Transient Recovery Voltage Following S.C.

Removal L , and C: natural

cap

Simplifying Assumptions CCT’s R, and Losses ignored After sep. of contacts, I flows in Arc I reach zero by controlling arc In Ac two I=0 in each cycle Current : Symm. & compl. Inductive At I=0 : VccT Max, VCB=Varc Assuming Varc=0 Time measu. from Inst. of Interr.

CCT Equation The KVL & Ic:L dI/dt+ Vc=Vm cosωtI=C dVc/dt (only I)

Physical InterPr.:- Ultimately: V supply- t=0, previous arc V=0- C charged, through L

& cause Osc.

2

2cosc c md V V V

tdt LC LC

Similar to LC ccT analysis Assuming:

ω0=1/LC applying L.T.:

Vc(0)=0 arc vol. Vc’(0)=I(0)/C=0

2 ' 20 0 2 2

( ) (0) (0) ( )c c c c m

ss v s sV V v s V

s

'

20 2 2 2 2 2 2 2 2

0 0 0

(0)( ) (0)

( )( )c

c m c

Vs sv s V V

s s s s

Time Respose Inv. Transf. just 1st

term Need:

Then:

Vc(t) is TRV Eq.:

2 2 2 2 2 2 2 2 2 20 0 0

1( )

( )( )

s s s

s s s s

20

2 2 2 2 2 20 0

( ) ( )c m

s sv s V

s s

20

02 20

( ) (cos cos )c mV t V t t

TRV Eq. (Park & Skeats) Discussion As ω0 » ω ,

Thus:

∆V(P.F.) very small

20

2 20

1.0

0( ) (cos cos )c mV t V t t

0( ) [1 cos ]c mV t V t

TRV Discussion Continued Fig : TRVp=2 x P.F. Vp TRV Osc.s damped out C.B. Ops follows Cap. Being charged VCB rises fast if: - L or C or both very small (ω0 large)

□ if RRRV>RDSB : Reignition

□ Then Switch passes If (another half Cycle)

TRV named “Restriking Voltage”

Experimental TRV reults

The TRV lasts 600 μs Decline of current TRV starts a small opp. polar.To ins. Vol.

due to: some “Current Chopping” Shows the H.F. Osc. Shows how H.F. Damped

r.r.r.v. factor A measure of severity of CCT for C.B. r.r.r.v.s high as Natural freq. higher air-cored reactor L=1 mH,C=400 pF F0=1/(2∏√10^-3x10^-10)=250KHz T0=4μs, in T0/2 TRV swing to 2Vp

In a 13.8KV CCT , r.r.r.v.=2x13.8√2/(2x√3)=11.3KV/μs

Beyond Capab. Most C.B.s Ex. Of very fact TRV: Kilometric Faults ch.9

Interruption of Asymmetrical If

Sw. closes at random,I likely to Asym And Degree of Asym. for If Now C.B. opens at I=0,V not at peak TRV now is not so High : Figure R.V. osc. Around Vinst.(nolonger at Peak)

TRV is not as high

TRV considerig C.B. Arc Voltage If arc vol. not negl. The inv. Trans. of

term:

Vc(0)=arc vol.,I=0 Increasing Sw.Tra Effect: I more into

Phase with supply voltage Fig.

102 2

(0) (0)cosc c

sV V t

s

Assignment No. 1 Question1 C1, 120 KV 1st S closed 45μs later G What is IR2?& Vc1? C1=5μF,C2=0.5μF R1=100Ω, R2=1000Ω

Solution of Question 1

C1V1(0)+C2V2(0)=(C1+C2)Vfinal

Vfinal=C1/(C1+C2).V1(0)=600/5.5= 109. KV With τ=100x0.4545=45.45μs V2(t)=109.(1-e^(-t/45.45))=68. KV Therefore IR2=68KV/1000=68A V1(t)=120-11(1-e^(-t/45.45))=113.KV

Question 2 C1 has 1.0 C, C2

discharged

C1=60, C2=40μF R=5 Ω Ipeak?

I(t=200μs) ? Eultimate in C2 ? Vc1(ultimate) ?

