Electric Circuits Assignment
Added on - 28 May 2020
Showing pages 1 to 3 of 7 pages
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. Thediagram should show the total earth fault loop path for the charging unit2.Suitable luminairesSpaceSuitable 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 resultfrom 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 anoverload on an electrical circuit, which may result from short-circuiting, failure of the device, oroverloading 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 installationreferenceReferenceMethod AReferenceMethod AReferenceMethod AReferenceMethod AReferenceMethod ARating factors applicable20 A20 A40 A20 A40 AMinimum csa of liveconductors13.5 mm213.5 mm215.5 mm213.5 mm215.5 mm2Actual voltage drop45.9245.9258.88045.9258.88Maximum permissibleconnection time0.4 s0.4 s5 s0.4 s5 sEarth fault loopimpedance1.96 ohms3.38 ohms3.38 ohms1.96 ohms3.38 ohmsMaximum earth loop2.87 ohms2.87 ohms1.84 ohms2.87 ohms1.84 ohms
impedance5.The minimum cross-sectional area required for the CPC of one of the circuits supplyingthe fridge is determined by the equations=√tI2kmm2where I= the fault current, t= the opening time of the protective device, and k thefactor 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 impedanceassumption 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 mostinstallations and assuming that the cable is made from copper, the value of k=1151s=√0.4×1.962115mm2=0.3554 mm226.CircuitDemandDiversity factorDemand following1Hall, F. (2015).Building Services Handbook.London: Routledge.2Steffy, G. (2012).Architectural Lighting Design.New York: John Wiley & Sons.