8/10/2019 ch400113141 answrs

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8/10/2019 ch400113141 answrs

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(e) Fault Tree

Q2.

(a) Cut sets: T = [(A+C)(B+C)+D+E]FG=(AB+AC+BC+C+D+E)FG = ABFG + ACFG + BCFG + CFG + DFG +

EFG

Minimal cut sets: T = ABFG + CFG + DFG + EFG

(b) Path sets: T= {[(A+C)(B+C)+D+E]FG} = [(AC + BC)DE]+F+ G = ACDE + BCDE + F+ G

Minimal path sets: T = ACDE + BCDE + F+ G

(c)

(i) Demand rate = (A+C)(B+C)+D+E = AB + C + D + E = 252.5 PMH

FDT = = FG= [(1/2)(5/1000000)(4000)][(1/2)(1/1000000)(8000) = 0.00004

(ii) Hazard rate = AB+ c+D+ E= 0.0101 PMH

(iii) Safe shutdown Rate = (Demand)[(1-F) + (demand)(F)(1-G) = 252.5 PMH

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Q3

(a) (10 marks)

(i) Reliability and availability both describe the probabilities of a component or system

works.

(ii) Reliability is an inherent property and depends solely on the design and manufacturingquality of the component or system.

(iii) Reliability is improved by improving the product design and manufacturing quality.

(iv) Reliability is use-time dependent (function of use-time).

(v) Reliability is affected by the environmental conditions in which the product is used.

(vi) Availability is the fractional operational time of a product and it is not a function of time.

(vii) Availability depends on the reliability of the product and its maintenance

regime/strategy.

(viii) Availability can be improved through employing higher reliability product and bettermaintenance strategy.

(b) (10 marks)

(b) Combustion Reaction Z

CH4+ 2O2--> CO2 + 2H2O 2

C2H6+ 3.5O2--> 2CO2 + 3H2O 3.5

C7H16+ 11O2--> 7CO2 + 8H2O 11

Mixture 4.0

Chemical Vol LEL

Methane 30% 5.0%

Ethane 20% 3.0%

Heptane 10% 1.0%

F. Mixture 60% 2.6%

Answer: LOC = zLEL 10.6%

Q4.

cm/s K = Ko(Mo/M)1/3

0.43

Trichloroethylene

TLV TWA,ppm

TLV STEL,ppm

MP

sat,

cmHgP

sat,

kPaRg, kPam

3/

kg-mol KWater

MoKo,

cm/s

10 25 131.40 68.52 91.3 8.3143 18 0.83

8/10/2019 ch400113141 answrs

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eqn 3-9 Cppm= (106)QmRgT/(kQvPM)

eqn 3-12 Qm= MKAPsat/(RgTL) 0.0104

eqn 3-14 Cppm= (106)KAPsat/(kQvP) 193

LogP A +B/T + CLogT

A 30.482609B -2936.227

C -7.999975

T oK 298

Log T 2.474216264

LogP 0.835830005

P, cmHg 6.85219959

P, mmHg 68.5219959

m2 A 0.5

cm/s K = Ko(Mo/M)1/3 0.43Answer

(a):kg/s Qm 0.0104

m3/s Qv 80

Answer(b):

eqn 3-9 Cppm= (106)QmRgT/(kQvPM) 193

END