Solution of Question 2

Ipeak=1x10^6/60/5=3.33KA Ceq=24μF, τ=5x24=120μs I(t=200μs)=3.33xe^(-t/120)=629.5 A Vfinal=1x10^6/(60+40)=10KV 1/2x40x10^-6x(10^8)=2000 Js

Question 3 Field coil of a machine,

S1 closes 1 s, Energy in coil? Energy dissipated? S.S. reached,S1 opened S2

closed 0.1 s. later Vs1? E dissipated in R2 ? L=2 H, R1=3.6Ω R2=10Ω

Soultion (Question 3) Ifinal=800/3.6=222.22 A I(t=1)=222.2x(1-e^(-R1t/L))=185.5A 1/2LI^2=34406. Js Esuppl.=∫VI dt=800^2/3.6∫(1-e^(-1.8t))dt =1.778x10^5[t+e^(-1.8t)/1.8]= =95338 Js Edissip.=95338-34406=60932 Js IR2(t=0.1 s)=222.2xe^(-13.6t/2)=112.6A VR2=1126 volts, Vs1=800+1126=1926 volts Es.s.=1/2*2*222.2^2=49380Js, ER2=49380X10/13.6=36310 Js

Double Frequency Transients Simplest case Opening C.B. Ind.Load, Unload T. L1,C1 source side L2,C2, load side open:2halves osc

indep. Deduction: Pre-open. Vc=L2/(L1+L2)

x V

D.F. Transients continued

Normally L2»L1

C1 & C2 charged to about V(t) of Sys. This V, at peak when I=0 C2 discharge via L2, f2=1/(2∏√(L2.C2))

C1 osc f1=1/(2∏√L1.C1)) about Vsys

Figures :Load side Tran.s & Source side Transients

Clearing S.C. in sec. side of Transf.

Another usual D.F. Transients

L1 Ind. Upto Trans. L2 Leak. Ind. Trans. C1&C2 sides cap. Fig. Two LC loop

Second D.F. Transients Eq. CCT. Vc1(0)=L2/(L1+L2) .V Vc1=V-L1dI1/dt=

Vc1(0)+1/C1∫(I1-I2)dt

Vc2=1/C2∫I2.dt Vc2=V-L1dI1/dt-

L2dI2/dt

Apply the L. T. to solve Eq.s V/s-L1si1(s)-L1I1(0)=Vc1(0)/s+1/

C1s[i1(s)-i2(s)] -i1(s)(L1s+1/C1s)+i2(s)/C1(s)=

Vc1(0)/s+L1I1(0)-L2.si2(s)+L2I2(0) Vc2(s)=i2(s)/sC2

Vc2(s)=V/s-L1si1(s)+L1I1(0)-

L2si2(s)+L2I2(0)

The D.F. CCT response Switch clear at t=0 I1(0)=I2(0)=I(0)=0 i1(s)=V/(L1s)-

[(L2C2s^2+1)/L1s].vc2(s)

In term of vc2(s):

4 2

1 1 2 2 2 1 1 1 2 2

21 2 2 1 1 2 2

1 1 1 1( )

1( ) ( )

( )c

s sLC L C L C LC L C

sv s V

L L C LC L C s

CCT Natural Frequencies (s^2+ω1^2)(s^2+ω^2)vc2(s)=

AV(1/s+Bs) A,B,ω1,ω2 function of : L1,C2,L2,C2

2 1 22 2 2 2 2 2 2 21 2 1 1 2 2 1 2

2 12 21 2

1 1 1( ) [ cos cos ( )

( ) ( )

(cos cos )]

V t AV t t

Bt t

Parameters of Eqs A=1/(L1C1L2C2) B=(L1+L2)/L1L2 C1

21,2

1 1 2 2 2 1

21 1 2 2

1 1 2 2 2 1

1 1 1 1( )2

1 1 1 1( ) 4 /( )

2

LC L C L C

LC L CLC L C L C

Damping

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

RLC CCTs & General Diff. Eqs Φ:I of branch or V

across the CCT

Ψ:V across a comp. or I in CCT

Typical Differential Eq. of RLC The Parallel RLC:

The Series RLC :

2

2

1( )

d dF t

dt RC dt LC

2

2( )

d R dF t

dt L dt LC

Parameters Continued Tp= RC = Parallel CCT time constant Ts= L/R = series CCT time constant TpTs=LC=T η= R/Z0=R√(C/L) η=Tp/Ts=RC/L Duality Relationship of Series&Parallel Transforms & their inverse plots in Dimensionless curves using: η

Basic Transform of RLC CCTs Closing s in C Parallel RLC