Improving Machinery Reliability 3 Episode 8 pdf

Life Cycle Cost Studies 289 resides in local computer files, but it has not been analyzed or put to effective use. Analysis can start with arithmetic analysis and grow to more complicated statistical analysis. Follow the guidelines for each step listed to work out a typical engineering problem. (Remember, a single right or wrong method/solution does not exist. Many methods and routes can be used to find LCC). If you disagree with the cost or life data, substitute your own values determined by local operating conditions, local costs, and local grades of equipment. Step I: Define the Problem. A pump is operating without an on-line spare. At pump failure, the process shuts down and financial losses are incurred as each hour of down-time results in a gross margin loss of US$4,000/hour of outage. Find an effective LCC alternative as the plant has an estimated 10 years of remaining life and is expected to be sold out during this interval. Step 2: Alternatives and acquisitions/sustaining costs. Consider three obvious alternatives for LCC (other alternatives exist for solving this problem, however, the list is pared for brevity): 1. Do nothing. Continue solo ANSI pump operations with a 100 horsepower, 1750 RPM, 250 psi, 500 gpm, 70% hydraulic efficiency, while pumping fluid with a specific gravity of 1. 2. Add a new, second ANSI pump in parallel (literally in redundant standby) that can be started immediately without the loss of production upon failure of the running pump. Alternate running of the parallel unit every other week to avoid typical failures incurred by nonoperating equipment. The capital costs for the second pump are $8,000 plus $3,000 for check/isolation valves, plus $2,500 for installation. 3. Remove the existing solo ANSI pump and replace it with a new solo API pump with the same performance as for the ANSI model. The API pump cost $18,000 plus $3,500 for installation and the installation will incur a four-hour loss of production for connecting the new pump. Step 3 Prepare cost breakdown structure/tree. For the do-nothing case, the cost breakdown structure will incur cost in the categories given in Figure 5-13; the cost breakdown structure depicted in Figure 5-14 refers to the redundant ANSI case. For the API pump case, the cost breakdown structure will incur cost in the categories given in Figure 5-15. The individual details for each case will become obvious in step 5. Step 4: Choose analytical cost model. The model used for this case is explained in an engineering spreadsheet. The spreadsheet merges cost details and failure details to prepare the NPV calculations. Failure costs are prorated into each year because the specific time for failure, because of chance events, is not known. (text continued on page 292) 290 Improving Machinery Reliability 1 1-1 r I Figure 5-13. Cost components for solo ANSI pump. Figure 5-14. Cost components for paralleVredundant ANSI pumps. Life Cycle Cost Studies 291 292 Improving Machinery Reliability (text continued from page 289) The same spreadsheet will be used with more details when statistical uncertainty is added in a section that follows. Step 5: Gather cost estimates and cost models. This is the complicated section where all the details are assembled. Of course, the more thorough the collection process, the better the LCC model. For this text, the details have been shortened with just enough information described to show the trends. Alternative #I : Do-nothing case-the datum. Use the following details from plant experience. Assume all the equipment follows the exponential distribution for reliability with constant failure rates. Note the reciprocal of failure rate is the mean time to failure. Since failure rates are