Electrical Power Systems

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Now, \ V1 = 1.198 14. 5º, \d1 = 14.5º V 2 = V r + I 2Z 2 = 1 0º + (3.2 – j0.8)(0.02 + j0.08) \ V 2 = 1.153 12º , \d 2 = 12º I g1 = (4 – j1) = 4.123 -14º , \a 1 = – 14º I g2 = I 2 + I 4 = (3.2 – j0.8) + (2 – j0.5) = 5.2 – j1.3 \ I g2 = 5.36 -14º , \a 2 = –14º \ cos(a 2 – a 1) = cos(0º) = 1.0 Generator station power factors are: cosf 1 = cos(14.5º + 14º) = 0.8788 cosf 2 = (12º + 14º) = 0.8988 Loss coefficients are: B 11 = \ B 11 = \ B 11 = K=1 2 1 V 4 å 2 K1 K A R bcos f g \ B 11 = 0.02485 pu. Similarly, \ B 22 = \ B 22 = 0.01755 pu. 2 1 2 2 2 2 11 21 31 41 RA 1 + RA 2 + RA 3 + RA 4 2 dV1cosf1i Optimal System Operation 441 2 2 2 2 0. 02( 1) + 0. 02( - 0. 2174) + 0. 01( 0. 7826) + 0. 01( 0. 2174) (. 1198 ´ 0. 8788) 2 2 2 2 12 22 32 42 RA 1 + RA 2 + RA 3 + RA 4 B22 = 2 dV2cosf 2i B 12 = 2 2 2 2 0. 02() 0 + 0. 02(. 0 7826) + 0. 01(. 0 7826) + 0. 01(. 0 2174) cos( a1 - a2) V V cosf cosf 1 2 1 2 (. 1153 ´ 0. 8988) 4 å K=1 A A R K1 K2 K 2 2
442 Electrical Power Systems \ B 12 = Note that cos(a 1 – a 2 ) = cos 0º = 1.0 \ B 12 = \ B 12 = 0.002928 Base is 100 MVA, \ B 11 = B 22 = B 12 = R ( A A ) + R ( A A ) + R ( A A ) + R ( A A ) 1 11 12 2 21 22 3 31 32 4 41 42 V1 V2 cosf cosf 1 2 0. 02( - 0. 2174 ´ 0. 7826) + 0. 01(. 0 7826) + 0. 01(. 0 2174) 0. 02485 100 0. 01755 100 0. 002928 100 (. 1198 ´ 1153 . ´ 0. 8788 ´ 0. 8988) = 0.02485 ×10 2 - MW -1 = 0.01755 ×10 2 - MW -1 = 0.002928 ×10 2 - MW -1 2 2 Example 16.10: Three generating stations supply powers of P 1 = 100 MW, P 2 = 200 MW, and P 3 = 400 MW respectively into the power network. Calculate the transmission loss in the network and the value of P L/P i. Given that B 11 = 0.01, B 22 = 0.03, B 33 = 0.04, B 12 = 0.001, B 23 = 0.0004 and B 31 = –0.001. Solution: rom eqn.(16.86) P L = m m åå p=1q=1 Number of generating plants m = 3 \ P L = 3 3 åå p=1q=1 PB P p pq q PB P p pq q \ P L = P 1 2 B11 + P 1B 12P 2 + P 1B 13P 3 + P 2B 21P 1 Since B 12 = B 21 , B 13 = B 31 , B 23 = B 32 Now On 100 MVA Base, + P 2 2 B22 + P 2B 23P 3 + P 3B 31P 1 + P 3B 32P 2 + P 3 2 B33 P L = B 11 P 1 2 + B 22 P 2 2 + B33 P 3 2 + 2B12P 1P 2 + 2B 23P 2P 3 + 2P 3P 1B 31 P 1 = 100 MW, P 2 = 200 MW, P 3 = 400 MW, P 1 = 1.0 pu, P 2 = 2.0 pu, P 3 = 4.0 pu 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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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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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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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
Page 420 and 421: Optimal System Operation 407 where
Page 422 and 423: Optimal System Operation 409 2. The
Page 424 and 425: Complete algorithm is given below:
Page 426 and 427: DP g1 = Î 2 DP g2 = Î 2 Using Tay
Page 428 and 429: Optimal System Operation 415 Equati
Page 430 and 431: \ 2 C2 = 0.18 Pg2 + 32 Pg2 + K2 Whe
Page 432 and 433: Optimal System Operation 419 Equati
Page 434 and 435: DP L = rom eqns (16.36) and (16.39)
Page 436 and 437: Optimal System Operation 423 Now we
Page 438 and 439: Optimal System Operation 425 The pr
Page 440 and 441: Example 16.6: Consider Ex-16.5, wit
Page 442 and 443: \ DP g (l) = 264.1657 MW rom eqn. (
Page 444 and 445: \ (4) 2 2 P = 0.00005 × (188.95) +
Page 446 and 447: Optimal System Operation 433 The te
Page 448 and 449: \ Also d P 3 d 2 P 3 3 =- G 32 si
Page 450 and 451: Let the current in line K be I K1.
Page 452 and 453: + HG I K Optimal System Operation
Page 456 and 457: Optimal System Operation 443 B 11 =
Page 458 and 459: Z 1 = (0.03 + j0.12)pu; I 1 = (1.3
Page 460 and 461: Objective Questions Objective Quest
Page 462 and 463: Objective Questions 449 26. or a de
Page 464 and 465: Objective Questions 451 56. Peak lo
Page 466 and 467: 82. Inductance of a conductor due t
Page 468 and 469: Objective Questions 455 108. In ter
Page 470 and 471: Objective Questions 457 137. In a p
Page 472 and 473: Objective Questions 459 (c) are var
Page 474 and 475: 203. A system is said to be effecti
Page 476 and 477: Answers of Objective Questions 1. (
Page 478 and 479: Bibliography Bibliography 465 1. O.
Page 480 and 481: Index accelerating (or decelerating
Page 482 and 483: methods for voltage control 115 mul
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