Capacitor switching continued
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Capacitor Switching …continued. A 13.8 KV, 5000KVAR, 3ph bank,NGr Source Gr, inductance:1 mH Restrike at V p : 1- c=5/(377x13.8)=69.64 μ F 2- Z=√1000/69.64=3.789 Ω 3-I p =2√2x13.8/(√3x3.789)=5.947 KA 4-f 0 =603 Hz. Subsequent Occurences. C.B. Interr. H.F. current at its zero:

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Capacitor switching continued l.jpg
Capacitor Switching …continued

  • A 13.8 KV, 5000KVAR, 3ph bank,NGr

  • Source Gr, inductance:1 mH

  • Restrike at Vp:

    1- c=5/(377x13.8)=69.64μF

    2- Z=√1000/69.64=3.789Ω

    3-Ip=2√2x13.8/(√3x3.789)=5.947 KA

    4-f0=603 Hz

Subsequent occurences l.jpg
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

Other restriking phenomena l.jpg
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=ω0/2Π= 1/[2Π√L2C] ;Fig.

  • Several KHz if C stray Cap. small

  • reigniting effect; heuristic approach:

  • Superposition of :Vs(0) effect + Vc(0) effect

  • I=Vm/[ω(L1+L2)] sinωt

  • I1(0)+I2(0)=I(0)≈I’t(during short duration)

  • I’= Vm/[L1+L2]

Restriking cct reignition after isolating inductive load l.jpg
Restriking cct: reignition after isolating inductive load

  • Equivalent CCT

  • Short interval ∆t0

  • source sub. Battery

  • I1&I2 rising ramp

    as current restab.

  • Ramp component1: Vs(0).t/[L1+L2]

  • f01=1/2Π x {√[L1+L2]/L1L2C}

Discussion continued l.jpg
Discussion Continued ….

Now superposing effect of the Capacitance

Initial Voltage Vc(0) at reignition:

Surge impedance:


  • The Osc_component2:


  • fraction L1/(L1+L2) of it pass L2

  • fraction L2/(L2+L2)of it pass L1

Formal solution l.jpg
Formal Solution

  • Two Loops Equation:

    L1 dI1/dt+Vc(0)+1/c∫(I1-I2)dt=Vs(0), (1)

    Differ.(1): dI1/dt+I1/L1C-I2/L1C=0, (2)

    Vc(0)+1/C∫(I1-I2)dt=L2 dI2/dt, (3)

  • L.T. of Eqs (2) & (3) respectively:

    (s+ω1)i1(s)-ω1 i2(s)=s I1(0)+I1’(0) (4)

    ω2i1(s)-(s+ω2)i2(s)=sI2(0)+I’2(0) (5)

  • I’1(0)=Vs(0)-Vc(0)/L1, I’2(0)=Vc(0)/L2

  • ω1=1/L1C, ω2=1/L2C

Discussion of formal solution l.jpg
Discussion of Formal Solution

  • Solving Eqs (4) & (5) simultaneously

  • yields current 2 comp.s:

    1- a ramp component

    2- a damped oscillating component with: √(ω1+ω2) =√[(L1+L2)/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

Arcing ground l.jpg

  • L-Gr fault:arc;ph & Gr stop&Reig.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; Ep=1pu

  • VA neg.peak at F. instant

  • Shift shown in phasor D.

Discussion of arcing gr results l.jpg
Discussion of Arcing Gr. results

  • C0 of A discharge;N rise to Ep & B,C rise

  • C1&C0 share charges at B and C

    not at once to Diagram values

  • Charge Conservation:


  • V=Ep/2[(C0+3C1)/(C0+C1)]

  • VB&VC rise to 3/2Ep osc. C0,C1&Ls

    CCT shown in Fig.

Discussion continued10 l.jpg
Discussion Continued …

  • Equivalent CCT

  • f0=1/{2Π√3L(c0+C1)}

  • Z0=√{3L/4(C0+C1)}

  • VB&VC above 3/2 Ep

  • Ip=(3/2Ep-V)/Z0

  • subs. for V&Z0:



    □ IB pass out of node Bdivided:Ic1,Ic0

    □ Ic1 pass the arc; its frac.:


    □Also a 60Hz current due to VA throughCN

What happens afterward l.jpg
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 level;N in -1pu

  • 3ph Phasors in next Fig a

  • N keeps -1pu and after 1/2cycle:

    VA rise to -2pu Fig b

Phasor diagram after interrupting a reignition current l.jpg
Phasor Diagram after interrupting a reignition current

  • Fig a & b

  • If Reignite again :

    similar shift ∆V=2pu

    Rather than 1pu

  • Transients higher

  • N -1pu 1pu

  • Swings of A & B

  • Then seq. can repeat

Continued on case 1 then case2 l.jpg
continued on case 1& then Case2…

  • after 1/2cycle A,Gr however at +peak

  • B&C 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,Vc2attain Vp

    2-Arc extinguish C1 s rise to VLL(VAC=VAB=1.5pu)

  • Vp=1.5Ep+(1.5Ep-V)=3Ep-V

Continued on case 2 l.jpg
Continued on Case 2

i.e. Vp-3/2Ep=3/2Ep-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 Vmax=VAG,

1/2cycle later, situation of figb

Discussion l.jpg

  • 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

Assignment no 3 l.jpg
Assignment No. 3

  • Ques. 1:

  • C1=2μF,C2=.38μF

  • L=800μH,R=5Ω

  • VC1(0)=75 KV

  • S closes, compute

    1-Max energy in L?

