Electrical Power Systems

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Capacitance of Transmission Lines 59 calculation with the exception that the radius r of each conductor is used; If d is the bundle spacing, then for two conductor arrangement, Ds = bg rd 1 2 ...(3.31) or three conductor arrangement (equilateral triangle) Ds = r. d or a four conductor (quadruplex) arrangement, 1 2 3 e j ...(3.32) 1 3 4 Ds = e 2 . r. dj ...(3.33) Considering the line to be transposed, the capacitance per phase is given as HG Can = 00242 . eq log D D s I KJ m/km ...(3.34) 3.6 CAPACITANCE O THREE PHASE DOUBLE CIRCUIT LINES igure 3.7 shows three phase double circuit line with the three sections of the transposition cycle. ig. 3.7: Three sections of three phase double circuit transposed line. Each phase conductor is transposed within its groups. The effect of ground and shield wires are considered to negligible. In this case per phase equivalent capacitance to neutral is Where HG Can = 00242 . eq log D D I KJ m/km ...(3.35) s b g 1 3 ...(3.36) D eq = D D D ab bc ca b g 1 D s = D D D sa sb sc 3 ...(3.37)
60 Electrical Power Systems b ab ab¢ a¢ b¢ a¢ bg 1 Dab = d . d . d . d 1 1 2 b 2 2g b 2g 4 = D. d . D. d 4 = Dd b 1 g b 1 g b 1 2 2g b ca . ca¢ . c¢ a¢ . 1 4 c¢ ag = b 5 . 1 . 5 . 1 4 1g = b 1 5g RST b 1 2 2g . b 1 2 2g . b 1 1 1 3 2 5g UVW Dbc = dbc . dbc . dbc . d 4 ¢ ¢ ¢ b¢ c = D. d2. D. d 4 2 = Dd D ca = d d d d d d d d d d \ D eq = Dd Dd d d 1 1 2 2 \ Deq = Dd2. d1. d5 1 1 1 1 \ Deq = D3 3 d2d6 1 6 d5 Now R S| T| D sa = rd 3 D sb = rd 4 D sc = rd 3 U V| W| 1 3 . . . ...(3.38) 1 b g2 1 b g2 1 b g2 1 1 RST b 2 2 3g . b 4g . b 3g \ D s = rd rd rd 1 1 \ Ds = rd . 3. r2 2 . d4 1 1 1 1 \ Ds = r3. d3 3 . r6. d6 4 1 1 1 \ Ds = r2d3 3 d6 4 HG I KJ 1 3 1 2 UVW 1 3 . . ...(3.39) Note that Deq and Ds will remain same for section-II and section-III of transposition cycle. Substituting Deq and Ds in eqn. (3.34), we have, 1 2
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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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Structure of Power Systems and ew O
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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 30 and 31: Structure of Power Systems and ew O
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 60 and 61: \ D m = D D D D Resistance and Indu
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
Page 110 and 111: or the reference phase a, E a = (Z
Page 112 and 113: 5.3 THE PER-UNIT (pu) SYSTEM Power
Page 114 and 115: Power System Components and Per Uni
Page 118 and 119: Z L (pu) = Z L (ohm) (. 08+ j 03 .)
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Power System Components and Per Uni
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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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