Electrical Power Systems

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Objective Questions Objective Questions 447 1. or constant transmission efficiency, voltage is increased n times, the size of the conductor would be 1 I (a) reduced to HG 2 n KJ times of the original (b) increased to (n2 ) times of the original I HG KJ times of the original (d) increased to n times of the original (c) reduced to 1 n 2. It the shunt admittance of the transmission line is ignored, the torque angle at which maximum power will occur is (a) p/4 (b) – p/2(c) p/2(d) p 3. If the torque angle d continuously increases, system will become (a) stable (b) unstable (c) oscillatory stable (d) asymptotic stable. 4. If V S = V R = 33 KV (three phase) and X = 13 ohm per phase, what will be the maximum power transmission per phase? (a)29 MW (b) 30 MW (c) 60 MW (d) 28 MW. 5. If the transmission voltage is increased, the percentage resistance drop will (a) increase (b) decrease (c) remain unchanged (d) none of the above. 6. The power transmission capacity of the transmission line is (a) proportional to the square of the voltage (b) inversely proportional to the voltage (c) directly porportional to the voltage (d) proportional to the cube of the voltage. 7. A transformer has a reactance of 0.04 pu. If the secondary of the transformer is short circuited with nominal voltage applied to the primary, the symmetrical short circuit current will be (a) 4 times of the full load current (b) 25 times of the full load current (c) 16 times of the full load current (d) 29 times of the full load current. 8. Three phase short circuit current of an alternator will be (a) zero (b) maximum (c) infinity (d) very small. 9. The material used for fuse wire must have (a) low melting point and low specific resistance (b) low melting point and high specific resistance (c) high melting point and low specific resistance (d) low melting point with any specific resistance. 10. The Buchholz relay is used to protect the (a) alternators against all internal faults (b) oil immersed transformers against all internal faults (c) synchronous motors against all internal faults (d) transmission line against all types of faults. 11. Third harmonics current flows from the generator terminals through the system and returns by way of comparatively low zero–sequence reactance path to the neutral terminals will give rise to
448 Electrical Power Systems (a) overheating of generator (b) overheating of generator and neutral resistors of earthing transformer (c) overheating of earthing transformer only (d) none of the above. 12. The effect of electric shock in the human body will depend on (a) line voltage (b) line current (c) current flowing in the body (d) none of the above. 13. To protect the power transformer (Y – Y, with neutral grounded) against fault, the CT will have (a) D – D connection (b) D – Y connection (c) Y – D connection (d) Y – Y connection. 14. The class C insulation can withstand (a) 180ºC (b) 150ºC (c) 200ºC (d) 105ºC 15. The location of the lightning arresters is (a) near the transformer (b) away from the transformer (c) near the circuit breaker (d) away from the circuit breaker. 16. Which of the following distribution systems gives the greater reliability (a) radial system (b) ring system (c) DC three wire system (d) open loop system. 17. In the transformers (a) high tension winding is used near the core (b) high tension winding is used away from the core (c) low tension winding is used near the core (d) low tension winding is used away from the core. 18. The effect of series capacitor on a transmission line is to improve the (a) steady state stability (b) transient stability (c) (a) and (b) both (d) none of the above. 19. The power intended to be available even under emergency condition is called (a) firm power (b) hot reserve (c) cold reserve (d) spinning reserve. 20. Two transmission lines of the surge impedance of 600 ohms each is linked by a cable. What should be the surge impedance of the cable if there were no reflection at the junction? (a) 300 ohms (b) 30 ohms (c) 600 ohms (d) 60 ohms. 21. or 11 KV underground system, it is possible to work upon a length of (a) 800 miles (b) 100 miles (c) 400 miles (d) 10 miles. 22. Series capacitors are used to (a) improve the line voltage (b) compensate for line inductive reactance (c) compensate for line capacitive reactance (d) none of the above. 23. The short circuit in any windings of the transformer is the result of (a) impulse voltage (b) insulation failure (c) mechanical vibrations (d) loose connection. 24. Split–phase relaying responds to faults between turns of a transformer (a) true (b) false. 25. Which of the following relays are used for phase fault on long line? (a) impedance relays (b) reactance relays (c) MHO relays (d) over current relays.
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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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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
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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 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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