Exam2_solution

# Exam2_solution - 57:020 Mechanics of Fluids and Transport...

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Unformatted text preview: 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM2 Solutions Problem 1: Linear momentum equation (Chapter 5) Information and assumptions • ρ = 999 kg/m3 • Vj = 30 m/s • Aj = 0.01 m2 • θ = 3 0° • Frictionless cart • Jet velocity magnitude remains constant along the vane surface Find • The restrain force Fx Solution For a control volume that surrounds the vane and cart, the continuity equation gives, Since , thus (+1 point) The linear momentum equation in the horizontal direction is (+2 points) or cos Thus, 1 cos (+5 points) For ρ = 999 kg/m3, Vj = 30 m/s, Aj = 0.01 m2, and θ = 30°, 999 kg m 0.01 m 30 m s 1 cos 30 1205 N (+2 points) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM2 Solutions Problem 2: Energy equation (Chapter 5) Information and assumptions • Q = 220 m3/h • ρ = 999 kg/m3 • Friction head loss hL = 5 m • Δz = 2 m • De = 5 cm • α=1 Find • Pump power in kW delivered to the water Solution Energy equation: 2 (+2 points) 2 Let “1” be at the reservoir surface and “2” be at the nozzle exit. Then, 0 (gage) 2m 0 220 m ⁄h h⁄3600s 0.025 m ⁄2 31.12 m s 5m (+4 points) The energy equation with α1 = α2 = 1 becomes 2 or 31.12 m⁄s 2 9.81 m⁄s 2m 5m 56.4 m (+2 points) The pump power is kg 999 m m 9.81 s m h s 3600 h 220 56.4 m 33,778 W 33.8 kW (+2 points) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM2 Solutions Problem 3: Exact solution (Chapter 6) Information and assumptions • Steady, paralle, viscous flow • No pressure gradient in the flow direction • U = 0.2 m/s • b = 5 mm Find • Velocity distribution across the plates and flow rate between the plates per unit depth Solution (a) For a steady flow (∂/∂t =0) with v = w = 0 and for a zero pressure gradient in the flow direction (∂p/∂x = 0), the Navier-Stokes equations (also by using the continuity equation, ∂u/∂x = 0) reduce to 0 (+5 points) So that By using the boundary conditions, u = 0 at y =0 and u = U at y = b, 0 Therefore 0.2 m⁄s 0.005 m m⁄s 40 (+3 points) (b) The flow rate per unit depth is Thus, 2 0.2 m⁄s 0.005 m 2 5 10 m ⁄s (+2 points) 57:020 Mechanics of Fluids and Transport Fall 2009 EXAM2 Solutions Problem 4: Dimensional analysis (Chapter 7) Information and assumptions • ,, • For model: D = 17 lb, d = 1 ft, V = 4 ft/s • For prototype: d = 30 ft • For water: ρ = 1.94 slugs/ft3 • For air: ρ = 2.38 × 10-3 slugs/ft3 Find • Pi parameter and the drag D of the prototype Solution (a) From the pi theorem, 4 -3 = 1 pi term required. Π (+2 points) or Π (+3 points) (b) For similarity between model and prototype, (+3 points) where, the subscript m stands for ‘model’. Then, slugs ft slugs 1.94 ft 2.38 10 ft s ft 4 s 10 30 ft 1 ft 17 lb 117 lb (+2 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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