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Unformatted text preview: (a) Since the ball diameter is the same for model and prototype, necessarily ω m D m U m = ω p D p U p = ⇒ ω m = ω p U m U p ρ m U m D m μ m = ρ p U p D p μ p = ⇒ μ m = μ p U m U p We are given U m = 140 ft/sec and U p = 210 ft/sec. Thus, U m U p = 140 ft / sec 210 ft / sec = 2 3 Hence, since ω p = 24 sec − 1 , we conclude that ω m = 16 sec − 1 and μ m = 2 3 μ p 2.84. CHAPTER 2, PROBLEM 84 283 (b) If we assume the viscosity follows a powerlaw, then the viscosity ratio is μ m μ p = X T m T p ~ . 68 = ⇒ T m T p = X μ m μ p ~ 1 . 47 Hence, for the conditions of Part (a), we must have T m = w 2 3 W 1 . 47 T p = 0 . 55 T p We are given T p = 75 o F = 534.67 o R. Therefore, T m = 0 . 55(534 . 67 o R) = 294 . 1 o R ≈ − 166 o F...
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 Spring '06
 Phares
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