A system has two failure modes. One failure mode, due to external conditions, has a constant failure rate of 0.07 failures per year. The second, attributed to wear out, has a Weibull distribution with a characteristic life of 10 yr and a shape parameter of 1.8. Determine the system reliability for a design life of 1 yr. (four decimals)

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yemmy yemmy · Answer 1 · 2020-03-28T05:46:36+01:00

Answer:

0.9177

Step-by-step explanation:

let us first represent the two failure modes with respect to time as follows

R₁(t) for external conditions

R₂(t) for wear out condition ( Wiebull )

Now,

[tex]R1(t) = e^{-nt} .....1[/tex]

where t = time in years = 1,

n = failure rate constant = 0.07

Also,

[tex]R2(t)=e^{-(\frac{t}{Q} )^{B} }......2[/tex]

where t = time in years = 1

where Q = characteristic life in years = 10

and B = the shape parameter = 1.8

Substituting values into equation 1

[tex]R1(t) = e^{-(0.07)(1)} \\\\R1(t) = e^{-0.07}[/tex]

Substituting values into equation 2

[tex]R2(t)=e^{-(\frac{1}{10} )^{1.8} }\\\\R2(t)=e^{-(0.1)}^{1.8} }\\\\R2(t)=e^{-0.0158}[/tex]

let the system reliability for a design life of one year be Rs(t)

hence,

Rs(t) = R1(t) * R2(t)

t = 1

[tex]Rs(1) = [e^{-0.07} ] * [e^{-0.0158} ] = 0.917713[/tex]

Rs(1) = 0.9177 (approx to four decimal places)