Simple Switching Transients ï¯ Considering AC Drives ï¯ Two Switching Cases: 1-...
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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 L 1-Source Imp. neglig. 2-source:V,f=50,60Hz P.F.= RL CCT & Sine Drive CCT. B. Closes Related O.D.E RI + L dI/dt=V= Rewriting i(s) If: A=(V m/ L).. Cos B=(V m /L). sin =R/L Necessary Inverse Transforms Time Response After Reformulation Noting: tan=L/R=/ Discussion on Response (RL Load) Two Components in Response: 1- The steady state (sin. Osc. Term) 2- The Transient (exp. Term) If = : Transient term=0 I(t) sym. If -=/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) :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 I s.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 100 s.steel Design resistors: i.e.:length, cross section Require: Rs Temp. riseRDSB : Reignition T hen Switch passes I f (another half Cycle) TRV named Restriking Voltage Experimental TRV results 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 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 F 0 =1/(210^-3x10^-10)=250KHz T 0 =4s, in T 0 /2 TRV swing to 2V p 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 Interruption of Asymmetrical I f Sw. closes at random,I likely to Asym And Degree of Asym. for I f Now C.B. opens at I=0,V not at peak TRV now is not so High : Figure R.V. osc. Around V inst.(nolonger at Peak) TRV is not as high TRV considerig C.B. Arc Voltage If arc vol. not negl. The inv. Transf. of term: V c (0)=arc volt.,I=0 Increasing Sw.Trans. Effect: I more into Phase with supply voltage Fig. Assignment No. 1 Question1 C 1, 120 KV 1 st S closed 45s later G What is I R 2 ?& V c 1 ? C 1 =5F,C 2 =0.5F R 1 =100, R 2 =1000 Solution of Question 1 C 1 V 1 (0)+C 2 V 2 (0)=(C 1 +C 2 )V final V final =C 1 /(C 1 +C 2 ).V 1 (0)=600/5.5= 109. KV With =100x0.4545=45.45s V 2 (t)=109.(1-e^ (-t/45.45) )=68. KV Therefore I R2 =68KV/1000=68A V 1 (t)=120-11(1-e^ (-t/45.45) )=113.KV Question 2 C 1 has 1.0 C, C 2 discharged C1=60, C 2 =40F R=5 I peak? I( t=200s ) ? E ultimate in C 2 ? V c1( ultimate ) ? Solution of Question 2 I peak =1x10^6/60/5=3.33KA C eq =24F, =5x24=120s I( t=200s )=3.33xe^( -t/120)= A V final= 1x10^6/(60+40)=10KV 1/2x40x10^-6x(10^8)=2000 Js Question 3 Field coil of a machine, S 1 closes 1 s, Energy in coil? Energy dissipated? S.S. reached,S1 opened S2 closed 0.1 s. later V s1? E dissipated in R 2 ? L= 2 H, R1=3.6 R 2 =10 Soultion (Question 3) I final =800/3.6= A I(t=1)=222.2x(1-e^(-R 1 t/L))=185.5A 1/2LI^2= Js E suppl. =VI dt=800^2/3.6(1-e^ (-1.8t) )dt =1.778x10^5[t+e^(-1.8t)/1.8]= =95338 Js E dissip. = =60932 Js I R2(t=0.1 s) =222.2xe^(-13.6t/2)=112.6A V R2 =1126 volts, V s1 = =1926 volts E s.s. =1/2*2*222.2^2=49380Js, E R2=49380X10/13.6=36310 Js Double Frequency Transients Simplest case Opening C.B. Ind.Load,Unload T. L 1,C 1 source side L 2, C 2, load side open:2halves osc indep. Deduction: Pre-open. V c = L 2 /(L1+L2) x V D.F. Transients continued Normally L 2 L 1 C 1 & C 2 charged to about V(t) of Sys. This V, at peak when I=0 C 2 discharge via L 2, f 2=1/(2(L2.C2)) C 1 osc f1=1/(2L1.C1)) about V sys Figures :Load side Tran.s & Source side Transients Clearing S.C. in sec. side of Transf. Another usual D.F. Transients L 1 Ind. Upto Trans. L 2 Leak. Ind. Trans. C 1 &C 2 sides cap. Fig. Two LC loop Second D.F. Transients Eq. CCT. V c1 (0)= L 2 /(L 1 +L 2 ). V V c1 =V-L 1 dI 1/ dt= V c1 (0)+ 1/C1(I 1 -I 2 )dt V c2 = 1/C 2 I 2.dt V c2 =V-L 1 dI 1/ dt- L 2 dI 2 /dt Apply the L. T. to solve Eq.s V/s-L 1 si 1 (s)- L 1 I 1 (0)=V c1 (0)/s+1/C 1 s[i 1 (s)-i 2 (s)] -i 1 (s)(L 1 s+ 1/C 1 s)+i 2 (s)/C 1 (s)= V c1 (0)/ s+L 1 I 1 (0)-L 2. si 2 (s)+L 2 I 2 (0) V c2 (s)=i 2 (s)/sC 2 V c2(s)= V/s-L 1 si 1 (s)+L 1 I 1 (0)- L 2 s i 2 (s)+ L 2 I 2 (0)