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RLC CCTs To Simulate Damping Φ:I of branch or V across the CCT Ψ:V across a comp. or I in CCT

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Typical Differential Eq. of RLC The Parallel RLC Eq(1): The Series RLC Eq(2):

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Load Switching Switch on & off loads : most Freq. RL, Low P.F. when Inductive High P.F. when Resistive C_ loadbus :role in After sw. off. Transient V 0 : V s (at instant I ceases) C charged to V 0, disch., In RL, Damp Os. Dis. A damped cosine wave of Fig. 4.6 As P.F. improve, Transient decrease

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The RL Load and Switching off

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Arc Furnace Example Low voltage & High Curent Fed by step down furn. Transformer Low P.F. & freq. switching Cap.s connected to HV bus impr. P.F. Delta & Wye Connections Example:Wye connection,Transf.60Hz 13.8 KV,20 MVA Y/Y solid Gr P.F. at Full Load;0.6,C corr. P.F.to 1.0 Transient?, sw.off fully loaded Transf.

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Eq. CCTs & Discussion Schematic & Eq. I load =20000/(13.8 √3)=836.7 A (rms) Z=13.8/(√3x836) =9.522 Ω φ=cos−0.6=53. R T +R L = 9.52cosφ=5.7 X T +X L = 9.52sin φ=7.6 L=20.2 mH

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Discussion Furn. Ex. continued open:I s (0)=0, required:I c(0) =-I(0) I c =-I=836.7sinΦ=669.4A (rms) I c is at peak since V c =0, and I c (t=0)=669.4√2=946.67A (text result should be corrected) V c (0)=0 X c =13.8/(√3x669.4)=11.9 Ω (please correct text book results) C=222.6 μF

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Discussion of Transient Resp. for I, the current: dI/dt+1/T s dI/dt+1/T=0 i(s)(s+s/T s +1/T)=(s+1/T s )I(0)+I’(0) Transient of series RLC CCT: L dI/dt+IR=V c LI’(0)+I(0)R=V c (0)=0 I’(0)=-I(0)R/L=-I(0)/T s i(s)=s/(s+s/T s +1/T). I(0)- Fig4.6

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Discussion Continued Z 0 =√L/C=√20.2/0.2228=9.52 Ω λ=Z 0 /R=9.52/5.713=1.6664 I, starts with –946.67 A, swing to +ve peak of 0.105 half cycle later. & -.06X946.67 after another half cycle (these values should be corrected in the text book) For V c : dV c /dt+1/T s dV c /dt +V c /T=0 v c (s)(s+s/T s +1/T)=(s+1/T s )V c (0)+V’ c (0) Vc(0)=0, V’ c (0)=-I(0)/c v c (s)=1/(s+s/T s +1/T). I(0)/c

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Transformer Terminal Voltage Fig 4.4 λ=1.66 peak reaches 65% undamped:[-I(0)/C]T=-I(0)Z 0 The first voltage peak: 0.65x946.67x9.52=5.85 KV (please correct the value in the text book) The time scale is T=√LC= 2.121 ms Reaches peak in 1.4T=2.97 ms Fast Transient and Corona Damping Always higher freq. Damped quicker

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Abnoraml Switching Normal : 2 pu Abnormal : mag. Far beyond this 1-current suppression 2- Capacitor Bank switching off 3-Other Restriking Phenomena 4-Transformer Mangnetizing Inrush 5-Ferroresonance

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Current Suppression N.,I ceases, arc current, periodic Zero Abn., arc suppression force current 0 Current Chopping trapped mag. Energy Abn. Voltage Ex: sw. off Transformer magnetizing current Energy stored:½L m I 0 L m very large

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Cur. Chop. ½ CV=1/2 L m I 0 V=I 0 √L m /C I 0 : Instant. current chopped i.e. 1000KVA, 13.8 KV Transformer 1- magnetizing current=1.5 A (rms) 2-L m =V/ωI m = 13800/(√3x377x1.5)=14 H eff.Cap. type of wind.&ins (1000-7000PF) If C=5000 PF, Z 0 =√[14/5x10^-9]= 52915Ω If C.B. chops I_peak, can be 2.5 A, V(peak)≈132KV Abnormal for 13.8 KV

