- **Calculus & Beyond Homework**
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- - **[SOLVED] failure of a system of components logically arranged in series and parallel**
(*http://www.physicsforums.com/showthread.php?t=227605*)

[SOLVED] failure of a system of components logically arranged in series and parallelThe performance of the valves in (V1 OR V2 OR V3) AND V4 AND V5 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 modelled 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] 2. Relevant equationsFT (t) = 1 − exp[−_λit] where i=1,2,3,4,5 3. The attempt at a solutionAssummed that V1,V2 AND V3 ARE IN SERIES AND ARE IN PARALLEL WITH V4 AND V5 = 1- exp [-λ1+λ2+λ3+λ4+λ5*3] Have tried to substitute .05,.267 and .189 for lambada, *3 for t in the given equation but my answer is still very different from the given answer of 0.08643[/QUOTE] |

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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