4097977994

4097977994 - Reliability Network Modeling Reliability...

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1 ME4214 Design for Product Safety Reliability – Part II: Reliability Network Modeling Dr. David Yuen Department of Mechanical Engineering The Hong Kong Polytechnic University 1 2 1 - understand how the system operates - identify the ways in which it can fail - deduce consequences of failure - derive models to represent these characteristics - then select appropriate reliability evaluation technique Reliability Evaluation Reliability Network Modeling Reliability Network Model (Reliability Block Diagram): - Series Network - Parallel Network (models the actual requirement of a system, and not its topological structure) 2 3 2 Series & Parallel Systems Series System Parallel System non-redundant system (all components must work for system success) fully redundant system (all components must fail for system failure) TR1, 10 kVA TR2, 10 kVA 15 kVA Load TR1, 20 kVA TR2, 20 kVA 15 kVA Load TR2 TR1 TR2 TR1 input output input output 3 4 3 Let R = P [Success] Q = P [Failure] R + Q = 1 = = = n 1 i i 2 1 s R R R R 2 1 2 1 2 1 2 1 S s Q Q Q Q ) Q )(1 Q - (1 - 1 R R 1 R 1 Q + = = = = 2 1 product rule of reliability Series System S R 1 R 2 S Q 1 Q 2 4
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5 4 Example: A system consists of 10 identical components, all of which must work for system success. What is the system reliability if each component has a reliability of 0.95? Component reliability, R = 0.95 Number of components, n = 10 Using Product Rule of Reliability, System Reliability, R s = R n = (0.95) 10 = 0.5987 5 6 5 Series System 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 1 02 03 04 05 06 07 0 Number of Components System Reliability 0.9 0.98 0.999 If each component has a reliability of 0.9. System Reliability decreases as the number of components increases in a Series System. The number on the curve is the reliability of each component. Number of Components Reliability 1 2 3 4 5 10 20 50 0.9 0.81 0.729 0.6561 0.59049 0.348678 0.121577 0.005154 6 7 6 product rule of unreliability Parallel Redundant System S Q 1 Q 2 S R 1 R 2 2 1 2 1 2 1 2 1 2 1 S s R R R R ) R )(1 R - (1 - 1 Q Q 1 Q 1 R + = = = = 7 8 7 Parallel System Reliability of each component is 0.7 Number of Components System Reliability 1 0.700000 2 0.910000 3 0.973000 4 0.991900 5 0.997570 6 0.999271 7 0.999781 0.6 0.7 0.8 0.9 1.0 12345678 Number of Components 8
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9 8 Example: A system is to be designed with an overall reliability of 0.999 using components having individual reliabilities of 0.7. What is the minimum number of components that must be connected in parallel? System Reliability, R s = 0.999 Component reliability, R = 0.7 Number of components, n = ? System Unreliability, Q s = 1 - R s = 1 – 0.999 = 0.001 Component Unreliability, Q = 1 – R = 1 – 0.7 = 0.3 Using Product Rule of Unreliability, Q s = Q n i.e. 0.001 = (0.3) n therefore, n = ln (0.001) / ln(0.3) = 5.74 since n is an integer, n = 6 9 10 9 Series/Parallel Systems Network Reduction Technique Telcom. Repeater Station Power Supply System 4 2 3 1 Battery Bank DG Rectifier Cable R X = R 2 .R 3 Q Y = Q X .Q 4 R Y = 1 - Q X .Q 4 = 1 – (1-R X ).(1-R 4 ) R S = R Y .R 1 4 1 X 1 Y 0.10 0.01 0.07 0.01 = 0.98245 10 DG= D iesel G enerator 11 10 Example: Derive a general expression for the unreliability of the model shown below, and hence evaluate the unreliability of the system if all components have a reliability of 0.8.
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4097977994 - Reliability Network Modeling Reliability...

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