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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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Presentation on theme: "RLC CCTs To Simulate Damping  Φ:I of branch or V across the CCT  Ψ:V across a comp. or I in CCT."— Presentation transcript:

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

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

3 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

4 The RL Load and Switching off

5 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. 1.0  Transient?, fully loaded Transf.

6 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

7 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

8 Discussion of Transient Resp.  for I, the current: dI/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

9 Discussion Continued  Z 0 =√L/C=√20.2/0.2228=9.52 Ω  λ=Z 0 /R=9.52/5.713=  I, starts with – A, swing to +ve peak of half cycle later. & -.06X after another half cycle (these values should be corrected in the text book)  For V c : dV 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

10 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= ms  Reaches peak in 1.4T=2.97 ms  Fast Transient and Corona Damping  Always higher freq. Damped quicker

11 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

12 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

13 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 ( PF)  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

14 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

15 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- dV/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

16 … continued  Chopping of Magnetizing current of a 13.8 kV

17 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.)

18 The response with cur. Chop. Practical Ex: Shown in Figure  1-chop only 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

19 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

20 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

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

22 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

23 Capacitance Switching off

24 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

25 Capacitance Switching with a Restrike at Peak of Voltage

26 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

27 Multiple Restrikes During Capacitance Switching

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