Capacitor Switching continued A 13 8 KV 5000

Capacitor Switching …continued A 13. 8 KV, 5000 KVAR, 3 ph bank, NGr o Source Gr, inductance: 1 m. H o Restrike at Vp: 1 - c=5/(377 x 13. 8�)=69. 64μF 2 - Z=√ 1000/69. 64=3. 789Ω 3 -Ip=2√ 2 x 13. 8/(√ 3 x 3. 789)=5. 947 KA 4 -f 0=603 Hz o

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

Other Restriking Phenomena o Seq. restrik. & clearing of C. B. O. V. s (even inductive load) low p. f. results in Vp at zero current Dominant f. TRV=ω0/2Π= 1/[2Π√L 2 C] ; Fig. o Several KHz if C stray Cap. small o o o o reigniting effect; heuristic approach: Superposition of : Vs(0) effect + Vc(0) effect I=Vm/[ω(L 1+L 2)] sinωt I 1(0)+I 2(0)=I(0)≈I’t(during short duration) I’= Vm/[L 1+L 2]

Restriking cct: reignition after isolating inductive load o Equivalent CCT Short interval ∆t 0 source sub. Battery o I 1&I 2 rising ramp o o as current restab. Ramp component 1: Vs(0). t/[L 1+L 2] f 01=1/2Π x {√[L 1+L 2]/L 1 L 2 C}

Discussion Continued …. Now superposing effect of the Capacitance Initial Voltage Vc(0) at reignition: Surge impedance: Z 0=√{L 1 L 2/[c(L 1+L 2)]} o The Osc_component 2: Ic=Vc(0)/√{L 1 L 2/[c(L 1+L 2)]} o fraction L 1/(L 1+L 2) of it pass L 2 o fraction L 2/(L 2+L 2)of it pass L 1

Formal Solution o o Two Loops Equation: L 1 d. I 1/dt+Vc(0)+1/c∫(I 1 -I 2)dt=Vs(0), (1) Differ. (1): d�I 1/dt�+I 1/L 1 C-I 2/L 1 C=0, (2) Vc(0)+1/C∫(I 1 -I 2)dt=L 2 d. I 2/dt, (3) L. T. of Eqs (2) & (3) respectively: (s�+ω1�)i 1(s)-ω1� i 2(s)=s I 1(0)+I 1’(0) (4) ω2�i 1(s)-(s�+ω2�)i 2(s)=s. I 2(0)+I’ 2(0) o o I’ 1(0)=Vs(0)-Vc(0)/L 1, I’ 2(0)=Vc(0)/L 2 ω1�=1/L 1 C, ω2�=1/L 2 C (5)

Discussion of Formal Solution o o o Solving Eqs (4) & (5) simultaneously yields current 2 comp. s: 1 - a ramp component 2 - a damped oscillating component with: √(ω1�+ω2�) =√[(L 1+L 2)/L 1 L 2 C] In first method assumed I 2(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

ARCING GROUND o o o o L-Gr fault: arc; ph & Gr stop&Reig. repeated Fig of simple model without sources C 1 : ph to ph cap. C 0 : ph to Gr cap. N at Gr potenial A to Gr shift V; Ep=1 pu VA neg. peak at F. instant Shift shown in phasor D.

Discussion of Arcing Gr. results o o o C 0 of A discharge; N rise to Ep & B, C rise C 1&C 0 share charges at B and C not at once to Diagram values Charge Conservation: V(C 0+C 1)=1/2 C 0 Ep+3/2 C 1 Ep V=Ep/2[(C 0+3 C 1)/(C 0+C 1)] VB&VC rise to 3/2 Ep osc. C 0, C 1&Ls CCT shown in Fig.

