Capacitor Switching

1
Capacitor Switching continued

• A 13.8 KV, 5000KVAR, 3ph bank,NGr
• Source Gr, inductance1 mH
• Restrike at Vp
• 1- c5/(377x13.8?)69.64µF
• 2- Zv1000/69.643.789O
• 3-Ip2v2x13.8/(v3x3.789)5.947 KA
• 4-f0603 Hz

2
Subsequent Occurences

• C.B. Interr. H.F. current at its zero
• 3Vp on Cap.at most 4Vp across C.B.
• If 2nd B.D. occur, 2nd Osc.
• I doubles, Vc from 3Vp to -5Vp
• If C.B. opens, at most 8Vp across C.B.
• further ESCALATION possible
• Rs seq.Restrikes, Cs subseq.Clearing
• Variation of VCB shown in FIG

3
Other Restriking Phenomena

• Seq. restrik. clearing of C.B.
• O.V.s (even inductive load)
• low p.f. results in Vp at zero current
• Dominant fTRV?0/2? 1/2?vL2C Fig.
• Several KHz if C stray Cap. small
• reigniting effect heuristic approach
• Superposition of Vs(0) effect Vc(0) effect
• IVm/?(L1L2) sin?t
• I1(0)I2(0)I(0)It(during short duration)
• I Vm/L1L2

4
Restriking cct
reignition after isolating inductive load

• Equivalent CCT
• Short interval ?t0
• source sub. Battery
• I1I2 rising ramp
• as current restab.
• Ramp component1 Vs(0).t/L1L2
• f011/2? x vL1L2/L1L2C

5
Discussion Continued .

• Now superposing effect of the Capacitance
• Initial Voltage Vc(0) at reignition
• Surge impedance
• Z0vL1L2/c(L1L2)
• The Osc_component2
• IcVc(0)/vL1L2/c(L1L2)
• fraction L1/(L1L2) of it pass L2
• fraction L2/(L2L2)of it pass L1

6
Formal Solution

• Two Loops Equation
• L1 dI1/dtVc(0)1/c?(I1-I2)dtVs(0),
  (1)
• Differ.(1) d?I1/dt?I1/L1C-I2/L1C0,
  (2)
• Vc(0)1/C?(I1-I2)dtL2 dI2/dt, (3)
  
• L.T. of Eqs (2) (3) respectively
• (s??1?)i1(s)-?1? i2(s)s I1(0)I1(0)
  (4)
• ?2?i1(s)-(s??2?)i2(s)sI2(0)I2(0)
  (5)
• I1(0)Vs(0)-Vc(0)/L1, I2(0)Vc(0)/L2
• ?1?1/L1C, ?2?1/L2C

7
Discussion of Formal Solution

• Solving Eqs (4) (5) simultaneously
• yields current 2 comp.s
• 1- a ramp component
• 2- a damped oscillating component with
  v(?1??2?) v(L1L2)/L1L2C
• In first method assumed I2(0)0
• true if Reignite at peak of TRV
• sys Gr Neutral, this fault High Cur. cause
  Damage
• All sys Gr directly or through some stray C
• So ARCING Gr Next Discussion

8
ARCING GROUND

• L-Gr faultarcph Gr stopReig.repeated
• Fig of simple model without sources
• C1 ph to ph cap.
• C0 ph to Gr cap.
• N at Gr potenial
• A to Gr ?shift V Ep1pu
• VA neg.peak at F. instant
• Shift shown in phasor D.

9
Discussion of Arcing Gr. results

• C0 of A dischargeN rise to Ep B,C rise
• C1C0 share charges at B and C
• not at once to Diagram values
• Charge Conservation
• V(C0C1)1/2C0Ep3/2C1Ep
• VEp/2(C03C1)/(C0C1)
• VBVC rise to 3/2Ep osc. C0,C1Ls
• CCT shown in Fig.

