Electrical Power Systems

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Objective Questions 457 137. In a power system, the purpose of an inter–connector is to (a) reduce the SC current (b) improve the stability margin (c) allow the flow of current from one generator to another (d) all of the above. 138. A thermal power plant has an efficiency of about (a)20% (b) 30% (c) 40% (d) 60% 139. The generating station with minimum running cost is (a) hydro power station (b) nuclear power station (c) thermal power station (d) diesel power station. 140. The conductor size of a feeder is determined on the basis of (a) the amount of current it has to carry (b) the voltage across the feeder (c) the length of the feeder (d) the power factor of the load. 141. Power plant having the maximum efficiency is (a) thermal (b) hydro (c) diesel (d) nuclear. 142. Maximum speed of modern turbo–alternator is (a) 1000 rpm (b) 3000 rpm (c) 1500 rpm (d) 1500 rpm. 143. The charging current in the cables (a) lags the voltage by 90º (b) leads the voltage by 90º (c) in–phase with the voltage (d) none of these. 144. Tappings in on–load tap changing transformer are provided on (a) LV side (b) HV side (c) both LV and HV side (d) all of these. 145. The cost of power generation will decrease when the load factor (a) decreases (b) increases (c) remains constant (d) none of these. 146. The permissible variation in supply frequency is (a)±1% (b) ±2% (c) ±3% (d) ±5% 147. A telephone line has a surge impedance of (a) 50 ohm (b) 60 ohm (c) 75 ohm (d) 200 ohm. 148. The dielectric strength of air at 76 cm of Hg and 25ºC is (a) 30 KV/cm (b) 21.1 KV r.m.s./cm (c) 21.1 KV r.m.s./m (d) 21.1 KV/cm. 149. In a transmission line, the conductors take the form of (a) parabola (b) hyperbola (c) catanary (d) semi circle. 150. Water is used only for cooling purposes in (a) nuclear power station (b) diesel power station (c) thermal power station (d) gas–turbine power station. 151. The type of power station at Bhakra is (a) nuclear (b) thermal (c) hydro (d) diesel. 152. Turbo–alternators usually have (a) 2 poles (b) 6 poles (c) 10 poles (d) 4 poles. 153. Alternators installed in a hydro power station, are of (a) low speed (b) high speed (c) medium speed (d) very high speed. 154. The protection from negative sequence currents is provided for (a) motors (b) generators (c) transformers (d) transmission line. 155. An ACSR conductor of size 48/7 means that (a) it has 48 strands of steel and 7 strands of aluminium (b) it has 7 strands of steel and 48 strands of aluminium
458 Electrical Power Systems (c) the conductor has 55 strands (d) the conductor has 41 strands. 156. A double circuit transmission line will have four conductors including one earth conductor (a) true (b) false. 157. Hydro power stations are used as base load stations (a) true (b) false. 158. In D.C. transmission there is no skin effect (a) true (b) false. 159. Water hammer is developed in turbine (a) true (b) false. 160. The reflection factor for a short–circuited transmission line is unity (a) true (b) false. 161. The value of diversity factor is (a) less than one (b) greater than one (c) equal to one (d) less than half. 162. The per capita consumption of electrical energy in India is (a) more than 150 kwh (b) less than 150 kwh (c) less than 50 kwh (d) none of these. 163. The percentage of electric energy consumed on a world wide basis is (a) 80% (b) 75% (c) 30% (d) none of these. 164. In India, the thermal generation is about (a) 80% (b) 40% (c) 65% (d) 50% 165. The turbine used for high head in hydro plant is (a) rancis turbine (b) Kaplan turbine (c) Pelton wheel (d) none of these. 166. The Pelton wheel is (a) axial flow impulse turbine (b) reaction turbine (c) pressure turbine (d) none of these. 167. The maximum head upto which a pelton–wheel can be easily designed is (a) 150 m (b) 1800 m (c) 1000 m (d) 300 m 168. If low head and large quantity of water are available, a power engineer should setect (a) Kaplan turbine (b) Pelton wheel (c) rancis turbine (d) none of these. 169. The electric energy is directly generated by the combustion of fuel on a stationary device by a process, called (a) Magnetohydro dynamic process (b) thermal process (c) nuclear fission process (d) all of these. 170. A gas turbine is (a) more economical to a conventional steam plant (b) unable to maintain flat voltage profile in the system even it is used as synchronous compensator (c) able to meet sudden peaks of system load (d) all of the above. 171. The conductors used in practice for transmission line is stranded for (a) providing necessary flexibility for stringing (b) reducing the effective resistance (c) preventing unwinding (d) all of the above 172. The per unit impedances of a transformer referred from the primary and secondary side (a) are not equal (b) are equal
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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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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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Page 424 and 425: Complete algorithm is given below:
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Page 430 and 431: \ 2 C2 = 0.18 Pg2 + 32 Pg2 + K2 Whe
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Page 434 and 435: DP L = rom eqns (16.36) and (16.39)
Page 436 and 437: Optimal System Operation 423 Now we
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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 454 and 455: Now, \ V1 = 1.198 14. 5º, \d1 = 14
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 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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