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Capacitor Switching …continuedPowerPoint Presentation

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

- 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

- 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

- 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

- 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 ….

Now superposing effect of the Capacitance

Initial Voltage Vc(0) at reignition:

Surge impedance:

Z0=√{L1L2/[c(L1+L2)]}

- The Osc_component2:
Ic=Vc(0)/√{L1L2/[c(L1+L2)]}

- fraction L1/(L1+L2) of it pass L2
- fraction L2/(L2+L2)of it pass L1

Formal Solution

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

Differ.(1): dI1/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)

ω2i1(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

- 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-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

- 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(C0+C1)=1/2C0Ep+3/2C1Ep

- V=Ep/2[(C0+3C1)/(C0+C1)]
- VB&VC rise to 3/2Ep osc. C0,C1&Ls
CCT shown in Fig.

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:
Ip=2Ep(C0/C0+C1)x

√{(C0+C1)/3L}

□ IB pass out of node Bdivided:Ic1,Ic0

□ Ic1 pass the arc; its frac.:

C0/(C0+C1)

□Also a 60Hz current due to VA throughCN

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

- 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…

- 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

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

- 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

- 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

3-Vc1(t0)?

4-Max of Vc2 ?

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 reversesfig4.4 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)

- series RLC CCT
- At this instant:
Vc1(0),Vc2(0) required

- CCT diff. Eqs employed
- Solved for Vc2

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:

dVc2/dt+R/LdVc2/dt+{(C1+C2)/LC1C2}Vc2=

-Vc1(0)/LC2

svc2(s)-sR/LVc2(0)-R/LV’c2(0)+sR/Lvc2(s)-√{(c1+C2)/LC1C2}xVc2(s)=-Vc1(0)/sLC2

Solution of Q1: ..continued

- I(0)=0V’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’=10fig4.7 : Vpeak=1.855

1pu=-T^2.Vc1(0)/LC2=

=-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

- 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

- 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+cosx-2cosx

Q2 continued

- ∫(1-cosωt) dt=
3/2t+1/(4ω) sin2ωt-2/ω sinωt=6Π/ω

t=0,4Π/ω

- Energy loss if TRV was a 1-cosine wave:
V/Rx6Π/ω=112.7/800x6Π/314=0.9526MJ

(952.6KJs)

- However (1-cosine) is deformed and assuming
to be halved : Energy loss≈952/2=476 KJs

Question 3

- 246 KW Load on a 3ph S.G., 13.8 KV
p.f.=0.6 lag(load parallel RL) Cpf=1

load+cap switch as a unit

- S opened, Vpeak across C.B.?
- Zs negligible

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 …

- CCT a parallel RLC
- -cdVc/dt=Vc/R+
1/L∫Vcdt+IL(0)

- dVc/dt+1/RCdVc/dt + 1/LC =0
- [s+s/RC+1/LC]vc(s)=(s+1/RC)Vc(0)+V’c(0)

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

- 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

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

if excited ~ Natural Freq.

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,
C=1/ωL=1.75μF

- 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

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