Reliability Engineering in Complex Systems - Assignment Example

RELIABILITY ЕNGINЕЕRING IN СОMРLЕХ SYSTEMS Name Institution Subject Instructor Date Question one A1: Determining minimal cut sets Minimal cut set consists of systems failure events that are sufficient and relevant in causing the top event. In this case, of fault tree for cooker fire, the approach used here considers the fault tree as a logic tree as well as a reliability block diagram. The extraction of minimal cut sets takes place from the fault tree diagram. Here it considers the minimal cut sets of the top events that bring about the failure of the system. After the identification of all the minimal has been achieved, the probability computations of associated with the fire cooker based on the given figures is computed (Smith and Mobley, 2008, 122). Cut sets: 1. (Non-heat resistant material falls into the burning stove) 2. (Nobody removed a non-heat resistant material from the stove) 3. (Non-heat resistant material falls into the burning stove; nobody removed a non-heat resistant material from the stove) 4. (Nobody attend Alarm from timer; somebody forget to turn off the stove; Timer is set and turn off but nobody turn of the stove; nobody set alarm when cooking unattended; Timer control system fails) 5. (Nobody attend Alarm from timer) 6. (Somebody forget to turn off the stove; Timer is set and turn off but nobody turn of the stove) 7. (Nobody set alarm when cooking unattended; Timer control system fails) 8. (Nobody set alarm when cooking unattended) 9. (Timer control system fails) 10. (Nobody attends to the alarm from the timer, Somebody forgets to switch off the stove, timer is set and turned off but nobody) Minimal cut sets and minimal tie sets: 1. (Non-heat resistant material falls into the burning stove) 2. (Nobody removed a non-heat resistant material from the stove) 3. (Somebody forget to turn off the stove; Timer is set and turn off but nobody turn of the stove) 4. (Timer control system fails) 5. (Nobody set alarm when cooking unattended; Timer control system fails) Probability of a fire from a cooker in one year Probability of a fire from a cooker in one year = 0.37 Question Two A2: Determining Mean Time Between Failures (MTBF) For the de-sulphonator tower, the Mean Time Between Failures (MTBF) seeks to compute basic reliability measure for the components of the system that are capable of being repaired. MTBF gives a description of the duration of time spent before the failure of the de-sulphonator tower is experienced with regard to the components or the assembly of its system (O'connor and Kleyner, 2012, 88). Mean Time Between Failures (MTBF) also describes the computation of variables that are mostly used in maintainability and reliability analyses. The calculation for Mean Time Between Failures for the de-sulphonator tower takes place as that of the inverse of failure rate λ for the system. However, MTBF finds its common usage for items that are both repairable and non-repairable despite the fact that it is basically meant for repairable items (Kececioglu, 2002, 187). Considering the following pdf for the de-sulphonator; ( t ) hours Where: λ is the failure rate for the system = 2.7 failures per 100,000 hours t is time in hours f(t) is the failure density the system The de-sulphonator system is subjected maintenance and testing on a regular basis. Repair time when the system is out of operation is described by the function: ( t ) hours Where: µ is the failure rate for the system = 24 t is time in hours m(t) is the pdf function Determining Mean Time to Repair (MTTR) For the de-sulphonator tower, the Mean time to repair (MTTR) involves the computation of the overall amount of time spent during the performance of all preventive and corrective repairing operations divided by the number of operations. It also involves the calculation of the overall time spent from shut down or failure to the completion of maintenance or repair (Birolini, 2010, 254). Mean Time to Repair (MTTR) is typically computed for systems that are repairable such as the de-sulphonator tower. Determining Availability and the Unavailability For the de-sulphonator tower system, the numerical value of both unavailability and availability are computed in terms of probability between 0 and 1. The Availability, A (t), of the de-sulphonator tower system involves the computation of the probability with regard to the operation of the system at a given time t considering that it the operation begun at time zero. The computation of unavailability, Q (t), of the de-sulphonator tower system involves the computation of the probability. The probability is associated with the failed state of the system at a given time t. Unavailability is equivalent to the probability associated with the failure state of the system or its components at a given time t and can also be equated to the number of components that have failed divided by the overall sample. Since the de-sulphonator tower system must be either in operation or not in operation at any one particular time t, the summation of availability and unavailability of the system is equal 1. A (t) + Q (t) = 1 Unavailability Q (t) = 1 - Availability A (t) Unavailability Q (t) = 1 - 0.016 Unavailability Q (t) = 0.984 Number of outage hours per day = 24 hours x 0.984 = 23.6 hours For systems that are repairable such the de-sulphonator tower system, availability computation takes reliability into consideration. In this case, the computation of reliability takes place as in the form of probability that failure does not take place during the given time period. In the event that there is more than one failure, the expression of reliability takes place as the rate of failure. With regard to systems that are repairable, the characterization of reliability happens through the involvement of Mean Time Between Failures (MTBF). However, this only takes place in the case of constant rate of failure. There are also