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Cur. Chop. Discussion Not So High: 1- damping, 2- fraction of Energy release shaded area< 30% stored energy I 0 √(0.3L m /C)= 55% V (transient) Dis. Transf. most vulnerable

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Continued… Air cored reactors (core of significant air-gap) 1-All energy recoverable 2-If as shunt compensator, protected by L.A. Formal Evaluation of RLC CCT 1- I C +I R +I L =0, sub. & Diff. 2- dV/dt+1/RCdV/dt+V/L m C=0 3-v(s)(s+s/RC+1/L m C)=(s+1/RC)V(0)+V’ 0 V’(0)=-I c (0)/c=-I 0 /C V(s)=sV(0)/(s+s/RC+1/L m c)+V(0)/Rc x 1/(s+s/RC+1/L m c) –I 0 /[c(s+s/ RC+1/L m C )] Transforms of Fig4.4 & Fig 4.6 first two normal Transient terms without chop

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… continued Chopping of Magnetizing current of a 13.8 kV

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The response with cur. Chop. 1st term Fig4.6, pu=V(0) 2nd term fig4.4,pu=TV c (0)/T p = Vc(0)/η ζ− I 0 /{c[s+s/T p +1/T]}=TI 0 /C 2η/(√4η- 1). exp(-t’/2η) sin[√(4η-1) t’/2η] □ TI 0 /C=Z 0 I 0 peak Amp. Chopping Term (exclude damp.)

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The response with cur. Chop. Practical Ex: Shown in Figure 1-chop only 0.5-0.6 A (I – to zero) TRV 20KV 2- chop occur instantaneously 3- in practice I declines on a measurable time 4-TRV and time-to- chop/period H.F. Osc. : Figure 5-TRV max if t c =0, TRV reduce as t c >T/4

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Discussion on CB performance small contact sep. dielectric fails Successive attempts raise Higher Voltages until isolation TRV of Cur.Chop. Limited by reignitions (Fig) G. Practice: a cable between C.B. and Transformer drastic reduction in TRV 100 ft of 15 KV cable ( 100PF/ft ) Transformer( 3000PF eff. Cap.) TRV halved Motors No risk: Noload inductance very small compare to transformer

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Semiconductor Devices Current Suppression Gen. OVs to destroy them end half cycle of diode conduction 1-carriers remained at junction region allow current to flow & reverses 2- then sweeps the carriers & returns device to Block state:I collapses fast inductive CCT Eng. Transf. to C, large V

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Current Suppresssion Silicon Diode CCT and Current H.F. Osc. L&C Protection : 1-snubber cap. In P. 2-additional series R

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Capacitance Switching Off Disconnect: C /unload Transmission lines Concerns: reignite/restrike in opening Chance low, Cap. Sw. frequent Cap fully charged Half Cycle VCB=2 V p

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Capacitance Switching off

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Discussion Cap. Sw. Off In fact Vc>Vsys Ferranti Rise Vsource_side decrease to Vsys There is a ∆V change (however,exist in weak systems) Discon. a C.B. in lower side of step down Transformer supplying an unloaded cable Current in Cap. Sw. is freq. small and it is possible to disconnect it In first zero -- with small contact sep., 2 V appear across contacts --- increased possibility of restrike (small separation) Oscillating to new voltage with f0=1/2Π√LC I(restrike)=2Vp/√L/C sinω0t Transient peak of 3 Vp

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Capacitance Switching with a Restrike at Peak of Voltage

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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)=69.64μF 2- Z=√1000/69.64=3.789Ω 3-I p =2√2x13.8/(√3x3.789)=5.947 KA 4-f 0 =603 Hz

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Multiple Restrikes During Capacitance Switching

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