Electrical Power Systems

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\ D m = D D D D Resistance and Inductance of Transmission Lines 47 2 2 Dbb¢ = Daa¢ = 3 m; Dab¢ = 3 + 1 = 3.162 m = Dba¢ b aa¢ ab¢ ba¢ bb¢ g 1 1 b g 2 = 3.08 m 4 = 3´ 3162 . 308 . I \ L = 0.921 log mH/km = 1.375 mH/km HG 0. 099KJ Example 2.16: A three phase transmission line has an equilateral spacing of 6 m. It is desired to rebuild the line with same D eq and horizontal configuration so that the central conductor is midway between the outers. ind the spacing between the outer and central conductor. Solution: Using eqn. (2.59), we get L = 0.4605 log D r¢ I mH/km HG KJ When conductors are placed in horizontal configuration (ig. 2.26 (b)). HG eq L = 0.4605 log D r¢ I KJ e d 1 3 j . D eq = 2 3 \ L = 0.4605 log d.2 r¢ HG or both the cases, inductance is equal, therefore, 0.4605 log d.2 r¢ 1 3 I KJ HG I HG KJ D = 0.4605 log r¢ 1 3 I K J mH/km. \ d = D 6 m = m = 4.762 m. 3 2 3 2 Example 2.17: Determine the inductance per km/phase of a double circuit three phase line as shown in ig. 2.27. The radius of each conductor is 1.5 cm. Solution: ig. 2.27 ig. 2.26 (a) ig. 2.26 (b)
48 Electrical Power Systems We know, L = 0.4605 log D rom the symmetry of the configuration, HG D eq s I KJ 1 2 b g m. Dsa = Dsb = Dsc = r¢ 3d b g 1 b ab. ab¢ . a¢ b. a¢ b¢ g 1 b bc. bc¢ . b¢ c. b¢ c¢ g 1 b ca. ca¢ . c¢ a. c¢ a¢ g 1 b ab . bc . cag 1 \ D s = D D D sa sb sc D ab = d d d d D bc = d d d d D ca = d d d d 1 r.3d 2 b g m 3 = Dsa = ¢ 1 4 = bd. 4d. 2d. dg 4 = d( 8) 1 b g 4 = d( 8) 4 = d. 4d. 2d. d 4 = 2dd . . 5d. 2d 1 4 1 4 1 1 b g 4 4 = d( 20) \ Deq = D D D 3 = d( 8´ 8´ 2 1 0) 12 =b12 80g d \ Deq = 1.815 d = 3.176 m \ L = 0 4605 1 2 Ds = 0. 7788 ´ 0. 015 ´ 3 ´ 175 . 1 12 b g m = 0.2476 m 3176 . I . log mH/km = 0.5103 mH/km HG 0. 2476KJ Example 2.18: Determine the inductance per km per phase of a double circuit three phase line as shown in ig. 2.28. The radius of each conductor is 1.5 cm. Solution: L = 0.4605 log D ig. 2.28 HG D = 0.539 mH/km D s = D D D sa sb sc D sa = ¢ eq s I KJ b g 1 3 1 r. 5d 2 b g ; Dsb = br¢ .3dg mH/km = 0.4605 log 1 2 333 . I mH/km HG 02246 . KJ
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To My Wife Shanta Son Debojyoti and
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Preface During the last fifty years
Page 10 and 11: Contents Preface vii 1. Structure o
Page 12 and 13: Contents xi 9. Symmetrical Componen
Page 14 and 15: Structure of Power Systems and ew O
Page 16 and 17: 1.2 REASONS OR INTERCONNECTION Stru
Page 20 and 21: Case-1: If P 1 = P 2 = P 3 = ... =
Page 22 and 23: Load factor, L = Maximum load = 3 M
Page 24 and 25: \ LD = 1 MW (c) rom eqn.(1.13), coi
Page 26 and 27: Case-2: Very short lasting peak. He
Page 28 and 29: 1.9 DISADVANTAGES O LOW POWER ACTOR
Page 32 and 33: Resistance and Inductance of Transm
Page 34 and 35: 2.4.1 Internal Inductance igure 2.1
Page 38 and 39: Therefore, we can write L 1 = L 11
Page 40 and 41: L a = 0.4605 n log R S| T| Resistan
Page 42 and 43: L = 0.4605 L Resistance and Inducta
Page 44 and 45: =(2Dd 1 · 2Dd 1) 1/4 = (2 Dd 1) 1/
Page 46 and 47: Solution. The distances from strand
Page 50 and 51: Solution: Using eqn. (2.14), Hx = I
Page 52 and 53: = 0.4605 log 6611 . 04 . Resistance
Page 58 and 59: Since from symmetry D b1= D b2, l 1
Page 62 and 63: dg 1 2 \ Dsa = 0. 7788 ´ 0. 015 ´
Page 66 and 67: Capacitance of Transmission Lines 5
Page 68 and 69: V Ki = 1 pÎ N å q m 2 o m = 1 3.3
Page 70 and 71: Similarly for the 2nd transposition
Page 74 and 75: C an = log R S| T| 00242 . 1 1 1 1
Page 76 and 77: \ V12 = q ln pÎo \ C 12 = q \ C 12
Page 78 and 79: I chg = jw C anV LN \ |I chg| = w C
Page 80 and 81: Line length is 200 km. \ Can (total
Page 82 and 83: Now applying eqn. (3.7), we have C
Page 86 and 87: = 0.008993 m/km. Capacitance of Tra
Page 92 and 93: Synchronous Machine: Steady State a
Page 98 and 99: \ V = 138 Using eqn. (4.18), . 3 =
Page 100 and 101: om ig. 4.6, Synchronous Machine: St
Page 106 and 107: \ i asy max = from which Synchronou
Page 108 and 109: 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
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Optimal System Operation 415 Equati
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\ 2 C2 = 0.18 Pg2 + 32 Pg2 + K2 Whe
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Optimal System Operation 419 Equati
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DP L = rom eqns (16.36) and (16.39)
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Optimal System Operation 423 Now we
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Optimal System Operation 425 The pr
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Example 16.6: Consider Ex-16.5, wit
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\ DP g (l) = 264.1657 MW rom eqn. (
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\ (4) 2 2 P = 0.00005 × (188.95) +
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Optimal System Operation 433 The te
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\ Also d P 3 d 2 P 3 3 =- G 32 si
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Let the current in line K be I K1.
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+ HG I K Optimal System Operation
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Now, \ V1 = 1.198 14. 5º, \d1 = 14
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Optimal System Operation 443 B 11 =
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Z 1 = (0.03 + j0.12)pu; I 1 = (1.3
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Objective Questions Objective Quest
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Objective Questions 449 26. or a de
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Objective Questions 451 56. Peak lo
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82. Inductance of a conductor due t
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Objective Questions 455 108. In ter
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Objective Questions 457 137. In a p
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Objective Questions 459 (c) are var
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203. A system is said to be effecti
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Answers of Objective Questions 1. (
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Bibliography Bibliography 465 1. O.
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Index accelerating (or decelerating
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methods for voltage control 115 mul
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