CHE218_solutions2

CHE218_solutions2 - 2.30 See Example 2.9 t = 1000 lbf in 2...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2.30* See Example 2.9. t = 1000 lbf in 2 ⋅ 1 ft 2 ⋅ 10,000 lbf in 2 = 0. From the larger table in Perr diameter, schedule 80 pipe ha selected. 2.31 The plot below shows al Clearly one should not fit a s You can get R 2 = 0.976 by fi the smaller pipes do not corre If, as shown above, A ≈ 0.136 written t- 0.1336 in = 0.024 the definition of schedule num Sch. 40. The difference is co 0.05 ft = 0.60 in = 1.52 cm ry's equivalent to Appendix A.3 we see that a 12 as a wall thickness of 0.687 inches. It would pro all the values from App. A.2. single equation to that set of data. If one considers only the values for 2.5 inch n diameter and larger, then t = 0.1366 in + 0.02 R 2 = 0.994. Thus, for these pipes, A ≈ 0.1366 inch and B However if one fits the sizes from 1/8 to 2 in nominal, one finds t = 0.0676 in + 0.0550 D in R 2 = 0.803. fitting a quadratic to this data set, showing that th respond very well to the linear model....
View Full Document

This homework help was uploaded on 04/18/2008 for the course CHE 218 taught by Professor Smith during the Spring '08 term at SDSMT.

Page1 / 2

CHE218_solutions2 - 2.30 See Example 2.9 t = 1000 lbf in 2...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online