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1. Determine the voltage drop at the motor frame for an earth .docx
- 1. 1. Determine thevoltage drop at the motor frame for an earth fault at the motor in the circuit shown in FIGURE 1. The phase voltage is 240 V. FIG. 1 240 V Motor Fault Motor control centre Switchgear Transformer Earth continuity conductor 0.01Ω 0.0002Ω 0.0002Ω 0.0002Ω 0.0002Ω
- 2. 0.0002Ω 0.0002Ω TXr winding Teesside UniversityOpen Learning (Engineering) © Teesside University 2011 2. An 11 kV motor is fed by cables from a transformer via switchgear, having a phase impedance of 0.3 + j0.3 ohm. The earth return path to the transformer neutral has a resistance of 0.42 ohm. Determine a suitable value of neutral resistance if the voltage rise at the motor in the event of an earth fault at the motor is not to exceed 430 V. Neglect transformer winding resistance. 3. Using TABLE 1, determine the resistance of a conductor and the volt drop per ampere metre at 70°C for a 95 mm2 pvc/swa/pvc copper cable. The temperature coefficient of resistance is 0.004 per °C.
- 3. TABLE 1 Resistancesof Copper Conductor and Steel Wire Armour for 3-core copper PVC Insulated Cables at 20°C. Copper Conductor Size Resistance/metre of Conductor (microhms per metre) Resistance/metre of Steel Armour (microhms per metre) 6 mm2 10 mm2 16 mm2 25 mm2 35 mm2 50 mm2 70 mm2 95 mm2
- 4. 120 mm2 150 mm2 185mm2 240 mm2 300 mm2 3080 1830 1150 727 524 387 268 193 153 124 99 75 60
- 5. 4600 3700 3200 2400 2100 1990 1400 1200 1100 740 680 600 540 Teesside University OpenLearning (Engineering) © Teesside University 2011 4. A load, 100 m from a 415 V Supply Authority point is connected by a
- 6. 35 mm2 pvc/swa/pvccopper cable. The current rating is 119 A and the volt drop per ampere metre is 1.1 mV. If the load current is 90 A, determine whether the volt drop at the load is within the 17th Edition of the IEE Wiring Regulations 5% limit. Using TABLE 1, calculate the earth loop impedance, minimum fault current (neglect the supply/earth impedance) and select an appropriate fuse for the circuit with reference to B.S. 88 part 2 from Figure 3 Below. Teesside University Open Learning (Engineering) © Teesside University 2011 FIG. 3 6 A 20 A 32 A
- 7. 50 A 80 A 125A 200 A 315 A 500 A 800 A 100 101 102 103 104 105 Time, s P ro sp ec ti ve
- 8. c u rr en t, r m s a m p er es 10–2 10–1 100101 102 103 104 Teesside University Open Learning (Engineering) © Teesside University 2011 1. (a) State two disadvantages of Time Graded Protection Systems. (b) State two advantages and a disadvantage of Differential Protection Systems.
- 9. (c) What isthe preferred secondary current rating of a CT for a Differential Protection System? Why is this rating preferred? (d) State the minimum time grading interval between two relays and give reasons for this interval. (e) Complete the following: 1. Desired relay operating time = TMS × .................... 2. Saturation can be a feature of CTs and .................... (f) State the two rules of successful time grading of relays.55 5 Calculate the primary setting current and TMS for an inverse- time overcurrent relay in a system where the maximum load is 400 A, the maximum fault current is 7000 A and CT ratio is 600/1, if the relay has to operate in 0.7 s. Assume settings are 50% to 200% in 25% steps. (110% of max. load has to be catered for.) Use the BS curve of an inverse - time overcurrent relay, FIGURE 1 for reference.
- 10. 2 Teesside University OpenLearning (Engineering) © Teesside University 2011 FIG. 1 B.S. Characteristic Curve of an Inverse - Time Overcurrent Relay 1 2 3 5 10 20 30 40 100 1 2 3 5 10 20 100 Multiple of setting T
- 11. im e (s ec s) Teesside University OpenLearning (Engineering) © Teesside University 2011 6. For the radial power system shown in FIGURE 2, the outgoing circuit at C is fused and will operate in 0.1 s for the fault level at C. The fault level at B is 6000 A. If the required discrimination time between the relay at B and the fuse at C is 0.3 s, calculate a suitable TMS setting for the relay at B. Assume CT settings are from 50% to 200% in 25% steps. 110% of maximum load is to be catered for. FIG. 2 7 FIGURE 3 shows the use of unit protection where differential
- 12. relays are protecting transformers.For each of the circuit breaker configurations: (i) CB A open and CB B closed (ii) CB A closed and CB B open (all other CBs closed) (a) To a base of 10 MVA determine initially the fault levels at the 11 kV and 3.3 kV busbars for a fault at the transformer, T1, output as shown. (b) Determine the current settings and TMSs for the relays shown if the operating time of the directional overcurrent relays on the 3.3 kV side of the transformers is 0.5 s. (Use a 0.4 s discrimination interval.) State with which of the CB configurations the directional overcurrent relays (if any) will not operate. BA 800/1 400/1 300 A
- 13. max load 300 A maxload C Teesside University Open Learning (Engineering) © Teesside University 2011 FIG. 3 G G B A 11 kV 3.3 kV 5 MVA 10% 5 MVA 10% T2T1 CT 250/1 CT 250/1 Non-directional
- 14. overcurrent relays CT 1000/1 Directional overcurrentrelays CT 1000/1 t = 0.5s Fault current Fault current Fault current Fault current G2G1 8MW, 0.8pf 20% F t = 0.5s 8MW, 0.8pf 20% Teesside University Open Learning (Engineering) © Teesside University 2011
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