1. 1 Availability Modeling of Cooling Water Pumps to Assess if a Replacement Option is Economically Feasible. Dr. Salman Mishari

2. 2 Presenter  Dr. Salman Mishari  Affiliation  Reliability Engineer in S. Aramco  20 years postgraduate experience  Vibration  Rotating equipment  Reliability

3. 3 Presenter  Education  B.S. in ME (USA)  M.S in Engineering. Mgt (KFUPM)  PhD in Reliability (UOB)

4. 4 Presentation Outline Some background on the problem: – Availability concern on a group of low lift pumps. How availability was modeled and estimated for both: – The current situation – The replacement alternative How availability and other factors were used in the replacement analysis. The Results of the Assessment. Lessons Learned

5. 5 Background Sea Inlet Canal 4 X 60,000 GPM pumps3 X 35,000 GPM pumps Excess Capacity Seven low left pumps running in parallel are used to pump sea water to inlet canal

6. 6 Background Sea Inlet Canal 4 X 60,000 GPM pumps3 X 35,000 GPM pumps Excess Capacity Water in inlet canal is picked up by high lift pumps to provide cooling for the whole plant

7. 7 Background Sea Inlet Canal 4 X 60,000 GPM pumps3 X 35,000 GPM pumps Excess Capacity Three have a flow rate of 35,000 GPM. Four are rated at 60,000 GPM.

8. 8 Background Sea Inlet Canal 4 X 60,000 GPM pumps3 X 35,000 GPM pumps Excess Capacity Two or three pumps are normally needed depending on the capacities of the available pumps.

9. 9 Background Sea Inlet Canal 4 X 60,000 GPM pumps3 X 35,000 GPM pumps Excess Capacity The normal mode of operation is one low and one high capacity pumps giving a combined flow rate of 95,000 GPM..

10. 10 Background Sea Inlet Canal 4 X 60,000 GPM pumps3 X 35,000 GPM pumps Excess Capacity The flow requirement is only about 80,000 GPM so the excess amount is diverted to the outlet canal, and it is considered a waste of energy.

11. 11 Background – The pumps are very old with some being over 50 years of age. – Time spent to repair has been very long due to major component failures and waiting for custom-made spare parts. – All of this raised concerns about their future capability to sustain the required level of availability.

12. 12 Data Gathering Data Gathering And Analysis – Six-year worth of failure data history was collected from the Company computerized maintenance management system (CMMS). – They all averaged about five-year MTBF and one-year MDT. – The current MTBF value of five years was considered acceptable. – The MDT of one year was, however, considered to be on the high side.

13. 13 Alternative – As per Company design practice, the replacement alternative, if found necessary, is three 50% pumps

14. 14 Analysis – One of the big-capacity pumps is not in operation so it’s not included in the analysis. – The required flow rate is 80,000 GPM so any scenario that is capable of providing the required flow is considered a success state.

15. 15 Analysis – Success states, All pumps being operable is an obvious success state. Two small pumps being operable is a failure state they can provide only 70,000 GPM (2X35,000) when the requirement is 80,000 GPM.

16. 16 Success Scenarios – There are 16 possible scenarios 12 of which are success states and 4 are failed states.

17. 17 State # No. of Small Pumps No. of Big Pumps State Label Success/ Fail 133P33Success 232P32Success 331P31Success 430P30Success 523P23Success 622P22Success 721P21Success 820P20Fail 913P13Success 1012P12Success 1111P11Success 1210P10Fail 1303P03Success 1402P02Success 1501P01Fail 1600P00Fail Success States

18. 18 State Probability The probability of being at any state depends on – Failure rate – Repair rates. Failure and repair rates are reciprocals MTBF and MDT – FR = 0.2 per year, – RR = 1 year

19. 19 Availability Solution Problem can not be easily solved for availability using simple engineering reliability methods. An approach usually reserved for complex systems is the Markov model.

