Exam EI2452/EI3364 Reliability analysis of power systems

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1 Exam EI2452/EI3364 Relablty analyss of power systems Exam date: 21 Aprl 2015 Tme: 13:00-15:00 Examner: Tools: Result: Instructons: Patrk Hlber Calculator 1 week after exam. State made assumptons and defntons. The solutons must be readable and easy to follow. Max: 20 ponts <16 Fal (New exam 7th of May) 16 Pass and 0 grade ponts 17 Pass and 0.5 grade ponts 18 Pass and 1 grade pont Kunglga Teknska högskolan KTH, Skolan för Elektro- och Systemteknk, Teknkrngen 33, Stockholm Tel: , Fax: , Epost: hlber@kth.se 1 (7)

2 Questons 1. (2p) Component A component does on average fal every 25th year and t takes 176h to repar t. Determne the expected avalablty. 2. (3p) Seres Two components n seres consttute a small system. Each component got an avalablty of Use an approxmatve method to determne unavalablty. What s the dfference compared to the exact calculaton (n percent)? 3. (3p) Parallel a) Determne the structural functon for the fgure above. (2p) b) The system got two mnmal paths, how much must the components be mproved n the less relable path n order to get the same performance as the better path? All components got an avalablty of (1p) 4. (5p) SAIFI, SAIDI.. Consder the system below, whch conssts of two breaker swtches and two feeders, each wth a load pont. Component data: B1: breaker swtch n average actve component falures/year L1: underground cable n average 0.02 falures/year L2: overhead lne 3 km n average 0.10 falures/km, year requre 20 hours restoraton/falure requre 12 hours restoraton/falure requre 8 hours restoraton/falure Kunglga Teknska högskolan KTH, Skolan för Elektro- och Systemteknk, Teknkrngen 33, Stockholm Tel: , Fax: , Epost: hlber@kth.se 2 (7)

3 Load pont data: Lp1 900 customers average annul energy consumton 8 GWh Lp2 250 customers average annul energy consumton 7.8 GWh a) Assumng the breaker swtch B2 to be n perfet condton, calculate SAIFI, CAIDI and AENS. (3p) b) Breaker swtch B2 was perfect n the prevous queston. Now consder that B2 has a stuck condton probablty of 2%. Ths means that out of all the stuatons where B2 need to operate, the breaker stays n a stuck condton and fal to operate for 2% of cases and hence B1 wll have to solate the fault. In such stuatons B2 s repared and restored frst and ths requre 5 hours. B2 has no other falure ssues. Calculate the total unavalablty and total ENS at Lp1 for ths case. (2p) 5. (5p) Markov 1 2 In a substaton there are two transformers of dfferent types workng n parallell. The system s operated dependng on the substaton s capacty and the capacty s assumed to be: 100% capacty f both transformers are workng, 60% capacty f only transformer 1 s workng and 40% capacty f only transformer 2 s workng (0% capacty f none are workng). Common cause events are neglected and reparaton s assumed to be performed ndependantly of each other (that s, f both are faled they are both beng repared and ether can be repared frst). The followng parameters are assumed: Transormer 1 Transformer 2 MTTF 20 years 30 years MTTR 7 days 4 days a) Defne the states for the system. (1p) b) Draw a transton state dagram for the system. Motvate your reasonng shortly. (2p) c) Formulate the transton rate matrx (Q-matrx). (1p) d) Formulate the system of equatons to solve the lmtng state probabltes (but don t solve t). (1p) (Hnt: MTTF = Mean Tme To Falure, MTTR = Mean Tme To Repar) 6. (2p) Dstrbutons For the falure rate functons A and B, fnd the matchng probablty densty functon among the fve canddates to the rght. (Please answer lke A-x and B-y, where x and y are numbers between 1 and 5). All unts are normalzed and standard. Note that the x axs for falure rate B s dfferent than the other graphs. Kunglga Teknska högskolan KTH, Skolan för Elektro- och Systemteknk, Teknkrngen 33, Stockholm Tel: , Fax: , Epost: hlber@kth.se 3 (7)

