4_ch 12 Mechanical Design budynas_SM_ch12

4_ch 12 Mechanical Design budynas_SM_ch12 - P = 1250 25 2 =...

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Chapter 12 307 12-5 c min = b min d max 2 = 2 . 0024 2 2 = 0 . 0012 in r . = d 2 = 2 2 = 1in , l / d = 1 / 2 = 0 . 50 r / c = 1 / 0 . 0012 = 833 N = 800 / 60 = 13 . 33 rev/s P = 600 2(1) = 300 psi Fig. 12-12: SAE 20, µ ± = 3 . 75 µ reyn S = (833 2 ) ± 3 . 75(10 6 )(13 . 3) 300 ² = 0 . 115 From Figs. 12-16, 12-18 and 12-19: h o / c = 0 . 23, rf / c = 3 . 8, Q / ( rcNl ) = 5 . 3 h o = 0 . 23(0 . 0012) = 0 . 000 276 in Ans . f = 3 . 8 833 = 0 . 004 56 The power loss due to friction is H = 2 π fWrN 778(12) = 2 π (0 . 004 56)(600)(1)(13 . 33) 778(12) = 0 . 0245 Btu/s Ans . Q = 5 . 3 rcNl = 5 . 3(1)(0 . 0012)(13 . 33)(1) = 0 . 0848 in 3 /s Ans. 12-6 c min = b min d max 2 = 25 . 04 25 2 = 0 . 02 mm r ˙= d / 2 = 25 / 2 = 12 . 5mm , l / d = 1 r / c = 12 . 5 / 0 . 02 = 625 N = 1200 / 60
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Unformatted text preview: P = 1250 25 2 = 2 MPa For µ = 50 mPa · s, S = (625 2 ) ± 50(10 − 3 )(20) 2(10 6 ) ² = . 195 From Figs. 12-16, 12-18 and 12-20: h o / c = . 52, f r / c = 4 . 5, Q s / Q = . 57 h o = . 52(0 . 02) = . 0104 mm Ans . f = 4 . 5 625 = . 0072 T = f Wr = . 0072(1 . 25)(12 . 5) = . 1125 N · m...
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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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