Failure of a system of components logically arranged in series and parallel

  1. [SOLVED] failure of a system of components logically arranged in series and parallel

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

    3. The attempt at a solution
    Assummed 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]
     
    Last edited: Apr 9, 2008
  2. jcsd
  3. 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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