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

  • Thread starter abba02
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[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. Homework 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:

Answers and Replies

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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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