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` COLLEGE OF ENGINEERING EXAMINATION MODEL ANSWER SHEET MAY/JUNE 2016 Module No: Paper Title: Examiner(s): Question No: EG-M36 Systems Monitoring, Control, Reliability, Survivability, Integrity and Maintenance K. Wada 25 Marks 1 Question 1 (a) In the field of engineering, what is engineering integri...

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COLLEGE OF ENGINEERING EXAMINATION MODEL ANSWER SHEET MAY/JUNE 2016 Module No:

EG-M36 Systems Monitoring, Control, Reliability, Survivability, Integrity and Maintenance

Paper Title: Examiner(s): Question No:

K. Wada 25 Marks

1

Question 1 (a)

In the field of engineering, what is engineering integrity in association with reliability, availability, maintainability and safety of a system? The overall combination of these four topics constitutes a methodology that ensures good engineering design with the desired engineering integrity. [2] (b)

Describe briefly, what is a system in terms of ‘structure’, ‘behaviour’ and ‘interconnectivity’? [Structure] It contains parts (or components) that are directly or indirectly related to each other. [1] [Behaviour] It exhibits processes that fulfill its function or purpose. [1] [Interconnectivity] Aspect of system reliability that takes equipment maintainability into account. [1] A system’s structure and behavior may be decomposed via subsystems and subprocesses to elementary parts and process steps. [1] (1 mark for each sensible/similar description to maximum of 4) (c)

[4]

For a complex system, why is it necessary to adopt a system

engineering approach to design? Because a system is dependable on the complexity of the functioning of the units. In other words, ‘System engineering ensures that all likely aspects of a project or system are considered, and integrated into a whole.’ [3] (d)

What is the fundamental difference between Quality and Reliability?

Reliability of a product is its ability to retain its quality as time progresses. [1] A product can only have high quality if it also has high reliability (initial quality is of little use if it is soon lost. [1] A product with high reliability does not necessarily have high quality. [1] Page 1 of 11

` This means that the product is merely retaining low quality over a long period of time. [1] (1 mark for each element with brief description to maximum of 4) (e)

[4]

Both reliability (R) and unreliability (F) vary with time t. Describe the characteristics of R(t) and F(t) with changing time, and write mathematical expressions of R and F with respect to t.

R(t) decreases with time. [1] F(t) increases with time. [1] R(t) + F(t) = 1. [1]

𝑅(𝑡) = exp(−𝜆𝑡) 𝐹(𝑡) = 1 − exp(−𝜆𝑡) (f)

[0.5] [0.5] [4]

There are 30 items of a repairable product subjected to a test interval of 24 hours. Mean down time (MDT) is measured to be 1.5 hours (where the total number of failures were 5). (i) Using the Equation (1-1), calculate the mean time between failures (MTBF). 𝑀𝑇𝐵𝐹 =

𝑁𝑇 − 𝑁𝐹 𝑀𝐷𝑇 𝑁𝐹

……………… (1-1)

MTBF = (30 x 24 – 5 x 1.5)/5 = 142.5hours

[3]

(ii) Using the calculated value of MTBF from (f)-(i), calculate the mean failure rate (per hour). Failure rate () = 1/MTBF = 7x10-3

[2]

(iii) Using the calculated values from (f)-(i), use Equation (1-2) to calculate the availability (A) and unavailability (U) of the product. 𝐴=

𝑀𝑇𝐵𝐹 𝑀𝑇𝐵𝐹 + 𝑀𝐷𝑇 ……………… (1-2)

A = 142.5/(142.5 + 1.5) = 0.99 U = 1 – A = 0.01

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[2] [1]

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COLLEGE OF ENGINEERING EXAMINATION MODEL ANSWER SHEET MAY/JUNE 2016 Module No: Question No:

EG-M36 25 Marks

2

Question 2 (a)

By and large, failures are classified as either a i) system failure or ii) a component failure. Give two examples each for system failures and component failures of a product. Example: Automobile [System failure] Brake, Engine, Gear box, Ignition, Fuel, etc. (1 mark for each sensible/similar element to maximum of 2) [2] [Component failure] Windshield wiper, Head lamps, Air-conditioning unit, Seat, etc. (1 mark for each sensible/similar element to maximum of 2) [2] [4] (b)

