Exaphpre 1 14.4 Tr For a 132 kV system, the reactance and capacitance up to the location of the circuit breaker is 3 ohms and 0.015 4F, respectively. Calculate the following: (a) The frequency of transient oscillation (b) The maximum value of restriking voltage across the contacts of the circuit breaker (c) The maximum value of RRRV Solution: (a) The frequency of transient oscillation “as f= 50, the system frequency = zs = 0,00954 H S 1 a 2nNLC * os 1 2n V0.00954 x 0.015 x 10° 10° 10° = = = 13.291 kHz 2m x 1.1962 — 7.5241 (b) The restriking voltage v. = V,, [1 — cos o,f]Circuit Breakers 541 m value of the restriking voltage = 2V,, = 2 BEF = 215.56 kV alue of RRRV = @,V,, se main y The maximum V: « = 2rfy x 22 x V2 x 1000 = 2n x 13.291 x 1000 x 132 x 2 x 1000 V/s 3 = 9010.45 x 10° V/s = 9.01045 kV/usee no \s* 2a VEC acai (14.18) Eaeus | Ina 220 kV system, the reactance and capacitance up to the locas limof circuit breaker is 8 © and 0.025 wR, respectively, A resistance of 600 ohms Sconnected across the contacts of the circuit breaker. Determine the following: (®) Natural frequency of oscillation () Damped frequency of oscillation ©) Critical value of Tesistance which will giv The value of Tesistance which will give damped frequency of oscillation, ne-fourth of the natural frequency of oscillation © no transient oscillation Suton; L=—8_- aw = 0.02544 H 2750 nt ©) Natural fr foe { v quency of oscill: on \Le ‘i, = 6.304 kHz re : “eney of damped oscillation is given by544 Power System Protection and Switchgear a —__—— i po Le - G2 2 0.02544 x 0,025 x 10% 4(0.025 * 10-°) x (600) iw ee pao 10" 5.413 kale 6.36 20 (iii) The value of critical resistance pated) [0.02544 [0.03544 = 504.35 2 “2 2 Vou2s x Se (iv) The damped frequency of oscillation is tx 6.304 kHz = 1576 Hz al iil i 1576 = + To ETe a 2x \0.02544 x 0.025 x 10° 4(0.025 x 10° 10, 16, or 1576 = 1 to _ 10" 2x \636 25K? Therefore, R= 52082. i 4087) CURRENT CUnnniAnHCircuit Breakers 545 Fe o14.3 | A circuit breaker interrupts, the magnetising current of a 100 MVA rendoemer 3 KV. The Magnetising current of the transformer is 5% of the full load current. Determine the maximum voltage which may appear across the spat the breaker when the magnetising current is interrupted at 53%_of its peak “ine, The stray capacitance is 2500 \r, The inductance is 30 H. ~~ value. The full load current of the transformer 6 = 10x10 agraga VB x 220 x 10 solution: Magnetising current =< x 262.44 = 34.44 A Current chopping occurs at 0.53 x 34.44 V2 = 25.83 A v = 2829 kV“Fxamplyhe 4 | A three-phase, 11 kV/132 kV, A-Y connected power transformer is protected by differential protection. The CT, on the LV side have a current ratio of 500/5. What must be the current ratio of the CT, on the HV side and how should they be connected. Solution: In order that circulating currents in the relay are in phase onposition, the ‘T.on the delta connected LV side of the transformer should be connected in star and the CT, on the star connected HV side of the transformer should be connected in delta, Connections of CT, on LV and HV sides are shown in Fig. 10.11 a374 Power System Protection and Switchgear a TOOT Pit | = = ay 05 Fig. 10.11 ~ 500/58 hy Let the line currents on the primary and secondary sides of the transformer be, and I, respectively. Then, VB x x I), = V3 x 132 x Ip u or la? LW AL 732 * Since the CT, on the LV side are connected in star, the current through the sec- ondary of the CT and the pilot wire will be 5 A. The CT, on the HV side being delta connected will have a current of 5/V3 A in the secondary. Hence CT ratio on HV side = 41.66/53 NB x 4166/5 = 72.1515 For I, of 500A, Jy x 500 = 41.66 A exephie 2 | A three-phase, 11 kV/33 kV, Y-A connected power transformer is protected by differential pratection, The CT, on the LV side have a current ratio of 400/5. What must be the ratio of CT, on the HV side. How the CT on both the sides of the transformer are connected. Solution: ‘The connections of the CT, on both the sides of the transformer are shown in Fig. 10.12. Let the line currents on the primary and secondary sides of the transformer be /y1 and /,3 respectively. j Then, NB x LL x My, = V3 x33 x Lp or 4a= Fela For 1,, of 400A, Iy= Hh x 400= 133.3.Exam) An IL kV, 100 MVA alternator is grounded through a resistance of 5 Q. The CTs have a ratio 1000/5. The relay is set to operate when there is an out of balance current of 1 A. What percentage of the generator winding will be protected by the percentage differential scheme of protection? Solution: Primary earth-fault current at which the relay operates 'y Op = <1 22004 Suppose p% of the winding from the neutral remains unprotected. » 3 The Fault current =

15.75 This means that 15.75% of the winding from the neutral is not protected. In other words, 100 — p = 100 - 15.75 = 84.25% of the winding from the terminal is Protected. Note: Near the neutral point voltage stress is less and therefore, phase to earth faults ate not likely to occur. Iebmleg Van 11 kV, 100 MVA alternator is provided with differential pro- tection. The percentage of winding to be protected against phase to ground fault 's 85%. The relay is set to operate when there is 20% out of balance current. “lermine the value of the resistance to be placed in the neutral to ground connection, Solution: . lutions (a) Primary earth-fault current at which the relay operates — 10010", 20 _ 949.759 V3x 11 100 Suppose that the percentage of winding which remains unprotected is p = 100-85 = 15%. Py dix 10° 100” V3 R, 'S the resistance in the neutral connection 1S. 1x10? 100" VR The fault current = Where g = 1049.759 a354 Power System Protection and Switchgear : g _ IS Hx 10! 100 x V3 % 1049,759 =0.919 (b) Protection against Stator Interturn Faults Longitudinal».eseyble 10.2 | A three-phase, 11 kV/33 KV, Y-A connected power transfa protected by differential pratection. The CT, on the LV side have of 400/5. What must be the ratio of CT, on the HV side. How sides of the transformer are connected. mers a Current rajg the CT, on both the Solution: The connections of the CT, on both the sides of the transformer ul ae shou in Fig, 10.12. Let the line currents on the primary and secondary sides of the transformer be/, and I, respectively. _ Then, VB x IL XI, = 13 x33 x py or I iz in For 1, 0f 400 A. yy a x 400 = 133.3.A beFY, Transformer and Buszone Protection 375 [OOO cy Fig. 10.12 ‘ . er ‘ough secondary of the CT on the primary side of the transformer eure ne the CT, on the primary side are connected in delta, the curres © 5A. Sinct s sii be > ‘i ire will be S V3 A. The CT, on the secondary side being star con- yah its pilot Wi nits pilot pout ili have a current of 5 V3 in the secondary, iwi : we io on HV side psee CT alo = 133.3/5 V3 =76.7/5