1. Clearing S.C. in sec. side of Transf.  Another usual D.F. Transients  L 1 Ind. Upto Trans.  L 2 Leak. Ind. Trans.  C 1 &C 2 sides cap.  Fig. Two LC loop

2. Second D.F. Transients  Eq. CCT.  V c1 (0)= L 2 /(L 1 +L 2 ). V  V c1 =V-L 1 dI 1/ dt= V c1 (0)+ 1/C1∫(I 1 -I 2 )dt  V c2 = 1/C 2 ∫I 2.dt  V c2 =V-L 1 dI 1/ dt- L 2 dI 2 /dt

3. Apply the L. T. to solve Eq.s  V/s-L 1 si 1 (s)- L 1 I 1 (0)=V c1 (0)/s+1/C 1 s[i 1 (s)-i 2 (s)]  -i 1 (s)(L 1 s+ 1/C 1 s)+i 2 (s)/C 1 (s)= V c1 (0)/ s+L 1 I 1 (0)-L 2. si 2 (s)+L 2 I 2 (0)  V c2 (s)=i 2 (s)/sC 2  V c2(s)= V/s-L 1 si 1 (s)+L 1 I 1 (0)- L 2 s i 2 (s)+ L 2 I 2 (0)

4. The D.F. CCT response  Switch clear at t=0  I 1 (0)=I 2 (0)=I(0)=0  i 1 (s)=V/(L 1 s)- [ (L 2 C 2 s^2+1)/L 1 s].v c2 (s)  In term of v c2 (s):

5. CCT Natural Frequencies  (s^2+ω 1 ^2)(s^2+ ω 2 ^2)v c2 (s)= AV(1/s+Bs) A,B,ω 1,ω 2 function of : L 1, C 2, L 2,C 2

6. Parameters of Eqs  A=1/(L 1 C 1 L 2 C 2 )  B=(L 1 +L 2 )/L 1 L 2 C 1

7. Damping Effects  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

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

9. Typical Differential Eq. of RLC  The Parallel RLC Eq(1):  The Series RLC Eq(2):

10. Parameters Continued  T p = RC = Parallel CCT time constant  T s = L/R = series CCT time constant  T p T s =LC=T  η= R/Z 0 =R√(C/L)  η=T p /T s =RC/L  Duality Relationship of Series&Parallel  Transforms & their inverse plots in Dimensionless curves using: η

11. Basic Transform of RLC CCTs  Closing s, Parallel RLC  C disch. In R & L  -C dV c /dt=I L +V c /R  Also: V c =L dI L /dt  dI L /dt+1/RC.dI L /dt+I L /LC =0  dI L /dt+1/T p. dI L /dt +I L /T=0 Is Gen. Eq(1), F(t)=0

12. L.T. application  (s+s/T p +1/T)i L (s)=(s+1/T p )I L (0)+ I L `(0) I L (0)=0, I L `(0)=V c (0)/L □ i L (s)=V c (0)/L. 1/(s+s/T p +1/T) □ 1/(s+s/T p +1/T)=1/[(s-s 1 )(s-s 2 )]= 1/(s 1 -s 2 ). [1/(s-s 1 ) – 1/(s-s 2 )] □s 1,2 =-1/(2T p) ±1/2√(1/T p  - 4/T)

13. Parallel RLC Time response  i L (s)=V c (0)/{L√(1/T p )-(4/T)}.[1/(s-s 1 ) – 1/(s-s 2 )]  i L (t)=V c (0)/{L√(1/T p )-(4/)}. [exp(s 1 t)-exp(s 2 (t)] Discussion : 1- 1/T p >4/T or (η<1/2), s 1 &s 2 real response O.D. No Oscillation(η=1/2, C.D.) 2- 1/T p  1/2), s 1 &s 2 complex reponse oscillatory: U.D.

