Final 2006

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Fluid Mechanics I MECH 331 Final Exam, Winter 2006

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1. [6 points] The wind blows across a field with an approximate velocity profile as shown in the figure. From the definition of the Reynolds Transport Theorem compute the (a) the volume flow rate across the vertical surface A-B. (b) the net rate of flow of momentum across the vertical surface A-B. (c) the net rate of flow of kinetic energy across the vertical surface A-B. The vertical surface A-B is of unit depth into the paper. (Note: Density, ρ = 2.38 × 10−3 slug/ft3 )

2. [6 points] A small round object is tested in a 1 m diameter wind tunnel. The pressure is uniform across sections (1) and (2). The upstream pressure is 196 N/m2 (gage), the downstream pressure is 98 N/m2 (gage), and the mean air speed is 10 m/s. The velocity profile at section (2) is parabolic and is defined as u(r) = Vmax

³ r ´2 R

,

where R is the radius of the wind tunnel and the velocity varies from zero at the tunnel centerline to a maximum at the tunnel wall. Calculate (a) the maximum velocity at section (2), and (b) the drag of the object and its supporting vane. Neglect viscous resistance at the tunnel wall. (Note: Density, ρ = 1.23 kg/m3 )

Fluid Mechanics I MECH 331 Final Exam, Winter 2006

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3. [10 points] The total mass of the helicopter-type craft shown is 1500 kg. The pressure of the air is atmospheric at the outlet. Assume the flow is steady and one-dimensional. Treat the air as incompressible at standard conditions and calculate (a) the speed of the air leaving the craft and (b) the minimum power that must be delivered to the air by the propeller if the craft is hovering. Assume that the velocity is uniform across the flow at the inlet and outlet. The inner diameter at the outlet is 3 m and the outer diameter is 3.3 m. (Note: Density, ρ = 1.23 kg/m3 )

4. [14 points] A pipe is titled by an angle θ and contains a fully developed flow. The fluid cylinder has diameter, D and contains liquid with density, ρ. The weight of the fluid cylinder must be taken into consideration. (a) Using a force balance on the fluid cylinder shown below derive an expression that relates the pressure drop, ∆p, to the shear stress, τ . (b) Formulate an expression for the velocity profile, u(r). (c) Derive an expression for the volume flowrate, Q, using the equation formulated in part (b). (d) From dimensional analysis, develop an expression for the volume flowrate, Q. Compare and comment, if any, on the differences between the expressions for the volume flowrate formulated in parts (c) and (d).

Fluid Mechanics I MECH 331 Final Exam, Winter 2006

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5. [14 points] A fire nozzle is attached to a 90 m long rubber-lined hose. The hose is 40 mm in diameter. Water from a hydrant is supplied to a booster pump on board the pumper truck at 345 kPa. At the present condition, the pressure at the nozzle inlet is 690kPa, and the pressure drop along the hose is 75 kPa per 10 m of length. Assume that the surface of the rubber-lined hose is smooth. Neglect minor losses. (Note: Density, ρ = 999 kg/m3 ; dynamic viscosity, µ = 1.12 × 10−3 Ns/m2 ; gravity, g = 9.81 m/s2 , Moody chart provided in the following page.) (a) Compute the flow rate in the pipe, (b) Determine the nozzle exit velocity, assuming no losses in the nozzle, (c) Calculate the power required to drive the booster pump, if its efficiency is 70%. (d) The nozzle in relation to the Booster pump can reach a maximum elevation of 20 m before the hose ruptures. If the flowrate is Q = 0.006 m3 /s, determine the maximum power that can be supplied to the booster pump before the hose ruptures. Use the friction factor computed from part (a).

90 m Nozzle Hose Booster Pump

Nozzle Inlet

Fluid Mechanics I MECH 331 Final Exam, Winter 2006

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Fluid Mechanics I MECH 331 Final Exam, Winter 2006

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USEFUL EQUATIONS 1. p2 − p1 = −γ(z2 − z1) 2. FR = γhcA,

yR =

Ixc yc A

+ yc

3. p + 12 ρV 2 + ρgz = constant 4. F = γV 5. τ = −µ du dr Du Dt

6. Material Derivative:

=

∂u ∂t

³

´

+ V~ · ∇ u

7. Reynolds Transport Theorem: R R DBsys ∂ ~ n dA Dt = ∂t cv ρb dV + cs ρbV · ~ 8. Conservation of Mass: R R ∂ ~ n dA = 0 Integral Form : ∂t ρ dV + cv cs ρV · ~ ³ ´ ∂ρ Vector Form : ∂t + ∇ · ρV~ = 0 9. Conservation of Momentum: R R ∂ ~ dV + ~ (V~ · ~n) dA = P F Integral Form : ∂t ρ V ρ V cv cs ~

Vector Form : ρ DDtV = ρ~g − ∇p 10. Conservation of Energy: Integral Form : ∂ ∂t

³ ˆ+ cv ρ u

R

V2 2

´

+ gz

³ dV + cs ρ uˆ + ρp + R

V2 2

One-Dimensional Form: p2 γ

+

V22 2g

+ z2 = 2

p1 γ

hL = f Dl V2g , hs =

+

V12 2g

˙ W γQ

+ z1 − hL + hs

´

˙ + gz V~ · ~n dA = Q˙ + W