1_ch 12 Mechanical Design budynas_SM_ch12

1_ch 12 Mechanical Design budynas_SM_ch12 - s = . 5(0 ....

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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 ² = 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 rcNl = 3 . 98(0 . 5)(0 . 000 75)(18 . 33)(1) = 0 . 0274 in 3 /s Q
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Unformatted text preview: s = . 5(0 . 0274) = . 0137 in 3 /s Ans . 12-2 c min = b min d max 2 = 1 . 252 1 . 250 2 = . 001 in r . = 1 . 25 / 2 = . 625 in r / c = . 625 / . 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 = . 585...
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