Estimation of Fundamental Natural Frequency, Damping Ratio and Equivalent Mass 523L (Session 4)
RLC CCTs To Simulate Damping
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Transcript of 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):
2
2
1( )
d dF t
dt RC dt LC
2
2( )
d R dF t
dt L dt LC
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
The RL Load and Switching off
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.
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
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
Discussion of Transient Resp. for I, the current: dI/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
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: dVc/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
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
Abnoraml Switching
Normal : 2 pu Abnormal : mag. Far beyond this1-current suppression2- Capacitor Bank switching off3-Other Restriking Phenomena4-Transformer Mangnetizing Inrush5-Ferroresonance
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
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
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
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- dV/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
… continued Chopping of Magnetizing current of a 13.8 kV
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.)
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
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
Semiconductor DevicesCurrent Suppression
Gen. OVs to destroy them end half cycle of diode conduction 1-carriers remained at junction region allow current to flow & reverses2- then sweeps the carriers & returns
device to Block state:I collapses fast inductive CCT Eng. Transf. to C,large V
Current Suppresssion Silicon Diode
CCT and Current
H.F. Osc. L&C
Protection : 1-snubber cap. In P. 2-additional series R
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
Capacitance Switching off
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
Capacitance Switching with a Restrike at Peak of Voltage
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
Multiple Restrikes During Capacitance Switching