1 / 35

Short Line Fault

Short Line Fault. Skeats Described: L:sourc.Inductance L 1 :line Induct. To F E :Instant. emf V PW =1/2 L 1 /[L 1 + L] E 2 travel B and F At F, sh. CCT. At C.B., open CCT. . Wedge-shaped Wave Formation . Initial Distribution After Wave Separation Each reflected α C.B. Ter.=1

willis
Télécharger la présentation

Short Line Fault

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Short Line Fault • Skeats Described: • L:sourc.Inductance L1:line Induct. To F E :Instant. emf VPW=1/2 L1/[L1+L] E • 2 travel B and F • At F, sh. CCT. • At C.B., open CCT.

  2. Wedge-shaped Wave Formation • Initial Distribution • After Wave Separation • Each reflected • α C.B. Ter.=1 • α F Ter. = -1 Voltage Distribution At different points • C.B. Ter.,middle,3/4to F

  3. Wave Reflections • At T and 2T • How 1 and 2 reflect • At Each end V found = V1(t)+V2(t) • At C.B. start: t=0,V=Vp t=T,V=0 t=2T,V=-Vp • T; travel time to F

  4. Wave Distribution • At C.B. Swing rate :2Vp/2T T:Very Small,atCB Fig1  Fast Rise of TRV Risk for C.B. • at Middle ----Fig2 • at ¾ to Fault--Fig3

  5. Solution of Sh. Line Fault application of Transmission Line Response • Vs= VR Coshjω√LC+IR Z0 sinhjω√LC • Is= IR coshjω√LC+ VR/Z0 sinhjω√LC • Zs= =

  6. Solution of Sh. Line Fault … • ZR=VR/IR • if a Sh. CCt. At Rec.ZR=0 • |Zsc|=Z0 tanh jω√LC • Substituting s for jω : • Zsc(s)=Z0 tanh√LC s • T=√LCtanhTs=sinhTs/coshTs= [1-exp(-2Ts)]/[1+exp(-2Ts)] • exp(-2Ts)=α  tanh Ts=(1-2α+2α^2- 2α^3+..) • Z(s)=Z0 [1-2exp(-2Ts)+2exp(-4Ts)-2…]

  7. Recovery Voltage Determination • Injecting minus fault current into CCT • I=-E/[L+L1] . T • i(s)=-E/[L+L1] . 1/s • response of Sh. Trans. Line : i(s)Z(s)=-E/[L+L1] Z0 1/s[1-2e^(-2Ts) +2e^(-4Ts)-…] • Inverse Transform results in TRV • 1st neg. ramp slope Z0 X current • Next terms ramp 2xslope of 1st delayed: 2T,4T,6T,… and alternate in Sign Sum a Saw tooth wave ∆t=4T, Vpp=E/[L+L1] Z0 2T Z0T=L1  Vpp=2EL1/[L+L1]

  8. CCT TRV • Response of shorted line Calculated • Response of source side: If source : by parallel L & C Rise a TRV of 1–cosine form T=2Π√LC & Amplitude EL/[L+L1] ,i.e.: Vsource =E L/[L+L1] {1-cos t/√LC} • VC.B.=Vsource-Vshort fault as in Figure

  9. Transformer Transient Model • Have the greatest exposure to electrical Transients (after Transmission line) • Different Model for specific studies • Transformer Model for: Switching on OPEN CCT • Im & Ih+e so a parallel RL CCT, & Cg capacitance of winding with length l R0=V/W0 • Cg/l per unit length & small fraction (Cg/l)∆x

  10. Transformer Equivalent CCT • As a load on Auto Transformer • Its Imp.:l/[ωCg∆x] • looking to A seen As : (l/x){l/[ωCg∆x]} • Related Cap.: Cg x ∆x /l • Ceff=Cg/l∫xdx • Integrating 0 to l Ceff=Cg/l[x/3]=Cg/3

  11. Other Models • Terminal Model & Detailed Model 1- Based on Name plate Information A paper, not discussed here 2-Software for Transformer Transient Study , not discussed here • Internal Model for a Transformer for special studies

