Simple Switching Transients Considering AC Drives Two Switching Cases:

1- Closing a CCT Breaker, Energizing Load

2- Opening a Breaker, Clearing Fault

The CCT Closing Transients Energizing an RL

CCT load:Series R and L1-Source Imp. neglig.2-source:V,f=50,60Hz

P.F.= 2 2 2( )R R

R L

Z

RL CCT & Sine Drive

CCT. B. Closes Related O.D.E RI + L dI/dt=V= ( )mV Sin t

(sin cos cos sin )mdIRI L V t tdt

2 2 2 2( ) ( ) (0) ( cos sin )msRi s Li s LI V

s s

2 2 2 2

1 cos sin( ) ( )( )

mV si s RL s ss L

Rewriting i(s) If: A=(Vm/L).ω. Cosθ B=(Vm/L). sinθ α=R/L

2 2 2 2( )( )( ) ( )( )

A Bsi ss s s s

Necessary Inverse Transforms

2 2 2 2 2 2 2 2

1 1 1( )( )( )

ss s s s s

12 2 2 2

1 1 ( cos sin )( )( ) ( )

te t ts s

12 2 2 2

1 ( sin cos 1 1 0)( )( )

ts e t ts s

Time Response After Reformulation

Noting: tanΦ=ωL/R=ω/α

12 2 2sin /( )

12 2 2cos

( )

2 2 2( ) [sin( ) sin( ) ]( )

tmVI t t eR L

Discussion on Response (RL Load) Two Components in Response: 1- The steady state (sin. Osc. Term) 2- The Transient (exp. Term)

If θ=Φ : Transient term=0 I(t) sym.If θ-Φ=±∏/2 : peak = 2 x (s.s. comp.)I(t) Max. Asym.

Breaker Capability (conclusion 1)

CCT.B. May Close into a S.C.(i.e. R,L relavent to source)

θ:Arbitrary In 3 ph a good chance of Full Asym.□ Designed Elec., Mech, & Thermally to withstand this duty □ Perform Subsequent open & Reclose

Serious Forces (conclusion 2) Buckling of parts can occur by Electro. Forces caused by current□Very high current, weld the contacts □If contact popping happens conditions get worse □Any weld formed must not impair the efficiency of CCT B.□Mechanism be able to break weld and rupture the fault current

Breaking Current (conclusion 3) time from Fa. initiation to contacts

parting,& instant. Current:

--Actual current to interrupt --C.B. of high decr., duty is less severe

□Damage, Thermal O.L. lines, system stability Fast. Opening of S.C.

Considering Transient & SubTransient reactances conclusion4 S.C. close to Generator T. & SubT. reactance Extra high first peak of Is.c.

After T. & sub. T., for several periods (1 Ph. fault): current zero is lacking

Example : 3ph Cap.Bs. 1st:3 X 60MVA,13.8 KV, star con. 2nd:3X30MVA,13.8KV,star con.●Be paralleled momen. by 100Ω s.steel●Design resistors: i.e.:length, cross section●Require: R’s Temp. rise<200 C sw. op. when one Bank. in pos. peak, other in

neg. peak

Stainless steel characteristics density:7.9 g/ cubic. cm S.Heat:0.5 J/g per C Resistivity:72μΩcmNote:no heat is lost during Sw. Op.

●What is weight of resistor●What will be the peak current

Example Solution R=ρ l/s l/s=R/ρ=100/(72x10^(-8))= 1.389x10^8 Reactance cap. Bank 1:(13.8)^2/160=1.058 Ω C1=1/(1.058x314.15)=3.0 mF C2=1.5 mF Peak voltage:13.8X√2/√3=11.27 KV Dissipated E.:1/2 Ceqx(V1(0)-V2(0))^2 = 2.54x10^5 joules Ceq=C1C2/(C1+C2)

Solution ……continued Dissip.Energy.=v.g.k.∆T= =lxsx7.9x0.5x200=790xlxs ls=2.54x10^5/790=321.52 l/sxls=l^2=1.389x10^8x321.52= =21.13x10^2 sq. m. l=45.96 m S=321.5/4596=6.99x10^-2 sq. cm. m=lsg=321.52x7.9=2.54 Kg I(peak)=∆V/R=22.54x10^3/100=225.4 A

Simple Sw.:Cap. Inrush current Shunt Capacitors :closing op. Assume a number of parallel units In Fig. there are two with sw:S1,S2 Source Inductance L, local CCT L1 I(t)=V(0)/Z0 sinω0t Z0=√L/C, V(0):inst. Voltage ω0=1/√LC

CCT Configuration

Consider 34.5 KV (a S.Gr.) system S.C. current at Bus:25 KA (rms), sym If C1=18 MVA, C2=10 MVA Next Fig: Bus B. Schematic diag. (dimen.)CALCULATION RESULTS(1st Trans. S1 close): I=1.8x10^7/(√3x34.5x10^3)=301.2A Xc1=34500/(√3x301.2)=66.13Ω C1=40.1 μF Xs=34.5/(√3x25)=0.797Ω , L=0.0021H Ip=(34.5√2/√3)(40.1/2100)^0.5=3.893KA f=1/2∏√LC1=548 Hz

Schematic Diagram of Cap. B.

Closing Switch S2 (2nd Transient) L1 depends geometry & Physical Dim. Inductance of Bus work prop:Cond.size Al. angles & channels used in Cap. Banks Ind. Param.:0.2 to 0.4 μH/ft(0.3 selected) L1=0.3X64(ft)=19.2 μH C2=22.28 μF C1C2/(C1+C2)=14.32μF

2nd Transient Results Z01=√(19.2/14.32)=1.158Ω Inrush from C1 to C2(C2discharged,C1Peak) Iinr=Vp/Z01=(34.5√2)/(√3x1.158)=22.33 KA Freqinr=10^6/2∏(√19.2x14.32) =9.6KHz Justifies negl. the Current from Source Extreme Case (Vc2=-Vp), results: Iinr.=48 KA

Discussion of H.F. Inrush Eff.s Severe Mechanical Stresses Transients in Nearby CCTs Cap. Bank Inrush Reduction Methods: Z increase, Extra L in form of Reactor Add R, damp Osc.s (and short it later) the 2nd switch cost much Synchronous Sw. (pre-strike limited app.)

3rd stage: Final Common Voltage of 2nd: C1V1(0)+C2V2(0)=(C1+C2).V(final)i.e.: Vfinal=(C1V1(0)+C2 V2(0))/(C1+C2)= (40.1x28.17-22.28x8)/(40.1+22.28)=15.25 KV Now two cap.s swing to sys. Peak Osc. Freq. = 1/[2∏√L(C1+C2)]=439.7 Hz V, rises: 28.17+(28.17-15.25)=41.09 KV Current (Vp-Vf)√[(C1+C2)/L].sinω1t= (28.17-15.25)√(62.38/2100) . Sinω1t= 2.227 sinω1t KA…………Fig. of OSC.

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 cosωt I=C dVc/dt (only I)

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

2

2 cosc 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 Vs

'20 2 2 2 2 2 2 2 2

0 0 0

(0)( ) (0)( )( )

cc m c

Vs sv s V Vs 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 ss s s s

20

2 2 2 2 2 20 0

( ) ( )c ms 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 results 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. Transf. of term:

Vc(0)=arc volt.,I=0 Increasing Sw.Trans. Effect: I more into Phase with

supply voltage Fig.

102 2(0) (0)cosc c

sV V ts

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)