15_ch 15 Mechanical Design budynas_SM_ch15

# 15_ch 15 Mechanical Design budynas_SM_ch15 -...

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Chapter 15 393 15-14 N W = 1, N G = 56, P t = 8 teeth/in, d = 1 . 5in , H o = 1hp, φ n = 20 , t a = 70 F, K a = 1 . 25, n d = 1, F e = 2in , A = 850 in 2 (a) m G = N G / N W = 56, D = N G / P t = 56 / 8 = 7 . 0in p x = π/ 8 = 0 . 3927 in, C = 1 . 5 + 7 = 8 . 5in Eq. (15-39): a = p x = 0 . 3927 = 0 . 125 in Eq. (15-40): b = 0 . 3683 p x = 0 . 1446 in Eq. (15-41): h t = 0 . 6866 p x = 0 . 2696 in Eq. (15-42): d o = 1 . 5 + 2(0 . 125) = 1 . 75 in Eq. (15-43): d r = 3 2(0 . 1446) = 2 . 711 in Eq. (15-44): D t = 7 + 2(0 . 125) = 7 . 25 in Eq. (15-45): D r = 7 2(0 . 1446) = 6 . 711 in Eq. (15-46): c = 0 . 1446 0 . 125 = 0 . 0196 in Eq. (15-47): ( F W ) max = 2 ± 2(7)0 . 125 = 2 . 646 in V W = π (1 . 5)(1725 / 12) = 677 . 4ft/min V G = π (7)(1725 / 56) 12 = 56 . 45 ft/min Eq. (13-28): L = p x N W = 0 . 3927 in, λ = tan 1 ² 0 . 3927 π (1 . 5) ³ = 4 . 764 P n = P t cos λ = 8 cos 4 . 764° = 8 . 028 p n = π P n = 0 . 3913 in Eq. (15-62): V s = π (1 . 5)(1725) 12 cos 4 . 764° = 679 . 8ft/min (b) Eq. (15-38): f = 0 . 103 exp ´ 0 . 110(679 . 8) 0 . 450 µ + 0 . 012 = 0 . 0250 Eq. (15-54): The efﬁciency is, e =
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