Test Rig - Maintenance Organisation - Exam, Exams of Organization Behaviour

Main points of this past exam are: Test Rig, Two Metrics, Reliability, Modelling Distributions, Typical Applications, Each Distribution, Reliability, Continuous Modelling Distributions, Constant Failure, Failure Times

Typology: Exams

2012/2013

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CORK INSTITUTE OF TECHNOLOGY
INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ
Semester 1 Examinations 2009
Module Title: Maintenance & Reliability
Module Code: MANU8003
School: Mechanical & Process Engineering
Programme Title: Bachelor of Engineering (Honours) in Sustainable Energy
Technology
Programme Code: ESENT_8_Y4
External Examiner(s): Prof. E. Coyle, Mr. R. Linger
Internal Examiner(s): Mr. Dan O’Brien
Instructions: Answer any 3 questions
Duration: 2 hours
Sitting: Winter 2009
Requirements for this examination: Probability Plotting Paper, Hazard Plotting
Paper, Statistical Tables
Note to Candidates: Please check the Programme Title and the Module Title to ensure that you are
attempting the correct examination.
If in doubt please contact an Invigilator.
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CORK INSTITUTE OF TECHNOLOGY

INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ

Semester 1 Examinations 2009

Module Title: Maintenance & Reliability

Module Code: MANU

School: Mechanical & Process Engineering

Programme Title: Bachelor of Engineering (Honours) in Sustainable Energy Technology

Programme Code: ESENT_8_Y

External Examiner(s): Prof. E. Coyle, Mr. R. Linger Internal Examiner(s): Mr. Dan O’Brien

Instructions: Answer any 3 questions

Duration: 2 hours

Sitting: Winter 2009

Requirements for this examination: Probability Plotting Paper, Hazard Plotting Paper, Statistical Tables

Note to Candidates: Please check the Programme Title and the Module Title to ensure that you are attempting the correct examination. If in doubt please contact an Invigilator.

Q. 1 Reliability Modelling Distributions:

(a) MTBF and MTTR are two metrics used in reliability. Differentiate between both. (3 marks) (b) Figure one beneath illustrates three continuous modelling distributions used in reliability. Describe each distribution and state typical applications. (12 marks)

Figure 1. Continuous modelling distributions. (c) Ten identical items with constant failure rate were put on test, their failure times ( in hours ) were observed to be: Table 1 – Failure Times 10 17 25 33 34 41 48 59 72 79

Calculate the MTBF and hence the reliability at 20hrs (8 marks) (d) The three parameter Weibull distribution may be expressed by the probability density function:

The time to fail for a flexible membrane follows the Weibull distribution with β=2 and θ = 300months. What is the reliability at 200 months? After how many months is 90% reliability achieved? (10 marks)

 −^ −

= ^ −

f ( x ) x exp x , for x

1

Q. 3 Reliability in Design

(a) What is the purpose of a FMEA and explain how it is performed, detailing what information is analysed and the significance of RPN? How does FMEA differ from FTA? (7 marks) (b) A control system consists of an electrical power supply, a standby battery supply which is activated by a sensor and switch if the main supply fails, a hydraulic power pack, a controller, and two actuators acting in parallel ( i.e. control exists if either or both actuators are functioning ) i. Draw the system reliability block diagram ii. Draw the fault tree appropriate to the top event “total loss of actuator control”. (12 marks)

(c) Describe the Robust Design Approach (8 marks)

(d) An experiment with three repetitions is shown in Table 3 beneath

Table 3. L4 with Results and Averages

Assuming that the bigger is better is the sought characteristic, which trial number should be selected?

(6 marks)

S/N for bigger is better

Q. 4 Maintenance Management / Inventory Managament

(a) Differentiate between TPM and RCM. (8 marks) (b) The acquisition cost of a plant was €30m. Running costs are €4m/year. €100m income is expected if the plant runs with an Overall Equipment Effectiveness (OEE) of 100%. A 1% reduction in OEE costs €1m/year. OEE = throughput % x availability % x quality% Throughput is 95%, Availability is 95% and Quality is 98% Life cycle costs are made up of acquisition costs + running costs + costs associated with running below 100% OEE. Use the above information to calculate the LCC of the plant for 25years. (6 marks)

(c) Spare parts management is divided into slow moving spares and fast moving spares. Describe both policies. (5 marks)

(d) A maintenance engineer carrying out the review of a life plan for a discharge system is concerned about the inventory policy for the main pump. He has gathered the following information. The pump is not repairable. On failure it is replaced with a new pump purchased from the manufacturer – the lead time is 4 months. The pump fails randomly with an average incidence of 0.5 failures per year. The estimated cost of holding the pump is €200 per year. The estimated cost of not being able to replace the pump from stock is €4000. Recommend an inventory policy for the pump. (7 marks)

(e) The failure distribution of a component is described by a 2-parameter Weibull distribution, with β = 2.5 and θ = 1000 hours. The cost for a corrective replacement is €5. The cost for a preventive replacement is €1. Estimate the optimum replacement age in order to minimize these costs given T = mθ + δ (7 marks)

Figure 2 – Weibull Distribution Function Plotting Paper

Figure 3 – Hazard Plotting Paper

Table 5 – Optimum Parts Replacement – Values for m

  • 1.5 2 2.5 Cf/Cp β
  • 2.0 2.229 1.091 0.893 0.810 0.766 0.761 0.775 0.
  • 2.2 1.830 0.981 0.816 0.760 0.731 0.733 0.755 0.
  • 2.4 1.579 0.899 0.764 0.720 0.702 0.711 0.738 0.
  • 2.6 1.401 0.834 0.722 0.688 0.679 0.692 0.725 0.
  • 2.8 1.265 0.782 0.687 0.660 0.659 0.675 0.713 0.
  • 3.0 1.158 0.738 0.657 0.637 0.642 0.661 0.702 0.
  • 3.3 1.033 0.684 0.620 0.607 0.619 0.642 0.687 0.
  • 3.6 0.937 0.641 0.589 0.582 0.600 0.627 0.676 0.
  • 4.0 0.839 0.594 0.555 0.554 0.579 0.609 0.662 0.
  • 4.5 0.746 0.547 0.521 0.526 0.557 0.591 0.648 0.
  • 5 0.676 0.511 0.493 0.503 0.538 0.575 0.635 0.
  • 6 0.574 0.455 0.450 0.466 0.509 0.550 0.615 0.
  • 7 0.503 0.414 0.418 0.438 0.486 0.530 0.600 0.
  • 8 0.451 0.382 0.392 0.416 0.468 0.514 0.587 0.
  • 9 0.411 0.358 0.372 0.398 0.452 0.500 0.575 0.
  • 10 0.378 0.337 0.355 0.382 0.439 0.488 0.566 0.