Electric Circuits Assignment - Desklib
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Added on 2020-05-28
Electric Circuits Assignment - Desklib
Added on 2020-05-28
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1.Draw a fully labeled circuit diagram illustrating the three-phase 400 V supply system andearthling arrangement from the substation transformer through to the charging unit. The diagram should show the total earth fault loop path for the charging unit2.Suitable luminairesSpace Suitable luminaireEateryCeiling up lighter, wall mounted double light, halogen lightServer/kitchenFluorescent lightHotel roomHalogen spot light, wall mounted lightSwimming poolCeiling down lighter3.Circuit protection devices
Circuit Breakers: This device would prevent the breakage of an electrical circuit that may result from an overcurrent, which is in most cases associated with either an overload or a short circuit. Circuit breakers disrupt the flow of current upon detection of a fault. Electronic Fuses: A fuse is a resistor of low resistance, which offers protection in case of an overload on an electrical circuit, which may result from short-circuiting, failure of the device, or overloading of the circuit. 4.Design current=InCorretionFactorZs=Ze+R1+R2Zs=Zt×UU210WhereZs =is the earth fault loop impedance required for safetyZt=is the tabulated value of earth fault loop impedanceU=is the actual supply voltageU240 =is the supply voltage assumed in the Table.Circuit230 V 16 A400 V 16 A400 V 32 A240 V 16 A400 V 32 ADesign current0.4571 A0.4571 A0.625 A0.4571 A0.625 ANominal rating16A16A32A16A32AMethod of installation referenceReference Method AReference Method AReference Method AReference Method AReference Method ARating factors applicable20 A20 A40 A20 A40 AMinimum csa of live conductors13.5 mm213.5 mm215.5 mm213.5 mm215.5 mm2Actual voltage drop45.9245.9258.88045.9258.88Maximum permissible connection time0.4 s0.4 s 5 s0.4 s5 sEarth fault loop impedance1.96 ohms3.38 ohms3.38 ohms1.96 ohms3.38 ohmsMaximum earth loop 2.87 ohms2.87 ohms1.84 ohms2.87 ohms1.84 ohms
impedance5.The minimum cross-sectional area required for the CPC of one of the circuits supplying the fridge is determined by the equations=√tI2kmm2 where I= the fault current, t= the opening time of the protective device, and k the factor which depends on the conductor of the material and the insulationI=Uo/Zs where Uo= the supply voltage and Zs=the earth-fault loop with zero impedance assumption for the faultZs=Ze+R1+R2=0.11+1.85=1.96 ohmsUo=240 VThe opening time, t, for the protective device would be 0.4 seconds as in the case for most installations and assuming that the cable is made from copper, the value of k=1151s=√0.4×1.962115mm2=0.3554 mm226.Circuit DemandDiversity factorDemand following 1Hall, F. (2015). Building Services Handbook. London: Routledge.2Steffy, G. (2012). Architectural Lighting Design. New York: John Wiley & Sons.
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