<p>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.</p><p>TABLE 1 Resistances of Copper Conductor and Steel Wire Armour for 3-core copper PVC Insulated Cables at 20°C.</p><p>Resistance of 95mm2 copper conductor = 193micro-ohm/meter at 200C.</p><p>Here α = 0.004 per degC </p><p>Hence, using Rt =R0(1+αt), we get</p><p>193 = R0(1+20α) = R0(1+0.004*20) ….(1)</p><p>R70=R0(1+70α) = R0(1+0.004*70) ….. (2)</p><p>Dividing (2) by (1), we get </p><p>R70 = 193*(1+0.004*70)/(1+0.004*20) = 228.74 micro-ohms/meter.</p><p>Hence, voltage drop = 228.74 micro-volt/amp-meter at 700C. </p><p>4. A load, 100 m from a 415 V Supply Authority point is connected by a 35 mm2 pvc/swa/pvc copper 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.</p><p>Voltage drop = 1.1*90*100 milli-volt</p><p>= 9.9 V</p><p>= 9.9/(415/√3) </p><p>= 0.0413 or, 4.13%</p><p>Hence, voltage drop is within permissible limits. Armour resistance = 2100 micro-ohms / meter at 200C for 35mm2 size as seen from the given table. </p><p>Length = 100m</p><p>Hence, armour resistance = 2100*10-6*100 = 0.21ohms.</p><p>Conductor resistance = V/I = 9.9/90 = 0.11ohms.</p><p>Minimum earth fault current will occur when fault takes place at the far end of the conductor. </p><p>Hence, earth loop impedance = 0.11+0.21 = 0.32ohms.</p><p>Hence, minimum earth fault current = (415/√3)/0.32 = 748.75A</p><p>Maximum earth fault current will occur when armor resistance is not coming in current flow path.</p><p>Hence, maximum earth fault current = (415/√3)/0.11 = 2178.19A.</p><p>From the figure, we see that 125A fuse will be suitable for this requirement. </p><p>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.</p><p>Primary setting current = ?</p><p>TMS = ?</p><p>Maximum load current = 400*1.1 = 440 considering 110% of maximum load.</p><p>Hence, required plug setting = 440/600 = 0.7333. Pick up setting of over current relay = 0.7333</p><p>Fault current = 7000A</p><p>CT secondary current under fault condition = 7000/600 = 11.67A</p><p>Hence, plug multiplier setting = 11.67/0.7333 = 15.914, say 15.91.</p><p>From given figure, we see that relay takes 2.5 secs to operate at PMS = 15.91 but we want that relay should operate in 0.7 secs.</p><p>Hence, TMS = 0.7/2.5 = 0.28.</p><p>We can calculate it from formula also.</p><p>Time to operate for IDMT relay </p><p>Hence, same as obtained from graph. </p><p>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.</p><p>Time for relay to operate at B = 0.1+0.3 = 0.4 secs </p><p>Maximum load = 300*1.1 = 330A considering 110% of maximum load.</p><p>Overload relay plug setting = 330/400 = 0.825 as CT ratio is 400/1. </p><p>Fault current = 6000A</p><p>CT secondary current under fault condition = 6000/400 = 15A</p><p>Hence, plug multiplier setting = 15/0.825 = 18.18</p><p>Putting in formula: Time to operate for IDMT relay, we get From given figure, we see that relay takes 2.3 secs to operate at PMS = 18.18 but we want that relay should operate in 0.4 secs.</p><p>Hence, TMS = 0.4/2.3 = 0.17 same as obtained using formula. </p><p>7 FIGURE 3 shows the use of unit protection where differential 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.</p><p>(i) (a) CB A opened and CB B closed. </p><p>At 10MVA base, transformer reactance = 10*(10/5) = 20%=0.2pu</p><p>Generator MVA = 8/0.8 = 10</p><p>Generator impedance = 20% or 0.2pu at 10MVA. </p><p>We can draw the following figure showing fault location in pu reactances. </p><p>Bothe generators will feed the fault. Reactance from G1 up to fault = 0.2+0.2 = 0.4 pu.</p><p>Same reactance is from G2 up to fault.</p><p>Hence, total reactance = 0.4/2 = 0.2 pu as 0.4 pu reactances are connected in parallel. </p><p>Hence, fault level = base MVA/pu reactance = 10/0.2 = 50MVA. (b) Fault contribution from G2 = 10/0.4 = 25MVA</p><p>Fault contribution From G1 = 10/0.4 = 25MVA</p><p>Rated MVA for T1 = 5MVA</p><p>Hence, PMS for relay on 11KV side of T1 = 25/5 = 5.</p><p>Time for relay to operate = 0.4+0.5 = 0.9 secs as relay on 3.3Kv side is operating in 0.5 secs and relay on 11KV side has to operate 0.5 secs after that. </p><p>Both relays on 11KV side will operate to clear the fault. </p><p>Hence, using formula, we get </p><p>Hence, TMS = 0.21 for 11KV side relays. </p><p>In this configuration, directional relay on 3.3KV side of T1 will operate. </p><p>(ii) (a) CB A closed and CB B opened.</p><p>We can draw the following figure showing fault location in pu reactances. </p><p> pu reactance = (0.2/2)+0.2 = 0.3pu ( generators’ reactances are connected in parallel and this equivalent is connected in series with transformer reactance.)</p><p>Hence, fault level = 10/0.3 = 33.33MVA </p><p>(b) Relay on 11KV side of T1 will operate. </p><p>Load on T1 = 5MVA</p><p>Fault level = 33.33 MVA</p><p>Hence, PMS = 33.33/5 = 6.67</p><p>Hence, using formula, we get </p><p>Actually, we can select operating time of this relay independently in this configuration as it need not be coordinated with any other relay. If we select its operating time as 0.5 secs, we will get </p>