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Chapter_17_Solutions

# Chapter_17_Solutions - Chapter 17 17-1 Given F-1 Polyamide...

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Chapter 17 17-1 Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, H nom = 2 hp, C = 9(12) = 108 in, velocity ratio = 0.5, K s = 1.25, n d = 1 V = d n / 12 = (2)(1750) / 12 = 916.3 ft/min D = d / vel ratio = 2 / 0.5 = 4 in Eq. (17-1): 1 1 4 2 2sin 2sin 3.123 rad 2 2(108) d D d C Table 17-2: t = 0.05 in, d min = 1.0 in, F a = 35 lbf/in, = 0.035 lbf/in 3 , f = 0.5 w = 12 bt = 12(0.035)6(0.05) = 0.126 lbf/ft (a) Eq. ( e ), p. 885: 2 2 0.126 916.3 0.913 lbf . 60 32.17 60 c V F Ans g w 1 2 63 025 63 025(2)(1.25)(1) 90.0 lbf · in 1750 2 2(90.0) 90.0 lbf 2 nom s d a H K n T n T F F F d Table 17-4: C p = 0.70 Eq. (17-12): ( F 1 ) a = bF a C p C v = 6(35)(0 . 70)(1) = 147 lbf Ans. F 2 = ( F 1 ) a [( F 1 ) a F 2 ] = 147 90 = 57 lbf Ans. Do not use Eq. (17-9) because we do not yet know f Eq. ( i ), p. 886: 1 2 147 57 0.913 101.1 lbf . 2 2 a i c F F F F A ns Using Eq. (17-7) solved for f ¢ (see step 8, p.888), 1 2 1 ( ) 1 147 0.913 ln ln 0.307 3.123 57 0.913 a c d c F F f F F   The friction is thus underdeveloped. (b) The transmitted horsepower is, with F = ( F 1 ) a F 2 = 90 lbf, Chapter 17, Page 1/39

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