This is a hand calculation problem, I have attached the file here for the questions and the answers are also made available for it. I just need the calculations so I can have a good understanding of the problem.
Microsoft Word - DT705-Protection_Qns-Feb18 Dr. J Kearney Power Sys Analysis 2 DT705 1 Tut Qns DT705 MEng. Power System Analysis 2 Protection. Ex 1 (a). On the Figure .. of the small power system circle the zones of protection. (b). State which circuit breakers trip for a fault at: (i) L1 (ii) Bus 3 (iii) Generator G2 (iv) Transformer T4 (v) Bus 2 (vi) Load (vii) M1 Ex 2 (a) An induction-disc overcurrent relay has a characteristic as shown in Table 1 and has a plug setting (PS) of 150% and a time multiplier of 0.2. The relay is supplied by a 200 A/5 A current transformer on a three-phase, 10 kV local feeder of reactance 3 Ω per phase. The feeder is supplied by a 100 MVA, 38 kV/10 kV transformer of 0.15 p.u. reactance. A three- phase solid earth occurs at the end of the 10 kV feeder. Calculate: (i) the fault current, (ii) the time of operation of the relay. Table 1 PSM 1 2 3 4 6 8 10 12 14 18 20 Time in Seconds 32 10 3 4.2 3.7 3.3 3.0 2.7 2.5 2.3 2.2 (G2) Generator 1 L1 3 B2 T1 B1 B3 B8 B4 B9 L2 L3 2 B5 B6 B7 B10 T3 B11 L4 B12 B13 (M1) Motor (G1) Generator T2 T4 (Transformer) r Load B1-B14 (Circuit Breakers) B14 Dr. J Kearney Power Sys Analysis 2 DT705 2 Ans. Actual relay time = 3.7 . (0.2)= 0.74 sec. Ex. 3 A 50 Hz, 20 kV radial system is protected be a circuit breaker A as shown in Figure..The maximum load on the line is 4 MVA. The fault level at the 20 kV bus is 400 MVA and the positive sequence reactance of the line is 3 Ω. Calculate the maximum load current and the fault current for a three-phase to ground fault at Bus 2. The CT for the relay has a ratio of 200:5. If the relay setting is such that the relay should trip when the current is at least twice the maximum load current and 1/5 of the minimum fault current, determine the appropriate current tap setting. The characteristic of the relay is given by: k I t . 1 014.0 02.0 − = Where t is the response time, I is the tap setting multiple and k is the time dial constant setting. Determine the constant k in the relay characteristics to ensure that the relay responds in 0.3 seconds. Fig. Ans CTS = 14.44, k=0.864 Ex 4 Differential relay protection for a single-phase transformer. Explain with aid of a sketch the operation of a differential method of protection for a single phase two-winding transformer. A single-phase two-winding, 10 MVA, 80kV/20kVtransformer has differential relay protection. Select suitable CT ratios, from below; Table 2 Current Transformer ratios 50:5 100:5 150:5 200:5 250:5 300:5 400:5 450:5 500:5 600:5 800:5 900:5 1000:5 Select k for a mismatch of 25%. Ans Similar in notes. CT1 = 150:5, CT2 =600:5, k=0.222 Ex 5 Load 20kV kV CT 200/5 A Line Grid A Bus 1 Bus 2 Relay Load CT 200/5 A 0.15 pu 3 Ω/ph CB 38/10 kV Dr. J Kearney Power Sys Analysis 2 DT705 3 Two-source system protection with directional and time-delay overcurrent relays. Explain how directional and time-delay overcurrent relays can be used to protect the system in the figure.Which relays should be coordinated for a fault (a) at P1, (b) at P2. (c) Is the system also protected against bus faults? 1 2 3 B1 B12 B21 B23 B32 B3 P2 P1 Load L1 Load L2 Load L3 Figure Ans. In notes Ex 6 A 40 MVA, 110 kV/38 kV delta/star transformer has its star point earthed through a resistor which limits the current, due to an earth fault on one of the line terminals, to rated value. Circulating-current protection is employed, with 5 A in the relay circuits at rated load. (a) Determine the current transformer ratios required. (b) Draw a circuit diagram showing the currents an all windings when a phase-to-phase fault of 1000A occurs on the 38kV side beyond the protected zone. (c) Find the maximum current setting of the relay such that it will trip if an earth fault occurs halfway along one of the 38 kV windings. Note: The diagram in (b) may also be used to obtain the solution for (c). Dr. J Kearney Power Sys Analysis 2 DT705 4 Ans Primary 200/5 CT’s , 40/1 ratio. Secondary 1000/5 CT’s ratio 200/1. Ans Relay R&T will operate with 0.75 A setting for fault ½ way along 38 kV winding. Ex 7 A three-phase delta-star connected 30 MVA 38/10kV transformer is protected by a differential relay. Calculate the relay current setting for faults drawing up to 200 per cent of the rated current. The CT current ratio on the primary side is 500:5 and on the secondary side is 2000:5. Ans I ` 1 = 4.56A, I ` 2 = 7.49A, Io (200%) = 5.862A Ex 8 A single-phase two-winding, 5 MVA 20/8.66 kV transformer is protected by a differential relay. Available relay tap settings are 5:5, 5:5.5, 5:6.6, 5:7.3, 5:8, 5:9 and 5:10. Giving tap ratios of 1.00, 1.10, 1.32, 1.46, 1.60,1.80 and 2.00. CT ratios, are: 50:5, 100:5, 150:5, 200:5, 250:5, 300:5, 400:5, 450:5, 500:5, 600:5, 800:5, 900:5. Select CT ratios and relay tap settings. Also determine the percentage mismatch for the selected tap setting. Relay Tap Settings 5:5 5:5.5 5:6.6 5:7.3 5:8 5:9 5:10 Relay Tap Ratios 1.00 1.10 1.32 1.46 1.60 1.80 2.00 CT Ratios 50:5 100:5 150:5 200:5 250:5 300:5 400:5 450:5 500:5 600:5 800:5 900:5 Ans Primary CT = 300:5, Sec= 600:5,Tap setting = 1.1 Mismatch =4.7 % If2 s/c Y B R Oper. Coil Restraining Coil 38kV 110kV Direction of Fault Current B Y R Dr. J Kearney Power Sys Analysis 2 DT705 5 G&S Ex. 9 Table below gives the positive-sequence line impedances and the CT and VT ratios at B12 for the 345 kV system as shown in the Fig above. (a) Determine the settings Zr1, Zr2 and Zr3 for the B12 three-zone, directional impedance relays connected as shown (Fig . below ) . Consider only solid, three-phase faults. (b) Maximum current for line 1-2 during emergency loading conditions is 1500 A at a power factor of 0.95 lagging. Verify that B12 does not trip during normal and emergency conditions. B12 B21 B13 B31 B32 B23 B24 B42 Zone 1 Zone 2 Zone 3 1 2 3 4 P1 P2 P3 Dr. J Kearney Power Sys Analysis 2 DT705 6 Table Line Positive-Sequence Impedance Ω 1-2 2-3 2-4 1-3 8 + j50 8 + j50 5.3 + j33 4.3+ j27 Breaker CT ratio VT Ratio B12 1500:1 3000:1 Ans Zr1 =4.05 ∠80.9 o Ω , Zr2 = =6.08 ∠80.9