Explanation Write the expression for the speed of the pump. m G = ω M ω P (I) Here, the angular velocity of the induction motor is ω M and the angular velocity of the pump is ω P . Write the expression for the tooth count of the sprocket. N 2 = m G N 1 (II) Here, the number of teeth for the driving sprocket is N 1 . Wite the expression for the design power. H d = ( H n o m ) K s n d Here, power is H n o m , service factor is K s and the design factor is n d . Write the expression for the allowable power. H a = ( K 1 K 2 ) H t a b Here, tabular power is H t a b . Write the expression for the tabulated power. H t a b = K s × n d × H n o m K 1 K 2 (III) Write the expression for the factor of safety. n f s = H t a b ( K 1 K 2 ) H n o m ( K s ) (IV) Write the expression for the approximate length of the chain in pitches. L p = 2 C p + N 1 + N 2 2 + ( N 2 − N 1 ) 4 π 2 ( C / p ) (V) Write the expression for the area. A = N 1 + N 2 2 − L p (VI) Write the expression for the center to center distance. C = p 4 [ − A + A 2 − 8 ( N 2 − N 1 2 π ) ] (VII) Conclusion: Substitute 700 rev / min for ω M and 140 rev / min for ω P in Equation (I). m G = 700 rev / min 140 rev / min = 700 140 = 5 Consider that the number of teeth in the driving sprocket ( N 1 ) is 17 . Substitute 17 for N 1 and 5 for m G in Equation (II). N 2 = 5 ( 17 ) = ( 5 × 17 ) = 85 ≈ 84 Refer to Table 17-22 “Tooth correction factors, K 1 ” to obtain tooth correction factor ( K 1 ) as 1 corresponding to the number of tooth in the driving socket ( N 1 ) as 17 . Refer to Table 17-23 “Multiple strand factors, K 2 ” to obtain multiple strand factors ( K 2 ) as 1 , ( K 2 ) as 1.7 , ( K 2 ) as 2.5 and ( K 2 ) as 3.3 corresponding to the number of strands 1 , 2 , 3 and 4 . For the condition of tooth correction factor ( K 1 ) as 1 . H d = H a (VIII) Substitute ( H n o m ) K s n d for H d and ( K 1 K 2 ) H t a b for H a in Equation (VIII). ( H n o m ) K s n d = ( K 1 K 2 ) H t a b H t a b ( K 1 K 2 ) = ( H n o m ) K s n d H t a b = K s × n d × H n o m K 1 K 2 Substitute 1.5 for K s , 1.1 for n d , 25 hp for H n o m , 1 for K 1 and 1 for K 2 in Equation (III). H t a b = 1.5 × 1.1 × 25 hp ( 1 ) ( 1 ) = 1.65 × 25 hp 1 = 41.3 hp Substitute 1.5 for K s , 1.1 for n d , 25 hp for H n o m , 1 for K 1 and 1.7 for K 2 in Equation (III). H t a b = 1.5 × 1.1 × 25 hp ( 1 ) ( 1.7 ) = 1

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

10th Edition

ISBN: 9780073398204

Author: Richard G Budynas, Keith J Nisbett

Publisher: McGraw-Hill Education

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Chapter 17, Problem 27P

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Chapter 17 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 17 - Example 17-2 resulted in selection of a 10-in-wide...Ch. 17 - A 6-in-widc polyamide F-l flat belt is used to...Ch. 17 - Prob. 3P Ch. 17 - A flat-belt drive is to consist of two...Ch. 17 - In solving problems and examining examples, you...Ch. 17 - Return to Ex. 17-1 and complete the following. (a)...Ch. 17 - Prob. 7P Ch. 17 - Prob. 8P Ch. 17 - Prob. 9P Ch. 17 - Two shafts 20 ft apart, with axes in the same...

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