    2-t0 instant current flow in c2


    4-Max of Vc2 ?

Solution of question 1 l.jpg
Solution of Question 1

  • Z0=√L/C1= √800/2=20Ω

  • S close: Ipeak(undamped)=75/20=3.75KA

  • λ=Z0/R=4(fig 4.4)

    Ip=0.83x3.75= 3.112 KA

  • Emax=0.0008x3.112^2/2=4.12KJ

  • I flows in C2 when reversesfig4.4 at t’=3.15

  • or t=3.15√LC1=126μs, (T=40μs) fig4.6  Vc1=-0.69x75=-51.75 KV

Eq cct diode goes off l.jpg
Eq. CCT (Diode goes off)

  • series RLC CCT

  • At this instant:

    Vc1(0),Vc2(0) required

  • CCT diff. Eqs employed

  • Solved for Vc2

Solution of question 1 continued l.jpg
Solution of Question 1 continued

  • 2nd1/2cycle,Vc1(0)=51.75KV,Vc2(0)=0,I(0)=0

  • C1&C2 now in series Ceq=C1C2/(C1+C2)

  • Z0’=√(800x2.38)/(2x0.38)=50.05Ωλ’=50/5=10

  • Now for series C1,C2,R,L we have:

    Vc1+Vc2=RI+L dI/dt (1)

    I=-C1 dVc1/dt=-C2 dVc2/dt (2)

    Vc1=Vc1(0)-1/C1∫Idt=Vc1(0)+C2/C1Vc2 (3)

    Solving Eqs 1,2,3 for Vc2:




Solution of q1 continued l.jpg
Solution of Q1: ..continued

  • I(0)=0V’c2(0)=0,T=√{Lc1c2/(c1+c2)

  • vc2(s)=

    -Vc1(0)/LC2 x 1/{s(s^2+Rs/L+1/LC)}

    where C=C1C2/(C1+C2)

    λ’=10fig4.7 : Vpeak=1.855


    =-C1/(C1+C2)Vc1(0)=-2/2.38 x (-51.7)=43.48KV

    Vpeak=1.855x43.48=80.7 KV

Question 2 c b opening resistor l.jpg
Question 2 , C.B. opening resistor

  • C.B. clear 28000 A sym fault

  • R=800Ω, Cbus=0.04μF,

  • Vsys=138KV,f=50Hz

  • 1-Peak of TRV?

  • 2-energy loss in R (it is 2 cycle in)?

Q2 solution l.jpg
Q2, Solution

  • X=138/(√3x28)=2.8455, L=9.1 mH

  • Without R, TRV=2x138x√(2/3) =225.4KV

  • Z0=√L/C=√{0.0091/0.04x10^-6}=477Ω, η=800/477=1.677

  • fig4.7 1.38x 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-2cosx

Q2 continued l.jpg
Q2 continued

  • ∫(1-cosωt) dt=

    3/2t+1/(4ω) sin2ωt-2/ω sinωt=6Π/ω


  • Energy loss if TRV was a 1-cosine wave:



  • However (1-cosine) is deformed and assuming

    to be halved : Energy loss≈952/2=476 KJs

Question 3 l.jpg
Question 3

  • 246 KW Load on a 3ph 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

Q3 solution l.jpg
Q3, solution

  • 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/(314.15x580.6)=5.48μF

  • switch open at Is=0, Vs=0, when IL&IC

    are at peak (opposite sign)

  • Ipeak=13.8√(2/3) /580.6=19.41 A

Q3 solution continued l.jpg
Q3, solution continued …

  • CCT a parallel RLC

  • -cdVc/dt=Vc/R+


  • dVc/dt+1/RCdVc/dt + 1/LC =0

  • [s+s/RC+1/LC]vc(s)=(s+1/RC)Vc(0)+V’c(0)

Q3 continued l.jpg
Q3 continued

  • 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 sinω0t

    =-IL(0) √L/C sinω0t

  • √L/C=√1.84/5.48=579.5Ω,

  • IL(0)=19.41 A Vpeak=11.27KV

  • η=R/Z0=1.333 fig 4.4 0.6x11.27=6.76KV

Transformer magnetizing current l.jpg
Transformer Magnetizing current

  • 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.

  • 1000KVA, 13.8KV: load 42 A, Inrush peak 150 A

  • A DC declines and finaly mag. current

Ferroresonance l.jpg

  • series Resonance

    Very H.V. across C&L (series LC)

    if excited ~ Natural Freq.

Example of transformer ferroresonance l.jpg
Example of Transformer Ferroresonance

  • 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=8H,B=1.76 or L=dΦ/dt=8exp(-I/1.76)

  • if 60Hz res. Occur in L=4H,


  • neglecting resistance:

    L dI/dt+1/C∫I dt=V sin(ωt+Φ)

    8 exp(I/1.76) dI/dt +

    10^6/1.75∫I dt=13.8√2/3 sin(377t+Φ)

  • A nonlinear Diff. Eq. solved for I and then V, Φ=0,45