constant, use one year time buckets to collect the cost of fail- ures per year as the literal failure date is unknown. Use the following assumptions based on an accounting principle that costs will follow activity-in this case it will follow failure activity. Capital cost are zero as the solo ANSI pump is currently a sunk cost and will not change. Lost gross margin occurs at US$4,000/hour when the process is down for repairs. Annual power cost for running the pump is US$l65/year per horsepower. The plant incurs 1.6 power outages each year for an average downtime of 0.5 hours, and this cost is charged into plant overhead rather than to individual pieces of equipment. Annual power costs are (US$165/hp-yr) * (100 hp) = US$16,500. Pump seals have a mean time to failure of three years. When seal failure occurs, eight hours of downtime is also lost production time. Maintenance crew costs for labor, incidental materials, and expense are US$lOO/hr. Seal replacement costs are US$1,500/seal plus US$300/incident for bearing replacements that occur as good maintenance practice while the pump is disassembled. Seal and bearing transporta- tion costs are usually expedited and cost US$l50 per incident. Annual seal costs are (1 yr/3 yeardfailure) * (US$( 1500 + 300 + 150) + (US$IOO/hr) * 8 hours + (US$4,000/hour * 8 hours)) = US$ll,583 Pump shafts have a mean time to failure of 18 years. When shaft failure occurs, ten hours of downtime is also lost production time. Maintenance crew costs for labor, incidental materials, and expense are US$lOO/hour. Shaft replacement costs are US$2,500/shaft plus US$1 ,XOO/incident for seal and bearing replacements that occur as good maintenance practice while the pump is disassembled. Shaft, seal, and bearing transportation costs are usually expedited and cost US$450 per incident. Annual shaft costs are (1 yeadl 8 yeadfailure) * (US$(2,500 + 1,800 + 450) + (US$lOO/hour) * 10 hours + (US$4,000/hour * 10 hours)] = US$2,542 Life Cycle Cost Studies 293 Pump impellers have a mean time to failure of 12 years. When impeller failure occurs, 8 hours of downtime is also lost production time. Maintenance crew costs are US$lOO/hr. Impeller replacement costs are US$3,000/impeller plus US$1,800/inci- dent for seal and bearing replacements that occur as good maintenance practice while the pump is disassembled. Impeller, seal, and bearing transportation costs are expedited and cost US$750 per incident. Annual impeller costs are (1 yeadl2 yeardfailure) * {US$(3,000 + 1,800 + 750) + Pump housings (scroll end) have a mean time to failure of 18 years. When hous- ing failures occur, 14 hours of downtime is also lost production time. Maintenance crew costs are US$lOO/hour. Housing replacement costs are US$3,000/housing plus US$l,800/incident for seal and bearing that occur as good maintenance practice while the pump is disassembled. Housing, seal, and bearing transportation costs are expedited and cost US$1,150 per incident. (US$lOO/hour) * 8 hours + (US$4,000/hour * 8 hours)) = US$3,171 Annual housing costs are (1 yeadl8 yeardfailure) * (US$(3,000 + 1,800 + 1,150) Pump bearing sets (a set = two bearings) have a mean time to failure of four years. When bearing failure occurs, eight hours of downtime is also lost production time. Maintenance crew costs are US$lOO/hour. Bearing replacement costs are US$300/bearing plus US$l,500/incident for seal replacement that occurs as good maintenance practice while the pump is disassembled. Bearing and seal transporta- tion costs are usually expedited and cost US$300 per incident. + (US$lOO/hour) * 14 hours + (US$4,000/hour * 14 hours)} = US$3,519 Annual bearing costs are (1 year/4 yeardfailure) * { US$(300 + 1,500 + 300) + Motors have a mean time to failure of 12 years considering all