10_ch 12 Mechanical Design budynas_SM_ch12

10_ch 12 Mechanical Design budynas_SM_ch12 - For a trial...

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Chapter 12 313 From Fig. 12-24, 0 . 120 ± T P = 0 . 349 + 6 . 009(0 . 358) + 0 . 0475(0 . 358) 2 = 2 . 5 ± T = 2 . 5 ± 0 . 469 0 . 120 ² = 9 . 8°C Discrepancy = 10°C 9 . 8°C = 0 . 2°C O.K. T a v = 65°C Ans . T 1 = T a v ± T / 2 = 65°C (10°C / 2) = 60°C T 2 = T a v + ± T / 2 = 65°C + (10°C / 2) = 70°C S = 0 . 358 From Figures 12-16, 12-18, 12-19 and 12-20: h o c = 0 . 68, fr / c = 7 . 5, Q rcN l = 3 . 8, Q s Q = 0 . 44 h o = 0 . 68(0 . 04) = 0 . 0272 mm Ans . f = 7 . 5 1000 = 0 . 0075 T = fWr = 0 . 0075(3)(40) = 0 . 9N · m H = 2 π TN = 2 π (0 . 9)(8) = 45 . 2W Ans . Q = 3 . 8(40)(0 . 04)(8)(80) = 3891 mm 3 /s Q s = 0 . 44(3891) = 1712 mm 3 /s Ans . 12-12 Given: d = 2 . 5in, b = 2 . 504 in, c min = 0 . 002 in, W = 1200 lbf,SAE = 20, T s = 110°F, N = 1120 rev/min, and l = 2 . 5in.
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Unformatted text preview: For a trial film temperature T f = 150°F T f µ ± S ± T (From Fig. 12-24) 150 2.421 0.0921 18.5 T a v = T s + ± T 2 = 110°F + 18 . 5°F 2 = 119 . 3°F T f − T a v = 150°F − 119 . 3°F which is not 0.1 or less, therefore try averaging ( T f ) new = 150°F + 119 . 3°F 2 = 134 . 6°F...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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