qm zg528-l6.pptx

BITS PilaniPilani Campus

QM ZG528 L-6Reliability Engineering

Gajanand GuptaDepartment of Mechanical Engineering

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BITS PilaniPilani Campus

Reliability of SystemsChapter - 5

BITS Pilani, Pilani Campus

• A system consist of number of different components• Components within a system may be related to one

another in two primary ways -Series configuration or -Parallel configuration • In series configuration all components must function for

the system to function. • However in a parallel or redundant configuration at least

one component must function for the system to function.

Introduction

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• In a series configuration, all assets are considered critical

Reliability block diagram for components in series

• E1=Event that component 1 does not fail• E2=Event that component 2 does not fail• Then P(E1)=R1 and P(E2)=R2 R1,R2=Reliabilities• Rs=P(E1∩E2)=P(E1) P(E2)= R1 R2

• Therefore generelizing system reliability as- Rs= (R1) x (R2) x (R3) x …………….x (Rn)

Serial Configuration

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If component has constant failure rate

MTTF=1/ = 1/

EXAMPLE- Consider 4 components of system in series are independent and identically distributed with CFR. If Rs(100)=0.95 , Find individual component MTTF

Serial Configuration

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• A system consists of three units which function reliability wise in series. The failure rates of these units are as follows:

λ1 = 3.5 fr/106 hr,λ2 = 16 fr/106 hr and λ3 = 1.2 fr/106 hr.Determine the following parameters of this system:1. Failure rate, λss(T).2. Reliability, Rss(t), for a 100-hr mission.3. Mean time between failures, MTBFSS.4. The series system's failure rate is

Example

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1. The series system's failure rate is

2. The series system's reliability is given by

where λss = 20.7 fr/106 hr, and t = 100 hr; hence

Rss = e-0.00207 = 0.9979.

Solution

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3. The series system's MTBF is

This means that on the average such identical systems, operating under identical application and operation stresses, would fail after 48,309 hr of operation.

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• Two exponential subsystems make up a system. Subsystem 1 has a reliability of 99.95% for a 100-hr mission, and Subsystem 2 has a reliability of 99.95% for a 10-hr mission. What is this system's reliability for a 100-hr mission when the two subsystems are reliability wise in series?

Example

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• For Subsystem 1 it is known thatR1(t = 100 hr) = 99.95% = 0.9995,and for Subsystem 2 it is known thatR2(t = 10 hr) = 99.95% = 0.9995.For exponential units it is known thatR(t) = e-λt,andThereforeIt is also known that

Solution

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Therefore, Rss(t = 100 hr) = e-(0.0000050013+0.000050013)(100) =

e-0.0055014,or Rss(t = 100 hr) = 0.9945 = 99.45%.

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• In parallel system all units must fail to for system to fail• Parallel units are represented by block diagram

• System reliability is found by taking 1 minus probability that all n components fail

Parallel Configuration

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• Rs(t) = 1 - [(1 - R1) (1 - R2) (1 - R3) …………. (1 - Rn)]

• For a redundant system consisting of all CFR components

EXAMPLE- For two component parallel system and having CFR derive expression for MTTF

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Parallel Configuration


• Two parallel, identical and independent components have CFR. If it is desired that Rs(1000) = 0.95, Find the component and system MTTF.

Example

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• A system consists of three units which function reliability wise in parallel . The failure rates of these units are as follows:

λ1 = 3.5 fr/106 hr,λ2 = 16 fr/106 hr and λ3 = 1.2 fr/106 hr. • Find the following for this system:• The reliability for a 100-hr mission starting the mission at

age zero.• The MTBF.

Example

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• The reliability of this system for t = 100 hr is

Rsp(100 hr) = 0.9999

• The MTBF of this system is

Solution

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MTBFsp = 909,600 hr.

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Thanks

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qm zg528-l6.pptx

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