causes. (Motors have many parts and can fail for many reasons. A thorough analysis would be more accurate than this overview approach taken by lumping all details into one MTBF number.) When motor failure occurs, eight hours of downtime is also lost production time as the motor is swapped for a similar unit in stores. Maintenance crew costs are US$lOO/hour. Motor replacement costs are US$3,000/motor. Motor transportation costs for expedited delivery use US$500. (US$lOO/hour) * 8 hours + (US$4,000/hour * 8 hours)} = US$8,688 Annual motor costs are (1 yeadl2 yeardfailure) * [US$(3,000 + 500) + (US$lOO/hour) * 8 hours + (US$4,000/hour * 8 hours)} = US$3,025 Couplings have a mean time to failure of eight years considering all causes. When coupling failure occurs, eight hours of downtime is also lost production time. Main- tenance crew costs for labor, incidental materials, and expense are US$lOO/hour. Coupling replacement costs are US$400. Coupling transportation costs for expedited delivery are US$300. Annual coupling costs are (1 yead8 yeadfailure) * [ US$(400 + 300) + (US$lOO/hr) * 8 hours + (US$4,000/hour * 8 hours)] = US$4,188 294 Improving Machinery Reliability Maintenance personnel visit the pump monthly for routine PM inspection, lube oil additionkhange out, and emissions tests. Maintenance cost is US$SO/hour for labor, incidental materials, and expense with 1 hour on the average charged per visit. No failure times are incurred during this activity. Annual maintenance PM costs are (12 visits * 1 hour/visit) * US$SO/hour = Operations visits the pump once per week for routine PM inspection and vibration logging. Operations cost is US$35/hour for labor and expense, with 0.2 hours charged for each visit. US$600 Annual operations PM costs are (52 visits * 0.2 houdvisit) * US$35/hour = The Reliability Group receives vibration data from operations by e-mail and scans the data weekly for abnormalities. Surveillance cost is US$SO/hour for labor and expense, and on the average, 0.2 hours is charged for each weekly visit. US$364 Annual vibrations PM costs are (52 visits * 0.2 houdvisit) * US$SO/hour = US$520 Maintenance and operations conduct a joint tailgate training session on good maintenance and operation practices for this pump once per year. Three people from maintenance attend at US$SO/hour-person and three people from operations attend at US$35/hour-person. The training session consumes and elapsed time of 0.5 hours. Annual training costs are (0.5 hour * (3 people * US$50 + 3 people * US$35)) = Disposal costs will occur as a lump sum at the end of the ten-year remaining life are expected to be US$500 for permits and legal costs associated with disposition, US$500 for wrecking/disposal costs, US$l,OOO for remediation costs, US$O for write-off/recovery costs, and US$ 1,000 estimated green/clean costs associated with disposal of the asset. These costs will occur in the final year. Table 5-1 1 shows non- annualized acquisition and sustaining costs for the existing solo ANSI pump, while Table 5-12 shows the annualized recurring costs. A quick cost review of the single ANSI pump shows lost gross margin from out- ages is the biggest annual cost problem as shown in Table 5-12 for a sustaining cost of US$54,827/year. The ANSI pump will consume 16.7 corrective and 35.8 preven- tive man-hours each year. Use of MTBFs and expected failures are based on the exponential distribution that is an acceptable first cut for costs, but this technique is not an accurate predic- tor of failures for wear-out phenomena expected for many of these components. An improved accuracy method will be described later using Weibull distributions for failures. US$128 Life Cycle Cost Studies 295 Table 5-1 1 Non-annualized Acquisition and Sustaining