Problem 2.48 - piston Mass SG Al ρ water ⋅ π D piston 2...

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Problem 2.48 [Difficulty: 2] Given: Flow data on apparatus Find: The terminal velocity of mass m Solution: Given data: D piston 73 mm = D tube 75 mm = Mass 2 kg = L 100 mm = SG Al 2.64 = Reference data: ρ water 1000 kg m 3 = (maximum density of water) From Fig. A.2:, the dynamic viscosity of SAE 10W-30 oil at 25 o C is: μ 0.13 Ns m 2 = The terminal velocity of the mass m is equivalent to the terminal velocity of the piston. At that terminal speed, the acceleration of the piston is zero. Therefore, all forces acting on the piston must be balanced. This means that the force driving the motion (i.e. the weight of mass m and the piston) balances the viscous forces acting on the surface of the piston. Thus, at r = R
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Unformatted text preview: piston: Mass SG Al ρ water ⋅ π D piston 2 ⋅ L ⋅ 4 ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ g ⋅ τ rz A ⋅ = μ r V z d d ⋅ ⎛ ⎝ ⎞ ⎠ π D piston ⋅ L ⋅ ( ) ⋅ = The velocity profile within the oil film is linear . .. Therefore r V z d d V D tube D piston − 2 ⎛ ⎜ ⎝ ⎞ ⎠ = Thus, the terminal velocity of the piston, V , is: V g SG Al ρ water ⋅ π ⋅ D piston 2 ⋅ L ⋅ 4 Mass ⋅ + ⎛ ⎝ ⎞ ⎠ ⋅ D tube D piston − ( ) ⋅ 8 μ ⋅ π ⋅ D piston ⋅ L ⋅ = or V 10.2 m s =...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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