p0832 - 8.32. CHAPTER 8, PROBLEM 32 941 8.32 Chapter 8,...

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8.32. CHAPTER 8, PROBLEM 32 941 8.32 Chapter 8, Problem 32 M 1 =1 . 5 p 2 =1a tm ............................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 8.2: Supersonic flow past a blunt object. For air , we can use the tables. So, note first that p t = p t p 1 p 1 p 2 p 2 = p 2 ( p 1 /p t )( p 2 /p 1 ) From the isentropic-flow tables, we find p 1 /p t =0 . 2724 . From the normal-shock tables, p 2 /p 1 =2 . 4583 .Thu s , p t = 1atm (0 . 2724) (2 . 4583) =1 . 49 atm For carbon dioxide , γ =1 . 30 W =1 . 4 ,soth a tw ec anno
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