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p0874 - M 2 = 0 7815 992 CHAPTER 8 ONE-DIMENSIONAL...

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8.74. CHAPTER 8, PROBLEM 74 991 8.74 Chapter 8, Problem 74 Figure 8.20: Flow through a duct of varying area. Station 1: We can solve this problem in terms of the area ratio. First, noting that M 1 = 1 . 7 , we can use the isentropic-flow tables to determine the area ratio at Station 1. That is, M 1 = 1 . 7 = A 1 A = 1 . 3376 Station 2: From the geometry, we can compute the area ratio just ahead of the shock at Station 2. Thus, A 2 A = A 2 A 1 A 1 A = 4 5 (1 . 3376) = 1 . 0701 and the isentropic-flow tables yield A/A M 1.0663 1.300 1.0701 1.309 1.0750 1.320 where the central row follows from linear interpolation. Hence, the Mach number ahead of the shock is M 1 = 1 . 309 Now, turning to the normal-shock tables, the Mach number behind the shock is M 1 M 2 1.300 0.7860 1.309 0.7815 1.320 0.7760 where linear interpolation has been used to arrive at
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Unformatted text preview: M 2 = 0 . 7815 . 992 CHAPTER 8. ONE-DIMENSIONAL COMPRESSIBLE FLOW Using the isentropic-flow tables again to compute the area ratio behind the shock, we find M 2 A 2 /A ∗ 0.7800 1.0471 0.7815 1.0464 0.8000 1.0382 Therefore, the area ratio is A 2 A ∗ = 1 . 0464 Station 3: Finally, from the geometry, the area ratio at Station 3 is A 3 A ∗ = A 3 A 2 A 2 A ∗ = 5 4 (1 . 0464) = 1 . 3080 Using the isentropic-flow tables we have A 3 /A ∗ M 3 1.3398 0.500 1.3080 0.517 1.3034 0.520 where we again use linear interpolation to arrive at the results in the central row. Therefore, the Mach number at Station 3 is M 3 . = 0 . 52...
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