Unformatted text preview: 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM3 Solutions
Problem 1: Linear momentum equation (Chapter 5)
Information and assumptions
• ρ = 998 kg/m3 for water at 20°C
• D1 = 0.1 m
• D2 = 0.03 m
• F = 70 N
• Steady, frictionless, onedimensional flow
• ρ = 13,600 kg/m3 for mercury
Find
• Water velocity at sections (1) and (2) and mercury manometer reading h. Solution
(a) Momentum of the jet striking the plate, 998 70 N 0.03 4 9.96 m⁄s
From continuity equation,
4 0.03 4 0.1 (b) From Bernoulli equation,
1
998 9.96
2 9.96 0.9 0.9 m⁄s (+4 points) 49,100 Pa (+3 points) 0.4 m (+3 point) By using a manometry equation, or,
49100
13,600 998 9.81 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM3 Solutions
Problem 2: Acceleration (Chapter 4)
Information and assumptions
• Inviscid, incompressible, steady flow
• a = 0.1 m
⁄
1
•
= 10 m/s
•
⁄
• Euler equation:
3
•
= 1.23 kg/m
Find
• Acceleration and pressure gradient ⁄ at = 2 = 0.2 m Solution
(a) Acceleration
(+4 points) where, 1 . Then, 1 At 3 3 0.1
0.2 0.1
0.2 1 = 2 ,
3 10 1 164.1 m
s (+4 points) (b) From Euler equation,
1.23 164.1 201.8 Pa⁄m (+2 points) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM3 Solutions
Problem 3: Friction loss (Chapter 8)
Information and assumptions
• Roughness, = 0.12 mm
• Diameter, = 40 mm
• Pressure drop Δ = 1.3 kPa per ℓ = 12 m
• Initial guess f = 0.0315 for iterative process
•
= 999 kg/m3; = 1.12 × 106 m2/s
Find
• Flow rate at = 10 years Solution
Energy equation
ℓ
2
Where, = and = 2 2 . Then,
ℓ Δ 2
or
Δ 2 Assume 0.09314 1 (+5 points) ℓ = 0.0315,
0.09314
√0.0315 0.5248 0.5248 0.04
1.12 10 ⁄ 18743 Check ,
1 1.8 log 0.003
3.7 . 6.9
18743 0.0315 5.6367 O. K . (+3 points) Flow rate
4 0.04 0.5248 6.6 10 ⁄ (+2 points) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM3 Solutions
Problem 4: Minor loss (Chapter 8)
Information and assumptions
• ℓ = 1200 m
• D = 0.05 m
• z1 = 400 m and z2 = 500 m
• Q = 0.005 m3/s
•
= 998 kg/m3; = 0.001 kg/m⋅s
•
= 0.0315
•
=
o 0.2 for 45° elbow
o 0.3 for 90° elbow
o 8.5 for open globe valve
o 1.0 for sharp exit
Find
• Gage pressure at point 1 Solution
Energy equation
ℓ
2
Where 0 and 2 2 0. Then,
ℓ
2 (+5 points) 2 From continuity
0.005
⁄4 0.05 2.55 m
s 1 11.1 Minor losses:
2 0.2 4 0.3 8.5 0.0315 1200
0.05 (+3 points) Then,
500 400 11.1 1 2.55
2 9.81 353 m or
998 9.81 353 3.46 MPa (+2 points) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM3 Solutions
Problem 5: Boundary layer (Chapter 9)
Information and assumptions
•
= 29.06 m/s
•
=8m
•
= 0.6 m
• ρ = 1.2 kg/m3
• μ = 1.8 × 105 kg/m⋅s
• Turbulent smoothwall flow from the leading
edge
Find
• Force on the sign Solution
Reynolds number
1.2 29.06 8
1.8 10 Friction drag coefficient1
0.031 1.55 1.55 10 (+3 points) 0.00291 0.031 10 (+3 points) Drag force 0.00291 1 1
2 1
2 2 1.2 29.06 0.6 8 2 (+4 points)
14 N Alternatively,
.
. . 0.0028 for turbulent smooth plate ( 0.0027for “tripped” turbulent boundary layer, 13.6 N)
≤ 107 ( 13.1 N) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM3 Solutions
Problem 6: Blunt body drag (Chapter 9)
Information and assumptions
• D = 1.69 in.
• U = 200 ft/s
• ρ = 0.00238 slugs/ft3
• ν = 1.57 × 104 ft2/s
Find
• Drag on a standard and smooth golf balls Solution
Reynolds number
200 1.69⁄12
1.57 10 1.79 10 Drag
1
2 (+4 points) 4 (a) For a standard golf ball
0.25
1
0.00238 200
2 4 1.69
12 0.25 0.185 lb (+3 points) (b) For a smooth golf ball
0.51
1
0.00238 200
2 4 1.69
12 0.51 0.378 lb (+3 points) ...
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 Fall '10
 FredrickStern
 Fluid Dynamics, Friction, Momentum, Transport Fall

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