Electrical Q&A

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  Guideline to Design Electrical Network for Building/Small Area.  (1) Calculate Electrical Load: ã Find out built up area in Sqft.of per flat per House/Dwellin unit. ã !ultipl area in Sqft. b Load/Sqft accordin to followin #able Type of LoadLoad/Sft $ndustrial1%% &att/SqftCo''ercial% &att/SqftDo'estic1 &att/Sqft ã *ppl t+e di,ersit factor and Co'pute t+e load of all dwellin units in t+e area. Type of LoadDi!ersity actor $ndustrial%.Co''ercial%.-Do'estic%. ã *dd t+e load of co''on ser,ices suc+ as *uditoriu' Street Li+ts Lifts and &ater 0u'ps etc. For si'plicit purpose %.&/dwellin units 'a be considered as co''on load. ã Co'pute t+e 2#otal Load3 of t+e area b addin load obser,ed at abo,e. ã *ppl t+e power factor of %.- to deter'ine t+e load in 4*. ã Co'pute t+e Load in 4*5 2#otal Load3/%.- ã Take transformer loading of #$%  considerin t+e networ arrane'ent 6in !ain Circuit.  &'( Decide !oltage grade for Electrical Load) ã $f load is equal to or 'ore t+an 7.%!4* t+e area s+all be fed t+rou+ 4 feeder. For suc+ loads t+e land space for /114 Sub8station s+all +a,e to be allocated b builder / Societ / *ut+orit . ã For load between 1 !4* to 7.!4* dedicated 114 feeder s+all be preferred. ã For load below 1 !4* e9istin 114 feed can be tapped t+rou+ 4C or 6!;.() Decide Si<e of #ransfor'er: ã Select #.C Si<e of 7 =4*> =4*1%% =4*7%% =4* or %% =4* accordin to our Load. ã #+e 'a9i'u' capacit of distribution transfor'er acceptable is %% 4* as a standard capacit . ã ?nl two8no of transfor'er at one location s+all be acceptable. $f t+ere is 'ore nu'ber of transfor'ers H# s+all be required to e9tend usin underround cables to locate additional transfor'er.() 6!; / L# 0anel: ã Eit+er 4C or 6in !ain Circuit s+all be used to control transfor'ers. #+ere cables s+ould +a,e 'eterin arrane'ent at 114. #+e protection s ste' at inco'in suppl s+all be usin nu'erical rela s.  ã ?n L# side of transfor'er L# 'ain feeder pillar s+all be pro,ided. #+e $nco'in s+all be protected b !CC/SF;. ã #+e distribution pillar8bo9 s+all be connected into 6in !ain ;nit. ã #+e inco'er of distribution pillar s+all +a,e !CC / SF;. #+e outoin s+all +a,e H6C fuses.() #+e L# cables fro' #.C to L# panel / !ain feeder pillar: ã Decide Si<e of L# Cable fro' #.C to L# 0anel as per followin #able. Transformer Si*e+a,le >%4* transfor'ers7 no 9 1C 9 >% Sq '' *l @L0E Cable%%4* transfor'ers1 no 9 1C 9 >% Sq '' *l @L0E7%4* transfor'ers A C 9 %% Sq '' *l @L0E1>%4* transfor'ers A C 9 %% Sq '' *l @L0E1%%4* transfor'ers A C 9 1% Sq '' *l @L0E(>) Considerin ,arious Factors B Lent+ of Cable: ã #+e factors for ca,le loading s-all ,e taken as %. ã #+e factor for multiplicity of ca,les from same ca,le trenc- s-all ,e 0%. ã #+e suested ma1imum lengt- of LT ca,le feeder s-all ,e '$ 2trs . ã #+e L# cables s+all be connected in rin 'ain circuit. ã #+e load on sub8feeder pillar s+all be restricted to 1%&.() L# cables fro' 'ain feeder pillars to distribution pillar bo9es: Load on distri,ution pillar   LT +a,le Si*e ;p to %& A C 9 1% sq'' *L @L0E;p to 1%%& A C 9 %% sq'' *L @L0E;p to 1% & A C 9 %%sq'' *L @L0E(-) Calculate 4oltae Drop and #BD Losses: ã #+e entire s ste' +as to be desined for a !oltage drop of '.% fro'   114 Side of transfor'er to 'eterin equip'ent at end consu'er pre'ises. ã #+e entire s ste' +as to be desined for T3D losses of ser!ice ma1imum '.% from 114 to end consu'er 'eter includin of ser,ice cable.  &4( 5-y EL+B can6t work if Neutral input of EL+B do not connect to ground7  ã ELC is used to detect eart+ leaae fault. ?nce t+e p+ase and neutral are connected in an ELC t+e current will flow t+rou+ p+ase and t+at 'uc+ current will +a,e to return neutral so resultant current is <ero. ã ?nce t+ere is a round fault in t+e load side current fro' p+ase will directl pass t+rou+ eart+ and it will not return t+rou+ neutral t+rou+ ELC. #+at 'eans once side current is oin and not returnin and +ence because of t+is difference in current ELC will trip and it will safe uard t+e ot+er circuits fro' fault loads. $f t+e neutral is not rounded fault current will definitel +i+ and t+at full fault current will co'e bac t+rou+ ELC and t+ere will be no difference in current. &'( 5-y in a t-ree pin plug t-e eart- pin is t-icker and longer t-an t-e ot-er pins7 ã $t depends upon 65rl/a w+ere area(a) is in,ersel proportional to resistance (6) so if (a) increases 6 decreases B if 6 is less t+e leaae current will tae low resistance pat+ so t+e eart+ pin s+ould be t+icer. ã $t is loner because t+e #+e First to 'ae t+e connection and Last to disconnect s+ould be eart+ 0in. #+is assures Safet for t+e person w+o uses t+e electrical instru'ent. &8(5-y Delta Star Transformers are used for Lig-ting Loads7 ã For li+tin loads neutral conductor is 'ust and +ence t+e secondar 'ust be star windin. and t+is li+tin load is alwa s unbalanced in all t+ree p+ases. ã #o 'ini'i<e t+e current unbalance in t+e pri'ar we use delta windin in t+e pri'ar . So delta / star transfor'er is used for li+tin loads. &9(5-at are t-e ad!antages of star:delta starter wit- induction motor7 ã (1)#+e 'ain ad,antae of usin t+e star delta starter is reduction of current durin t+e startin of t+e 'otor. Startin current is reduced to 8 ti'es ?f current of Direct online startin. ã (7) Hence t+e startin current is reduced  t+e ,oltae drops durin t+e startin of 'otor in s ste's are reduced. &$(5-at is meant ,y regenerati!e ,raking7 ã &+en t+e suppl is cut off for a runnin 'otor it still continue runnin due to inertia. $n order to stop it quicl we place a load (resistor) across t+e ar'ature windin and t+e 'otor s+ould +a,e 'aintained continuous field suppl . so t+at bac e.'.f ,oltae is 'ade to appl across t+e resistor and due to load t+e 'otor stops quicl . #+is t pe of breain is called as  26eenerati,e reain3. &#(5-en !oltage increases t-en current also increases t-en w-at is t-e need of o!er !oltage relay and o!er current relay7 +an we measure o!er !oltage and o!er current ,y measuring current only7  ã o. &e cant sense t+e o,er ,oltae b ust 'easurin t+e current onl because t+e current increases not onl for o,er ,oltaes but also for under ,oltae (*s 'ost of t+e loads are non8linear in nature).So t+e o,er ,oltae protection B o,er current protection are co'pletel different. ã ?,er ,oltae rela 'eant for sensin o,er ,oltaes B protect t+e s ste' fro' insulation brea down and firin. ?,er current rela 'eant for sensin an internal s+ort circuit o,er load condition eart+ fault t+ereb reducin t+e s ste' failure B ris of fire. So for a better protection of t+e s ste'. $t s+ould +a,e bot+ o,er ,oltae B o,er current rela . &(;f one lamp connects ,etween two p-ases it will glow or not7 $f t+e ,oltae between t+e two p+ases is equal to t+e la'p ,oltae t+en t+e la'p will low. ã &+en t+e ,oltae difference is bi it will da'ae t+e la'p and w+en t+e difference is s'aller t+e la'p will low dependin on t+e t pe of la'p. &0( 5-at are <=+ fuses and w-ere it is used7 ã H6C stand for 2+i+ rupturin capacit 3 fuse and it is used in distribution s ste' for electrical transfor'ers &>(2ention t-e met-ods for starting an induction motor7 ã #+e different 'et+ods of startin an induction 'otor ã D?L: direct online starter ã Star delta starter ã *uto transfor'er starter ã 6esistance starter ã Series reactor starter &4(5-at is t-e difference ,etween eart- resistance and eart- electrode resistance7 ã ?nl one of t+e ter'inals is e,ident in t+e eart+ resistance. $n order to find t+e second ter'inal we s+ould recourse to its definition: ã Eart+ 6esistance is t+e resistance e9istin between t+e electricall accessible part of a buried electrode and anot+er point of t+e eart+ w+ic+ is far awa . ã #+e resistance of t+e electrode +as t+e followin co'ponents:(*) t+e resistance of t+e 'etal and t+at of t+e connection to it.() t+e contact resistance of t+e surroundin eart+ to t+e electrode. &44(5-at is t-e power factor of an alternator at no load7
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