Electrical Power Systems

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Load low Analysis 159 are permitted to vary during iterative process. This is must as voltage magnitude and phase angle are specified at slack bus. With this data, net bus injected power (Pi + jQi) at all buses are known other than slack bus. Step-2: ormation of YBus Matrix With the line and shunt admittance data, form YBus matrix. Step-3: Iterative Computation of Bus Voltage To start the iterative computation, a set of initial voltage values is assumed. Since in a power system, the voltage variation is not too wide, it is usual practice to use a flat voltage start, i.e. initially all voltages are set equal to (1 + j0) except the voltage of the slack bus which is specified and fixed. It should be noted that (n – 1) voltage equations are to be solved iteratively for finding (n – 1) complex voltages V2 , V3 , ..... Vn . The iterative computation is continued till the change in maximum magnitude of bus voltage, (DV) is less than a certain tolerance for all bus voltages, i.e. (p+1) i (p) i DV = max| V - V | < Î , i = 23,... n Step-4: Computation of Slack Bus Power Slack bus power can be computed using eqns. (7.15) and (7.16), i.e. n P1 = å| V1|| Vk|| Yik|cos( q1k - di + dk) k = 1 n Q1 = - V1VkY1k q1k - d1+ dk k = 1 å| || || |sin ( ) ...(7.41) ...(7.42) Step-5: Computation of Line lows This is the last step in the load flow analysis. The power flows on the various lines are computed using eqns. (7.34) and (7.35). Real and reactive power loss can be computed using eqns. (7.39) and (7.40) respectively. Example 7.2: ig. 7.7 shows the single line diagram of a sample 3- bus power system. Data for this system are given in Table 7.2 and 7.3. (a) Using the Gauss- Seidel method, determine the phasor values of the voltage at buses 2 and 3. (Perform only two iterations). (b) ind the slack bus real and reactive power after second iteration. (c) Determine the line flows and line losses after second iteration. Neglect line charging admittance. ig. 7.7: 3-bus sample power system.
160 Electrical Power Systems Table 7.2: Scheduled generation and loads and assumed bus voltage for sample power system Generation Load Bus code Assumed MW MVAr MW MVAr i bus voltage 1 1.05 + j0.0 – – 0 0 (slack bus) 2 1 + j0.0 50 30 305.6 140.2 31 + j0.0 0.0 0.0 138.6 45.2 Solution: Step-1: Initial computations Convert all the loads in per-unit values PL 2 = 305. 6 100 Table 7.3: Line impedances Bus code Impedance i – k Z ik 1-2 0.02 + j0.04 1-30.01 + j0.03 2-30.0125 + j0.025 = 3. 056 pu; QL2 = 140. 2 100 = 1402 . pu 138. 6 45. 2 PL3 = = 1386 . pu; QL3 = = 0. 452 pu 100 100 Convert all the generation in per-unit values. P g2 = 50 100 = 050 . pu; Qg2 = 30 100 Compute net-injected power at bus 2 and 3. = 030 . pu P 2 = P g2 – P L2 = (0.5 – 3.056) = –2.556 pu Q 2 = Q g2 – Q L2 = (0.3 – 1.402) = –1.102 pu P 3 = P g3 – P L3 = 0 – 1.386 = –1.386 pu Q 3 = Q g3 – Q L3 = 0 – 0.452 = –0.452 pu Step-2: ormation of Y BUS matrix 1 1 y12 = y21 = = = ( 10 - j20) Z 002 . + j004 . 12 Base MVA = 100
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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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Page 122 and 123: Power System Components and Per Uni
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 146 and 147: as: \ |V S| = 145.13 kV. \ Sending
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 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
Page 196 and 197: Assuming q ik - d i + d k » d ik,
Page 198 and 199: Table 7.4: Line impedances BUS code
Page 200 and 201: Symmetrical ault 187 is drawing 10
Page 202 and 203: \ X TH = ig. 8.4: Thevenin equivale
Page 204 and 205: ig. 8.5: Circuit diagram of Example
Page 206 and 207: = | Vo| | Vo| | Z| ´ × (MVA) Base
Page 208 and 209: \ 1 - V 1f = j0.14 × (-j4.165) \ V
Page 210 and 211: Base current I B = Per unit reactan
Page 212 and 213: Symmetrical ault 199 Example 8.9: T
Page 214 and 215: Solution: Let Base MVA = 12 Base Vo
Page 216 and 217: Solution: Let Base MVA = 100 Base V
Page 218 and 219: Therefore, current to be interrupte
Page 220 and 221: 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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