FInal - Irtiza

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New England College *

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5330

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Mechanical Engineering

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Dec 6, 2023

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DoE Online course. MECH 5750 Final Exam. – Quiz Option (A-E). Name_________________ Please answer questions Quiz version A, B, C, D or E depending on your student ID last (end) digit 1, 2 answer Quiz A | 3 answer Quiz B | 4, 5, 6 answer Quiz C | 7, 8 answer Quiz D | 9, 0 answer Quiz E 1. If you start with the bicycle hill climb problem in week 9 assignment, factor 6 was folded over to remove the confounding of other factors to factor 6, as shown in the notes. Note you do not have to calculate any values just note down the resulting interactions. Problem 1, Part 1. Please only answer one version depending on your last student ID digit (use right chart above showing the isolated interaction[s]). Please only write down the confounding interactions (no calculations values are required). A. If the DoE was folded with two factors at the same time (2 and 6), what is the confounding (aliasing) table? Average (1+1’) = Average (1-1’) = ______________ Average (2+2’) = Average (2-2’) = ______________ Average (3+3’) = Average (3-3’) = _______________ Average (4 +4’) = Average (4-4’) = _______________ Average (5+5’) = Average (5-5’) = _______________ Average (6+6’) = Average (6-6’) = ________________ Average (7+7’) = Average (7-7’) = ________________

Problem 1 Part 2 . All students Please answer this question for all versions (A, B, C, D, E) . If the DoE design team folded all factors at the same time (1-7), what is the confounding (aliasing) table? Show all factors and all interactions confounding (aliasing). Average (1+1’) = (2*3), (4*5), (6*7) Average (1-1’) = No confounding Average (2+2’) = (1*3), (4*6), (5*7) Average (2-2’) = No confounding Average (3+3’) = (1*2), (4*7), (5*6) Average (3-3’) = No confounding Average (4 +4’) = (1*5), (2*6), (3*7) Average (4-4’) = No confounding Average (5+5’) = (1*4), (2*7), (3*6) Average (5-5’) = No confounding Average (6+6’) = (1*7), (2*4), (3*5) Average (6-6’) = No confounding Average (7+7’) = (1*6), (2*5), (3*4) Average (7-7’) = No confounding Problem 2. (All Students answer All questions a – j) In an industrial waste treatment plant, metals are precipitated by mixing lime in a tank to produce metal hydroxides. Design an experiment to maximize the metal precipitation @Five factors at two levels 2 lime flows [FLOW] (10, 20 L/m) 2 mixing speeds [SPEED] (20 or 25 rpm) 2 tank levels [LVL] (50% or 100%) 2 mixing Times [MIX] (10, 20 rpm) 2 Effluent Temperatures [TEMP] (ambient, 50’F) Select best fit/smallest orthogonal array experiments and show column assignments (in numbers) only: a. Considering all interactions for the five factors at two levels: Array: L32 , Factor Assignments: Flow 1 , Speed 2 , LVL 4 , MIX 8 , TEMP 16 b. Ignore all interactions for the five factors at two levels: Array: L8 , Factor Assignments: Flow 1 , Speed 2 , LVL 4 , MIX 5 , TEMP 3 c. Considering 2 way interactions only among the five factors and ignoring all other interactions: Array: L16 Factor Assignments: Flow 1 , Speed 2 , LVL 4 , MIX 8 , TEMP 15 d. Considering only these two (two–way interactions) for the five factors, and ignoring all other interactions. Answer your versions only (A, B C D or E – Please circle your letter below ) A. [TEMP] x [MIX] and [TEMP] x [FLOW] B. SPEED x TEMP and SPEED x FLOW C. FLOW x MIX and FLOW x LVL D. MIX x LVL and Mix x TEMP E. TEMP x FLOW and TEMP x SPEED Array: L8 Factor Assignments: Flow: 4 , Speed: 1 , LVL: 6 , MIX: 7 , TEMP: 2

e. Considering only four factors at two levels: Flow, Speed, LVL and MIX, and consider all interactions Array: L16 , Factor Assignments: Flow: 1 , Speed: 2 , LVL: 4 , MIX: 8 f. Considering only four factors at two levels Flow, Speed, LVL and MIX, and ignore all interactions Array: L8 Factor Assignments: Flow 1 , Speed 2 , LVL 3 , MIX 4 g. Considering only four factors at two levels: Flow, Speed, LVL and MIX, and consider only two-way interactions while ignoring all others Array: L12 Factor Assignments: Flow 1 , Speed 2 , LVL 4 , MIX 7 h. Considering only three factors at two levels: Flow, Speed and LVL, and consider all interactions Array: L8 Factor Assignments: Flow 1 , Speed 2 , LVL 4 i. Considering only three factors at two levels: Flow, Speed and LVL, and ignore all interactions Array: L4 Factor Assignments: Flow 1 , Speed 2 , LVL 3 j. Considering only three factors at two levels: Flow, Speed and LVL, and consider only two- way interactions while ignoring all others Array: L8 Factor Assignments: Flow 1 , Speed 2 , LVL 4 Problem 3. (All Students answer All questions a – k). In an industrial waste treatment plant, metals are precipitated by mixing lime in a tank to produce metal hydroxides. Design an experiment to maximize the metal precipitation: Five factors at three levels. 3 lime flows [FLOW] (10, 20 or 30 L/m) 3 mixing speeds [SPEED] (20 or 25 or 35 rpm) 3 tank levels [LVL] (50% 75% or 100%) 3 mixing Times [MIX] (10, 20 or 30rpm) 3 Effluent Temperatures [TEMP] (ambient, 25 or 50’F) Select best fit/smallest orthogonal array experiments and show column assignments (in numbers) only: a Considering all 5 factors have three levels and ignore all interactions: Array: L27 Factor Assignments: Flow 1 , Speed 2 , LVL 3 , MIX 7 , TEMP 5 b. Considering all five factors with three levels and selecting 2 way interactions only among three selected factors (Flow Speed and LVL), while ignoring all other interactions: Array: L27 Factor Assignments: Flow 1 , Speed 2 , LVL 3 , MIX 7 , TEMP 8

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