Electrical Power Systems

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as: \ |V S| = 145.13 kV. \ Sending end line-to-line voltage Voltage regulation = Characteristics and Performance of Transmission Lines 133 |V S| L–L = 3 × 145.13 = 251.37 kV 25137 . - 220 220 Per phase real power loss in the line Per phase receiving end power = 14.26% P Loss = |I| 2 R = (787.4) 2 × 9 × 10 –6 MW = 5.58 MW. P R = 300 3 \ Per phase sending end power Transmission line efficiency is ´ 08 . = 80 MW P S = (80 + 5.58) = 85.58 MW. R h = P P (b) 3|V R ||I R | = 300 \ |V R| = 220 kV S = 80 85. 58 \ |I R| = 787.4 Amp = 93.47%. Load is at 0.8 power factor leading. or leading power factor load, eqn. (6.6) can be written |V S| = |V R|+ |I| (R cos d R – X sin d R) \ |V S| = 127 + 787.4 (9 × 0.8 – 26.39 × 0.6) = 120.2 kV \ |V S| L–L = 3 × 120.2 = 208.2 kV Voltage regulation = 208. 2 - 220 220 = –5.36% Per phase real power loss = (787.4) 2 × 9 × 10 –6 = 5.58 MW Per phase receiving end power, P R = 300 3 Per phase sending end power, Transmission line efficiency ´ 08 . = 80 MW P S = (80 + 5.58) = 85.58 MW R h = P P S = 80 85. 58 = 93.47%
134 Electrical Power Systems Example 6.3: Determine the efficiency and regulation of a 3-phase, 150 km long, 50 Hz transmission line delivering 20 MW at a power factor of 0.8 lagging and 66 kV to a balanced load. Resistance of the line is 0.075 W/km, 1.5 cm outside dia, spaced equilaterally 2 meters between centres. Use nominal p method. Solution: R = 0.075 × 150 = 11.25 W diameter of the conductor = 1.5 cm \ radius r = 15 . = 0.75 cm 2 d = 2 mt = 200 cm \ L = 2 × 10 –7 × (150 × 1000) ln 200 HG I Henry 075 . KJ \ L = 0.1675 Henry \ X = 2 × p × 50 × 0.1675 = 52.62 W. HG I KJ 2 ´ p ´ 8. 854 ´ 10 The capacitance per phase = 200 ln 075 . -12 \ Y = jwC = j 2p × 50 × 1.49 × 10 –6 mho \ Y = j 468.1 × 10 –6 mho \ Now Y 2 = j 234 × 10–6 mho 3 × |I R|× 66 × 0.8 = 20 × 1000 Z = (11.25 + j52.62) = 53. 809 77. 9° W. \ |I R | = 218.7 Amp. at 0.8 pf lagging Receiving end phase voltage |V R | = 66 rom eqn. (6.15), we have 3 V S = 1 = 38.104 kV ZYI + V + ZI HG 2 KJ R R IR = 218. 7 - 36. 87° VR = 38. 104 0° ZY 2 = 53.809 × 234 × 10–6 77. 9°+ 90° ´ ( 150 ´ 1000) = 1.49 m. = 0. 01259 167. 9° = (–0.0123 + j0.00264)
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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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Synchronous Machine: Steady State a
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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 .)
Page 126 and 127: \ VL (pu) = 0. 2033 - 76. 68° pu P
Page 128 and 129: Base impedance for the load is ( )
Page 130 and 131: Now we can write, \ |I| cos f = P |
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Page 140 and 141: Eqns (6.15) and (6.17) can be writt
Page 142 and 143: Characteristics and Performance of
Page 144 and 145: om eqn. (6.6), Characteristics and
Page 150 and 151: Sending end power factor angle = 8.
Page 152 and 153: \ I R = V R = 208 Characteristics a
Page 154 and 155: 6.6 VOLTAGE WAVES Characteristics a
Page 158 and 159: 6.9 ERRANTI EECT Characteristics an
Page 160 and 161: Load low Analysis 7.1 INTRODUCTION
Page 162 and 163: Applying KCL to the independent nod
Page 164 and 165: y 10 = y 20 = y 30 = y13 ¢ y12 ¢
Page 166 and 167: \ V i = 1 Y ii L NM P - jQ i i * Vi
Page 168 and 169: n 2 i ii ii ik i k ik k i \ Pi - jQ
Page 170 and 171: Load low Analysis 157 ig. 7.6: P-re
Page 172 and 173: Load low Analysis 159 are permitted
Page 174 and 175: 1 1 y13 = y31 = = = ( 10 - j30) Z (
Page 176 and 177: (1) \ V3 = 10011 . - 2. 06° After
Page 178 and 179: Load low Analysis 165 Using eqn. (7
Page 180 and 181: Load low Analysis 167 \ Q 2 = -|V 2
Page 182 and 183: |V 1| = 1.05,|V 2| = 1.0,|V 3| (1)
Page 184 and 185: \ L NM e j ( 0) ( 0) ( 0) 1 1 1 2 n
Page 186 and 187: ( p) p The terms DPi and DQi Load l
Page 188 and 189: Load low Analysis 175 P2 = -35.77
Page 190 and 191: (1) P2( cal ) = -2.62 (1) P3( cal )
Page 192 and 193: Load low Analysis 179 Use deoupled
Page 194 and 195: 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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