Calculating Probability of Valve Failure in Q5: Exponential Distribution Model

AI Thread Summary
The discussion focuses on calculating the probability of valve failure in Q5 using an exponential distribution model. The useful life of the valves is defined as up to 5 years, with specific failure rates assigned to each valve. The probability of a loss of flow from the manifold within 3 years is derived from the exponential distribution function. The correct calculation yields a probability of 0.08643, confirming the model's accuracy. This analysis emphasizes the importance of understanding valve performance and failure probabilities in engineering applications.
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[SOLVED] probability of failure

The performance of the valves in Q5 has been assessed in more detail under conditions

closer to those experienced in-service and the distribution functions of the random time

to failure have been quantified. The useful life period, prior to wear-out, occurs from

installtion to 5years. During this period, all of the distribution functions are modeled

using an exponential distribution function of the form:

FT (t) = 1 − exp[−_λit] where i=1,2,3,4,5

If _λ1 = λ_2 = _λ3 = 0.05; λ_4 = 0.267; λ_5 = 0.189 (all in years−1), calculate the probability

of a loss of flow from the manifold sometime in the period (0,3)years.

ANSWER[P[F]=0.08643]

ATTEMPT
Have tried to substitute .05 for lambada and 3 for t in the given equation but my answer is still very different from the given answer of 0.08643
 
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The attempt at a solution:

V1,V2 AND V3 ARE IN SERIES AND SS1 ARE IN PARALLEL TO V4 AND V5

(PV1 OR PV2 OR PV3) AND PV4 AND PV5

FT (t) = 1 − exp[−_λit] where i=1,2,3,4,5

If _λ1 = λ_2 = _λ3 = 0.05; λ_4 = 0.267; λ_5 = 0.189 (all in years−1),

For PSS FT (t) =1- exp [-λ1+λ2+λ3*3] where t=3 = .36237

For PV4, FT (t) =1- exp [-λ4*3]=.55112

For PV5,FT (t) =1- exp [-λ5*3]= .43278

Therefore the probability of loss of flow from the manifold at time 3 years is

PSS1 AND PV4 AND PV5= .36237*.55112*.43278= .08643
 
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