After 1 year of operating a pump, it has been observed that the flow rate of a pump fluctuates unexpectedly from time to time. In the category of ‘sudden and gradual failures’ such a circumstance is classified as what type of failure? Depending on the size of a system, this type of failure can be classified as “Catastrophic Failures”, which could takes place suddenly and cannot be anticipated in advance. [2] (c)

Explain briefly, how improper selection of materials can lead to a permanent failure? Improper selection of materials mean choosing a component that has insufficient strength to withstand the loads (tension, compression, torsion, shear, bending, random loading conditions, etc.) in operation. Uncontrolled manufacturing process, for example, could potentially lead to processing a component with inhomogeneity of material strength to withstand the loads, thus leading to a permanent failure. [2]

(d)

Explain briefly, how manufacturing process can lead to a premature failure on the production line? - In actual manufacturing, uncontrolled manufacturing process could lead to producing a component outside the defined tolerances. Using this uncontrolled component for assembly further leads to a high possibility of having unexpected stress raisers, leakage, thermal stress, etc. - In the course of machining, exposure to increased surface temperature will change the surface characteristics (inhomogeneous material streng th).

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` - Improper heat-treatment can cause surface irregularities, weakness in the material strength due to not successful in removing the dislocations (e.g. from the prior coldrolling process), brittleness causing premature failure due to impact loading, etc. (1 mark for each statement or similar to maximum of 2) (e)

[2]

A desktop computer monitor is designed to withstand sudden impact loading applied to the polymer exterior rim. The rim’s fracture toughness is normally distributed with the mean of 90 MPa and the standard deviation () of 20 MPa. The stress generated as a result of impact load is also normally distributed with the mean of 45 MPa and the standard deviation () of 15 MPa. The rim’s probability of failures per application of stress is 0.0015. The safety margin (SM) and loading roughness (LR) are represented by Equations (2-1) and (2-2), respectively. 𝑆𝑀 = 𝐿𝑅 =

𝑦−𝑥

………. (2-1)

𝜎𝑥

………. (2-2)

√(𝜎𝑥2 +𝜎𝑦2 ) √(𝜎𝑥2 +𝜎𝑦2 )

(i) Using the Equation 2-1 calculate the SM of a product. SM = (90 – 45)/(152 + 202) = 1.8 [1] (ii) Using the Equation 2-2 calculate the LR of a product. LR = 15/(152 + 202) = 0.6 [1] (iii)

Using Figure 2-1 as a guide and together with the calculated results from (e)-(i) and (e)-(ii), comment on the product’s degree of reliability.

Log10 (0.0015) = -2.8 SM = 1.8 (calculated value) LR = 0.6 (calculated value) The product’s degree of reliability is said to be “unreliable” (in the Region (c)), as indicated by the mark X in the figure. Log10 (failures per application 0 of stress) -2

X

-4

LR

-6 -8

Region (b)

Region (c)

Region (a)

-10 -12

0

1

2

3

4

5

6

Safety Margin, SM

Figure 2-1. Variation of failure rate with SM

[5] Page 4 of 11

(iv)

` Suggest ways to improve its reliability more to ‘Region (a)’, as

shown in Figure 2-1, which is known as ‘intrinsically reliable region’. - Change material to increase its strength (e.g. hardness, ductility). - Change design to reduce (or absorb) the impact load more effectively. - Change the design feature of the rim, adding more reinforcements (e.g. ribs). - Add shock absorbing internal features. (2 marks for each statement or similar to maximum of 8)

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[8]

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COLLEGE OF ENGINEERING EXAMINATION MODEL ANSWER SHEET MAY/JUNE 2016 Module No: Question No:

EG-M36 25 Marks

3

Question 3 (a)

Draw block diagrams and highlight the advantages and disadvantages of a parallel system reliability in comparison to a series system reliability. Example: Parallel system

Series system

(0.5 marks each for the block diagrams to maximum of 1)

[1]

Advantages: - have backup (redundancy) in a system so that if one element/subsystem survives the overall system survives (in series system, all the element/subsystem needs to survive in order to avoid system failure) - probability of failure in a system is minimized - minimum maintenance Disadvantages; - complexity of a system - high cost due to having more backup (redundant) element/subsystem - total mass of a system gets heavier (1 mark for each statement or similar to maximum of 4)

[4]

(b) Briefly describe with an example, what is full active redundancy? All elements/subsystems are normally in continuous operation. Can be an expensive solution. [1] Example includes, power generation systems (having two compressor groups). [1] [2] Page 6 of 11

(c)

` Figure 3-1 shows a fire alarm standby system consisting of three identical subsystems (A, B, C) and a switching system S.

