PS4 - MECH 411 Introduction to Fluid Mechanics American...

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MECH 411 Introduction to Fluid Mechanics Dec 11, 2004 American University of Beirut, Fall 2004-2005 Handout # PS4 Problem Set 4 Solution Textbook: 4.17, 4.21, 4.37, 4.41, 4.43, 4.53, 4.60, 4.83 1
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Problem 4.17 (a) The fow is two-dimensional, incompressible, and steady. Given u ( x,y ) we are asked to Fnd v ( x,y ). We employ the continuity (conservation o± mass) equation in di²erential ±orm Dt + u ·∇ u = 0 ³or incompressible 2D fow, we get ∂u ∂x + ∂v ∂y = 0 ∂v ∂y = ∂u ∂x = ∂u ∂δ ∂δ ∂x ∂v ∂y = U p 2 y δ 2 + 2 y 2 δ 3 P 1 2 Cx - 1 / 2 Integrating ±rom y = 0 to y and noting that δ = δ ( x ) and using the no slip condition: v ( y = 0) = 0, then i y 0 ∂v ∂y dy = i y 0 U p 2 y δ 2 + 2 y 2 δ 3 P 1 2 Cx - 1 / 2 dy v ( x,y ) = U p y 2 δ 2 + 2 y 3 3 δ 3 P 1 2 Cx - 1 / 2 v ( x,y ) = U p y 2 δ 2 2 y 3 3 δ 3 P 1 2 δ x It can be shown that v is maximum where ∂u/∂y = 0 leading to y * = δ so that v max = U 6 δ x ³or U = 3 m/s, δ = 1 . 1 cm, and x = 1 m, we get v max = 0 . 0055 m/s. 2
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Problem 4.21 The fow is steady and 1D. The speed oF sound is a s = 340 m/s. The objective is to ±nd ( D e /D 0 ) min to neglect compressibility e²ects For (a) V 0 = 10 m/s and (b) V 0 = 30 m/s. The incompressibility assumption is valid For M < 0 . 3, where the Mach number is M ≡ V/a s . We assume that the fow is incompressible and later we ±nd the condition that has to be satis±ed. So For an incompressible, steady, 1D fow the continuity equation reduces to ρ 0 V 0 A 0 = ρ e V e A e with ρ 0 = ρ e , A 0 = πD 2 0 / 4, A e = πD 2 e / 4, then V 0 D 2 0 = V e D 2 e Assume that the speed oF sound does not change (what does that mean For an ideal gas knowing that a s = γRT ?) then p D e D 0 P 2 = V 0 a s M e Incompressibility implies that M e < 0 . 3 so that p D e D 0 P 2 > V 0 0 . 3 a s ThereFore the minimum value oF D e /D 0 is p D e D 0 P min = p V 0 0 . 3 a s P 1 / 2 (a) With a s = 340 m/s, V 0 = 10 m/s, we get ( D e /D 0 ) min = 0 . 313 (b) With a s = 340 m/s, V 0 = 30 m/s, we get ( D e /D 0 ) min = 0 . 542 So For higher inlet speed, the outer diameter must be larger.
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PS4 - MECH 411 Introduction to Fluid Mechanics American...

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