Chapter8 - 57:020 Mechanics of Fluids and Transport...

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57:020 Mechanics of Fluids and Transport Processes Chapter 8 Professor Fred Stern Fall 2007 1 Chapter 8 Flow in Conduits Entrance and developed flows Le = f(D, V, ρ , μ ) Π i theorem Le/D = f(Re) Laminar flow: Re crit 2000, i.e., for Re < Re crit laminar Re > Re crit turbulent Le/D = .06Re from experiments Le max = .06Re crit D 138D maximum Le for laminar flow
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57:020 Mechanics of Fluids and Transport Processes Chapter 8 Professor Fred Stern Fall 2007 2 Turbulent flow: D Le 6 / 1 Re 4 . 4 from experiment Laminar vs. Turbulent Flow laminar turbulent spark photo Reynolds 1883 showed difference depends on Re = ν VD Re Le/D 4000 18 10 4 20 10 5 30 10 6 44 10 7 65 10 8 95 i.e., relatively shorter than for laminar flow
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57:020 Mechanics of Fluids and Transport Processes Chapter 8 Professor Fred Stern Fall 2007 3 Shear-Stress Distribution Across a Pipe Section Continuity: Q 1 = Q 2 = constant , i.e., V 1 = V 2 since A 1 = A 2 Momentum: ( ) ρ = A V u F s = ( ) ( ) 2 2 2 1 1 1 A V V A V V ρ ρ = ( ) 0 V V Q 1 2 = ρ ( ) 0 ds r 2 sin W A ds ds dp p pA = π τ α Δ + Ads W γ = Δ ds dz sin = α ( ) 0 ds r 2 ds dz Ads dsA ds dp = π τ γ ÷ Ads ( ) γ + = τ z p ds d 2 r ( ) 0 2 W r d p z ds τ γ = + τ varies linearly from 0.0 at r = 0 (centerline) to τ max (= τ w ) at r = r 0 (wall), which is valid for laminar and turbulent flow.
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57:020 Mechanics of Fluids and Transport Processes Chapter 8 Professor Fred Stern Fall 2007 4 no slip condition Laminar Flow in Pipes ( ) γ + = μ = μ = τ z p ds d 2 r dr dV dy dV y = wall coordinate = r o r dy dV dr dy dy dV dr dV = = ( ) γ + μ = z p ds d 2 r dr dV ( ) C z p ds d 4 r V 2 + γ + μ = ( ) ( ) γ + μ = = z p ds d 4 r C 0 r V 2 o o ( ) ( ) 2 2 2 0 1 4 o C r r d r V r p z V ds r γ μ = + = where = A d V Q = ( ) π o r 0 rdr 2 r V ( ) π = θ = 2 rdr rdrd dA Exact solution to Navier-Stokes equations for laminar flow in circular pipe ( ) 2 4 o C r d V p z ds γ μ = +
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57:020 Mechanics of Fluids and Transport Processes Chapter 8 Professor Fred Stern Fall 2007 5 ( ) γ + μ π = z p ds d 8 r Q 4 o ( ) 2 8 2 o C r V Q d V p z A ds γ μ = = + = For a horizontal pipe, ( ) 2 2 4 4 o o C r r d p V p z ds L γ μ μ Δ = + = where L = length of pipe = ds ( ) ( ) 2 2 2 2 0 0 1 4 4 o r p r p V r r r L r L μ μ Δ Δ = = ( ) 0 4 2 2 0 0 2 128 r p D p Q r r rdr L L π μ μ Δ Δ = = Energy equation: L 2 2 2 2 1 2 1 1 h z g 2 V p z g 2 V p + + + γ = + + γ 1 2 1 2 p p h z z γ γ Δ = + +
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