HW5sol - ME 363 Homework#5 1(a h0 h f = 2R First roll = 2...

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ME 363 Homework #5 1. (a) 2 0 hh R 21 . 5 First roll: = 0.183 15 1.5 1.5 0.75 Second roll: = 0.158 15 1.125 1.125 0.75 Third roll: = 0.137 15 minimum 0.183 f μ −= = −× = = ∴= (b) () 00 011 122 233 000 333 30 3 3 33 0 3 3 0.75 0.75 0.75 1.03 12 26.1 0.75 1.03 40 86.8 ft min 0.46 0.75 1.03 VwtVwtVwtVwt wtl wtl tt ww wt ll i n v v ×× = =×× ××= ×× = × = = × × == = × 3 1
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(c) () 01 0 0 0 0.08 - 15 0.5 2.74 0.75 1.03 1.147 45.9 / min 22 1 45.9 12 / 60 1 ln ln 0.964 0.75 2.74 0.75 24,000(0.964) 23,929 2 ' 23,929 27,631 3 '1 2 2.74 10 2 r r m avg avg ev LR h h i n V V VV V L Cp Y L FLWY h ε σε μ == × = + + × = = × = =× = = ⎛⎞ =× × + ⎜⎟ ⎝⎠ × ± ± f t s i 3 0.183 2.74 7,631 1 865 10 21 . 7 5 lbs + × += × × 2
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2. (a) () 22 00 2 0 0 1 0 0.2 1 1 6- 1 180 a n=0.2 π 44 1 2 0.5 . 2 0.5 ln ln -1.386 2 180 1.386 192 0.3 2 1 192 1 3 3 0.5 269 269 10 2 25.4 10 545 ff f f n f f dh d hh i n d h h a d a h a Fa d π ε σε μ σ ππ Κ= ΜΡ = = = == = = = ⎛⎞ × Ρ= + = + ⎜⎟ × ⎝⎠ Ρ =Ρ × = × × × × Ν 6 (b) 0.07 2 500 39.37 0.5 25.4 80 39.37 103.4 0.3 2 103.4 1 144.8 30 . 5 144.8 10 2 25.4 10 4 293 m f a h CM PM F KN 6 P a P a υ + = × =× = = × =+= × × × × × = ± ± 3
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3. () 0 00 2 0 33 01 20 3 3 100 20 100 ln 2ln 3.22 20 6t a n - 6 200 (100) tan30 3.22 100 - 20 22.49 f ff dm m m d d d dd ε υα εε == Α = = Α ×× = ± 0 0.1 0 0 0 2 6- 6 @900 0.1 150 150 (22.49) 204 2 1.7ln 2 100 204 1.7 3.22 100 1,524 1,524 10 100 10 4 12 Neither is sufficient. One must reduced the speed by e e Cm C M P a MPa L R D MPa FA MN
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This note was uploaded on 12/25/2011 for the course ME 300363 taught by Professor Staff during the Fall '10 term at Purdue.

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HW5sol - ME 363 Homework#5 1(a h0 h f = 2R First roll = 2...

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