10
Discussion Continued

• Equivalent CCT
• f01/2?v3L(c0C1)
• Z0v3L/4(C0C1)
• VBVC? above 3/2 Ep
• Ip(3/2Ep-V)/Z0
• subs. for VZ0
• Ip2Ep(C0/C0C1)x
• v(C0C1)/3L
• ? IB pass out of node
  B?dividedIc1,Ic0
• ? Ic1 pass the arc its frac.
• C0/(C0C1)
• ? Also a 60Hz current due to VA through CN

11
What Happens Afterward?

• Depends on Arc behavior
• 1-Iarc till next p.f. zero in 1/2cycle
• 2-Iarc immed. Ceased in 1st zero of the IHF
• If 1, occur after ½ cycle
• since A is still at Gr levelN in -1pu
• 3ph Phasors in next Fig a
• N keeps -1pu and after 1/2cycle
• VA rise to -2pu Fig b

12
Phasor Diagram after interrupting a reignition
current

• Fig a b
• If Reignite again
• similar shift ?V2pu
• Rather than 1pu
• Transients higher
• N -1pu? 1pu
• Swings of A B
• Then seq. can repeat

13
continued on case 1 then Case2

• after 1/2cycle A,Gr however at peak
• BC instant. -1.5pu to Gr
• If now arc interrupts VN change -1pu
• After ½ cycle exactly as last fig b
• Case 2 If interrupt at 1st zero of IHF
• occur at point P of curves
• 1-Vc1,Vc2 attain Vp
• 2-Arc extinguish C1 s rise to VLL(VACVAB1.5pu)
  
• Vp1.5Ep(1.5Ep-V)3Ep-V

14
Continued on Case 2

• i.e. Vp-3/2Ep3/2Ep-V across arc path
• ? N has corresponding Disp.
• ?Phasor Diag.Fig.a?
• ?different from 1st fig
• ?Then arc interruption
• at p.f. current zero
• when VmaxVAG,
• 1/2cycle later, situation of figb

15
Discussion

• transient following next reignit. Is greater
• Arc is interrupted in 1st H.F. zero I
• neutral displacement increases
• increase in energy trapped on zero seq Cap
  escalate the voltage
• How to suppress arcing Gr OVs
• 1- an appropriate reactance in neutral
• 2- Peterson coil , sensitive for fault
  detection

16
Assignment No. 3

• Ques. 1
• C12µF,C2.38µF
• L800µH,R5O
• VC1(0)75 KV
• S closes, compute
• 1-Max energy in L?
• 2-t0 instant current flow in c2
• 3-Vc1(t0)?
• 4-Max of Vc2 ?

17
Solution of Question 1

• Z0vL/C1 v800/220O
• S close Ipeak(undamped)75/203.75KA
• ?Z0/R4(fig 4.4)?
• Ip0.83x3.75 3.112 KA
• Emax0.0008x3.1122/24.12KJ
• I flows in C2 when reverses?fig4.4 at t3.15
• or t3.15vLC1126µs, (T40µs)
  fig4.6 ? Vc1-0.69x75-51.75 KV

18
Eq. CCT (Diode goes off)

• series RLC CCT
• At this instant
• Vc1(0),Vc2(0) required
• CCT diff. Eqs employed
• Solved for Vc2

19
Solution of Question 1 continued

• 2nd1/2cycle,Vc1(0)51.75KV,Vc2(0)0,I(0)0
• C1C2 now in series CeqC1C2/(C1C2)
• Z0v(800x2.38)/(2x0.38)50.05O ?50/510
• Now for series C1,C2,R,L we have
• Vc1Vc2RIL dI/dt
  (1)
• I-C1 dVc1/dt-C2 dVc2/dt (2)
• Vc1Vc1(0)-1/C1?IdtVc1(0)C2/C1Vc2 (3)
• Solving Eqs 1,2,3 for Vc2
• d?Vc2/dt?R/LdVc2/dt(C1C2)/LC1C2Vc2
• 
  -Vc1(0)/LC2
• s?vc2(s)-sR/LVc2(0)-R/LVc2(0)sR/Lvc2(s)-v(c1C2
  )/LC1C2xVc2(s)-Vc1(0)/sLC2

20
Solution of Q1 ..continued

• I(0)0?Vc2(0)0,TvLc1c2/(c1c2)
• vc2(s)
• -Vc1(0)/LC2 x 1/s(s2Rs/L1/LC)
• where CC1C2/(C1C2)
• ?10?fig4.7 Vpeak1.855
• 1pu-T2.Vc1(0)/LC2
• -C1/(C1C2)Vc1(0)-2/2.38 x (-51.7)43.48KV
• Vpeak1.855x43.4880.7 KV

21
Question 2 , C.B. opening resistor

• C.B. clear 28000 A sym fault
• R800O, Cbus0.04µF,
• Vsys138KV,f50Hz
• 1-Peak of TRV?
• 2-energy loss in R (it is 2 cycle in)?

