Exam1_solution - 57:020 Mechanics of Fluids and Transport...

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Unformatted text preview: 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM1 Solutions Problem 1: Shear stress (Chapter 1) Information and assumptions • • • • • Find • 1 = 0.04 lb⋅s/ft2 V = 2 ft/s h = 0.2 in A = 2 ft2 The shear stress at the midplane and the shear-force acting on the bottom wall Solution (+4 points) (a) Along the midplane where y = 0, 3 0 Thus, 0 (+2 points) (b) At the bottom wall where y = -h 3 3 Thus, 3 0.04 lb s⁄ft 3 2 ft⁄s 0.2 in. 1 ft⁄12 in. 14.4 lb⁄ft (+2 points) Then, the shear-force is 14.4 lb⁄ft 2 ft 28.8 lb (+2 points) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM1 Solutions Problem 2: Hydrostatic force (Chapter 2) Information and assumptions • = 9.81 kN/m3 • Wall width = 150 m • Water depth = 25 m • Wall slant angle = 60° • Find • The force acting on the wall and its line of action Solution Let 150 m and 25 m⁄sin 60 face underwater, respectively. 28.87 m be the width and the length of the wall sur- (a) 9.81 kN⁄m 25 m 150 m 2 5.31 10 N 28.87 m (+ 4 points) where, 12 150 m 28.87 m 300780.98 m 12 28.87 m 14.435 m 2 2 150 m 28.87 m 4330.5 m Thus, 300780.98 14.435 m 4330.5 m 14.435 m 19.25 m (+5 points) (b) sin 60 4.60 10 N (+1 point) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM1 Solutions Problem 3: Bernoulli equation (Chapter 3) Information and assumptions • Width b = 6.0 m • Height h = 5.0 m • Gate opening height a = 0.8 m • Contraction coefficient Cc = 0.61 Find • Flow rate Solution From Bernoulli equation, 2 where, 0, 2 5 m, and 0.61 0.8 m 0.488 m. (+4 points) From the continuity equation, 5m 0.488 m 10.25 (+2 points) Then, the Bernoulli equation becomes, 2 9.81 m⁄s 5m 10.25 2 9.81 m⁄s 0.488 m or, 0.922 m⁄s (+3 points) Hence 6 m 5 m 0.922 m⁄ 27.66 m ⁄ (+1 point) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM1 Solutions Problem 4: Acceleration (Chapter 4) Information and assumptions • Streamline coordinate . • 0.6 • Particle velocity • Fluid velocity Find • • • at = 1 sec at = 1 ft Fluid acceleration 0.5 at = 1 ft Solution (a) . 0.3 0.182 ft⁄s (+3 points) (b) | 0.5 1 0.5 ft⁄s (+2 points) (c) 0.5 1 0.5 0.25 ft⁄s (+5 points) ...
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This note was uploaded on 12/08/2011 for the course MECHANICS 57:020 taught by Professor Fredrickstern during the Fall '10 term at University of Iowa.

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