p0826 - U p U = 1 M √ M 2 = 1 934 CHAPTER 8...

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8.26. CHAPTER 8, PROBLEM 26 933 8.26 Chapter 8, Problem 26 8.26(a): For isentropic flow, the total pressure is p t = p w 1+ γ 1 2 M 2 W γ / ( γ 1) So, we have p t p ρ = p ρ ^ w 1+ γ 1 2 M 2 W γ / ( γ 1) 1 ± But, we know that a 2 = γ p/ ρ and M = U/a .Thu s , p ρ = 1 γ a 2 = U 2 γ M 2 Thus, p t p ρ = U 2 γ M 2 ^ w 1+ γ 1 2 M 2 W γ / ( γ 1) 1 ± Hence, the velocity measured by a Pitot tube is U p = ² 2( p t p ) ρ = U M > ³ ³ : 2 γ ^ w 1+ γ 1 2 M 2 W γ / ( γ 1) 1 ± Therefore, we conclude that U p U = 1 M > ³ ³ : 2 γ ^ w 1+ γ 1 2 M 2 W γ / ( γ 1) 1 ± 8.26(b): For M U 1 , expanding in Taylor series yields w 1+ γ 1 2 M 2 W γ / ( γ 1) =1+ γ 2 M 2 + ··· wherefore 2 γ ^ w 1+ γ 1 2 M 2 W γ / ( γ 1) 1 ± = 2 γ } 1+ γ 2 M 2 + ··· 1 ] = M 2 +
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Unformatted text preview: U p U = 1 M √ M 2 + · · · = 1 + · · · 934 CHAPTER 8. ONE-DIMENSIONAL COMPRESSIBLE FLOW Therefore, dropping the higher-order terms, we see that U p → U as M → 8.26(c): Using the formula derived in Part (a), the velocity ratio varies with Mach number as follows. M U p /U 0.0 1.000 0.2 1.005 0.4 1.020 0.6 1.046 0.8 1.082 1.0 1.129 Inspection of this table shows that the difference is less than 2% for M ≤ . 4 ....
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p0826 - U p U = 1 M √ M 2 = 1 934 CHAPTER 8...

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