shi20396_ch12-1

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

This preview shows pages 1–6. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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).

### Page1 / 21

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online