Question

1 a) Please briefly explain the basic idea of analyzing asymmetric short circuit. b) Is the symmetrical component method suitable for asymmetrical systems? Why? c) How to list the six equations for analyzing the Asymmetric Short Circuit in power system (without writing the specific equation)? What are the six unknowns? d) Please start with the meaning of the parameters of the transformer equivalent circuit, explain the similarities and differences of the transformer sequence parameters. e) Please briefly explain in text why the zero-sequence excitation reactance of the transformer is related to the iron core structure? f) What is the relationship between the positive sequence, negative sequence and zero sequence reactance of the line? why? (Without formula deduction, with refined language) g) Why does the integrated load generally not need to establish a zero sequence equivalent circuit?

Answer 1

b) answer

Yes we use symmetrical component method for asymmetrical system
SYMMETRICAL COMPONENTS load voltage. The performance of given diagram can be analyse by knowing the Balanced load : 120 + 8y L120 + 48 (240 co. In = o. The load voltage can be calculated by using per ph. reactance deagram along with po valveg. m -> (less severa) upu = Epo- & Teq pu ter = (x+x+x) AU (single ph. deogram is sufficient to cale ladie mast balance. internal con £10% fl current). unbalance External (In>10% 1. faulta (Requiel Symon.compa to cale food vot. ). (Severe) of load is externally unbalance due to occurence. of faulty then the electrical quantitics which are associate in each ph. are severely onbalance and for evaluate toad volt consider individual ph. diagram and 3 110 eq.e to be solved. (for Varu, you, sou). In order to reduce while evaluating unbalanced electrical quantities it iç preferred to expressed by a set of 3 balanced electrical quantities namely (Symmetrical quartificą . they are 0 +ve seq. componente (1) ③ -ve seg. Componentę (2), 3. zero seq. compo oco). • VR = Ro + VR + time taking

IR not conect Vy Vyo.+ y + y2 VB VBO & VBI + VB2 Here VK, Vy, ve } unbalanced quantities VRO, VR, VR, Vyo.... } balanced quantities IR PRo & & Rit But by deo + 8 + ву. PR Prothetke B IBO & & BI & B2 in ph. manner Adr. of sym. compos is: reduce the no. of calces so that time taking is less. AS sm. comp.g balance quantities so B&Y сотря are expressed in a ph. by using k. 'k is defined as unity magretude, andersoek 120 ph. desplacement in anticw. direction. k = 1 (120 power is not calco arc = - 0.5 + JO.867 K² = 1 (240 > K²+k²+ k = 0. 1+k'+k=0. kt = ksk = 1k & 14126. ko kek² = 1.4² 11240 FAULTS 1 olc faulty (scrieç faults) (less frequent Saulde) sic faulty e shunt faulty) (frequent faults)
a) answer

most Cidomon LG OM → more everton All LLL Forsyningtrical olc faults (serieg faulte ) 11). Metting of one of the phage fuge ob opening of CB (ii). Melting of fuse in two phs or opening of CB in two ph.s. These are unsymmetrical of un balanced faulte. sl faulty (shunt faulty ) Sources: on Symonpetricat (ii). LL falting of tree branches, un balance (iii). LLG flashover of string of insulators collapsing of line supports. w). LLLG in the LG occurs at alter- balance nator terminale severe Most common - LG. both in Alternator Sub transient (ist Cycle when fault occurs). у 2 of failty can be characterized as, 41. incrcasing nh. voltages 12). falling of this corrent. 13). Slight improvement in it 14). The rige in the volt-e of this will increage the working field intensity which will result a failure of dictectric strength of insulation. - most B & Th. E B 11 of treg of poxy. Supply

13). 44). so otc fault means study of ve of pha and they are to be expressed in peak value 6'coz behaviour of fault is sub transient. at a sic fault takes, then resulty ag 1. fall in phie volt. (volt-s of phages ). (2) Rice in corrents of the phagee. Reduction of it Reduction of free of supply Due to rige in the corrents of this more temp will be generated in the insulation 35 / 137 and if operating temp. is more than temp. of insulation then insulation will fail. so ste fault is the study of currents of phases and expressed in terme me, even though it is subtransient period (Hect-tR+) the other way to say fault coment is high 6lcoj sob. transient reactance for the How of fault i is very less when compared to steady state reactance of the system, o xd" = sub transient x 11x dampli xa ха 1st cycle.