considerations with regard to the computation of availability A (t) whose determination is equivalent to that of the determination of the probability associated with the operable state of a system at any given time (Kececioglu, 2002, 96). The categorisation of systems either as repairable or non-repairable considers the rate of failure, which is loosely applied to non-repairable systems. Question Three B1: In determining the Rs function of the structure above, it is otherwise known as the System Reliability. The reliability of the above structure involves the determination of the probability associated with success under certain condition within a given duration of time. The reliability of a system Rs for the above system under consideration considers the reliabilities of the individual components in such a manner as described below: Where R1 = the reliability of individual component 1 R2 = the reliability of individual component 2 R3 = the reliability of individual component 3 Rn = the reliability of individual component n In this determination, the assumption is that the individual components are interdependent. This means that the failure or non-failure of a single component does not bring about any changes to the reliabilities of the other components in the system structure (Levitin, 2007, 103). In determining the Rs function, which is the overall reliability of the system, components are categorized and analysed as either components in series or components in parallel (Birolini, 2010, 186). For system components in series: Rs = (R1) x (R2) x (R3) x …………….x (Rn) Whereas for system components in parallel: Rs (t) = 1 - [(1 - R1) x (1 - R2) x (1 - R3) …………. (1 - Rn)] Therefore for the above ACME widgets structure whose functional details are described in tabulation as indicated below: System components Component label Probability of failing Detectors A 0.035 Alarm F 0.007 Cables B 0.008 Computer G 0.020 Personnel Response C 0.031 Personnel Access H 0.015 Manual Override D 0.001 System for Control Rod I 0.021 System for cooling E 0.038 System for Inerting J 0.006 Structure function Rs: Rs (t) = 1 - [(1 - R1) (1 - R2) (1 - R3) …………. (1 - Rn)] Rs (t) = 1 - (1 – e-3t) (1 – e-3t) (1 – e-3t) System reliability Rs: Rs1= 1 - [(1 - 0.038) x (1 - 0.008)] = 0.046 Rs2 = (0.035) x (Rs1) = (0.035) x (0.046) = 0.002 Rs3 = 1 - [(1 - 0.001) x (1 – Rs2)] = 1 - [(1 - 0.001) x (1 – 0.002)] = 0.997 Therefore, structure function Rs = (Rs3) x (0.031) = (0.997) x (0.031) = 0.03 Importance of component ‘D’ The component D whose definition is expfressed as ID=dRs/dRD is significance in the sense that it shows the relationship the output and input for the system under consideration. It also forms the basis for analysis and computation of probabilities for various system components and the overall system reliability (Elsayed, 2012, 221). if all the components have the same reliability R: The polynomial structure of the function reduces to; Rs (t) = 1 - [(1 - R1) (1 - R2) (1 - R3) …………. (1 - Rn)] Rs (t) = 1 - [(1 - x) (1 - x) (1 - x) …………. (1 -xn)] Rs (t) = 1 - [(k) (k) (k) …………. (kn)] Rs (t) = 1 – kn If the same values of R are used, then this value relates to Rs = (Rs3) x (0.031) = (0.997) x (0.031) = 0.03 Question Three B3: Determining Mean Time To Failure MTTF for Bionic processor for Windows In computation of Mean Time To Failure MTTF for Bionic processor for Windows in the system structure shown above, assumptions are made regarding the components of the system. It is assumed that the individual components of the system are at a constant rate of hazard with regard to the phase of bathtub curve. In the event that the rate of failure is equal to one, then the rate of hazard is constant and failures do not depend in time. In this case, there is no decrease or increase in the age of a system component with time. A negative exponential forms an expression for the function’s failure density. For this particular system, the computation of Mean Time To Failure MTTF can be equated to MTBF only if the failure function has an exponential distribution. For failure rate of one, the rate of arrival is also equal to one (Smith and Mobley, 2008, 180). Failure rate could be stated as the failure’s arrival rate. The failure rate with time is equal to constant rate of hazard when the failure distribution follows an exponential function. Cimmulative rate of failure, F (t) = (1 – e- λ 3t) Reliability, R (t) = e- λ 3t Mean Time To Failure MTTF for Bionic processor for Windows is the inverse of rate of failure. This condition holds for a non-repairable system component with an average operating performance (Elsayed, 2012, 321). A = 1/15, B = 1/25, C = 1/10, D = 1/8, E = 1/35 per year. AB = (1/15) x (1/25) = 1/250 ABC = 1 - [(1 – (1/15)) x (1 – (1/250))] = 0.0704 ABCE = (1/35) x (0.0704) = 0.002 Therefore, Mean Time To Failure (MTTF) ABCDE = 1 - [(1 – (0.002)) x (1 – (1/8))] = 0.127 Rs = 1 – 0.127 = 0.873 The number of minimal cuts in this particular component is four. Considering the alternative circuit, the reliability, Rs is given by: DA = (1/8) x (1/15) = 0.0083 CE = (1/10) x (1/35) = 0.0029 ABCDE = 1 - [(1 – 0.0083) x (1 – 0.0029) x (1 – 0.04)] = 0.05 Therefore Rs = 1 – 0.05 = 0.95 Ratio of difference in reliability = (0.95 – 0.873) / 0.873 = 8.8 % = 88:1000 References Birolini, A. (2010). Reliability engineering theory and practice. Berlin, Springer. Elsayed, E. A. (2012). Reliability engineering. Hoboken, John Wiley & Sons. Kececioglu, D. (2002). Reliability engineering handbook. Lancaster, Pa, DEStech Publications. Levitin, G. (2007). Computational intelligence in reliability engineering: evolutionary techniques in reliability analysis and optimization. Berlin, Springer. O'connor, P. D. T., & Kleyner, A. (2012). Practical reliability engineering. Chichester [u.a.], Wiley. Smith, R., & Mobley, R. K. (2008). Rules of thumb for maintenance and reliability engineers. Amsterdam, Elsevier/Butterworth-Heinemann. Read More