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solu0004

# solu0004 - CIVL 252 Hydraulics Problem 1 Solution a...

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CIVL 252- Hydraulics Solutions for Problem Set Spring 2002 1 Problem 1 Solution: a. Dimensionless velocity Laminar flow: 2 2 max 1 ) ( R r u r u - = ; Turbulent flow: 7 / 1 max ) 1 ( ) ( R r u r u - = Dimensionless shear stress: Laminar: 2 max 2 R r u r u μ μ τ = - = R r R u ma 2 / = μ τ Turbulent: 7 / 6 max ) 1 ( 7 1 - - = - = R r R u r u μ μ τ 7 / 6 ) 1 ( 7 1 / - - = R r R u ma μ τ (Please remember the total shear stress should be ' ' v u r u ρ μ τ - - = , since it is difficult to model the turbulent shear stress, (refer Problem 2) , so we just calculate the viscous part of the total shear stress!) The velocity profile and shear distribution are plotted in the figure: 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 r/R u(r)/u max or τ /( μ u max /R) - u lam inar : u turbulent -- τ lam inar -. τ turbulent Comment: velocity distribution across sectional area of turbulent flow is more uniform that of laminar flow except the region near the pipe wall.

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CIVL 252- Hydraulics Solutions for Problem Set Spring 2002 2 Shear stress for turbulent flow has infinity value at the wall, which means Although the power law approximation for velocity is good fitting for the Velocity profile, the derivative of the power law is not that good Approximation. b. Average velocity Laminar flow: 2 2 ) 1 ( 2 ) 1 ( 1 2 ) ( 1 1 max 0 2 4 2 2 max 0 2 2 2 2 max 0 2 2 max 2 0 2 u R r r R u dr R r R u rdr R r u R rdr r u R udA A V R R R R A = - = - = - = = = π π π π Dimensionless 2 1 max = u V Turbulent flow: - = - = = = R R R A rdr R r R u rdr R r u R rdr r u R udA A V 0 7 / 1 2 max 0 2 2 max 2 0 2 ) 1 ( 2 2 ) 1 ( 1 2 ) ( 1 1 π π π π Integral by part, max 0 7 / 15
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solu0004 - CIVL 252 Hydraulics Problem 1 Solution a...

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