QUALyTEK QUALIDADE, TECNOLOGIA E SISTEMAS LTDA.

WEIBULL - CÁLCULO MATEMÁTICO

	A hundred identical pumps are operating continually untill to failure. It has been anoted the times to failure of each one, and we obtain the following table: Observed Data Time To Failure (hours) Observed Frequency 1000 => 1100 1100 => 1200 1200 => 1300 1300 => 1400 1400 => 1500 1500 => 1600 1600 => 1700 1700 => 1800 1800 => 1900 2 6 16 14 26 22 7 6 1 I- Calculate: The minimal life or intrinsic reliability "t0" => failure free time). The characteristic life or scale parameter (h). The shape parameter (b) and failure characterisitic. The correlation coefficient (r). The failure probability for a operating interval of 1350 hours. The reliability to a operating interval of 1400 hours. The pump's MTTF (Mean Time To Failure). The standard deviation (s). The variation coefficient (s/m). II- Trace the reliability graphic. III- The corrective maintenance cost per intervention (Ccm) is US$ 600.00 and the preventive maintenance cost per intervention (Cpm) is US$ 250.00. Is there a optimal period to effectuate preventive maintenance? In affirmartive case, what is this period? Solution: Pumps Data Time to Failure (hours) Obs. Freq. Observed Simple Rel. Freq. Observed Cumul. Rel. Freq. 1000 => 1100 1100 => 1200 1200 => 1300 1300 => 1400 1400 => 1500 1500 => 1600 1600 => 1700 1700 => 1800 1800 => 1900 Total: 2 6 16 14 26 22 7 6 1 100 0.02 0.06 0.16 0.14 0.26 0.22 0.07 0.06 0.01 1.00 0.02 0.08 0.24 0.38 0.64 0.86 0.93 0.99 1.00 --- Item I.1- To determine "t0", there are three methods: by trial graphic computer program Trial: by means of selecting arbitrary values to "t0". The value giving the best correlation coefficient, will be the more adequated one. Graphic: by means of utilization of graphics paper (Team Chartwell, by example), and use of below formula. Computer program: by means of trials, where several values to "t0" are tested, being choosed the one presenting the best correlation coefficient. In this case, the best option is "t0 = 900 hours". Item I.2 and I.3- We know that the failure cumulative frequency to the Weibull distribution, is given by: , transforming to the linear form, we obtain: Therefore, we can construct the following table: t F(t) Y=Ln{-Ln[1-F(t)]} X=Ln(t-to) to=900 h 1100 1200 1300 1400 1500 1600 1700 1800 1900 0.02 0.08 0.24 0.38 0.64 0.86 0.93 0.99 1.00 -3.9019 -2.4843 -1.2930 -0.7381 0.0214 0.6761 0.9780 1.5272 ------ 5.2983 5.7038 5.9915 6.2146 6.3969 6.5511 6.6846 6.8024 ------ Now, we can apply linear regression to determine "b" and "h": Table to Facilitate Calculus Ord. "Yi" "Xi" "Yi2" "Xi2" "Xi x Yi" 1 2 3 4 5 6 7 8 S -3.9019 -2.4843 -1.2930 -0.7381 0.0214 0.6761 0.9780 1.5272 -5.2146 5.2983 5.7038 5.9915 6.2146 6.3969 6.5511 6.6846 6.8024 49.6432 15.2251 6.1719 1.6719 0.5447 0.0005 0.4571 0.9566 2.3323 27.3601 28.0722 32.5331 35.8976 38.6214 40.9207 42.9167 44.6840 46.2726 309.9183 -20.6737 -14.1701 -7.7472 -4.5868 0.1370 4.4289 6.5379 10.3885 -25.6855 Determination of the angular coefficient (b): * typical failures by wear-out Determination of the linear coefficient (- b . Ln h) Therefore: h = 594.28 hours (characteristic life - scale parameter) Item I.4- Determination of the correlation coefficient (r): Item I.5- Pump failure probability (t=1350 hours): Thus, to "t=1350 hours", we have: Item I.6- Pump reliability (t=1400 hours) Thus, to "t=1400 hours", we have: Item I.7- Pump's MTTF: Item I.8- Standard deviation: Item I.9- Variation coefficient: Item II- Reliability graphic: Item III- Preventive maintenance interval: Values: The following equations will be used: # Exist a finite time to execute systematics preventive maintenance, when: We also can use the graphic below: # If the equation above is true, then the optimal time interval to execute preventive maintenance, is given by: Entering with the values, we obtain: Condition: Optimal interval: