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Transient Recovery Voltage

Transient Recovery Voltage. Following S.C. Removal L , and C: natural cap. Simplifying Assumptions. CCT’s R, and Losses ignored After sep. of contacts, I flows in Arc I reach zero by controlling arc In Ac two I=0 in each cycle Current : Symm. & compl. Inductive

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Transient Recovery Voltage

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  1. Transient Recovery Voltage • Following S.C. Removal • L , and C: natural cap

  2. Simplifying Assumptions • CCT’s R, and Losses ignored • After sep. of contacts, I flows in Arc • I reach zero by controlling arc • In Ac two I=0 in each cycle • Current : Symm. & compl. Inductive • At I=0 : VccT Max, VCB=Varc • Assuming Varc=0 • Time measu. from Inst. of Interr.

  3. CCT Equation • The KVL & Ic: L dI/dt+ Vc=Vm cosωt I=C dVc/dt (only I) • Physical InterPr.: - Ultimately: V supply • t=0, previous arc V=0 • C charged, through L & cause Osc.

  4. Similar to LC ccT analysis • Assuming: ω0=1/LC applying L.T.: • Vc(0)=0 arc vol. • Vc’(0)=I(0)/C=0

  5. Time Respose • Inv. Transf. just 1st term • Need: • Then: • Vc(t) is TRV Eq.:

  6. TRV Eq. (Park & Skeats) Discussion • As ω0 » ω , • Thus: • ∆V(P.F.) very small

  7. TRV Discussion Continued • Fig : TRVp=2 x P.F. Vp • TRV Osc.s damped out • C.B. Ops follows Cap. Being charged • VCB rises fast if: - L or C or both very small (ω0 large) □ if RRRV>RDSB : Reignition □ Then Switch passes If(another half Cycle) • TRV named “Restriking Voltage”

  8. Experimental TRV reults • The TRV lasts 600 μs • Decline of current • TRV starts a small opp. polar.To ins. Vol. • due to: some “Current Chopping” • Shows the H.F. Osc. • Shows how H.F. Damped

  9. r.r.r.v. factor • A measure of severity of CCT for C.B. • r.r.r.v.s high as Natural freq. higher • air-cored reactor L=1 mH,C=400 pF • F0=1/(2∏√10^-3x10^-10)=250KHz • T0=4μs, in T0/2 TRV swing to 2Vp • In a 13.8KV CCT , r.r.r.v.=2x13.8√2/(2x√3)=11.3KV/μs • Beyond Capab. Most C.B.s • Ex. Of very fact TRV: Kilometric Faults ch.9

  10. Interruption of Asymmetrical If • Sw. closes at random,I likely to Asym • And Degree of Asym. for If • Now C.B. opens at I=0,V not at peak • TRV now is not so High : Figure • R.V. osc. Around Vinst.(nolonger at Peak) • TRV is not as high

  11. TRV considerig C.B. Arc Voltage • If arc vol. not negl. • The inv. Trans. of term: • Vc(0)=arc vol.,I=0 • Increasing Sw.Tra • Effect: I more into Phase with supply voltage Fig.

  12. Assignment No. 1 • Question1 • C1, 120 KV • 1st S closed • 45μs later G • What is IR2?& Vc1? • C1=5μF,C2=0.5μF • R1=100Ω, • R2=1000Ω

  13. Solution of Question 1 • C1V1(0)+C2V2(0)=(C1+C2)Vfinal • Vfinal=C1/(C1+C2).V1(0)=600/5.5= 109. KV • With τ=100x0.4545=45.45μs • V2(t)=109.(1-e^(-t/45.45))=68. KV • Therefore IR2=68KV/1000=68A • V1(t)=120-11(1-e^(-t/45.45))=113.KV

  14. Question 2 • C1 has 1.0 C, C2 discharged • C1=60, C2=40μF • R=5 Ω • Ipeak? • I(t=200μs) ? • Eultimate in C2 ? • Vc1(ultimate) ?

