Chapter_17_Solutions

Chapter_17_Solutions - Chapter 17 17-1 Given: F-1...

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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): 11 42 2sin 3.123 rad 2 2(108) d Dd 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: 22 0.126 916.3 0.913 lbf . 60 32.17 60 c V F Ans g     w  12 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 HK n T n T FF 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: 147 57 0.913 101.1 lbf . a ic F FA n s 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 ac dc 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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Eq. (j), p. 887: ( ) 90(916.3) 2.5 hp . 33 000 33 000 FV H Ans  nom 2.5 1 2(1.25) fs s H n HK  Eq. (17-1): 11 42 2sin 3.160 rad 2 2(108) D Dd C       Eq. (17-2): L = [4 C 2 ( D d ) 2 ] 1/2 + ( D D + d d )/2 = [4(108) 2 (4 2) 2 ] 1/2 + [4(3.160) + 2(3.123)]/2 = 225 . 4 in Ans. (c) Eq. (17-13): 22 3 3(108 / 12) (0.126) dip 0.151 in . 2 2(101.1) i C Ans F w Comment : The solution of the problem is finished; however, a note concerning the design is presented here. The friction is under-developed. Narrowing the belt width to 5 in (if size is available) will increase f . The limit of narrowing is b min = 4 . 680 in, whence 1 2 12 0.0983 lbf/ft ( ) 114.7 lbf 0.713 lbf 24.7 lbf 90 lbf · in (same) 0.50 ( ) 90 lbf a c a F FF Tf F    w dip 0.173 in 68.9 lbf i F f Longer life can be obtained with a 6-inch wide belt by reducing F i to attain Prob. 17-8 develops an equation we can use here 0.50. f 1 21 1 2 2 ( ) exp( ) exp( ) 1 2 1 ln 3 dip 2 cc ic c dc i f F F f F f C F        w which in this case, d = 3.123 rad, exp( f ) = exp[0.5(3.123)] = 4.766, w = 0.126 lbf/ft, F = 90.0 lbf, F c = 0.913 lbf, and gives Chapter 17, Page 2/39
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 1 0.913 90 4.766 0.913 114.8 lbf 4.766 1 F   F 2 = 114.8 90 = 24.8 lbf F i = (114.8 + 24.8)/ 2 0.913 = 68.9 lbf 1 114.8 0.913 ln 0.50 3.123 24.8 0.913 f    2 3 108 / 12 0.126 dip 0.222 in 2(68.9) So, reducing F i from 101.1 lbf to 68.9 lbf will bring the undeveloped friction up to 0.50, with a corresponding dip of 0.222 in. Having reduced F 1 and F 2 , the endurance of the belt is improved. Power, service factor and design factor have remained intact. ______________________________________________________________________________ 17-2 Double the dimensions of Prob. 17-1. In Prob. 17-1, F-1 Polyamide was used with a thickness of 0.05 in. With what is available in Table 17-2 we will select the Polyamide A-2 belt with a thickness of 0.11 in. Also, let b = 12 in, d = 4 in with n = 1750 rev/min, H nom = 2 hp, C = 18(12) = 216 in, velocity ratio = 0.5, K s = 1.25, n d = 1.
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This note was uploaded on 11/20/2011 for the course MCE 325 taught by Professor Ahm during the Spring '11 term at American Dubai.

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Chapter_17_Solutions - Chapter 17 17-1 Given: F-1...

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