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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....
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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.
 Spring '08
 Smith

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