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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 shearforce 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 shearforce 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.
 Fall '10
 FredrickStern
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