Exam3_solution

Exam3_solution - 57:020 Mechanics of Fluids and Transport...

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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, one-dimensional 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 × 10-6 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 × 10-5 kg/m⋅s • Turbulent smooth-wall 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 × 10-4 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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