Discussion Continued … Equivalent CCT o o o f 0=1/{2Π√ 3 L(c 0+C 1)} Z 0=√{3 L/4(C 0+C 1)} VB&VC above 3/2 Ep Ip=(3/2 Ep-V)/Z 0 subs. for V&Z 0: Ip=2 Ep(C 0/C 0+C 1)x □ □ √{(C 0+C 1)/3 L} IB pass out of node B divided: Ic 1, Ic 0 Ic 1 pass the arc; its frac. : C 0/(C 0+C 1) □ Also a 60 Hz current due to VA through CN

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

Phasor Diagram after interrupting a reignition current o o o Fig a & b If Reignite again : similar shift ∆V=2 pu Rather than 1 pu Transients higher N -1 pu Swings of A & B Then seq. can repeat

continued on case 1& then Case 2… o o o after 1/2 cycle A, Gr however at +peak B&C instant. -1. 5 pu to Gr If now arc interrupts VN change -1 pu After ½ cycle exactly as last fig b Case 2 : If interrupt at 1 st zero of IHF occur at point P of curves 1 -Vc 1, Vc 2 attain Vp 2 -Arc extinguish C 1 s rise to VLL(VAC=VAB=1. 5 pu) o Vp=1. 5 Ep+(1. 5 Ep-V)=3 Ep-V

Continued on Case 2 i. e. Vp-3/2 Ep=3/2 Ep-V across arc path □ N has corresponding Disp. □Phasor Diag. Fig. a □different from 1 st fig □Then arc interruption at p. f. current zero when Vmax=VAG, 1/2 cycle later, situation of figb

Discussion o o o transient following next reignit. Is greater Arc is interrupted in 1 st 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

Assignment No. 3 o Ques. 1: o C 1=2μF, C 2=. 38μF L=800μH, R=5Ω o VC 1(0)=75 KV o S closes, compute 1 -Max energy in L? 2 -t 0 instant current flow in c 2 3 -Vc 1(t 0)? 4 -Max of Vc 2 ? o

Solution of Question 1 o o o Z 0=√L/C 1= √ 800/2=20Ω S close: Ipeak(undamped)=75/20=3. 75 KA λ=Z 0/R=4(fig 4. 4) Ip=0. 83 x 3. 75= 3. 112 KA Emax=0. 0008 x 3. 112^2/2=4. 12 KJ I flows in C 2 when reverses fig 4. 4 at t’=3. 15 or t=3. 15√LC 1=126μs, (T=40μs) fig 4. 6 Vc 1=-0. 69 x 75=-51. 75 KV

Eq. CCT (Diode goes off) o o series RLC CCT At this instant: Vc 1(0), Vc 2(0) required o o CCT diff. Eqs employed Solved for Vc 2

Solution of Question 1 continued o o 2 nd 1/2 cycle, Vc 1(0)=51. 75 KV, Vc 2(0)=0, I(0)=0 C 1&C 2 now in series Ceq=C 1 C 2/(C 1+C 2) Z 0’=√(800 x 2. 38)/(2 x 0. 38)=50. 05Ω λ’=50/5=10 Now for series C 1, C 2, R, L we have: Vc 1+Vc 2=RI+L d. I/dt (1) I=-C 1 d. Vc 1/dt=-C 2 d. Vc 2/dt (2) Vc 1=Vc 1(0)-1/C 1∫Idt=Vc 1(0)+C 2/C 1 Vc 2 (3) Solving Eqs 1, 2, 3 for Vc 2: d�Vc 2/dt�+R/Ld. Vc 2/dt+{(C 1+C 2)/LC 1 C 2}Vc 2= -V c 1(0)/LC 2 s�vc 2(s)-s. R/LVc 2(0)-R/LV’c 2(0)+s. R/Lvc 2(s)√{(c 1+C 2)/LC 1 C 2}x. Vc 2(s)=-Vc 1(0)/s. LC 2

Solution of Q 1: . . continued I(0)=0 V’c 2(0)=0, T=√{Lc 1 c 2/(c 1+c 2) o vc 2(s)= -Vc 1(0)/LC 2 x 1/{s(s^2+Rs/L+1/LC)} where C=C 1 C 2/(C 1+C 2) λ’=10 fig 4. 7 : Vpeak=1. 855 1 pu=-T^2. Vc 1(0)/LC 2= =-C 1/(C 1+C 2)Vc 1(0)=-2/2. 38 x (51. 7)=43. 48 KV o Vpeak=1. 855 x 43. 48=80. 7 KV

Question 2 , C. B. opening resistor o o o C. B. clear 28000 A sym fault R=800Ω, Cbus=0. 04μF, Vsys=138 KV, f=50 Hz 1 -Peak of TRV? 2 -energy loss in R (it is 2 cycle in)?

