Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

Ninety prototypes of the new compressor have been developed All units tested either ran to failure or were terminated after an equivalent 1200 days of

Ninety prototypes of the new compressor have been developed All units tested either \"ran to failure\" or were terminated after an equivalent 1200 days of operation had Each failed compressor was subsequent repaired and its repair time recorded Time to failure and repair times in days Failure days to failure Repair time Failure days to failure Repair time 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 77 129 135 136 171 199 241 269 272 289 309 310 314 367 398 409 424 429 450 461 463 466 474 490 505 511 522 550 554 557 557 560 574 580 589 635 639 2.4 0.9 1.9 1.1 1.9 1.8 1.4 1.2 1.1 1.6 1.6 1 1 1.5 1.6 0.8 1.4 1.3 1.2 1.2 1.3 1.3 1.3 0.8 1.5 2.1 2.2 1.9 1.2 1.4 1.6 1.1 1.3 1.1 1.6 1.4 2.1 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 653 661 667 695 701 709 745 747 748 750 759 761 791 791 805 838 839 842 869 879 884 937 949 958 969 983 1019 1026 1054 1055 1099 1105 1124 1132 1133 1144 1.7 1.2 1.4 1.3 1.3 1.2 1.4 2.1 1.3 1.4 1.4 1.3 1.6 1.2 1.6 1.5 1.4 0.9 1.3 1.4 1.2 1.8 1.7 2 1.7 1.4 1.1 1.3 1 1.5 0.9 1.6 1.2 2.1 1.2 1.7 0 days of operation had elapsed Section 10.1.1 and Section 11.1 Preventive Maintenance Repair Distribution Chapter 10 MPMT = Exponential 10 Tpm = 200 Weibull PM Crew size = 1 Normal design or economic life (td) Lognormal m b q m 5 2 5 5 5 MTTR 5 4.431 5 25 % repairs 1.438 2.682 1.628 50% (median) repairs 3.466 4.163 5.000 75% repairs 6.931 5.887 8.372 90% repairs 11.513 7.587 11.408 Renewal Process Mean downtime (M-bar) 7.500 7.216 7.500 MH/OH 0.1 0.094 0.1 Minimal Repair Process Mean downtime (M-bar) 7.814 7.565 7.814 MH/OH 0.089 0.084 0.089 Chapter 11 Steady-State Availability (Renewal process) Inherent Availability 0.9756 0.9783 0.9756 MTBM 100.00 100.00 100.00 Achieved Availability 0.9302 0.9327 0.9302 Chapter 11.1.5 Availability with minimal repair over the design life (power law process) w/o PM 0.9809 0.9831 0.9809 with PM 0.9351 0.9370 0.9351 t At2 t1 5000 2 t2 t1 where m(t1 , t2 ) (t ) dt t2 t1 m(t1 , t2 ) MTTR t1 m s 5 0.8 6.886 2.915 5.000 8.577 13.939 8.443 0.119 8.638 0.103 0.9667 100.00 0.9221 0.9740 0.9287 failure distribution Crew size 2 Renewal Power law process - MTBF process -a b 200 25 0.00001 = m(t) = 1.7 19.420 MTBF Ainh MTBF + MTTR MTBM Aa td m td td / Tpm MTBM MTBM + M Section 11.2 Exponential Model intermediate MTBF = MTTR = t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 200 Steady-state Avail start time 0 10 0.952381 stop time 10 point availability interval availability = 0.98186 A(t) 0.995254 0.99098 0.987133 0.983669 0.98055 0.977742 0.975215 0.972939 0.97089 0.969045 0.967384 0.965888 0.964542 0.96333 0.962238 0.961256 0.960371 0.959575 0.958858 0.958212 0.957631 0.957108 0.956637 0.956212 0.95583 0.955487 0.955177 0.954898 0.954647 0.954422 0.954218 0.954035 0.95387 0.953722 0.953588 0.953468 0.953359 0.953262 0.953174 0.953095 0.953024 0.95296 0.952902 0.045351 1 0.349938 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 0.95285 0.952803 0.952761 0.952723 0.952689 0.952658 0.952631 0.952606 0.952584 0.952563 0.952545 0.952529 0.952514 0.952501 0.952489 0.952478 0.952468 0.95246 0.952452 0.952445 0.952438 0.952433 0.952428 0.952423 0.952419 0.952415 0.952412 0.952409 0.952406 0.952403 0.952401 0.952399 0.952397 0.952396 0.952394 0.952393 0.952392 0.952391 0.95239 0.952389 0.952388 0.952387 0.952387 0.952386 0.952386 0.952385 0.952385 0.952384 0.952384 0.952384 94 95 96 97 98 99 100 0.952383 0.952383 0.952383 0.952383 0.952383 0.952382 0.952382 intermediate l 0.005 r = 0.1 A(t) 1 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0 20 40 60 time 80 100 120 Section 10.2.4 Replacement Model and Section 10.2.5 Preventive Maintenance Model Power law process unit of time: oper hr 0.000001 a = b= 2.8 replacement cost Cu = $21,000 cost of a failure Cf = $500 cost of a PM = $200 time to replace = 428.0 oper hr min cost per oper hr $76.33 time between PM = 81.2 oper hr min