Solution_HW4 - Solutions for homework set #4 Problem #1: 1m...

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Solutions for homework set #4 Problem #1: 0.15 m 1 m 0.3 m ground tube 1.1 m long and 0.005 m in diameter a. Determine the initial volumetric flow rate Q of oil through the tube The volumetric flow rate equation is given as BSL 2.3-21 4 wP R Q 8L πΔ == ρμ In this example 32 P gh 1000kg / m *9.81m / s * 0.45m 4414.5Pa Δ= ρ = = Note that h is the vertical distance between the oil-air interface (at z=0.45m) and the tube exit (z=0). Therefore, 4 53 *4414.5Pa *(0.0025m) Q6 8*0.001Pa s*1.1m π . 1 6 x 1 0 m / s b. Justification of laminar flow For the assumption of laminar flow to be justified, Re < 2100. ( )( ) () ( ) 3 2 2 6.16 10 / 1000 / 22 Re 15676 2100 0.001 0.0025 xm s k g m Du DQ RQ Q ARR P a s m ρ ρρ μ μπ π = = = = > Therefore, because Re > 2100, the assumption of laminar flow cannot be justified.
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Problem #2: Newtonian fluid inside tube R z cylinder wall moves upward with velocity V a. The last equation you could use is equation 3.6-7 in the example problem, or equation 2.3-16 in Chapter 2. Note that everything is identical up to the point where the last boundary condition (u z = V at r = R in this case). b. Velocity distribution Starting with equation 2.3-16 in the textbook: 2 2 4 z P ur L μ Δ =− + C BC: At r = R, u z = V 2 2 2 2 4 4 P
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This note was uploaded on 04/17/2008 for the course PGE 322K taught by Professor Dicarlo during the Fall '08 term at University of Texas at Austin.

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Solution_HW4 - Solutions for homework set #4 Problem #1: 1m...

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