Electrical Power Systems

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Corona 357 air around the conductor becomes conducting due to electrostatic stress. If the potential difference is raised still further, the glow and the noise will increase in intensity until eventually a sparkover will take place. If the conductors are perfectly uniform and smooth, the glow will be uniform along their length, otherwise the rough points of the conductor will appear brighter. With conductors only a short distance apart in comparison with their diameter, the spark-over may take place before there is any luminous glow is observed. If the applied potential difference is D.C. instead of A.C., the positive conductor having a uniform glow while the negative conductor has a more patchy glow and often it is accompanied by streamers if there are any rough places. An important point in connection with corona is that it is accompanied by loss of power, which is dissipated in the form of heat, light, sound and chemical action. In case of A.C. system, current flow due to corona is non-sinusoidal and in practice this non-sinusoidal current and non-sinusoidal voltage drop by corona may be more important than the power loss. The effects of corona can be summarized as follows: (i) It is accompanied by power loss (ii) A luminous violet glow is observed around the conductor (iii) This glow is much brighter over rough and dirty surfaces of the conductor (iv) It produces a hissing noise (v) It produces ozone gas which can be readily detected by its characteristic odour 14.3 POTENTIAL GRADIENT OR SINGLE-PHASE LINE igure 14.1 shows a single phase line. Radius of each conductor is r. ig.14.1: Single phase transmission line. The voltage gradient at a point P from conductor 1 is given by G = G1 – G2 Where G1 = Voltage gradient at point P due to a charge + q on conductor 1. G2 = Voltage gradient at point P due to a charge – q on conductor 2. We Know, G 1 = G 2 = q 2pÎ0 x ...(14.1) volt/m ... (14.2) -q volt/m ...(14.3) p Î ( D-x) 2 0
358 Electrical Power Systems \ Therefore, G = q + 2pÎ0 x L NM q pÎ ( D-x) 2 0 O QP \ G = q 2 pÎ 0 1 1 + volt/m x D - x ... (14.4) The voltage gradient is a maximum at the surface of the conductor, where x = r. Substituting x = r in eqn. (14.4), we get q L1 Gmax = + 2 pÎ 0 NM r 1 O volt/m D - rQP ... (14.5) Since D is large compared with r, we can write D – r » D. Therefore, Gmax = q volt/m 2pÎ 0 r ...(14.6) The R.M.S. value of voltage gradient is given by as: G rms = G max 2 ...(14.7) Potential difference V 12 between the conductors of a single phase line (chapter-3) is written I volt ...(14.8) HG KJ V12 = q pÎ0 ln D r Voltage from conductor to neutral is given as: Vn = V12 2 = q D ln 2pÎ0 r Eqn. (14.9) can be written as: \ I HG KJ Vn q D = ln r 2pÎ 0 r r q 2pÎ 0 r = r \ G max = r Vn D r ln HG I KJ Vn D r ln HG I KJ I volt ... (14.9) HG KJ volt/m ... (14.10)
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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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Page 322 and 323: 12.4 ISOCHRONOUS GOVERNOR Automatic
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Page 374 and 375: Similarly, G b = G c = r r V bn ln
Page 376 and 377: Corona 363 In the above expressions
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Page 380 and 381: Solution: rom eqn. (14.38), air den
Page 382 and 383: \ V 0 = 110.15 KV(rms) V n = 220 3
Page 384 and 385: \ W = 55 ´ b69. 28 - 57g 2 635 . -
Page 386 and 387: Analysis of Sag and Tension 15.1 IN
Page 388 and 389: where Dl = l0. DT MA DT = T1 - T0 T
Page 390 and 391: when x = L l , s = 2 2 , \ l 2 = H
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Page 394 and 395: Squaring eqn. (15.37), we get, s 2
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Page 398 and 399: \ T max = 572.59 kg (d) rom eqn. (1
Page 400 and 401: ig. 15.6: Case of negative x 1. Ana
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Page 406 and 407: (f ) Vertical sag = dcos q cos q =
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Page 414 and 415: Total weight of the conductor = wl
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Page 418 and 419: 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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