shi20396_ch12-1 - Chapter 12 12-1 Given dmax = 1.000 in and...

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Chapter 12 12-1 Given d max = 1 . 000 in and b min = 1 . 0015 in, the minimum radial clearance is c min = b min d max 2 = 1 . 0015 1 . 000 2 = 0 . 000 75 in Also l / d = 1 r ˙= 1 . 000 / 2 = 0 . 500 r / c = 0 . 500 / 0 . 000 75 = 667 N = 1100 / 60 = 18 . 33 rev/s P = W / ( ld ) = 250 / [(1)(1)] = 250 psi Eq. (12-7): S = (667 2 ) · 8(10 6 )(18 . 33) 250 ¸ = 0 . 261 Fig. 12-16: h 0 / c = 0 . 595 Fig. 12-19: Q / ( rcNl ) = 3 . 98 Fig. 12-18: fr / c = 5 . 8 Fig. 12-20: Q s / Q = 0 . 5 h 0 = 0 . 595(0 . 000 75) = 0 . 000 466 in Ans . f = 5 . 8 r / c = 5 . 8 667 = 0 . 0087 The power loss in Btu/s is H = 2 π fWrN 778(12) = 2 π (0 . 0087)(250)(0 . 5)(18 . 33) 778(12) = 0 . 0134 Btu/s Ans . Q = 3 . 98 = 3 . 98(0 . 5)(0 . 000 75)(18 . 33)(1) = 0 . 0274 in 3 /s Q s = 0 . 5(0 . 0274) = 0 . 0137 in 3 /s Ans . 12-2 c min = b min d max 2 = 1 . 252 1 . 250 2 = 0 . 001 in r . = 1 . 25 / 2 = 0 . 625 in r / c = 0 . 625 / 0 . 001 = 625 N = 1150 / 60 = 19 . 167 rev/s P = 400 1 . 25(2 . 5) = 128 psi l / d = 2 . 5 / 1 . 25 = 2 S = (625 2 )(10)(10 6 )(19 . 167) 128 = 0 . 585
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Chapter 12 313 The interpolation formula of Eq. (12-16) will have to be used. From Figs. 12-16, 12-21, and 12-19 For l / d =∞ , h o / c = 0 . 96, P / p max = 0 . 84, Q rcNl = 3 . 09 l / d = 1, h o / c = 0 . 77, P / p max = 0 . 52, Q = 3 . 6 l / d = 1 2 , h o / c = 0 . 54, P / p max = 0 . 42, Q = 4 . 4 l / d = 1 4 , h o / c = 0 . 31, P / p max = 0 . 28, Q = 5 . 25 Equation (12-16) is easily programmed by code or by using a spreadsheet. The results are: l / dy y 1 y 1 / 2 y 1 / 4 y l / d h o / c 2 0.96 0.77 0.54 0.31 0.88 P / p max 2 0.84 0.52 0.42 0.28 0.64 Q / 2 3.09 3.60 4.40 5.25 3.28 h o = 0 . 88(0 . 001) = 0 . 000 88 in Ans . p max = 128 0 . 64 = 200 psi Ans . Q = 3 . 28(0 . 625)(0 . 001)(19 . 167)(2 . 5) = 0 . 098 in 3 /s Ans . 12-3 c min = b min d max 2 = 3 . 005 3 . 000 2 = 0 . 0025 in r . = 3 . 000 / 2 = 1 . 500 in l / d = 1 . 5 / 3 = 0 . 5 r / c = 1 . 5 / 0 . 0025 = 600 N = 600 / 60 = 10 rev/s P = 800 1 . 5(3) = 177 . 78 psi Fig. 12-12: SAE 10, µ 0 = 1 . 75 µ reyn S = (600 2 ) · 1 . 75(10 6 )(10) 177 . 78 ¸ = 0 . 0354 Figs. 12-16 and 12-21: h o / c = 0 . 11, P / p max = 0 . 21 h o = 0 . 11(0 . 0025) = 0 . 000 275 in Ans . p max = 177 . 78 / 0 . 21 = 847 psi Ans .
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314 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fig. 12-12: SAE 40, µ 0 = 4 . 5 µ reyn S = 0 . 0354 µ 4 . 5 1 . 75 = 0 . 0910 h o / c = 0 . 19, P / p max = 0 . 275 h o = 0 . 19(0 . 0025) = 0 . 000 475 in Ans . p max = 177 . 78 / 0 . 275 = 646 psi Ans . 12-4 c min = b min d max 2 = 3 . 006 3 . 000 2 = 0 . 003 r . = 3 . 000 / 2 = 1 . 5in l / d = 1 r / c = 1 . 5 / 0 . 003 = 500 N = 750 / 60 = 12 . 5rev/s P = 600 3(3) = 66 . 7 psi Fig. 12-14: SAE 10W, µ 0 = 2 . 1 µ reyn S = (500 2 ) · 2 . 1(10 6 )(12 . 5) 66 . 7 ¸ = 0 . 0984 From Figs. 12-16 and 12-21: h o / c = 0 . 34, P / p max = 0 . 395 h o = 0 . 34(0 . 003) = 0 . 001 020 in Ans . p max = 66 . 7 0 . 395 = 169 psi Ans . Fig. 12-14: SAE 20W-40, µ 0 = 5 . 05 µ reyn S = (500 2 ) · 5 . 05(10 6 . 5) 66 . 7 ¸ = 0 . 237 From Figs. 12-16 and 12-21: h o / c = 0 . 57, P / p max = 0 . 47 h o = 0 . 57(0 . 003) = 0 . 001 71 in Ans . p max = 66 . 7 0 . 47 = 142 psi Ans .
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Chapter 12 315 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 0 = 3 . 75 µ reyn S = (833 2 ) · 3 . 75(10 6 )(13 . 3) 300 ¸ = 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 = 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 = 20 rev/s P = 1250 25 2 = 2MPa For µ = 50 MPa · s, S = (625 2 ) · 50(10 3 )(20) 2(10 6 ) ¸ = 0 . 195 From Figs. 12-16, 12-18 and 12-20: h o / c = 0 . 52, fr / c = 4 . 5, Q s / Q = 0 . 57 h o = 0 . 52(0 . 02) = 0 . 0104 mm Ans . f = 4 . 5 625 = 0 . 0072 T = fWr = 0 . 0072(1 . 25)(12 . 5) = 0 . 1125 N · m
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316 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The power loss due to friction is H = 2 π TN = 2 π (0 . 1125)(20) = 14 . 14 W Ans .
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This note was uploaded on 04/30/2008 for the course MAE 4900 taught by Professor Khanna during the Fall '06 term at Missouri (Mizzou).

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shi20396_ch12-1 - Chapter 12 12-1 Given dmax = 1.000 in and...

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