ECE 54: Session 6; Page / Spring 08 ECE 54: Reducing Capacitor Switching Transients Define units: MW 000kW MVA MW MVAr MVA Example : f 60Hz ω πf ω 76.99 rad s t 0 0.00000sec 60 sec Add inductive reactance to limit the worst case inrush current to 000A when closing into C with C open, with an X/R ratio of 0 in the source reactance and an X/R ratio of for the added reactor jx s Assume source and capacitors are Y connected. V LL 4.5kVRMS C C Capacitor Banks: Q Q L 8MVAr 0MVAr 9.μH Source I sc 5kA at 4.5kV at 4.5kV between caps
ECE 54: Session 6; Page / Spring 08 Determine circuit parameters: Find capacitance using: Q = V LL X c V LL V LL X c X c 66. Ω X c X c 9.0 Ω Q C C ωx 40.μF C C c ωx.9μf c Q V m V LL V m 8.7kV ϕ 0deg v s () t V m cos( ωt ϕ) Case : approximate with resistances ignored. V ln Find source impedance using: I sc = V LL X s X s X I s 0.797 Ω L s sc ω L s.mh X s Z 0 L lim L s C L lim Z 0 ( 0mH) 7.6 Ω
ECE 54: Session 6; Page / Spring 08 ω n L lim L s L lim C f n L lim ω n L lim f π n ( 0) 546.6Hz Initial conditons: V c_0 0V capacitor is discharged i Ls_0 0A no load current Homogeneous solution: i h_ tl lim V m V c_0 Z 0 L lim sin ω n L lim t If we ignore the 60Hz current we can solve for L lim V m V c_0 Z 0 L lim = 000A or Z 0 L lim = V m V c_0 000A Substitute: Z 0 L lim L s C L lim V m V c_0 L Lim C L 000A s L Lim 9.7mH Now find the resulting natural frequency 40.85Hz f n L Lim
ECE 54: Session 6; Page 4/ Spring 08 One quarter cycle at that frequency takes: t topeak 4f n L t topeak.77 ms Lim t topeak 60Hz60deg 8.4deg So we proably can't neglect the particular solution, lets check R s 0 0 ohm atan η L lim L lim ω L s R s ωc 90 η L Lim deg i p_ tl lim ωc V LL L lim ω L s cos ωt ϕ ηl lim.85 i p_ t topeak L Lim A So, the actual first positive peak current is: i p_ t topeak L Lim i h_ t topeak L Lim 677.5 A
ECE 54: Session 6; Page 5/ Spring 08 But now look at first negative peak at the natural frequency. t tonegpeak 4f n L t tonegpeak 5. ms Lim i p_ t tonegpeak L Lim i h_ t tonegpeak L Lim 47.6 A bigger than 000A, need larger L lim Now try a simultaneous solution, note that the time to the peak value varies, and is not necessarily at /4 cycle of the natural frequency. L lim 0mH Given i h_ 4f n L lim L Lim L Lim L lim i p_ 4f n L lim L lim Find L lim L Lim.7mH 56.9mH Check the results: 0.98Hz f n L Lim = 000A This is still an approximation, since we are assuming the peak happens at /4 cycle.
ECE 54: Session 6; Page 6/ Spring 08 Three quarters cycle at that frequency takes: t tonegpeak 4f n L t tonegpeak 7. ms t tonegpeak 60Hz60deg 55.8deg Lim i h_ t tonegpeak L Lim i p_ t tonegpeak L Lim 000A 0 i p_ tl Lim i h_ tl Lim 500 0 500 0 0 0.0 0.0 0.0 t
ECE 54: Session 6; Page 7/ Spring 08 i () t i h_ tl Lim i p_ tl Lim i () t 0 0 0 0 0 0 0.0 0.0 0.0 t Note that later peaks exceed 000A. Including the resistance will lower this. First peak is actually 008A Close enough, since inductor cannot be designed so precisely in any case. X lim ωl Lim X lim. Ω Note that the particular solution current increases Without the inductor: I P_pk V LL I P_pk 4.9 A ωl s ωc
ECE 54: Session 6; Page 8/ Spring 08 With the inductor: I P_pk V LL I P_pk 68.66 A ωl s L Lim ωc Bigger current, effectively raises voltage across the capacitor Now look at the effect of lowering the inrush current on the transient voltage on the capacitor v cap () t I P_pk cos( ωt ϕ) X c V m V c_0 cosω n L Lim t V cappk I P_pk X c V m V c_0 V cappk 70.4kV Higher than result without added inductor, but only happens when both peaks coincide, which may not happen for some time. v cap () t 0 5 50 4 0 50 4 0 5 0 0.0 0.0 0.0 t First negative peak: -67.5 kv
ECE 54: Session 6; Page 9/ Spring 08 If the resistance of the source impedance and the resistance of the reactor are included, it is a little more complicated, but the same general approach works: Z s V LL I sc Z s = R X X = X X = 0 0 Z s X s X s X s 0.79 Ω L s L ω s.0mh 0 R s X s R 0 s 0.079 Ω Assume it will still be underdamped, since X/R of the reactor is large. αl lim ω d L lim ωl lim R s L s L lim α ( 0 ) 8.85 s ωl lim R s L s L lim ω L s L d ( 0) 44.9 rad lim s C
