21_ch 16 Mechanical Design budynas_SM_ch16

21_ch 16 Mechanical Design budynas_SM_ch16 - + 2690 . 4 lbf...

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416 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is T L = ± ± ± ± 1300(12) 10 ± ± ± ± = 1560 lbf · in The rated motor torque T r is T r = 63 025(3) 1125 = 168 . 07 lbf · in For Eqs. (16-65): ω r = 2 π 60 (1125) = 117 . 81 rad/s ω s = 2 π 60 (1200) = 125 . 66 rad/s a = T r ω s ω r =− 168 . 07 125 . 66 117 . 81 =− 21 . 41 b = T r ω s ω s ω r = 168 . 07(125 . 66) 125 . 66 117 . 81 = 2690 . 4 lbf · in The linear portion of the squirrel-cage motor characteristic can now be expressed as T M =− 21 . 41 ω
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Unformatted text preview: + 2690 . 4 lbf · in Eq. (16-68): T 2 = 168 . 07 ² 1560 − 168 . 07 1560 − T 2 ³ 19 One root is 168.07 which is for infinite time. The root for 10 s is wanted. Use a successive substitution method T 2 New T 2 0.00 19.30 19.30 24.40 24.40 26.00 26.00 26.50 26.50 26.67 Continue until convergence. T 2 = 26 . 771 Eq. (16-69): I = − 21 . 41(10 − . 5) ln(26 . 771 / 168 . 07) = 110 . 72 in · lbf · s/rad ω = T − b a...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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