Costs for Solo ANSI Pump ANSI Pump: CoslE,emenl Year Year Year Year Year Year Year Year Year Year Year 0 1 2 34 5 6 7 8 8 10 Aquisdion Cos& Picgram Managemenl 0 Engineering Design 0 Engineernng Data 0 Spare pans B Lagisbu 0 Faulilies 8 ConsVucLon 0 lnieal Training 0 Technical Data 0 Capital Equlpmenl 0 Suseining Costs DOwmenlaliOn Coas 0 Disposal Costs 3000 TOM 0 0 0 0 00 0 0 0 0 3000 Alternative #2: Add redundant ANSI pump. Use the following details from plant experience. This case results in pumps installed in parallel but operated as a standby redundant system as the redundant components are not energized but are literally standing by waiting to be used when failure of the operating system is detected. Of course, the detectiodswitching device is very important for calculating overall system reliabili- ty, and for this case the reliability is assumed to be 100%. Also for simplicity, the reliability of the system is calculated as if the redundant pumps are operating in par- allel. Furthermore, experience in most chemical plants and refineries shows impend- ing failure is usually detected and redundant systems are usually started in a timely manner to avoid lost production from the failing device; therefore, assume no loss of production by use of redundant pumps. Capital costs for the redundant ANSI pumps are $8,000 plus $3,000 for checkhso- lation valves, plus $2,500 for construction and installation along with US$l,OOO for program management, US$1,500 for engineering design, and US$l,OOO for docu- mentation. Likewise, the plant maintenance organization will incur US$l,OOO for engineering documentation costs to put the equipment into the paperwork system. Lost gross margin occurs at US$4,000/hour when the process is down for repairs, Annual power cost for running the pump is US$165/year per horsepower. Remem- ber. either the old pump runs or the new pump (not both at the same time). The plant incurs 1.6 power outages each year for an average downtime of 0.5 hours, and this cost is charged into plant overhead rather than to individual pieces of equipment. Annual power costs are (US$165/hp-year) * (100 hp) = US$16,500 Assume no lost production time by use of the redundant pumps. Keep all other costs as described for the single ANSI pump and depreciate the assets over the ten year project life. Table 5-12 Annual Sustaining Cost for Single ANSI Pump cost El- Seal Shafl tmpeller Houung PumpBeanngs Motors Coupling Mainbnanw PM nil& Openbons PM nMs Vibrabon bpt Training costs TOhl Failurns Elapwd Cost For MTBF. per year Repair or bb.,E.xp, Pan Cos yam oraclwb' AcbW US,,,r &Man USS perv horn uss - - - 3 03333 8 100 s 267 5 800 18 005s 10 100 S 56 S 239 12 0.0833 8 1W S 67 S 400 18 005s 14 1w s 78 s 267 4 025W 8 100 5 2w s 450 12 00833 8 100 5 67 S 29 8 0 1250 8 1w s 100 s 50 12 50s 6w 10 4 35s 364 10 4 so s 520 05 255 0 128 S 2.446 S 2,266 - Logirticr cost US! per illcidant $50 S 25 $38 $64 s38 S 42 538 - - s 293 - Lo11 Gms! Margin uss - S 10.667 s 2.222 5 2.667 S 3.111 s 8.m S 2.667 s 4.m S 33,333 I $ 16.600 System farlure me= 0 9861 oood Nlnl.MW System MTBF= 1 01408 Seal replacement = seals+baanngt Bsanng mplscemmFssals+beanngs SM mplsgment=shan+seala+bsanngs ImpeHer mplscama~im~ler*aaals+~a~gs 1 yr Availakllty. A= 0.999 HwMng rsplsasmant=hwamg+seals+beanngs 1 yr reliablify. R= 37% 1 yr unmhably. UR= 63% Total Cos1 S 3.519 5 3.025 128 S 64.127 Item Cost Logistics 1 US$ IC.stusI seal coas Bearings cost= shan cost= Impeller cos* Pump housing= Motor cost= Coupling cost= os1 gross margin USmr = ower cost(USSl65~pyr)= Motor rize(hp)= s 1.5w s 75 S300S 75 S 2.5W S 3M) s 3.000 s 300 s 3.000 s 1,ooo 5 3.000 t 500 S4M)S300 5 4.000 S 165 100 Life Cycle Cost Studies 297 Disposal costs occur as a lump sum at the end of the ten-year