ECE 54: Session 6; Page 0/ Spring 08 f d L lim ω d L lim f π d ( 0) 547.96Hz Particular solution: atan η L lim L lim ω L s η( 0) 89.9deg R s ωl lim ωc i L_p tl lim R s ωl lim V LL ωc ωl s L lim cos ωt ϕ ηl lim Boundary conditions: i L ( 0) = 0 Switch was open. 0 = e 0 k k 0 Therefore k 0 d v L ( 0) = V m 0 = L i L_h d t
ECE 54: Session 6; Page / Spring 08 k V m = L s L lim e α0 αsin ω d 0 ω d cos ω d 0 k L lim i L_h tl lim V m ω d L lim k ( 0 ) 890.6 A L s L lim k L lim i L tl lim e α L limt sinω d L lim t i L_h tl lim i L_p tl lim L lim 5mH Given i L 4f d L lim L lim = 000A L Lim Find L lim L Lim 8.6mH Check the results: L Lim 5.9085mH 07.Hz f d L Lim One quarter cycle at that frequency takes: t topeak 4f d L t topeak 7ms Lim
ECE 54: Session 6; Page / Spring 08 t topeak 60Hz60deg 5.5deg 000 i L t topeak L Lim A Same approximation as above. 0 i L_p tl Lim i L_h tl Lim 500 0 500 0 0 0 0.0 0.0 0.0 t 0 i L tl Lim 0 0 0 0 00 00 00
ECE 54: 0 0.0 0.0 0.0 t Session 6; Page / Spring 08 Capacitor voltage v c () t = C i L () t dt V c ( 0) v c_ωd () t k L Lim C α L Lim e α L Limt ω d L Lim αl Lim sinω d L Lim t ω d L Lim cosω d L Lim t v c_ω () t ωc R s ωl Lim V LL ωc ωl s L Lim sin ( ωt ϕ) η L Lim
ECE 54: Session 6; Page 4/ Spring 08 v c () t v c_ω () t v c_ωd () t 60 4 40 4 v c_ωd () t v c_ω () t 0 4 0 0 4 40 4 60 4 0 0.0 0.0 0.0 t
ECE 54: Session 6; Page 5/ Spring 08 50 4 v c () t 0 First peak is now 54.804kV 50 4 0 0.0 0.0 0.0 t
ECE 54: Session 6; Page 6/ Spring 08 Switch in capacitor through a resistance: Eliminate the source resistance again. Suppose the bank is energized through a resistor with bank already in the system. Determine the resistance needed to limit the peak line to ground voltage on either bank to.5kv C C C eq C C C eq 4.μF Simulation Results With R = 5.0 ohm, switching at t = 5.6ms. Results identical with L ignored.
ECE 54: Session 6; Page 7/ Spring 08 40 *0 0 0 0 0-0 -0-0 0 0 0 0 40 50 *0-60 (f ile ExamplR.pl4; x-v ar t) t: VMAX v:c - v:c - Analytical Solution: C 40.μF C.9μF L 9.μH V m 8.7kV Actual value with C in the system Undamped resonant frequencies: f 0 f 0 49.5 π L s C C s
ECE 54: Session 6; Page 8/ Spring 08 f 0 f 0 9596.5 C C s π L C C Likely to be damped severly by a resistance that can damp the lower frequency, so neglect this resonant circuit. Driving voltage for f 0 : From charge balance: C V C ( 0) = C C V( ) Assume near voltage peak when switch C V infinity V m V infinity 8.kV C C V drivingpoint V m V infinity V drivingpoint 0.06kV Damped resonant voltage will be: V f0 = V drivingpoint e αt sinω d t Where: α = R C C C C
ECE 54: Session 6; Page 9/ Spring 08 ω d = πf 0 α Find maximum voltage that can be added to source: V maxadded.5kv V m V maxadded 5.kV So we want the positive peak of the damped sine wave to be: V maxadded = V drivingpoint e αt sinω d t Or with the numbers added: 5.kV = ( 0.06kV) e αt sin ω d t Positive maximum occurs when: sinω d t = This implies that: t pk πf 0 α = π π or: t pk = Equation # πf 0 α The damped amplitude requires:
ECE 54: Session 6; Page 0/ Spring 08 5.kV 0.06kV = e α t pk take natural log of both sides Exp_αt pk ln V maxadded Exp_αt pk 0.65 s therefore V drivingpoint sec αt pk = 0.65 Equation # Combining these two equations: π α = 0.65 πf 0 α Square the equation, and rearrange: α 9π = Exp_αt pk 4 π Collecting terms α Exp_αt pk 4 π f0 = 9π Exp_αt pk 4 f0 α
ECE 54: Session 6; Page / Spring 08 Then, taking a square root α Exp_αt pk 4πf 0 9π sec α 68.7 Exp_αt pk 4 sec Recall α = R C C R α C C R.7ohm ω d πf 0 α ω d π 45.4Hz However, there is a problem with this result. This result assumes that the transient happens so fast that the 60Hz waveform is not varying. It turns out that the transient will be worse the switching transient occurs prior to the 60 Hz peak with a tradeoff between the driving point voltage and the sum of the resonant response plus the 60 Hz response.