  12. Internal Model • for our purpose Detailed one not desirable • Divide the time after surge into: 1- extremely short (about fraction of μs) 2- Next in range of m S and seconds 3- the steady state • In the fraction of μs no current penetrate to inductance only displacements currents to Capacitances

  13. Initial Distribution • Eq. CCT. • Cg:Cap to Gr Cs:Cap from end to end of winding • Gr Cap./unit length=Cg/l SeriesCap./length=Cs l E:voltage at any point on winding Ig=total current in Gr Cap Is=current in series Cap

  14. Elementary length of Winding • The Gr current: ∆Ig=Cg ∆x ωE/l (1) • Also : ∆Ig=dIs/dx ∆x (2) • With (1) & (2): dIs/dx=CgωE/l (3) • Series Cap. Of Elem. Length: l Cs/∆x

  15. Deriving Eeuations • Voltage across Element=dE/dx ∆x • series Cap. current=C dE/dt • Is=l Csω dE/dx (4) • Differentiating above Eq. : dIs/dx=lCsω dE/dx (5) Equate (3) & (5): dE/dx-1/l Cg/CsE=0 (6)

  16. Fast Response of winding to switching • The General solution of Eq.(6): E=A1 e^px + A2 e^-px • where p=1/l √Cg/Cs • A1& A2 from Boundary Conditions • NEUTRAL GROUNDED: x=0, E=0 and x=l, E=V • V amplitude of step surge

  17. Coefficients for Neutral Gr • A1+A2=0, A1e^pl+A2e^-pl=V A1=-A2=V/[ e^pl- e^-pl ] • If pl=√Cg/Cs=α • A1=-A2=V/[2sinhα] • E=V/[2sinhα] . {e^px – e^-px}= = V sinh(αx/l)/sinhα (7) • If Neutral Isolated: x=l , E=V & x=0 , Is=0 or dE/dx=0

  18. Coefficient of Neutral Isolated • Two Boundary Eqs: p(A1-A2)=0 A1e^pl + A2e^-pl=V • Coefficients are : A1=A2=V/[2cosh pl]=V/[2coshα] • The response : • E= V/2coshα [e^px + e^-px]= V cosh(αx/l) / coshα (8) Eqs (7) & (8) are Initial Voltage Distribution

  19. Discussion on responses • both depend on α=√Cg/Cs • More Nonuniform as α increases • α=10% to 60% voltage across the 1st 10% of winding at Line end • The Gradient at line terminal by differentiating: • a- Gr Neutral dE/dx=α/l V cosh(αx/l)/ sinhα; x=l  dE/dx=αV/l cothα • B-Neutral Isolated dE/dx=α/l V sinh(αx/l)/coshα x=l  dE/dx= αV/l tanhα

  20. Discussion Continued • For large values of α, cothα=tanhα=1 ; dE/dx=αV/l or is α times mean Gradient • Under Surge; Turns Insulation at Line End Severly Stressed • Slower fronted surges, capacitance not governing Voltage distribution • Winding Cap.:Geometery Dependent

  21. Discussion Continued • Different Winding Styles : Vary in α • Source of steep fronted surges: Flashover of an insulator, closing of a switch or C.B.,reignition in a switch,Fast Trans. in opening GISswitches,Lightning • Remedial Methods: strengthen End turn insulation; Failed placing metallic shields, Design interleaved Winding

  22. Initial and Approximate limit of Excursion • Ceff of a winding, excited by V at ω • I=ωCeffV , Is=ωCsl dE/dx [Is]l=ωCsl (dE/dx)l; (dE/dx)l=α(V/l) cothα • Thus: ωCeffV=ωCsαV cothα or : Ceff=αVscothα=√Cg/Cs Cs cothα • Typical α, cothα=1, therefore:Ceff=√CgCs

  23. A sample Transformer • Cg=3000 PF, Cs=30 PF • α=√3000/30=10 Ceff=√90000=300PF • For a Transmission Line Z0=400Ω charging Time const.=400x3x 10^-10=0.12μs • Points with transients: starts at initials, and finally Osc. About some Final value: Swing almost as far above steady value as starts below it In winding Final Dis. Uniform or it is the α=0 line Therefore excursions of any point lie within the envelope