A

S B Switching system C

Subsystems

Figure 3-1. Fire alarm standby system (i)

What are the three possible combinations of overall system failures?

All A, B, C fails (S survives).

[1]

Or two subsystems fail and one switching failure.

[1]

Or one subsystem fails and two switching failures.

[1] [3]

(ii)

Considering that the subsystem failure and switching failure are represented by F and Fs, respectively, the expression for overall system failure (FSYST) is represented by Equation 3-1. If each of the subsystem has failure rate  = 5 (based on 105 operating hours) and switching system  = 1 (based on 105 operating hours), calculate the FSYST after 1 year of operation.

𝐹𝑆𝑌𝑆𝑇 = 𝐹3 (1 − 𝐹𝑆 ) + 𝐹 2 𝐹𝑆 + 𝐹𝐹𝑆2 𝐹 = 1 − 𝑒 −𝜆𝑡 = 1 − 𝑒

(24×365) 105

−5 ×[

𝐹𝑆 = 1 − 𝑒 −𝜆𝑡 = 1 − 𝑒

(24×365) 105

−1×[

…………… (3-1)

]

= 0.35

[2]

]

= 0.08

[2]

𝐹𝑆𝑌𝑆𝑇 = 0.353 (1 − 0.08) + 0.352 × 0.08 + 0.35 × 0. 082 = 0.05

[1] [5]

(d)

A standby boiler feed water system consists of n identical pumps and a switch. The switch enables the discharge from any pump to be connected to the output pipe so that the system is running with one pump operational and the remainder pump(s) on standby. The aim is to Page 7 of 11

` have overall system reliability better than 0.99 over a period of four months. Assuming a constant failure rate and using the data given below calculate the required value of n. [Data] Failure rate for a single pump = 0.35/year Failure rate for a single switching operation = 0.17/year Failure probability of a single pump 𝐹 = 1 − 𝑒 −𝜆𝑡 = 1 − 𝑒 −0.35×[

4 ] 12

= 0.110

[1]

Failure probability of a single switching operation 𝐹𝑆 = 1 − 𝑒 −𝜆𝑡 = 1 − 𝑒

−0.17×[

4 ] 12

= 0.055

[1]

(a) Reliability of a single pump (n = 1, no standby) 𝑅𝑆𝑌𝑆𝑇 = 1 − 𝐹 = 0.89, not adequate

[2]

(b) Reliability of standby system with two pumps (n = 2) 𝐹𝑆𝑌𝑆𝑇 = 𝐹2 (1 − 𝐹𝑆 ) + 𝐹𝐹𝑆 = 0.018

𝑅𝑆𝑌𝑆𝑇 = 1 − 𝐹𝑆𝑌𝑆𝑇 = 0.982, still not adequate

[3]

(c) Reliability of standby system with three pumps (n = 3) 𝐹𝑆𝑌𝑆𝑇 = 𝐹3 (1 − 𝐹𝑆 ) + 𝐹2 𝐹𝑆 + 𝐹𝐹𝑆 2 = 2.26 × 10−3 𝑅𝑆𝑌𝑆𝑇 = 1 − 𝐹𝑆𝑌𝑆𝑇 = 0.998, meets the requirement

[3] [10]

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COLLEGE OF ENGINEERING EXAMINATION MODEL ANSWER SHEET MAY/JUNE 2016 Module No: Question No:

EG-M36 25 Marks

4

Question 4 (a)

When dealing with the design for reliability (DFR) activities at the ‘Analysis’ phase, various types of analysis can be done after the first design draft is created. List the four analysis activities. - Finite Element Analysis (FEA) - Physics of Failure (POF) - Warranty Data Analysis - Design Review Based on Failure Mode (DRBFM) - Study on the lessons learned on previous programmes - Design of Experiments (DOE) (1 mark for each statement or similar to maximum of 4)

(b)

[4]

Explain the difference between Accelerated Life Testing (ALT) and Highly Accelerated Life Testing (HALT).

