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SolutionHW_C5_1

# SolutionHW_C5_1 - Homework Set#5 Due date by 11:00 am...

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Homework Set #5: Due date: 10/20/06, by 11:00 am. Problem #1 : A fluid with viscosity 18.3 cp and density 1.32 g/cm3 is flowing in a long horizontal tube of radius 1.05 in (2.67 cm). For what pressure gradient will the flow become turbulent? Answer: The minimum value of Re 4 /( ) wD π μ = needed to produce turbulent flow in a long, smooth tube is about 2100. Poiseuille’s law holds until this critical Re value, giving () 4 0 critical Re<Re 8 L PP R w L πρ = Hence, the pressure gradient needed to initiate the laminar-turbulent transition is 2 5 44 3 88 4 Re 2100 1.1 10 Pa/km 4 critical dp w D dz R R R μμ πρπ ρ == Problem #2: Water is flowing through a long, straight, level run of smooth 6.00 in inner diameter pipe, at temperature 68 0 F. The pressure gradient along the length of the pipe is 1 psi/mile. (a) Determine the wall shear stress τ 0 in Pa? (b) Assume the flow to be turbulent and determine the radial distance from the pipe wall at which , using Fig. 5.5-3 in the textbook (second edition), also available in the lecture note. Pay attention at the definition of dimensionless variables. ,max 0.0 0.2 uu = , 0.1, , 0.4, 0.7, 0.85, 1.0 / zz (c) Plot the complete velocity profile, versus r/R. ,max / (d) Is the assumption of turbulent flow justified? Answer: (a) Wall shear stress τ 0 ( ) 0 0 0.1633 Pa 2 L pp R L τ (b) The radial distances from the pipe wall at which ,max 0.0, 0.1, 0.2, 0.4, 0.7, 0.85, 1.0 / = Physical properties of water at temperature 68 0 F ρ = 0.9992 10 3 kg/m 3 , μ = 1.0019 10 -3 Pa-s, ν =1.0037 10 -6 m 2 /s

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Friction velocity: * 0.01278 m/s o u τ ρ == , Thus u * / ν = 1.273 10 4 m -1 , R = 0.0762 m At the tube center * 970 Ru y ν + , and Fig. 5-5.3 in the BSL gives 22.7 m/s yR u + = = As a result ,max * 0.290 m/s z uu u + = =
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SolutionHW_C5_1 - Homework Set#5 Due date by 11:00 am...

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