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... = 1/1.28× 10−4= 7812.5 hours MTTFs= MTTFc/4=1953.125 hours 3. A space vehicle requires three out four ... 1 Reliability of Technical Systems Tutorial # 3 Solution 1. Consider the probability density ... = 1/1.28× 10−4= 7812.5 hours MTTFs= MTTFc/4=1953.125 hours 3. A space vehicle requires three out four ... 1 Reliability of Technical Systems Tutorial # 3 Solution 1. Consider the probability density ... 1 ...
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... ash on a landfill for stabilized residues: 2 2 260'295.9) 5 2 1(%2000'110 COkg t COkg t ... landfilling the fly ash on a landfill for stabilized residues: t COkg t COkg 22 3.135.9) 5 2 1( total: t ... ash on a landfill for stabilized residues: 2 2 260'295.9) 5 2 1(%2000'110 COkg t COkg t ... landfilling the fly ash on a landfill for stabilized residues: t COkg t COkg 22 3.135.9) 5 2 1( total: t ... 1 ...
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... a multi-output process, which produces 10 kg of product A and 5 kg of product B per minute. 1 kg of ... product A is worth 20 CHF, 1 kg of B is worth 5 CHF. The process emits 120 kg CO2 per hour. Product B can ... a multi-output process, which produces 10 kg of product A and 5 kg of product B per minute. 1 kg of ... product A is worth 20 CHF, 1 kg of B is worth 5 CHF. The process emits 120 kg CO2 per hour. Product B can ... 1 ...
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... (flammable, non-flammable and poisonous) Class 3 Flammable liquids Class 4 Flammable solids Class 5 Oxidizing ... 1 www.lsa.ethz.ch/education/vorl Methods of Technical Risk Assessment in a Regional Context Methods ... (flammable, non-flammable and poisonous) Class 3 Flammable liquids Class 4 Flammable solids Class 5 Oxidizing ... 1 ... 1 ...
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... 1 Reliability of Technical Systems Tutorial # 3 Due : October 12, 2010 1. Consider the probability ... density function. ≥ = − otherwise te tf t 0 0 002.0 )( 002.0 for t is in hours Calculate reliability ... 1 Reliability of Technical Systems Tutorial # 3 Due : October 12, 2010 1. Consider the probability ... 1 ... 1 ...
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... the system, while the failure likelihood of the components is given qi = 0.01, i = 1, 2, 3, 4. Q2 ... qj = 0.01, j = 1, 2, 3, 4. ... the system, while the failure likelihood of the components is given qi = 0.01, i = 1, 2, 3, 4. Q2 ... qj = 0.01, j = 1, 2, 3, 4. ... 1 ...
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... [ 1/hour] • Failure rate of the spare tire: 𝜆𝑆= 10- 3 [ 1/hour] • Repair duration (time to install the ... tire (you may assume repair rate is about 6⋅ 10- 3 ( 1/hour) ). In this case, what is the probability that ... [ 1/hour] • Failure rate of the spare tire: 𝜆𝑆= 10- 3 [ 1/hour] • Repair duration (time to install the ... tire (you may assume repair rate is about 6⋅ 10- 3 ( 1/hour) ). In this case, what is the probability that ... 1 ...
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... 1 1 Reliability of Technical Systems Tutorial # 5 (FTA Cont) Due: October 26th, 2010 The cooling ... : 9.999E- 5 Transducer: 5E- 5 Electric Valve: 4.182E- 3 Electric Pump: 7.937E- 3 Power supply: 1E-4 Backup ... 1 1 Reliability of Technical Systems Tutorial # 5 (FTA Cont) Due: October 26th, 2010 The cooling ... : 9.999E- 5 Transducer: 5E- 5 Electric Valve: 4.182E- 3 Electric Pump: 7.937E- 3 Power supply: 1E-4 Backup ... 1 ...
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... 1 1 Reliability of Technical Systems Tutorial #4 (FTA) Due: October 19th, 2010 The cooling system ... 1 ... 1 1 Reliability of Technical Systems Tutorial #4 (FTA) Due: October 19th, 2010 The cooling system ... 1 ...
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... 1 1 Reliability of Technical Systems Tutorial #8 Due: November 23th, 2010 Q1) Below is a system ... distributed with sane CFR. The failure rate for C1 and C2 is 0.000253 ( 1/hour). Component C3 is also ... 1 ... 1 1 Reliability of Technical Systems Tutorial #8 Due: November 23th, 2010 Q1) Below is a system ... 1 ...
