p0276 - ρ p F p = ρ m F m 2.76(b): We are given H p = 12...

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2.76. CHAPTER 2, PROBLEM 76 213 2.76 Chapter 2, Problem 76 The parameters for this problem are the same as those of the sphere example in the text. Anticipating that the flow will lie in the high Reynolds number range, we thus expect that the force on the tractor-trailer will vary according to F = ρ U 2 H 2 f ( Re H ) ,R e H ρ UH μ 2.76(a): To achieve dynamic similitude, we must match Reynolds number. Since the temperature is the same for model and prototype and both fluids are air, necessarily μ m = μ p .Thu s , ρ m U m H m μ m = ρ p U p H p μ p = U m = U p ρ p ρ m H p H m So, the force scales according to F m F p = ρ m ρ p X U m U p ~ 2 X H m H p ~ 2 = ρ m ρ p X ρ p ρ m ~ 2 w H p H m W 2 X H m H p ~ 2 = ρ p ρ m
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Unformatted text preview: ρ p F p = ρ m F m 2.76(b): We are given H p = 12 H m . Also, the densities are ρ p = 0.00234 slug/ft 3 and ρ m = 0.00585 slug/ft 3 . So, the velocity ratio is U m U p = ρ p ρ m H p H m = X . 00234 slug / ft 3 . 00585 slug / ft 3 ~ (12) = 4 . 8 Since the speed for flow past the model is given to be 240 ft/sec, we find U p = U m 4 . 8 = 50 ft sec Also, the force on the model is 70 lb, so that F p = F m ρ m ρ p = (70 lb) X . 00585 slug / ft 3 . 00234 slug / ft 3 ~ = 175 lb...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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