14. Parallel RLC Time response Inductor Current in a parallel RLC cct

15. Discussion of Osc. Responses  complex roots rewritten as: S 1,2 =-1/2T p [1±j√(4η -1)]  final reponse in term of sine func.: I L (t)=V c (0)/L. 2T p exp(-t/2T p )/(4η -1) X sin√(4η-1). t/2T p  Power sys. More concern about this  In Gen.: 1 st Useful Transform is

16. General LTs for RLC CCT response  Eq (3) : ζ- 1/[s+s/T p +1/T]= 2T p. exp(-t/2T p )/√(4η-1). Sin[√(4η-1) t/(2T p )]  variable substitution for General Plots  If t’=tη/T p =t/T (T=1/ω 0 ), In term of t’ plot only function :η and t’  I L (t)=V(0)/L.T.2η.exp(-t’/2η)/√(4η- 1). Sin[√(4η-1) t’/2η] …Eq(4)

17. Discussion on family of curves  V(0)/L. T=V(0) √LC/L=V(0)/√(L/C)=V(0)/Z 0  Eq4,Except top term is dimensionless  as follows :2η.exp(-t’/2η)/√(4η-1). sin[√(4η-1)t’/2η] Which is ζ− 1/(s+s/T p + 1/T) □ Fig 4.4, plot : 1/T. ζ− 1/(s+s/T p + 1/T) □ no damping η -->∞,I L (0)=V(0)/Z 0.sint’

18. Discussion on family of curves Fig 4.4

19. Discussion of Fig 4.4 continued  Vertical axis in P.u. +1 to -1  time scale T, time variable :t’  curves as function of D.F. : η  The same plots for series RLC λ(damping factor)=z 0 /R=1/η  Ex:V c =20 KV,C=0.1μF,L=8mH,R=430  S closed in C branch of p. RLC CCT

20. P. RLC Eample Solution  Z 0 =√L/C=√8. 100=283 Ω  η=R/Z 0 =430/283=1.52  I L_peak =V c (0)/Z 0 =20000/283=70.7 A  Fig 4.4 & η=1.5 gives a factor 0.65  => I L_p =70.7x0.65=46.0 A  T=√LC=√8 x 10^(-5)=28.3 μs  Acc. Fig 4.4 t’(peak)=1.35T=38 μs  freq. D.Osc. ≈5306 Hz

21. Discussion of Response η<1/2  s 1,2 =-1/2T p [1±√(1-4η)] : O.D.  ζ− 1/(s+s/T p +1/T)= 2T p. exp(-t/2T p )/[√(1-4η)] sinh√(1- 4η). t/2T p  For η=1/2 or Critically Damped:  η  1/2, a=√(1-4η2)  0 => sinh√a x / √a  x  ζ−1{1/[s+(1/2T p )]^2}=t exp(-t/2T p )

22. Other Transfer Functions (2 nd Type)  Now in P. RLC if V c is looked for,  dV c /dt+1/C dI L /dt+1/RC dV c /dt=0  sub. For dI L /dt=V c /L: dV c /dt+1/T p dV c /dt + V c /T=0 □ (s+s/T p +1/T)v c (s)=(s+1/T p )V c (0)+ V c ’(0) □ V c ’(t)=-I L (t)/C – V c (t)/RC, I L (0)=0  V c ’(0)=-I L (0)/c-V c (0)/RC=-V c (0)/T p

23. L.T. Eq. for V c P. RLC  v c (s)=V c (0). s/(s+s/T p +1/T)  knowing: ζ− s/(s+s/T p +1/T)= d/dt{ζ− 1/(s+s/T p +1/T)}+ F(0)  η>1/2: Fig 4.6 cosine forms exp(-t/2T p ){cos√(4η-1).t/2T p - sin[√(4η-1)t/2T p ]/√(4η-1)}  η=1/2 : exp(-t/2T p )(1-t/2T p )

24. Fig 4.6 cosine forms

25. The 3 rd Type of Response  I ramp inj. P. RLC  the KCL Eq.: V/R+ CdV/dt+1/L∫Vdt=I’t  L.T. : (s+s/T p +1/T)v(s)= I’/sC+(s+1/T p )V(0)+ V’(0)  V(0)=0, V’(0)=0

