LV generator protection - engineering site - Schneider Electric

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3.2. Downstream LV network protection3.2.1. Priority circuit protectionChoice of breaking capacityThis must be systematically checked with the characteristics of the main source(HV/LV transformer).Choice and setting of the Short Time Delay releasesb subdistribution boardsthe ratings of the protection devices for the subdistribution and final distributioncircuits are always lower than Generator Set rated current. Consequently, exceptin special cases, conditions are similar to supply by the transformer.b main LV switchboardv the sizing of the main feeder protection devices is normally similar to that of theGenerator Set. Setting of the STD must allow for the short-circuit characteristic ofthe Generator Set (see 3.1.2.).v discrimination of protection devices on the priority feeders must be provided ingenerator set operation (it can even be compulsory for safety feeders).It is necessary to check proper staggering of STD setting of the protectiondevices of the main feeders with that of the subdistribution protection devicesdownstream (normally set for distribution circuits at 10 ln).Note: when operating on the Generator Set, use of a low sensitivity RCDenables management of the insulation fault and ensures very simplediscrimination.3.2.2. Safety of peopleIn the IT (2 nd fault) and TN grounding systems, protection of people againstindirect contacts is provided by the STD protection of circuit-breakers. Theiroperation on a fault must be ensured, whether the installation is supplied by theMain source (Transformer) or by the Replacement source (Generator Set).Calculating the insulation fault currentZero-sequence reactance formulated as a % of Uo by the manufacturer x’o.The typical value is 8 %.The phase-to-neutral single-phase short-circuit current is given by:The insulation fault current in the TN system is slightly greater than the threephasefault current: for example, in event of an insulation fault on the system inthe previous example, the insulation fault current is equal to 3 kA.19
Protection and Monitoringof a LV Generator Set3.3.The monitoring functionsDue to the specific characteristics of the generator and its regulation, the properoperating parameters of the Generator Set must be monitored when specialloads are implemented.The behaviour of the generator is different from that of the transformer:b the active power it supplies is optimised for a power factor = 0.8b at less than power factor 0.8, the generator may, by increased excitation,supply part of the reactive power.3.3.1. Capacitor bankAn off-load generator connected to a capacitor bank may self-arc, consequentlyincreasing its overvoltage.The capacitor banks used for power factor regulation must therefore bedisconnected. This operation can be performed by sending the stopping setpointto the regulator (if it is connected to the system managing the source switchings)or by opening the circuit-breaker supplying the capacitors.If capacitors continue to be necessary, do not use regulation of the power factorrelay in this case (incorrect and over-slow setting).3.3.2. Motor restart and re-accelerationA generator can supply at most in transient period a current of between 3 and 5times its nominal current.A motor absorbs roughly 6 ln for 2 to 20 s during start-up.If Σ Pmotors is high, simultaneous start-up of loads generates a high pick-upcurrent that can be damaging: large voltage drop, due to the high value of theGenerator Set transient and subtransient reactances (20 % to 30 %), with a riskof:b non-starting of motorsb temperature rise linked to the prolonged starting time due to the voltage dropb tripping of the thermal protection devices.Moreover, the network and the actuators are disturbed by the voltage drop.ApplicationA generator supplies a set of motors.Generator short-circuit characteristics: P N = 130 kVA at a power factor of 0.8,ln = 150 AX’d = 20 % (for example) hence lsc = 750 A.b the Σ Pmotors is 45 kW (45 % of generator power)Calculating voltage drop at start-up:Σ Motors = 45 kW, l M = 81 A, hence a starting current ld = 480 A for 2 to 20 s.Voltage drop on the busbar for simultaneous motor starting:∆U≈UI N - I dI cc - I Nen %∆U ≈ 55 %which is not supportable for motors (failure to start).b the Σ Pmotors is 20 kW (20 % of generator power)Calculating voltage drop at start-up:Σ Motors = 20 kW, l M = 35 A, hence a starting current ld = 210 A for 2 to 20 s.Voltage drop on the busbar:∆U≈UI N - I dI cc - I Nen %∆U ≈ 10 %which is supportable but high.20
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