remaining life are expected to be US$500 for permits and legal costs associated with disposition, US$500 for wreckingldisposal costs, US$l,OOO for remediation costs, US$O for write-offlrecovery costs, and US$l,OOO estimated greedclean costs associated with disposal of the asset. These costs will occur in the final year. Table 5-13 shows non- annualized acquisition and sustaining costs for the parallel ANSI pumps; Table 5-14 shows the annualized recurring costs for parallel/redundant ANSI pumps. A quick cost review of the redundant ANSI pump shows electrical power costs are the biggest annual cost problem shown in Table 5-14 for a sustaining cost of US$21,493lyear. Alternative #3 Replace Solo ANSI Pump with Solo API Pump. Use the follow- ing details from plant experience. Capital costs are $18,000 for a solo API pump plus $3,500 for construction and installation along with US$l,OOO for program management, US$1,5OO for engineer- ing design, US$l,OOO for documentation, and US$500 for technical data. Likewise, the plant maintenance organization will incur US$l,OOO for engineering documenta- tion costs to put the equipment into the paperwork system and US$1,500 for training costs associated with the new class of equipment. Spare parts for the new equipment will be increased by $2,900 for a new set of seals and bearings. Lost gross margin occurs at US$4,000/hour when the process is down for repairs. Costs will be charged for each specific case using the accounting principle that cost follows activity. Annual power cost for running the pump is US$165/year per horsepower. The plant incurs 1.6 power outages each year for an average downtime of 0.5 hours, and this cost is charged into plant overhead rather than to individual pieces of equipment. Annual power costs are (US$165/hp-year) * (100 hp) = US$16,500 See Table 5-13 for other failure details, and plan to depreciate the assets over the 10-year project life. Table 5-13 Non-annualized Acquisition and Sustaining Costs for Parallel ANSI Pumps PanlleURedundant ANSI Pumps: Year Year Year Year Year Year Year Year Year Year Year 0 1 2 3 4 5 6 7 B 9 10 Cost Element Acqulsnion Cosls Program Managemenl io00 Engineering Dala 0 Spare parts B Logsl~cs 0 Faulilles 8 Construction 0 Inibal Training 0 Technical Dala 0 Engineering Design 1500 Capital Equipmenl 13500 Sustaining Costs Documenlalion Costs 1000 0 [Pisposal costs 3000 0 3000 0 0 0 0 0 0 0 0 ~otai= 17000 CIV'L.3 s IOoo'SL s I- 1 e21 s I I OZS s WES 009s Bns 639s BOtt boss 6LE s ea s [...]... 0 -32 011 Met Present Value I (zll3.175)~using 12% a 1 3 1 5 782 7 0 5 782 7 0 0 -5 782 7 21974 -35 8 53 0’ -35 8 53 140 -25519 -5 782 7 21974 -35 8 53 0 -35 8 53 125 - 285 82 0 4 [ 5 782 7 0 5 1 6 1 7 1 8 1 5 782 7 5 782 7 5 782 7 5 782 7 5 782 7 0 0 0 0 0 0 0 0 -35 8 53 157 -22 785 -35 8 53 176 -2 034 4 0 -5 782 7 21974 -35 8 53 0 -35 8 53 197 - 181 54 214 93 0 135 0 -2 28 43 86 80 -141 63 135 0 -1 28 13 1.57 -81 43 214 93 214 93 0 0 0 0 135 0 -2 28 43 86 80... -141 63 135 0 -1 28 13 1.76 -7270 135 0 4 28 43 86 80 -141 63 135 0 -1 28 13 1.97 -6491 135 0 -2 28 43 86 80 -141 63 135 0 -1 28 13 2.21 -5796 135 0 -2 28 43 86 80 -141 63 135 0 -1 28 13 2. 48 -5175 44444 44444 44444 0 180 0 -46244 175 73 - 286 71 180 0 -2 687 1 1.97 - 136 14 44444 0 180 0 -46244 175 73 - 286 71 18W *2 687 1 2.21 -12155 44444 0 180 0 -46244 175 73 - 286 71 180 0 -2 687 1 2. 