1. Product life expectancy MTTF refers to the expected time when 63.2% of the product is defective, or the time when the reliability is 36.8%. The product life expectancy L10 refers to the expected time when the product is 10% defective, or the reliability is 90 % Of time
MTTF=t1+(t2-t1)*0.632L10=t1+(t2-t1)*0.1
t1: the shortest time to reach the defective rate value t2: the longest time to reach the defective rate value
2. There is no certain standard for the definition of badness requested, for example:
(1) A batch of products can be used for testing. After 2000, 3000, 5000, 10000, and 18000 hours, test whether the speed and current consumption exceed the specifications as the definition of bad.
(2) Or use a precision measuring instrument to measure the residual amount of lubricating oil on the axis as a poor definition.
3. The more samples tested, the more reliable the data will be, at least three.
4. The test time can be 2000, 3000, 5000, 10000, 18000 hours as the norm. That is to say, the test time point i is 1,2,….n, because the test point has five points, so n=5, I=1,2,3,4,5 total five points.
| i | 1 | 2 | 3 | 4 | 5 |
| t (hour) | 2000 | 3000 | 5000 | 10000 | 18000 |
5. Assume that an existing fan product is deemed defective when the shaft oil is consumed to 50%. Because it takes a long time for the oil to reach 50%, it is necessary to find the time by the heterodyne method.
If the fuel loss curve approaches linearity, the fuel consumption formula can be written as Y=Ax+b
Y: Oil finger residual amount after i time elapses
X: Time (unit: hour)
a: The slope of the curve. Equal to Σ(Xi-X)*(Yi-Y)/ Σ(Xi-X)2
b: Constant (unit: %) equal to Y-[Σ(Xi-X)*(Yi-Y)/ Σ(Xi-X)2]*X=Y-Ax
Note: When calculating, if b>100% is calculated as 100%
X: = average value of each test time point
=(2000+3000+5000+10000+18000)/5=7600 hours
Y: The average value of grease residues at each test time point. Assuming that there are three samples S1, S2, and S3, their grease residues after i time are as follows:
| Xi (hour) | 2000 | 3000 | 5000 | 10000 | 18000 |
| S1(%) | 99.9 | 99.8 | 99.6 | 99.1 | 98 |
| S1(%) | 99.5 | 99.2 | 97 | 93 | 87 |
| S1(%) | 99.5 | 99.3 | 98 | 96 | 93 |
X=7600, Y=(99.9+99.8+99.6+99.1+98)/5=99.28%
a=[(2000-7600)(99.9-99.28)+(3000-7600)(99.8-99.28)+(5000-7600)(99.6-99.28)+ (10000-7600)(9931-99.28)+(18000- 7600) (98-99.28)]/[(2000-7600)2+(3000-7600)2+(5000-7600)2+(10000-7600)2+(18000-7600)2]=[(-3472 )+(-2392)+(-832)+(-432)+ (-13312)]/[(31360000)+(21160000)+(6760000)+(108160000)]=-20440/167440000=-0.000122
b=Y-aX=99.28-(-0.000122*7600)=99.28+0.927=100.2(%), calculated as 100(%) according to the fuel consumption formula Y=aX+b, we know that the S1 fan is in the lubricating oil quantity (Y) 50% time left X
X=(Y-b)/a=(50-100)/(-0.000122)=-50/(-0.000122)=409836 hours
X=7600, Y=(99.5+99.2+97+93+87)/5=95.14%
a=[(2000-7600)(99.5-95.14)+(3000-7600)(99.2-95.14)+(5000-7600)(97-95.14)+ (10000-7600)(9.-95.14)+(18000 -7600) (87-95.14)]/ [(2000-7600)2+(3000-7600)2+(5000-7600)2+(10000-7600)2+(18000-7600)2]=[(- 24420)+(-18680)+(-4840) +(-5140)+(-84660)]/[(31360000)+(21160000)+(6760000)+(108160000)]=-137740/167440000=-0.0008226< br /> b=Y-aX=95.14-(-0.0008226*7600)=95.14+6.25=101.39(%), calculated as 100%
According to the fuel consumption formula Y=aX+b, the time X when the S2 fan has 50% of the lubricating oil (Y) remaining:
X=(Y-b)/a=(50-100)/(-0.0008226)=-50/(-0.0008226)=60782 hours
X=7600, Y=(99.5+99.3+98+96+93)/5=97.16%
a=[(2000-7600)(99.5-97.16)+(3000-7600)(99.3-97.16)+(5000-7600)(98-97.16)+ (10000-7600)(96-97.16)+(18000- 7600) (93-97.16)]/ [(2000-7600)2+(3000-7600)2+(5000-7600)2+(10000-7600)2+(18000-7600)2]=[(-13104 )+(-9844)+(-2184) +(-2784)+(-43264)]/[(31360000)+(21160000)+(6760000)+(108160000)]=-71180/167440000=-0.0004251
b=Y-aX=0.90-(-0.0004251*7600)=97.14+3.23=100.37(%), calculated as 100%
According to the fuel consumption formula Y=aX+b, the time X when the S3 fan has 50% of the lubricating oil (Y) remaining:
X=(Y-b)/a=(50-100)/(-0.0004251)=-50/(-0.0004251)=117619 hours
6. According to the above three samples: The fan that consumes 50% of the oil first is S2, the time is 60,782 hours, the slowest is S1, the time is 409,836 hours, so we know that t1=60782 hours, t2=409836 hours
7.MTTF=t1+(t2-t1)*0.632=60782+(409836-60782)*0.632=60782+220602=281384 hours
L10=t1+(t2-t1)*0.1=60782+(409836-60782)*0.1=60782+34905=95687 hours
