p0194 - 2 π RL Thus F = 2 π RL τ R = 2 π RL w − p 1...

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102 CHAPTER 1. INTRODUCTION 1.94 Chapter 1, Problem 94 1.94(a): For the pipe-flow solution given in the text, we know that the maximum velocity is related to the pressure difference by u m = ( p 1 p 2 ) 4 μL R 2 = u m = ( p 1 p 2 ) 16 ρν L D 2 where ρ is density, ν is kinematic viscosity and D =2 R is diameter. Thus, the maximum velocity in the tube is u m = p 1 . 44 lb / ft 2 Q (0 . 02 ft) 2 16 p 1 . 94 slug / ft 3 Qp 1 . 41 · 10 5 ft 2 / sec Q (5 / 6ft) =1 . 58 ft sec 1.94(b): Hence, the Reynolds number is Re R = u m R ν = (1 . 58 ft / sec) (0 . 01 ft) 1 . 41 · 10 5 ft 2 / sec =1121 1.94(c): Because the Reynolds number is low enough ( Re R < 2300 ) to guarantee laminar flow, we can compute the force on the tube as the product of the surface shear stress, τ ( R ) , and the total surface area of the tube, which is
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Unformatted text preview: 2 π RL . Thus, F = 2 π RL τ ( R ) = 2 π RL w − p 1 − p 2 L R 2 W = − π ( p 1 − p 2 ) R 2 We can eliminate the pressure difference in favor of the maximum velocity by noting that u m = ( p 1 − p 2 ) 4 μL R 2 = ⇒ ( p 1 − p 2 ) R 2 = 4 μu m L Therefore, the total viscous force on the tube is F = − 4 π μu m L Finally, for the given conditions, the force is F = − 4 π p 1 . 94 slug / ft 3 Q p 1 . 41 · 10 − 5 ft 2 / sec Q (1 . 58 ft / sec) (5 / 6 ft) = − 4 . 53 · 10 − 4 lb...
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