22
Q2, Solution

• X138/(v3x28)2.8455, L9.1 mH
• Without R, TRV2x138xv(2/3) 225.4KV
• Z0vL/Cv0.0091/0.04x10-6477O,
  ?800/4771.677
• fig4.7? 1.38x 225.4/2155.5 KV
• Energy dissipated in 2 cycles
• VTRV112.71-exp(-t/2?)x sin.. cos
• Integrating PlossVTRV?/R over ?t0 to ?t4?
• for a (1-cosine) wave(1-cosx)?1cos?x-2cosx

23
Q2 continued

• ?(1-cos?t)? dt
• 3/2t1/(4?) sin2?t-2/?
  sin?t6?/?
• t0,4?/?
• Energy loss if TRV was a 1-cosine wave
• V?/Rx6?/?112.7?/800x6?/3140.9526MJ
• 
  (952.6KJs)
• However (1-cosine) is deformed and assuming
• to be halved Energy loss952/2476 KJs

24
Question 3

• 246 KW Load on a 3ph S.G., 13.8 KV
• p.f.0.6 lag(load parallel RL) C?pf1
• loadcap switch as a unit
• S opened, Vpeak across C.B.?
• Zs negligible

25
Q3, solution

• V?/R246 KW, R(13.8/v3)?/246/3774O
• p.f.0.6, F53.13
• tanF1.333R/X? X580.6O,L1.84 H? Xc580.6
• C1/(314.15x580.6)5.48µF
• switch open at Is0, Vs0, when ILIC
• are at peak (opposite sign)
• Ipeak13.8v(2/3) /580.619.41 A

26
Q3, solution continued

• CCT a parallel RLC
• -cdVc/dtVc/R
• 1/L?VcdtIL(0)
• d?Vc/dt1/RCdVc/dt 1/LC 0
• s?s/RC1/LCvc(s)(s1/RC)Vc(0)Vc(0)

27
Q3 continued

• Vc(0)-Vc(0)/RC-IL(0)/C, Vc(0)0
• ? vc(s)-IL(0)/cs2s/RC1/LC
• without R ?-IL(0)/c x 1/(s2?02)
• and Vc(t)-IL(0)/C?0 sin?0t
• -IL(0) vL/C sin?0t
• vL/Cv1.84/5.48579.5O,
• IL(0)19.41 A? Vpeak11.27KV
• ?R/Z01.333? fig 4.4 0.6x11.276.76KV

28
Transformer Magnetizing current

• Mag. Inrush ? transient
• Im0.5 to 2 rated, non-sinusoidal
• Distortion B in core
• Instant of energizing Residual Flux can
  cause Inrush
• Continue several sec, small Transf.
• several min, large Transf.
  
• 1000KVA, 13.8KV load 42 A, Inrush peak 150 A
• A DC declines and finaly mag. current

29
Ferroresonance

• series Resonance
• Very H.V. across CL (series LC)
• if excited Natural Freq.

30
Example of Transformer Ferroresonance

• simulate nonlinearity of core
• LdF/dtA exp(-I/B)I0, LA IB, LA/e
• A 13.8 KV step down T. resonance
• with cables in primary CCT, A8H,B1.76 or
  LdF/dt8exp(-I/1.76)
• if 60Hz res. Occur in L4H,
• C1/??L1.75µF
• neglecting resistance
• L dI/dt1/C?I dtV sin(?tF)
• 8 exp(I/1.76) dI/dt
• 106/1.75?I dt13.8v2/3 sin(377tF)
• A nonlinear Diff. Eq. solved for I and then V,
  F0,45