48 -1 08 53 0 W827 21974 -35 8 53 0 I 9 -5 782 7 21974 -35 8 53 0 -5 782 7... -35 8 53 0 -35 8 53 0 -5 782 7 21974 -35 8 53 0 -5 782 7 21974 -35 8 53 0 221 -162 18 -35 8 53 2 48 -14 480 -35 8 53 277 -12929 214 93 214 93 214 93 0 135 0 -2 28 43 10 6 082 7 0 0 -6 082 7 231 14 -37 7 13 0 -37 7 13 311 -12142 discount rale Alternallve #&Add ParallellRedundanlANSI Pump Capital 135 00 cost 35 00 214 93 214 93 214 93 Savings 0 0 0 Depreclalron 135 0 135 0 135 0 Profi! W4 (axes -2 28 43 -2 28 43 -2 28 43 Tax Provision 86 80 85 80 86 80... Provision 175 73 175 73 175 73 Ne1 Income - 286 71 - 286 71 - 286 71 Add Back Depreuation 180 0 180 0 I800 Cash Flow -30 900 -2 687 1 -2 687 1 -2 687 1 Discount Factors 1.00 1.12 1.25 1.40 Present Value -30 900 - 239 92 -21422 -19126 Net PresenlValuel S (1 83 , 3 28) ~using 12% discount rale a - 0 0 180 0 -46244 175 73 - 286 71 180 0 -2 687 1 1.57 -17077 180 0 -46244 175 73 - 286 71 180 0 -2 687 1 1.76 -15247 86 80 -141 63 135 0 -1 28 13 2.77 -4620... 02222 00455 00625 00455 01667 00 83 3 00500 1 yr reliability, R= 51% 1 yr unrslisbili UR= 49% 1 yrAvaila3ility.A= 0,999 8 10 8 14 8 8 a 100 5 100 t 100 5 100 s 100 t 100 S 100 5 0.6756 1. 480 1 1 78 45 50 64 133 s 67 5 S s 5 4 83 s 250 s 6 0 s t s ussm 7.111 E 1 .81 8 S 2.m 2.545 5 ,33 3 2.667 S s s 7,967 2.175 2.4 78 2.9 98 5.975 3. 025 1.715 6 0 0 50s 600 t S 5 35 5 50 s 255 S 36 4 520 s 3 6 4 t 520 t Total System failure... 175 73 - 286 71 180 0 -2 687 1 1.57 -17077 180 0 -46244 175 73 - 286 71 180 0 -2 687 1 1.76 -15247 86 80 -141 63 135 0 -1 28 13 2.77 -4620 44444 0 180 0 -46244 175 73 - 286 71 180 0 -2 687 1 2.77 -9690 244 93 0 135 0 -2 58 43 982 0 -160 23 135 0 -146 73 3.11 -4724 47444 0 IBOO -49244 187 13 -30 531 180 0 - 28 731 3. 11 -9251 tives are shown in Figure 5-16 for a quick grasp of how the break-even points compare to the base case Cumulative present... 86 80 85 80 86 80 Net lmome -141 63 -141 63 -141 63 Add Back Deprecrauon 135 0 135 0 135 0 Cash Flow -17000 -1 28 13 -1 28 13 -1 28 13 Discount Faclors 100 112 125 140 PresentValue -17000 -11440 -10214 -9120 Net Present Value I I (89 .9 93) lusinpa 12% discounlrate Allernative #3- ReplaceANSI Pump With Solo API Pump Capital I8000 Cor1 12900 44444 44444 44444 Savlngs 0 0 0 Depredation iaoo 180 0 180 0 Prolit b/4 laxer -46244... 40 1 28 2, 188 t 2 486 S Good ~lUlnbMW c k : m t S t 1.600 It 216 I 23, 076 I S 16,600 It 1 28 44.444 Motor size(hp)= 30 1 Life Cycle Cost Studies Table 5-17 Summary of Cost Profiles for Each Alternative Year 0 1 1 1 2 Alternative #l-Exlsting Solo ANSI Pump capluai 0 cost 5 782 7 Sanngs 0 Depreuatr0n 0 0 Profit W4 laxes -5 782 7 Tax Provision 21974 Ne( Income -35 8 53 Add Back Depteuahon 0 Cash Flow 0 -35 8 53 Discount... Electric Company, St Louis, Missouri Based on a presentation at the 5th International Process Plant Reliability Conference, Houston, Texas, October 1996 31 3 31 4 Improving Machinery Reliability TYPICAL TORQUE VS SPEED CURVES ‘050% NEMA DESIGNS A, I), C, D, E 30 0% 250% 3 200% u E 2 k! 150% r; K 8 100% 8 50% Figure 6-1 Typical torque vs speed curves for NEMA designs A, B, C, D,E Squirrel Cage Motors Are... Publishers, 169 Vista Drive, Marlton, NJ 080 53, 1996 MIL-HDBK-259, Military Handbook, Life Cycle Cost in Navy Acquisitions, 1 April 19 83 , available from Global Engineering Documents, phone 1 -80 0 -85 4-7179 MIL-HDBK-276- 1, Military Handbook, Life Cycle Cost Model for Defense Material Systems, Data Collection Workbook, 3 February 1 984 , Global Engineering Documents, phone 1 -80 0 -85 4-7179 MIL-HDBK-276-2, Military . -35 8 53 -35 8 53 -35 8 53 -35 8 53 -35 8 53 -37 7 13 Cash Flow 0 -35 8 53 -35 8 53 -35 8 53 -35 8 53 -35 8 53 -35 8 53 -35 8 53 -35 8 53 -35 8 53 -37 7 13 Discount Factors 1W 112 125 140 157 176 197 221 2 48 277 31 1. rate 214 93 0 135 0 -2 28 43 86 80 -141 63 135 0 -1 28 13 1.57 -81 43 214 93 0 135 0 -2 28 43 86 80 -141 63 135 0 -1 28 13 1.76 -7270 214 93 0 135 0 4 28 43 86 80 -141 63 135 0 -1 28 13 1.97 -6491. 1.97 -6491 214 93 0 135 0 -2 28 43 86 80 -141 63 135 0 -1 28 13 2.21 -5796 214 93 214 93 0 0 135 0 135 0 -2 28 43 -2 28 43 86 80 86 80 -141 63 -141 63 135 0 135 0 -1 28 13 -1 28 13 2. 48 2.77 -5175