4 Formulas N UN UN SAIFI, SAIDI, CAIDI, N N N 8760 N UN UN ASAI, ASUI, 8760 N 8760 N ENS AENS, ENS ENS N Kunglga Teknska högskolan KTH, Skolan för Elektro- och Systemteknk, Teknkrngen 33, Stockholm Tel: , Fax: , Epost: hlber@kth.se 4 (7)

5 Solutons 1. A=1-U=1-(1/25)*(176/8760)= or wth exact method 25/(25+176/8760) = Correct answer: U~= =0.04 U=1-0.98*0.98= Dfference: ( )/0.0396= 1% 3. a) 1-(1-x1x2)(1-x3) = x1x2+x3-x1x2x3 b) 3,1,2 or 3,2,1 c){3} got U=0.001, {1,2} got An mprovement of s needed, the dstrbuton of mprovement s rrelevant, but equal shares reasonable/realstc. A1= and A2= Assumptons: Usng approxmatve methods. 4. a) B1: actve component falures/year; 20 hours/falure L1: 0.02 falures/year; 12 hours/falure L2: 0.30 falures/year (3 km * 0.1 falures/(km and year); 8 hours/falure Lp1 Lp2 λ[f/yr] r[hrs/f] U[hrs/yr] λ[f/yr] r[hrs/f] U[hrs/yr] B L L Sum , , b) Kunglga Teknska högskolan KTH, Skolan för Elektro- och Systemteknk, Teknkrngen 33, Stockholm Tel: , Fax: , Epost: hlber@kth.se 5 (7)

6 Lp1 Lp2 λ[f/yr] r[hrs/f] U[hrs/yr] λ[f/yr] r[hrs/f] U[hrs/yr] B L L (2% of λ L2 ) Sum a) Snce the transformers are not dentcal we need to have four states: State 1: Both transformers are workng. State 2: Transformer 1 s not workng and transformer 2 s workng. State 3: Transformer 1 s workng and transformer 2 s not workng. State 4: Both transformers are not workng. b) Let and be the falure rates for transformer 1 and 2 respectvely. Let and be the repar rate for transformer 1 and 2 respectvely. Snce common cause events are neglected and all events are ndependent, only transtons where one transformer change workng state are possble. We get the followng transton state dagram: Wth the followng parameter values: 1 1 0,05 / ,033 / / 7 Kunglga Teknska högskolan KTH, Skolan för Elektro- och Systemteknk, Teknkrngen 33, Stockholm Tel: , Fax: , Epost: hlber@kth.se 6 (7)

7 / c) d) Wth the lmtng state probablty vector, the equatons are gven by: 1 Or lkewse: A-1 and B-5. Motvaton: Frst of all densty functon 3 s not a realstc densty functon snce t does not converge to zero and clearly the area under the graph s above 1. It can be ruled out. Falure rate A s constant, whch correspond to an exponental dstrbuton. Densty functon 1 and 4 are canddates. At tme zero the densty functon has the same value as the falure rate, thus A corresponds to 1. Also, you could note that for densty functon 4, t seems not totally unlkely that the component fals after tme 10, whch s qute unlkely for a component wth falure rate 0,5 (and MTTF 2). Falure rate B s ncreasng, and falures are more lkely at later tmes. Densty functons 2 and 5 are canddates.densty functon 2 has most of ts probablty around 8 and t seems unlkely wth a falure before tme 3. The falure rate B has value 1 already around tme 1,5 whch means that at that moment the MTTF s 1. Thus the probablty of falures before 3 must be qute lkely and densty functon 2 can be ruled out. Densty functon 5 fts the falure rate snce most of the probablty around s around tme 2. Kunglga Teknska högskolan KTH, Skolan för Elektro- och Systemteknk, Teknkrngen 33, Stockholm Tel: , Fax: , Epost: hlber@kth.se 7 (7)