ECE 524: Lecture 15 Reducing Capacitor Switching Transients. jx s C 2 C 1. Define units: MW 1000kW MVA MW MVAr MVA. rad s

Size: px

Start display at page:

Download "ECE 524: Lecture 15 Reducing Capacitor Switching Transients. jx s C 2 C 1. Define units: MW 1000kW MVA MW MVAr MVA. rad s"

Lynne Neal
5 years ago
Views:

1 ECE 54: Session 5; Page / Spring 04 ECE 54: Lecture 5 Reducing Capacitor Switching Transients Define units: MW 000kW MVA MW MVAr MVA Example : f 60Hz ω πf ω rad s t sec 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

2 ECE 54: Session 5; Page / Spring 04 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 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

3 ECE 54: Session 5; Page / Spring 04 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

4 ECE 54: Session 5; Page 4/ Spring 04 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 peak current is: i p_ t topeak L Lim i h_ t topeak L Lim A

5 ECE 54: Session 5; Page 5/ Spring 04 So the inductor is oversized... 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. The maximum will occur when the derivative of the total current is zero. d t i p_() t d ωc ω V LL L lim ω L s sin ωt ϕ ηl lim d t i h_() t d ω n L lim V m V c_0 Z 0 L lim cosω n L lim t Define variables for these derivatives d p_ tl lim ωc ω V LL L lim ω L s sin ωt ϕ ηl lim ω n L lim d h_ tl lim L lim 8.66mH V m V c_0 Z 0 L lim cosω n L lim t

6 ECE 54: Session 5; Page 6/ Spring 04 Given i h_ 4f n L lim L Lim L lim i p_ 4f n L lim L lim Find L lim L Lim 8.7mH 000A This is still an approximation, since we are assuming the peak happens at /4 cycle. Check the results: 75.58Hz f n L Lim One quarter cycle at that frequency takes: t topeak 4f n L t topeak.4 ms Lim t topeak 60Hz60deg 0.75deg i p_ t topeak L Lim i h_ t topeak L Lim 000A But: d h_ t topeak L Lim d p_ t topeak L Lim A s So not at a local minima

7 ECE 54: Session 5; Page 7/ Spring 04 0 i p_ tl Lim i h_ tl Lim i () t i h_ tl Lim i p_ tl Lim t i () t 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.

8 ECE 54: Session 5; Page 8/ Spring 04 X lim ωl Lim X lim 6.9 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 With the inductor: I P_pk V LL I P_pk 48. 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 60.06kV Higher than result without added inductor, but only happens when both peaks coincide, which may not happen for some time.

9 ECE 54: Session 5; Page 9/ Spring v cap () t 0 First peak: 44.90kV t 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 Ω

10 ECE 54: Session 5; Page 0/ Spring 04 Assume it will still be underdamped, since X/R of the reactor is large. αl lim ω d L lim f 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 ω d L lim f π d ( 0) Hz 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

11 ECE 54: Session 5; Page / Spring 04 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 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 ) 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 7.76mH

12 ECE 54: Session 5; Page / Spring 04 Check the results: 78.7Hz f d L Lim One quarter cycle at that frequency takes: t topeak 4f d L t topeak.4 ms Lim t topeak 60Hz60deg 0.9deg i L t topeak L Lim 000A Same approximation as above. 0 i L_p tl Lim i L_h tl Lim t

13 ECE 54: Session 5; Page / Spring i L tl Lim 0 Capacitor voltage t 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

14 ECE 54: Session 5; Page 4/ Spring 04 v c_ω () t ωc R s ωl Lim V LL ωc ωl s L Lim sin ( ωt ϕ) η L Lim v c () t v c_ω () t v c_ωd () t v c_ωd () t v c_ω () t t

15 ECE 54: Session 5; Page 5/ Spring v c () t First peak is now kV t

16 ECE 54: Session 5; Page 6/ Spring 04 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. 40 * *0-60 (f ile ExamplR.pl4; x-v ar t) t: VMAX v:c - v:c -

17 ECE 54: Session 5; Page 7/ Spring 04 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 π L s C C s f 0 f 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 C C infinity 8.kV V drivingpoint V m V infinity V drivingpoint 0.06kV

18 ECE 54: Session 5; Page 8/ Spring 04 Damped resonant voltage will be: V f0 V drivingpoint e αt sinω d t Where: α R C C C C 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

19 ECE 54: Session 5; Page 9/ Spring 04 This implies that: t pk πf 0 α π π or: t pk Equation # πf 0 α The damped amplitude requires: 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

20 ECE 54: Session 5; Page 0/ Spring 04 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 Then, taking a square root α Exp_αt pk 4πf 0 9π sec α 68.7 sec Exp_αt pk 4 Recall α R C C R α C C R.7ohm ω d πf 0 α ω d π 45.4Hz

21 ECE 54: Session 5; Page / Spring 04 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.

ECE 524: Reducing Capacitor Switching Transients

ECE 524: Reducing Capacitor Switching Transients 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