Q 2, Solution o o o X=138/(√ 3 x 28)=2. 8455, L=9. 1 m. H Without R, TRV=2 x 138 x√(2/3) =225. 4 KV Z 0=√L/C=√{0. 0091/0. 04 x 10^-6}=477Ω, η=800/477=1. 677 fig 4. 7 1. 38 x 225. 4/2=155. 5 KV Energy dissipated in 2 cycles: VTRV=112. 7{1 -exp(-t’/2η)x sin. . +cos…} Integrating Ploss=VTRV�/R over ωt=0 to ωt=4Π for a (1 -cosine) wave: (1 -cosx)�=1+cos�x 2 cosx

Q 2 continued o o o ∫(1 -cosωt)� dt= 3/2 t+1/(4ω) sin 2ωt-2/ω sinωt=6Π/ω t=0, 4Π/ω Energy loss if TRV was a 1 -cosine wave: V�/Rx 6Π/ω=112. 7�/800 x 6Π/314=0. 9526 MJ (952. 6 KJs) However (1 -cosine) is deformed and assuming to be halved : Energy loss≈952/2=476 KJs

Question 3 o o o 246 KW Load on a 3 ph S. G. , 13. 8 KV p. f. =0. 6 lag(load parallel RL) C pf=1 load+cap switch as a unit S opened, Vpeak across C. B. ? Zs negligible

Q 3, solution o o o V�/R=246 KW, R=(13. 8/√ 3)�/246/3=774Ω p. f. =0. 6, Φ=53. 13 tanΦ=1. 333=R/X X=580. 6Ω, L=1. 84 H Xc=580. 6 C=1/(314. 15 x 580. 6)=5. 48μF switch open at Is=0, Vs=0, when IL&IC are at peak (opposite sign) o Ipeak=13. 8√(2/3) /580. 6=19. 41 A

Q 3, solution continued … o CCT a parallel RLC o -cd. Vc/dt=Vc/R+ 1/L∫Vcdt+IL(0) o d�Vc/dt+1/RCd. V c/dt + 1/LC =0 o [s�+s/RC+1/LC] vc(s)=(s+1/RC)Vc (0)+V’c(0)

Q 3 continued o o o o V’c(0)=-Vc(0)/RC-IL(0)/C, Vc(0)=0 vc(s)=-IL(0)/{c[s^2+s/RC+1/LC]} without R -IL(0)/c x 1/(s^2+ω0^2) and : Vc(t)=-IL(0)/Cω0 sinω0 t =-IL(0) √L/C sinω0 t √L/C=√ 1. 84/5. 48=579. 5Ω, IL(0)=19. 41 A Vpeak=11. 27 KV η=R/Z 0=1. 333 fig 4. 4 0. 6 x 11. 27=6. 76 KV

Transformer Magnetizing current o o o o Mag. Inrush transient Im=0. 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. 1000 KVA, 13. 8 KV: load 42 A, Inrush peak 150 A A DC declines and finaly mag. current

Ferroresonance series Resonance Very H. V. across C&L (series LC) o if excited ~ Natural Freq.

Example of Transformer Ferroresonance o o simulate nonlinearity of core: L=dΦ/dt=A exp(-I/B): I=0, L=A; I=B, L=A/e A 13. 8 KV step down T. resonance with cables in primary CCT, A=8 H, B=1. 76 or L=dΦ/dt=8 exp(-I/1. 76) o if 60 Hz res. Occur in L=4 H, C=1/ω�L=1. 75μF o o neglecting resistance: L d. I/dt+1/C∫I dt=V sin(ωt+Φ) 8 exp(I/1. 76) d. I/dt + 10^6/1. 75∫I dt=13. 8√ 2/3 sin(377 t+Φ) A nonlinear Diff. Eq. solved for I and then V, Φ=0, 45