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RLC CCTs To Simulate Damping

RLC CCTs To Simulate Damping. Φ :I of branch or V across the CCT Ψ :V across a comp. or I in CCT. Typical Differential Eq. of RLC. The Parallel RLC Eq(1): The Series RLC Eq(2):. Load Switching. Switch on & off loads : most Freq. RL, Low P.F. when Inductive

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RLC CCTs To Simulate Damping

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  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 V0 : Vs (at instant I ceases) C charged to V0, 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. P.F.to 1.0 • Transient?, sw.off fully loaded Transf.

  6. Eq. CCTs & Discussion • Schematic & Eq. • Iload=20000/(13.8√3)=836.7 A (rms) • Z=13.8/(√3x836) =9.522 Ω • φ=cos−0.6=53. • RT+RL=9.52cosφ=5.7 • XT+XL=9.52sinφ=7.6 • L=20.2 mH

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

  8. Discussion of Transient Resp. • for I, the current: dI/dt+1/Ts dI/dt+1/T=0 i(s)(s+s/Ts+1/T)=(s+1/Ts)I(0)+I’(0) • Transient of series RLC CCT: L dI/dt+IR=Vc LI’(0)+I(0)R=Vc(0)=0 I’(0)=-I(0)R/L=-I(0)/Ts • i(s)=s/(s+s/Ts+1/T) . I(0)-Fig4.6

  9. Discussion Continued • Z0=√L/C=√20.2/0.2228=9.52 Ω • λ=Z0/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 Vc: dVc/dt+1/Ts dVc/dt +Vc/T=0 • vc(s)(s+s/Ts+1/T)=(s+1/Ts)Vc(0)+V’c(0) • Vc(0)=0, V’c(0)=-I(0)/c • vc(s)=1/(s+s/Ts+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)Z0 • 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

  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:½LmI0 • Lm very large

  13. Cur. Chop. • ½ CV=1/2 LmI0 V=I0 √Lm/C • I0: Instant. current chopped • i.e. 1000KVA, 13.8 KV Transformer 1- magnetizing current=1.5 A (rms) 2-Lm=V/ωIm=13800/(√3x377x1.5)=14 H • eff.Cap.type of wind.&ins(1000-7000PF) • If C=5000 PF, Z0=√[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 • I0√(0.3Lm/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- IC+IR+IL=0, sub. & Diff. 2- dV/dt+1/RCdV/dt+V/LmC=0 3-v(s)(s+s/RC+1/LmC)=(s+1/RC)V(0)+V’0 • V’(0)=-Ic(0)/c=-I0/C • V(s)=sV(0)/(s+s/RC+1/Lmc)+V(0)/Rc x 1/(s+s/RC+1/Lmc) –I0/[c(s+s/RC+1/LmC)] • 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=TVc(0)/Tp = Vc(0)/η • ζ− I0/{c[s+s/Tp+1/T]}=TI0/C 2η/(√4η-1) . exp(-t’/2η) sin[√(4η-1) t’/2η] □ TI0/C=Z0I0 peak Amp. Chopping Term (exclude damp.)

  18. 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 tc=0, TRV reduce as tc>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 DevicesCurrent 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 Vp

  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 Vp: 1- c=5/(377x13.8)=69.64μF 2- Z=√1000/69.64=3.789Ω 3-Ip=2√2x13.8/(√3x3.789)=5.947 KA 4-f0=603 Hz

  27. Multiple Restrikes During Capacitance Switching

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