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\ cosd cr = 0.654 Power System Stab
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Similarly, the change in the power
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d (2) = d (1) + Dd (2) = 17.855 + 1
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ig. 11.19: Sample network of 11.4.
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Automatic Generation Control: Conve
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12.4 ISOCHRONOUS GOVERNOR Automatic
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Where D = PL = constant. f Theref
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\ T p = 32 3 sec. Automatic Generat
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om ig. 12.20, state-variable equati
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The area control error for area-1 a
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Automatic Generation Control in a R
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Corona 357 air around the conductor
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14.4 POTENTIAL GRADIENT OR THREE-PH
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Similarly, G b = G c = r r V bn ln
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Corona 363 In the above expressions
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P c = 2.1f HG log 10 Vn DI HG r K
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Solution: rom eqn. (14.38), air den
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\ V 0 = 110.15 KV(rms) V n = 220 3
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\ W = 55 ´ b69. 28 - 57g 2 635 . -
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Analysis of Sag and Tension 15.1 IN
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where Dl = l0. DT MA DT = T1 - T0 T
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when x = L l , s = 2 2 , \ l 2 = H
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or approximately, 2 d = wL 8H ig. 1
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Squaring eqn. (15.37), we get, s 2
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under the action of T, H and wx. or
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\ T max = 572.59 kg (d) rom eqn. (1
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ig. 15.6: Case of negative x 1. Ana
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where Analysis of Sag and Tension 3
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or 1-meter length of conductor, Ana
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(f ) Vertical sag = dcos q cos q =
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Analysis of Sag and Tension 395 = 0
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Analysis of Sag and Tension 397 Ass
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Analysis of Sag and Tension 399 Dis
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Total weight of the conductor = wl
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Analysis of Sag and Tension 403 Whe
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Optimal System Operation 16.1 INTRO
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Optimal System Operation 407 where
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Optimal System Operation 409 2. The
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Complete algorithm is given below:
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DP g1 = Î 2 DP g2 = Î 2 Using Tay
Page 428 and 429: Optimal System Operation 415 Equati
Page 430 and 431: \ 2 C2 = 0.18 Pg2 + 32 Pg2 + K2 Whe
Page 432 and 433: Optimal System Operation 419 Equati
Page 434 and 435: DP L = rom eqns (16.36) and (16.39)
Page 436 and 437: Optimal System Operation 423 Now we
Page 438 and 439: Optimal System Operation 425 The pr
Page 440 and 441: Example 16.6: Consider Ex-16.5, wit
Page 442 and 443: \ DP g (l) = 264.1657 MW rom eqn. (
Page 444 and 445: \ (4) 2 2 P = 0.00005 × (188.95) +
Page 446 and 447: Optimal System Operation 433 The te
Page 448 and 449: \ Also d P 3 d 2 P 3 3 =- G 32 si
Page 450 and 451: Let the current in line K be I K1.
Page 452 and 453: + HG I K Optimal System Operation
Page 454 and 455: Now, \ V1 = 1.198 14. 5º, \d1 = 14
Page 456 and 457: Optimal System Operation 443 B 11 =
Page 458 and 459: Z 1 = (0.03 + j0.12)pu; I 1 = (1.3
Page 460 and 461: Objective Questions Objective Quest
Page 462 and 463: Objective Questions 449 26. or a de
Page 464 and 465: Objective Questions 451 56. Peak lo
Page 466 and 467: 82. Inductance of a conductor due t
Page 468 and 469: Objective Questions 455 108. In ter
Page 470 and 471: Objective Questions 457 137. In a p
Page 472 and 473: Objective Questions 459 (c) are var
Page 474 and 475: 203. A system is said to be effecti
Page 476 and 477: Answers of Objective Questions 1. (
Page 480 and 481: Index accelerating (or decelerating
Page 482 and 483: methods for voltage control 115 mul
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