20. 20 Markov Solution 3 2323 1313 0303 3030 2020 1010 0 3232 3131 2 2121 1212 0101 1 0202 0.2, 1 0.2, 2 0.2, 3 0.2, 1 0.2, 2 0.2, 3 0.2, 2 0.2, 1 0.2, 3 0.2, 2

21. 21 = = Markov Solution P33 P32 P31 P30 P23 P22 P21 P20 P13 P12 P11 P10 P03 P02 P01 P00

22. 22 Availability Solution Solution of current system – Availability is the sum of the success states and it was equal to 99.98% for the subject system. Solution of alternative – Similar availability analysis was done for the replacement alternative; i.e., the 3X50% pumps.

23. 23 Feasibility Assessment The result was a similar level of availability (99.98%). It was, therefore, concluded that replacement on the basis of availability was not economically feasible

24. 24 Economic Replacement Analysis So, Availability is not a feasible basis for replacement What about other factors – Unnecessary Extra Capacity – Maintaining 3 pumps instead of 7 All this savings need to be assessed against Capital investment. The engineering economic approach is used to evaluate the economic feasibility.

25. 25 Economic Replacement Analysis Estimated replacement capital investment – This cost was estimated to be $1000,000 Return on investment – Return on investment is realized through savings of: Unnecessary Extra Capacity Maintaining 3 pumps instead of 7

26. 26 Unnecessary extra capacity The current mode of operation is – One small and one big giving a combined flow rate of 95,000 GPM – Requirement is only about 80,000 GPM. Excess Energy: PSI X GPM 7.58 psi X 15,000 GPM HP = ------------------ = -------------------------------- = 77 hp = 58KW 1714 X eff. 1714 X 0.86 This translates to an annual cost savings of about – Annual Cost = 58 X 24 hrs X 365 days X $.026 =$13,000

27. 27 Difference in maintenance costs – The replacement alternative is three pumps while the current set up is seven. – Maintaining three pumps should be cheaper than maintaining seven. – Maintenance history indicates that the average maintenance cost per pump has been about $1000,000 for seven pumps over 18 years. Therefore,

28. 28 Difference in maintenance costs – Annual maintenance cost = ($1,000,000 ÷ 7 pumps) ÷ 18 years = $8,000/pump – Alternative is four pumps less, Maintenance savings = 4 X $8,000 = $32,000 – Total return on investment = $13,000 + 32,000 = $45,000 What do you think? Is this attractive?

29. 29 The Minimum Attractive Return – The estimated return is compared to the minimum attractive return – This is calculated using Compound Interested Tables. Project life window of 20 years Minimum attractive rate of return of 9%, – This corresponds to 0.11282.

30. 30 Interest Table

31. 31 The Minimum Attractive Return For an investment capital of $1,000,000, The minimum attractive return = $1,000,000 X 0.11282 = $112,820 Estimated is less than attractive Replacement of current system is not economically feasible.

32. 32 Lessons Learned Many lessons can be learned from this case study. However, I would like to focus on the area of assessing the maturity of an organization from a Reliability Management Perspective.

33. 33 Lessons Learned There are many reliability improvement opportunities. Maturity is needed, however, to be able to capture these opportunities. I suggest the following questions should be emphasized if you conduct a self assessment at your facility.

34. 34 Lessons Learned Does your facility keep good failure and maintenance records: – Bad data  badly estimated parameters  bad model  bad analysis  wrong decision

35. 35 Lessons Learned What triggers your reliability assessment and studies? – strictly SME judgment? Like in the example. – Certain level of loss? – Is it a Bad Actor Program? – Is it KPI’s? if yes, Is it accurate data? (for example MTBF <> MTBWO) Is it accurate and consistent calculation methodologies?

36. 36 Lessons Learned Who does your reliability and replacement analysis? – Do you have qualified reliability engineers Are they well trained in reliability modeling? Do they have knowledge in sampling and Statistics (estimating model parameters)? – Are they qualified in using Engineering Economics principles?

37. 37 Lessons Learned How good is your root cause analysis program? – What triggers an RCA study? – Who facilities the study? Is he trained in RCA? – Who participates in the study? – Where do you document the results and how easy can you retrieve them?

38. 38 Lessons Learned How good is your PM program? – How often do you review your PM program? – How do you set the PM tasks and their frequencies? – What methodology do you use? RCM; PMO?

39. 39 Lessons Learned How is your recommendation tracking system? – Who ensures that all recommendations are implemented? – Who ensures the credibility of the implemented recommendations?

40. 40 Any Questions