11_ch 12 Mechanical Design budynas_SM_ch12

# 11_ch 12 Mechanical Design budynas_SM_ch12 - 10(0 . 002 / 1...

This preview shows page 1. Sign up to view the full content.

314 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Proceed with additional trials Trial New T f µ ± S ± TT a v T f 150.0 2.421 0.0921 18.5 119.3 134.6 134.6 3.453 0.1310 23.1 121.5 128.1 128.1 4.070 0.1550 25.8 122.9 125.5 125.5 4.255 0.1650 27.0 123.5 124.5 124.5 4.471 0.1700 27.5 123.8 124.1 124.1 4.515 0.1710 27.7 123.9 124.0 124.0 4.532 0.1720 27.8 123.7 123.9 Note that the convergence begins rapidly. There are ways to speed this, but at this point they would only add complexity. Depending where you stop, you can enter the analysis. (a) µ = 4 . 541(10 6 )reyn , S = 0 . 1724 From Fig. 12-16: h o c = 0 . 482, h o = 0 . 482(0 . 002) = 0 . 000 964 in From Fig. 12-17: φ = 56° Ans . (b) e = c h o = 0 . 002 0 . 000 964 = 0 . 001 04 in Ans . (c) From Fig. 12-18: fr c = 4 . 10, f = 4 .
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 10(0 . 002 / 1 . 25) = . 006 56 Ans . (d) T = f Wr = . 006 56(1200)(1 . 25) = 9 . 84 lbf in H = 2 T N 778(12) = 2 (9 . 84)(1120 / 60) 778(12) = . 124 Btu/s Ans . (e) From Fig. 12-19: Q rcNl = 4 . 16, Q = 4 . 16(1 . 25)(0 . 002) 1120 60 (2 . 5) = . 485 in 3 /s Ans . From Fig. 12-20: Q s Q = . 6, Q s = . 6(0 . 485) = . 291 in 3 /s Ans . (f) From Fig. 12-21: P p max = . 45, p max = 1200 2 . 5 2 (0 . 45) = 427 psi Ans . p max = 16 Ans . (g) p = 82 Ans . (h) T f = 123 . 9F Ans . (i) T s + T = 110F + 27 . 8F = 137 . 8F Ans ....
View Full Document

## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online