Presentation on theme: "Arrester Application bringing it all together!"— Presentation transcript:
1Arrester Application bringing it all together! Selection (see EMSection 14 SI 1.00 andIEEE COvervoltageEnergyMarginsCoordinating across the transformer
2Arrester Selection – Step 1 a. Select an MCOV rating that is just greater than the maximum continuous operating voltage of the systemb. Check the system maximum temporary 60hz over-voltage (magnitude and duration) with the arrester temporary over-voltage capability from the manufacturer’s curvec. Choose the higher rating from a or b above
3Arrester Selection – Step 2 a. Select the arrester class based on the available fault duty. The available fault duty should be less than the arrester pressure relief ratingb. Check the energy dissipation capability of the arrester. This is given in kilo-joules per kv of rating (either MCOV or duty cycle rating). Usually this is determined from switching studies of the transmission system and is not a concern on the distribution system.
4Arrester Selection – Step 3 Calculate the protective margins and check they are adequate:PM1> 20% Front-of-WavePM2> 15% LightningPM3> 15% Switching
5Arrester Selection – example: Let’s pick arresters for the 230kv system. Step one is to select the MCOV rating. 1a. Maximum voltage for the 230kv system is 242kv from ANSI C84.1 – 1995 so the MCOV must be at least = 242kv = kv Standard ratings available above 139.7kv are:IEEE C
6Arrester Selection – example cont: 1b. The next step is to check the maximum possible system 60hz over-voltageFirst let’s check the maximum voltage rise on the un-faulted phase fora phase to ground fault. System analysis shows a worst case maximum voltage rise of 197kv for a phase to ground fault on the system (COG of 81.4%). Assume a clearing time of 5 cycles.Next check for the possibility of back-feed during clearing of a 230kv fault.Consider a kv delta-wye transformer. A high side fault is also back-fedfrom the 34.5kv system:
7Neutral shift And Back-feed: The 230kv breakers clearin 3-5 cyclesNeutral shiftAnd Back-feed:Phase to groundfault197kv242kvThe 230kv line fault isback-fed from the 34.5kvsystem until the reversepower relaying clears(up to 1 sec).Arrester voltagekv140kvNeutral shifts duringclearing of a phase toground faultThe xfmr high side arresters must withstand the increasedphase to ground voltage on the unfaulted phases until thefault clears
8Arrester Selection – example cont: 1c. The highest stress for the back-feed condition could be a maximum of242kv for 1 second. The minimum MCOV = 242/1.46 = 165.7kv1.46Pick a 228kvduty cycle:
9Arrester Selection – example cont: 2a. Next select the arrester class:The worst case phase to ground fault duty on the 230kv system is atDoubs and is around 50ka.Need to select a station class arrester since the fault duty is > 16ka.2b. Check transmission line discharge duty. Transmission studies show the maximum switching duty on the 230kv system is within the 7KJ/kv rating for standard station class metal oxide arresters.
10Arrester Selection – example cont: 3a. Calculate the protective margins and check they are adequate:PM1> 20% Front-of-WavePM2> 20% LightningPM3> 15% SwitchingNormally margins are calculated for the transformers since theyusually have reduced insulation (less than full BIL)
11Protective Ratios for Liquid Filled Transformers CWW= CWWFOWPR160 HZBIL= BILLPLPR2BSL20KAVoltage= BSLSPLPR3FOW10KA5KASPLDischarge voltage atAppropriate systemImpulse current1.010100100010000.1Time in u-seconds
16Protective Margins PM1= (PR1-1) x 100% > 20% PR1 = CWW/FOW PR2 = BIL/LPLPR3 = BSL/SPLPM1= (PR1-1) x 100% > 20%PM2= (PR2-1) x 100% > 20%PM3= (PR3-1) x 100% > 15%Example: kv transformer high side:PR1 = CWW/FOW=(825x1.1)/(559) = 1.62PM1= (PR1-1) x 100%=(1.62-1)x100%= 62%PR2= BIL/LPL= 825/(525) =1.57PM2= (PR2-1) x 100% =(1.57-1)x100%= 57%PR3 = BSL/SPL=(825x.83)/(451)= 1.52PM3= (PR3-1) x 100%=(1.52-1)= 52%
17Why are the low side margins so high? Protective MarginsExample: kv transformer low side:PR1 = CWW/FOW=(200x1.1)/(95.8) = 2.30PM1= (PR1-1) x 100%=(2.30-1)x100%= 130%PR2= BIL/LPL= 200/(84) =2.38PM2= (PR2-1) x 100% =(2.38-1)x100%= 238%PR3 = BSL/SPL=(200x.83)/(66)= 2.52PM3= (PR3-1) x 100%=(2.52-1)= 252%Why are the low side margins so high?
19Coordination across the transformer: Should you worry aboutMixing Silicon Carbide and metal oxide arresters?Consider a temporary 60hz over-voltage on the 230kv side of 200kv ph-gnd:SiC Duty cycle high side = 228kvWill the high side arrester conduct??V low side = 200 x (34.5/230)V low side = 30.0kvV low side = 30.0 kvArrester Duty cycle = 228 kvMOV low side arresters wouldbe stressed since MCOV= 24.4 kvUse MOVs on BOTH sides of the transformer
23The OMU (optical metering unit) What is the BIL? BSL? 60 hertz withstand?How do you know? What impulse testing has been done? Is there a design test report?Should it be protected by an arrester? Is the insulation self-restoring? Will it flashover on the outside or the inside?What happens when it loses SF6? What is the withstand at reduced gas pressure? Can we add gas if its energized?Should we trip it for low gas? What is the minimum pressure that is acceptable?When its switched out could the switching surge cause a flashover? (could the porcelain explode?)
24Thanks for your attention…… Your substation support staff! Luxor substation