cost per oper hr $3.83 increment 14 replacement 3 time cost time enter starting value 100 $212 10 2 114 $187 13 3 128 $167 16 4 142 $152 19 5 156 $139 22 6 170 $129 25 7 184 $120 28 8 198 $113 31 9 212 $107 34 10 226 $102 37 11 240 $97 40 12 254 $93 43 13 268 $90 46 14 282 $87 49 15 296 $85 52 16 310 $83 55 17 324 $81 58 18 338 $80 61 19 352 $79 64 20 366 $78 67 21 380 $77 70 22 394 $77 73 23 408 $76 76 24 422 $76 79 25 436 $76 82 26 450 $77 85 27 464 $77 88 28 478 $77 91 29 492 $78 94 30 506 $78 97 31 520 $79 100 32 534 $80 103 33 548 $81 106 34 562 $82 109 35 576 $83 112 36 590 $84 115 37 604 $85 118 38 618 $87 121 39 632 $88 124 PM cost $20.03 $15.44 $12.57 $10.63 $9.22 $8.16 $7.34 $6.69 $6.17 $5.74 $5.38 $5.09 $4.84 $4.63 $4.46 $4.31 $4.19 $4.10 $4.02 $3.95 $3.90 $3.87 $3.85 $3.83 $3.83 $3.84 $3.85 $3.88 $3.91 $3.95 $3.99 $4.04 $4.10 $4.16 $4.23 $4.30 $4.38 $4.46 $4.54 cost per unit of time $25.00 $20.00 $15.00 $10.00 $5.00 $0.00 0 50 100 15 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 646 660 674 688 702 716 730 744 758 772 786 800 814 828 842 856 870 884 898 912 926 940 954 968 982 996 1010 1024 1038 1052 1066 1080 1094 1108 1122 1136 $90 $91 $93 $95 $96 $98 $100 $102 $104 $106 $108 $110 $113 $115 $117 $119 $122 $124 $127 $129 $132 $135 $137 $140 $143 $146 $149 $152 $155 $158 $161 $164 $167 $170 $173 $176 127 130 133 136 139 142 145 148 151 154 157 160 163 166 169 172 175 178 181 184 187 190 193 196 199 202 205 208 211 214 217 220 223 226 229 232 $4.64 $4.73 $4.83 $4.93 $5.04 $5.15 $5.26 $5.38 $5.50 $5.63 $5.76 $5.89 $6.02 $6.16 $6.30 $6.45 $6.59 $6.74 $6.90 $7.05 $7.21 $7.37 $7.54 $7.70 $7.87 $8.05 $8.22 $8.40 $8.58 $8.76 $8.95 $9.14 $9.33 $9.52 $9.72 $9.92 Maintenance Model cost per unit of time $250 $200 $150 replacement time $100 $50 $0 0 200 r unit of time PM Interval 50 100 150 200 250 400 600 800 1000 1200 me 1200 Section 11.4 Inspect and Repair Availability Model Exponential Case l Inspect time - t1 Repair time - t2 T 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300 310 320 330 340 350 360 370 380 390 400 410 420 enter values 0.0002 16 units 48 hours A(T) 0.382818953 0.551510411 0.646187714 0.706610178 0.748384799 0.77888786 0.802058727 0.820192752 0.834717868 0.846568969 0.856383799 0.864612317 0.871581079 0.877532824 0.882651746 0.88708014 0.890929676 0.894289214 0.897230348 0.899811403 0.902080374 0.904077116 0.905835 0.907382177 0.908742565 0.909936612 0.910981903 0.911893648 0.912685065 0.9133677 0.913951679 0.914445925 0.914858325 0.915195882 0.915464832 0.915670748 0.915818624 0.915912949 0.915957772 0.915956748 0.915913193 0.915830115 Maximum Time between Availability Inspections - T* 0.915958 390 hours 1 e lT A(t ) l[T t1 t2 (1 A(T) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 inspection int 0.2 0.1 0 0 200 400 600 800 430 440 450 460 470 480 490 500 510 520 530 540 550 560 570 580 590 600 610 620 630 640 650 660 670 680 690 700 710 720 730 740 750 760 770 780 790 800 810 820 830 840 850 860 870 880 890 900 910 920 0.915710254 0.915556109 0.915369963 0.91515391 0.914909868 0.914639602 0.914344738 0.914026773 0.913687092 0.913326978 0.912947616 0.912550109 0.912135479 0.91170468 0.911258599 0.910798063 0.910323845 0.909836668 0.909337208 0.908826098 0.908303933 0.90777127 0.907228634 0.906676518 0.906115386 0.905545677 0.904967801 0.90438215 0.90378909 0.90318897 0.902582118 0.901968847 0.901349452 0.900724212 0.900093392 0.899457246 0.898816013 0.89816992 0.897519185 0.896864012 0.8962046 0.895541134 0.894873793 0.894202747 0.893528159 0.892850184 0.89216897 0.891484658 0.890797385 0.890107279 930 940 950 960 970 980 990 1000 0.889414464 0.88871906 0.888021181 0.887320934 0.886618425 0.885913755 0.885207019 0.884498309 1 e lT A(t ) l[T t1 t2 (1 e lT )] inspection interval 600 800 1000 1200 \f

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

Introduction to graph theory

Authors: Douglas B. West

2nd edition

131437372, 978-0131437371

More Books

Students also viewed these Mathematics questions

Question

1.. Distinguish between vertical and horizontal differentiation

Answered: 1 week ago