  15. Solution of Question 2 • Ipeak=1x10^6/60/5=3.33KA • Ceq=24μF, τ=5x24=120μs • I(t=200μs)=3.33xe^(-t/120)=629.5 A • Vfinal=1x10^6/(60+40)=10KV • 1/2x40x10^-6x(10^8)=2000 Js

  16. Question 3 • Field coil of a machine, S1 closes • 1 s, Energy in coil? • Energy dissipated? • S.S. reached,S1 opened S2 closed • 0.1 s. later Vs1? • E dissipated in R2 ? • L=2 H, R1=3.6Ω • R2=10Ω

  17. Soultion (Question 3) • Ifinal=800/3.6=222.22 A • I(t=1)=222.2x(1-e^(-R1t/L))=185.5A • 1/2LI^2=34406. Js • Esuppl.=∫VI dt=800^2/3.6∫(1-e^(-1.8t))dt =1.778x10^5[t+e^(-1.8t)/1.8]= =95338 Js • Edissip.=95338-34406=60932 Js • IR2(t=0.1 s)=222.2xe^(-13.6t/2)=112.6A • VR2=1126 volts, Vs1=800+1126=1926 volts • Es.s.=1/2*2*222.2^2=49380Js, • ER2=49380X10/13.6=36310 Js

  18. Double Frequency Transients • Simplest case • Opening C.B. • Ind.Load, Unload T. • L1,C1 source side • L2,C2, load side • open:2halves osc indep. • Deduction: • Pre-open. Vc=L2/(L1+L2) x V

  19. D.F. Transients continued • Normally L2»L1 • C1 & C2 charged to about V(t) of Sys. • This V, at peak when I=0 • C2 discharge via L2, f2=1/(2∏√(L2.C2)) • C1 osc f1=1/(2∏√L1.C1)) about Vsys • Figures :Load side Tran.s & Source side Transients

  20. Clearing S.C. in sec. side of Transf. • Another usual D.F. Transients • L1 Ind. Upto Trans. • L2 Leak. Ind. Trans. • C1&C2 sides cap. • Fig. Two LC loop

  21. Second D.F. Transients • Eq. CCT. • Vc1(0)=L2/(L1+L2) .V • Vc1=V-L1dI1/dt= Vc1(0)+1/C1∫(I1-I2)dt • Vc2=1/C2∫I2.dt • Vc2=V-L1dI1/dt- L2dI2/dt

  22. Apply the L. T. to solve Eq.s • V/s-L1si1(s)-L1I1(0)=Vc1(0)/s+1/C1s[i1(s)-i2(s)] • -i1(s)(L1s+1/C1s)+i2(s)/C1(s)= Vc1(0)/s+L1I1(0)-L2.si2(s)+L2I2(0) • Vc2(s)=i2(s)/sC2 • Vc2(s)=V/s-L1si1(s)+L1I1(0)- L2si2(s)+L2I2(0)

  23. The D.F. CCT response • Switch clear at t=0 • I1(0)=I2(0)=I(0)=0 • i1(s)=V/(L1s)-[(L2C2s^2+1)/L1s].vc2(s) • In term of vc2(s):

  24. CCT Natural Frequencies • (s^2+ω1^2)(s^2+ω^2)vc2(s)= AV(1/s+Bs) A,B,ω1,ω2 function of : L1,C2,L2,C2

  25. Parameters of Eqs • A=1/(L1C1L2C2) • B=(L1+L2)/L1L2 C1

  26. Damping • Observation of RLC CCT LC CCT assumed lossless loss simulated by resistance in CCT □ Resistance has damping effect □ The two important CCTs: 1- Parallel RLC CCT 2- Series RLC CCT

  27. RLC CCTs & General Diff. Eqs • Φ:I of branch or V across the CCT • Ψ:V across a comp. or I in CCT

  28. Typical Differential Eq. of RLC • The Parallel RLC: • The Series RLC :

  29. Parameters Continued • Tp= RC = Parallel CCT time constant • Ts= L/R = series CCT time constant • TpTs=LC=T • η= R/Z0=R√(C/L) • η=Tp/Ts=RC/L • Duality Relationship of Series&Parallel • Transforms & their inverse plots in Dimensionless curves using: η

  30. Basic Transform of RLC CCTs • Closing s in C Parallel RLC

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