specialtopics_3

8/3/2019 specialtopics_3

1/30

Transient Recovery Voltage Following S.C.

Removal

L , and C: naturalcap


2/30

Simplifying Assumptions CCTs 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.


3/30

CCT Equation The KVL & Ic:L dI/dt+ Vc=Vm costI=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 m

d V V V t

dt LC LC [ !


4/30

Similar to LC ccT analysis Assuming:

0=1/LC applyingL.T.:

Vc(0)=0 arc vol.

Vc(0)=I(0)/C=0

2 ' 2

0 0 2 2( ) (0) (0) ( )

c c c c m

s s v s sV V v s V

s[ [

[

!

'

2

0 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[

[ [ [ [!


5/30

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[ [ [ [ [ [!

2

0

2 2 2 2 2 2

0 0

( ) ( )c m

s sv s V

s s

[

[ [ [ [

!

2

002 2

0

( ) (cos cos )c m

V t V t t [

[ [

[ [

!


6/30

TRV Eq. (Park & Skeats) Discussion As 0 ,

Thus:

V(P.F.) very small

2

0

2 2

0

1.0[

[ [

}

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

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


7/30

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 (0 large)

if RRRV>RDSB : Reignition

Then Switch passes If (another half Cycle) TRV named Restriking Voltage


8/30

Experimental TRV reults The TRV lasts 600 s

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


9/30

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/(210^-3x10^-10)=250KHz

T0=4s, in T0/2 TRV swing to 2Vp

In a 13.8KV CCT ,

r.r.r.v.=2x13.82/(2x3)=11.3KV/s Beyond Capab. Most C.B.s

Ex. Of very fact TRV: Kilometric Faults ch.9


10/30

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


11/30

TRV considerig C.B. Arc Voltage If arc vol. not negl.

The inv. Trans. ofterm:

Vc(0)=arc vol.,I=0

Increasing Sw.Tra

Effect: I more intoPhase with supplyvoltage Fig.

102 2

(0) (0) cosc c

sV V ts

^ [[

!


12/30

Assignment No. 1 Question1

C1, 120 KV

1st

S closed 45s later G

What is IR2?& Vc1?

C1=5F,C2=0.5F

R1=100,

R2=1000


13/30

Solution of Question 1 C1V1(0)+C2V2(0)=(C1+C2)Vfinal

Vfinal=C1/(C1+C2).V1(0)=600/5.5=

109. KV With =100x0.4545=45.45s

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


14/30

Question 2 C1 has 1.0 C, C2

discharged

C1=60, C2=40F

R=5 Ipeak?

I(t=200s) ?

Eultimate in C2 ? Vc1(ultimate) ?


15/30

Solution of Question 2 Ipeak=1x10^6/60/5=3.33KA

Ceq=24F, =5x24=120s

I(t=200s)=3.33xe^(-t/120)=629.5 A Vfinal=1x10^6/(60+40)=10KV

1/2x40x10^-6x(10^8)=2000 Js


16/30

Question 3 Field coil of a machine,

S1 closes 1 s, Energy in coil? Energy dissipated?

S.S. reached,S1 opened S2closed

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


17/30

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


18/30

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


19/30

D.F. Transients continued Normally L2L1

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/(2L1.C1)) about Vsys

Figures :Load side Tran.s & Sourceside Transients


20/30

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


21/30

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/C2I2.dt

Vc2=V-L1dI1/dt-L2dI2/dt


22/30

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)


23/30

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

2

1 2 2 1 1 2 2

1 1 1 1( )

1( ) ( )

( )c

s sL C L C L C L C L C

sv s V

L L C L C L C s

- !


24/30

CCT Natural Frequencies (s^2+1^2)(s^2+^2)vc2(s)=

AV(1/s+Bs)

A,B,1,2 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

[ [

[ [ [ [ [ [ [ [

[ [

[ [

!


25/30

Parameters of Eqs A=1/(L1C1L2C2)

B=(L1+L2)/L1L2 C1

2

1,2

1 1 2 2 2 1

2

1 1 2 2

1 1 2 2 2 1

1 1 1 1( )2

1 1 1 1( ) 4 /( )

2

L C L C L C

L C L C

L C L C L C

[ ! s


26/30

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 CCT2- Series RLC CCT


27/30

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

across the CCT

:V across a

comp. or I in CCT


28/30

Typical Differential Eq. of RLC The Parallel RLC:

The Series RLC :

2

2

1( )

d dF t

dt RC dt LC

J J J !

2

2( )d R d F t

dt L dt LC

] ] ] !


29/30


30/30

Basic Transform of RLC CCTs Closing s in C

Parallel RLC

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