  24. Lightning • Lightning greatest cause of outages: 1- 26% outages in 230 KV CCTs & 65% of outages in 345 KV Results of study on 42 Companies in USA & CANADA • And 47% of 33 KV sys in UK study of 50000 faults reports Also Caused by Lightning • Clouds acquire charge& Electric fields within them and between them

  25. Development • When E excessive:space Insulation Breakdown or lightning flash occur • A high current discharge • Those terminate on or near power lines • similar to:close a switch between cloud & line or adjacent earth • a direct con. Or through mutual coupling

  26. Lightning surge • Disturbance on a lineTraveling wave • Travel both Direction, 1/2IZ0 I: lightning current Z0=Surge Imp. Line • The earth carries a net negatve charge of 5x10^5 C, downward E=0.13KV/m • An equivalent pos. charge in space • Upper Atmosph. Mean potential of 300 KV relative to earth

  27. Lightning • Localized charge of thunder clouds superimposes its field on the fine weather field, freq. causing it to reverse • As charges within cloud & by induction on earth below, field sufficient Breakdown(30 KV/cm) • Photographic evidence:a stepped leader stroke, random manner &short steps from cloud to earth • Then a power return stroke moves up the ionized channel prepared by leader

  28. Interaction Between Lightning & Power System

  29. Assignment N0.4 (Solution) • Question 1 • 13.8 KV, 3ph Bus L=0.4/314=1.3 mH Xc=13.8/5.4=35.27 Ω, C=90.2μF • Z0=10√1.3/9.02=3.796Ω • Vc(0)=11.27KV • Ipeak=18000/3.796= 4.74 KA

  30. Question 1 • 1- Vp=2x18-11.27=24.73 KV Trap • 2- Assuming no damping, reaches Again the same neg. peak and 11.27KV trap • 3- 1/2 cycle later –(18-11.27)=-6.73 Vp2=-(24.73+2x6.73)=-38.19 KV

  31. Question 2 • C.B. reignites during opening&1st • Peak voltage on L2 L2=352,L1=15mH, C=3.2nF So reigniting at Vp, 2 comp.: • Ramp:Vs(0).t/[L1+L2]= 138√2x10/[√3(352+15)x10-]=0.307x10^6 t • Oscill.of : f01=1/2Π x {√[L1+L2]/L1L2C} • Z0=√{L1L2/[c(L1+L2)]} • component2:as Sw closes Ic=[Vs(0)-Vc(0)] /√{L1L2/[c(L1+L2)]} ≈2Vp√C/L1=104.1 A

  32. Question 2 continued • Eq of Reignition current I’ t + Im sinω0t which at current zero: sinω0t=-I’t/Im , ω0=1/√LC1=1.443x10^5 • Sin 1.443x10^5t=-0.307x10^6t/104.1=-2.949x10^3t • Sin 1.443x10^5t =-2.949x10^3t t(μs): 70 68 67 66.7 66.8 -0.6259 -0.3780 -0.2409 -0.1987 -0.1959 -0.2064 0.2005 -0.1376 -0.1967 -0.1966

  33. Question 2 • t=66.68μs I1=0.307x66.68=20.47 A • Vp=I1√L2/C=20.47x10.488=214.7 KV

  34. Question 3 • 69 KV, 3ph Cap. N isolated, poles interrupt N.Seq. • 160◦ 1st reignite • Xc=69/30=158.7 C=20μF,CN=0.02μF Vs-at-reig=69√2/3cos160 =-52.94 KV • Trap Vol.: • V’A(0)=56.34KV V’B(0)=20.62KV,V’C(0)= -76.96KV,VCN(0)=28.17KV • Vrest=56.34+28.17+52.94=137.45 KV

  35. Question 3 continued • Z0=√L/CN=√5.3x0.2 x100=514Ω • Ip-restrike=137.45/514=0.267KA=267A • F0=1/[2Π√LCN]=10^6/{2Π√53x2}=15.45 KHz • Voltage swing N=2x137.45=274.9 KV • VN=28.7-274.9=-246.73 KV • VB’=-246.73+20.6=-226.13 KV • VC’=-246.73+-76.96=-323.69 KV

More Related