[ALT] Testing a product by subjecting it to conditions in excess of its normal service parameters (uncover faults and potential modes of failure). [1] [HALT] Stresses applied to a product well beyond normal shipping, storage and in-use levels. [1] [2] (c)

In view of design verification, what sort of tests are required for a wind turbine (e.g. 5000 kW power generation, diameter 124 m, height 114 m)? - Continuous cyclic loading test (fatigue behavior) - Structural integrity tests (bending, torsion, compression, combined stresses) - Weather tests (rain water, sea water, dust, etc.) - Vibration test (e.g. considering the earthquake conditions) - Thermal cycling test (high and low temperatures) - Airflow test (aerodynamic performance) - Lightning test (in case of lightning strike) (1 mark for each statement or similar to maximum of 5)

Page 9 of 11

[5]

` (d) In the context of DFR activities, what is the goal of design validation and how does it differ from verification? Goal of validation is to successfully resolve design and manufacturing issues in case they had been overlooked at the previous DFR phases. [1] Verification is more to do with uncovering design faults and potential modes of failure. [1] [2] (e) What is the difference between fail-danger-failure and fail-safe-failure? Fail-danger-failure: any system or component failure which prevents, or tends to prevent, the plant being tripped when potentially hazardous fault/condition occurs. [1] Fail-safe-failure: any system or component failure which produces a plant trip when a plant trip is not required. [1] [2] (f)

A protective system (pressure measurement) is to have maximum faildanger probability (PD) not exceeding 6 x 10-3 and maximum fail-safe probability (PS) not exceeding 6 x 10-2. The system is tested and proved to be working at 4 weeks intervals. Use the components listed in Table 4-1, together with the given information in Figures 4-1, 4-2, and 4-3 to answer the following questions. (i)

Without the use of majority voting equipment, what are the overall system fail-danger and fail-safe probabilities in the cases of single, two in parallel, and three in parallel configurations? Comment on the results in view of meeting the fail-danger and fail-safe requirements.

Without majority voting equipment: i) Single channel system: FSmax = FDmax = 6.7 x 10 -2 (too high)

[1]

ii) 2 in parallel: PD = FD2 = 4.49 x 10-3 (OK); PS = 2FS = 13.4 x 10-2 (too high) [1] iii) 3 in parallel: PD = FD3 = 3 x 10-4 (OK); PS = 3FS = 20.1 x 10-2 (getting worse) [1] None of the above cases will meet the fail-danger or fail-safe probability requirements. Need to use majority voting equipment. [1] [4] (ii)

With the use of majority voting equipment, what are the overall system fail-danger and fail-safe probabilities that meets the requirement?

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` With majority voting equipment: [HIVE] FS = FD = 3.84 x 10-3 (OK)

[1]

[HITI] single channel:  = 0.6 + 0.1, FS = FD = 5.2 x 10-2 (need to reduce) 2oo3 voting: PD = 3FD2 = 8.24 x 10-3 (a bit high), PS = 3FS2 = 8.24 x 10-3 (OK) 2oo4 voting: PD = 4FD3 = 5.6 x 10-4 (OK), PS = 6FS2 = 1.6 x 10-2 (OK) [1] [HISS] single channel:  = 0.1 + 0.1 + 0.1, FS = FD = 2.3 x 10-2 (need to reduce) 2 channels: PD = FD2 = 5.3 x 10-4 (OK); PS = 2FS = 4.6 x 10-2 (OK) [1] [Overall System] Fail-danger: (PD)HITI + (PD)HIVE + (PD)HISS = 4.93 x 10-3 (OK) Fail-safe: (PS)HITI + (PS)HIVE + (PS)HISS = 6.24 x 10-2 (slightly high, not OK) [1] Choose 3oo5 system instead: Referring to the table: PD = 10FD3, PS = 10FS3 [HITI] 3oo5 voting: PD = 10FD3 = 1.4 x 10-3 (OK), PS = 10FS3 = 1.4 x 10-3 (OK) [1] [Overall System] Fail-danger: (PD)HITI + (PD)HIVE + (PD)HISS = 5.77 x 10-3 (OK) Fail-safe: (PS)HITI + (PS)HIVE + (PS)HISS = 5.12 x 10-2 (OK)

[1] [6]

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