Xi' - Fransient reactance eesee xd having the equal and cycle Steady state reactance xa. 4 cycle. xdcxdcxd & ve seq. components: These are and magnitude 120 ph.displace- ment and ph. seq. is same that of oreginal ph. seq. of the lo -ve sea components: These are the components haung equal magnitute and 128 ph displacement and ph. seq. iq opposite to that of original ph. seq. of the oleo. zero feq. components: These are having equal magnitude without any ph displacement. If there is no ph. displacement then no phace sequence. Due to rotor airgap flux cwhich is assumed in ca, ent induced in stator wag, then this will able to deliver current to load the corr. field produced by its is assumed in the same dire. as that of field il cu.

1) at the dire. of rotation of stator field is same ag rotor field then such seq. is called ave seg. The reactance offered for the flow of i's is steady state reactance Ya Before fault JBL VR - VRI Even during the fault, there is no change in air gas flux of rotor, so it will induce an emt. in the stator wig, which results the corr-i iç delivered to the point in the same dire as that of original dire. but it is offered with sub trans sient reactance xd", so +ve scq docg existe before fault as well as during fault Ver 1 VR₂ Ivy v 1B gero B2 bakmad sea Vyz B1 RYB unbalanced RYB. balanced tve RBY balanced - ve Before fault +ve seq. compo. does exists. and also during any type of fault tve sea comp. also existing and it will helpful in order to set
answe c)

Due to rotor field, the volt cwhich induced in stator wag is only the seq. voltage. The mag. of -ve -ve scq volt at the fault point is always more than the corr stator. adg the cor corrent flows from the sault point towards source of generator. The field produced by - ve seg. comp.s. that of the field produced by the seq in the stator. However worot rotor et is having relative speed of 2Ns and it will as a current is induced in rotor at double the free and rotor field 19 opp. to YCjult

udg over heated an order to protect rotor field due to over bating -ve. seq. relay is employed in the stator wag. ZSC will exists provided that, (a) fault is ground fault (6). Neutral of the system is grounded of the fault is associated with ground & neutral of the system is grounded and fault is flowing to ground and enter into the system through N- grounding. Ground can be provide mag onthout any ph. displacement. VRO + VRI+ VR2 Vy Vyo & Vyl + y2 V Ro tku VB VBO + VBI + VB2 VRO + K. VR + K². VR - 1 k² k VR Vy VB - k k? Vre R 1 K² k - dy &B k K² dre -

Я ее 1 - - de IRI 1/3 IK ly 3 SRT k² k
answer d)

Zero seq. nols of THE To include no effect, the Tit can be made equivalent of mech. Switches ie serieg - paralel Switches on both side of TIF. 5000002 Xro po series switch - y - open wag Till switch shunt way (closed). Y scries switch open (3x = 0) 4 close (3x =o) Y close (3x). - scrieq switch series switch there is shunt switch. always cloge no closed path for the gero - seq. corrente. YY T000000 XTO 3x, o Xoeq = xoto

Co000007 Д- 1 Хто 3xp end Хое = xъ+ 3x, . 0000 3х, об Хто у - Д elecer v 1-4 3ҡ, =0 wolle — А 00 зхо 000 Хто Д-д toeg х, + об

911 01000 ) , 3x = a XTO 3xpo Хогу = х +0 Wells Y 3х, то 00 3ко Xory=x+, 3х, МУ 0000 Хто 3х, но 3x = Teele МУ 0) ЗХ, хто 3xp = 00 47 / 137 4) 0000 Хто 4 Хоер = xo+o Igleo Islo =xo ter eller Хто 30 хь +3х, 1st Islao relle жто 3х, со ДУ Xoeg хто relleb Хто sx=0 ДИ Sea = x +0 хто -

( short pitch with Relations among reactances : among seq tve ove Jero 11 Purbo mlc (P-26 Unform salient pole mic Non-uniform x" x < x, XGO < xg" air gap flux Certy air gap Hlux Gransformer Xx2 = X Хто хт Transmission line tu X 1₂ = x4 X - 3XL
subscript 0 ,1,2indicates zero, posstive, negative sequence components