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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 NavierStokes 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 × 103 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.
 Fall '10
 FredrickStern
 Friction, Momentum

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