26. Ramp function response con…  V(s)=1/{s[s+s/T p +1/T]}. I’/C  1/{s[s+s/T p +1/T]} =T{1/s- s/[s+s/T p +1/T]- 1/T p [s+s/T p +1/T] }  for η>1/2: ζ−….= T{1-exp(-t/2T p )[sin(√(4η- 1)t/2T p )/√(4η-1) +cos√(4η-1) t/2T p ]}

27. Critical and O.D. response  for η=1/2 : ζ−…=(2T p ){1-exp(-t/2T p ). [1+1/2T p ]}  for η<1/2 : ζ−…=T{1-T{1-exp(- t/2T p )[sinh(√(4η-1)t/2T p )/√(4η-1) +cosh√(4η-1) t/2T p ]}  Fig 4.7: has 1-cosine envelope

28. Fig 4.7: has 1-cosine envelope

29. Series RLC CCT  V (dc) applied to a series RLC  C is discharged  IR+LdI/dt+1/C∫Idt =v  Diff. & Rearrang.  dI/dt+R/L. dI/dt+1/LC=0  dI/dt+1/T s. dI/dt+ 1/T=0

30. L.T. of Eq.  After arrang.: (Remind;λ=1/μ=Z 0 /R)  (s+s/T s +1/T)i(s)=(s+1/T s )I(0)+ I’(0) I(0)=0, I’(0)=V/L(at t=0 dc v across L) □ i(s)=V/L. 1/{s+s/T s +1/T} □Resp. similar to Parallel as: □I(t)=V/L. 2T s /[√4λ-1]. exp(-t/2T s ). sin√(4λ-1) t/2T s for:λ>1/2

31. The other Responses  I=V/L. t. exp(-t/2T s ) for λ=1/2  I=V/L. 2T s /[√4λ-1]. exp(-t/2T s ). sinh√(4λ-1) t/2T s for:λ<1/2 □ Fig. 4.4 can employed if : 1 – T used for time scaling 2 – t’=t/T=Z 0 t/L=tλ/T s 3 – or t/2T s =t’/2λ

32. Res istance Sw itching  CB use R during OP. 1- employed during closing 2- Also when open a switch  In multi break C.B. ; dist. TRV uniformly  reduce severity of TRV  high Ω for closing<react(1/ω 0 C break )  ω 0 : frequency of recovery transient  TRV reduction into a low value

33. Eq. CCT of Res. Switching  Opening a Fault  Single Ph. employed: 1- apply –I(t) pre-open 2- Add the voltages  Interest: short dur. Post-open, I~ ramp  I=V/ωL.cosωt  dI/dt=-V/L sinωt  I inj = V/L. t (I’=V/L)  V(s)=1/{s[s+s/T p +1/T]}. I’/C

34. Results of a Res. Switching  3ph air blast C.B. 345 KV, 25000 MVA  Interrupt. Cap.: 40KA (2 breaks in series)  Assuming 25000 PF for 345 KV bus  R s?, reduce TRV to 70% of undamped  Solution: 1- TRV p =2x345√2/√3=562 KV 2-X s =345000/40000√3=5Ω, L=5/377=13.2 mH 3-Z 0 =√L/C=√13.2x10-/2.5x10-=725 Ω 4-from Fig4.7(70%)~η=1.8,R=ηZ 0=1.8X725=1300 Ω

35. Detail Information  each interrupter, 650Ω resistor  Resistor has a very low L  Made from thin ribbon stainless steel  Wound in circle, alternate layers in opp. Dir. : to reduce inductance  Interrupting R’s current,a 2 nd Transient starts  C’ across its break

36. 2 nd Transient of R sw, Eq. CCT  Interrupt. R’s switch  L’: induct. of Local loop to fault  to be solved by Sup. Pos.