Transport processes. 7. Semester Chemical Engineering Civil Engineering

Size: px

Start display at page:

Download "Transport processes. 7. Semester Chemical Engineering Civil Engineering"

Britney Bryant
5 years ago
Views:

1 Transport processes 7. Semester Chemical Engineering Civil Engineering

2 1. Elementary Fluid Dynamics 2. Fluid Kinematics 3. Finite Control Volume Analysis 4. Differential Analysis of Fluid Flow 5. Viscous Flow and Turbulence 6. Turbulent Boundary Layer Flow 7. Principles of Heat Transfer 8. Internal Forces Convection 9. Unsteady Heat Transfer 10. Boiling and Condensation 11. Mass Transfer 12. Porous Media Flow 13. Non-Newtonian Flow Course plan

3 Today's lecture Turbulent Boundary layer flow Blasius solution Momentum Integral Analysis The regions of the turbulent boundary layer

4 Boundary layer characteristics Boundary layer thickness ( ) δ = y for u y = 0.99U 5 Re = cr 2 10

5 Boundary layer thickness Standard boundary layer thickness ( ) δ = y for u y = 0.99U Displacement thickness δ 1 u = dy 0 U Momentum thickness u u θ = 1 dy 0 U U

6 Wall shear stress Given your knowledge of velocity profiles, explain the development of the wall shear stress Use two minutes to discuss with the person next to you

7 Boundary layer equations Simplified Navier-Stokes equations Flow is parallel to the plate Flow is 2 dimensional Fluid is convected downstream more quickly then it diffuses across the streamlines Only the largest terms are retained u v + = 0 x y µ u + v = x y ρ y 2 u u u 2 From continuity From x-,y-ns-equations, pressure eliminated

8 Boundary conditions The boundary condition between the b.layer and the wall: u = v = 0 for y = 0 no slip BC The boundary condition between the b.layer and the free flow: u = 0.99U for y = δ

9 Blasius solution The mathematical details of the solution are quite tedious and complex and will not be given here - Geankoplis. In general, the solutions of nonlinear partial differential equations are extremely difficult to obtain. - Munson et al.

10 Blasius solution the result! The boundary thickness grows with the square root of the distance from the leading edge δ = 5 µ x ρu The wall shear stress decreases with increasing boundary layer thickness τ = 0.332U w 32 µρ x δ 5 δ = = x Re x x Re x θ = x Re x

11 Momentum integral analysis Blasius exact result, but only for laminar boundary layers Momentum Integral analysis approximate, you have to guess an velocity profile Works for turbulent boundary layers Typical guesses for a laminar velocity profile

12 Momentum integral analysis No flow/stress across top streamline, steady flow Fx = VρdV t + ρ uvn da τ plate w ρ Vn ρ (1) (2) Drag da = u da + uvn da

13 Momentum integral analysis #2 τ plate w ρ ρ (1) (2) da = uvn da + uvn da Conservation of mass: 2 ( ) ρ Drag = ρ U U da + u da (1) (2) 2 2 Drag = ρu bh ρb δ u dy 0 0 ( ) δ = ρb u U u dy Flow rate ρ = Flow rate 1 2 δ Uhb = b udy δ 2 U bh = ρb uudy 0 0

14 Momentum integral analysis #3 δ 0 ( ) Drag = ρb u U u dy Drag ddrag dx 2 = ρbu θ = ρbu 2 dθ dx Conservation of momentum: momentum rate ( ) 2 bu = b u U u dy 0 ρ θ ρ = momentum rate 1 2 d 2 θ τw = ρu dx ddrag d d = τ da = b τ dx = bτ plate w plate w w dx dx dx The Momentum integral equation

15 Example Derive the boundary layer growth δ=f(x) for laminar flow assuming a linear velocity profile: The assumed velocity profile: y u = U δ The Momentum integral equation : 2 dθ τw = ρu dx For laminar flow applies: du U linear profile τ = w µ µ dy δ Definition of θ: u u δ y y δ θ = 1 dy 1 dy 0 = U U = 0 δ δ 6 Combining all of the above!: µ 2 U U d 6 = ρ δ δdδ = µ dx δ 6 dx ρu Integration from 0 to δ and 0 to x, rearranging: µ x δ = 3.46 ρu Even when starting with a poor approximation of the velocity profile the momentum integral analysis gives us a good approximation of the boundary layer growth

16 Example Derive the boundary layer growth δ=f(x) for turbulent flow assuming a 1/7 th velocity profile: The assumed velocity profile: For turbulent flow applies: (empirically) τ y u = δ 17 U ρU µ ρuδ 2 w = 14 Combining all of the above!: The Momentum integral equation : 2 dθ τw = ρu dx Definition of θ: θ u u y y 7 1 dy 1 dy 0 U U δ = = = 0 δ δ 72 δ µ 7 2 dδ 14 µ ρu = ρu δ dδ = dx ρuδ 72 dx Uδ Integration from 0 to δ and 0 to x, rearranging: µ x δ = U 15 x 45

17 Structures of the turbulent b.layer The Turbulent velocity profile using one set of Nondimensional units (Flipped!) The Turbulent velocity profile using another set of Nondimensional units

18 Wall bound Turbulent stresses + mean pressure gradient τ = ρ uv Turbulent stresses dominates du τ = ρ uv + µ x dy du τ = µ x dy Laminar stresses dominates M.Mandø Aalborg University

19 An illustration

20 The viscous sublayer Very close to the wall eddies can t exists, thus the laminar stress-strain relationship applies: du τ w = µ dy Hence, the region very close to the wall is named the viscous sub-layer We introduce a new grouping of scalars: τ w u = ρ This has units of m/s, and is thus named the friction velocity If we combine we get: u u = yρu µ This is known as the law of the wall Notice here the velocity varies linear with y inner coordinates u y + u = u + = yρu µ

21 The log law regime Further out from the wall the turbulent shear stress becomes dominant. If this is true the velocity should vary with the logarithm of y, the following empirical fit is given: u u yρu = 2.5ln µ This is known simply as the log law The region where it applies is named the log law region

22 The buffer layer Between the viscous sub-layer and the log law region both laminar and turbulent shear stresses are significant Hence, the name Buffer layer Spalding offers an empirical fit of the entire inner layer: + + ( ) ( ) κu κu + + κb κb + y = u + e e 1 κu The inner layer is the region where inner coordinates applies, that is: the viscous sublayer, the buffer layer and the log law layer

23 The outer layer Far away from the wall mean flow pressure gradients become important. When the flow deviate from the log law we are in the outer layer LAYER Viscous sublayer Buffer layer Log law layer Outer layer EXTEND 0<y+<5 5<y+<30 30<y+< <y+<

24 Some measurement of the turbulent b.layer Universal?

25 Flow over a golf ball

26 Flow over a golf ball Boundary layer separation

27 Flow over a golf ball Tripping the boundary layer

28 Excercises

7.6 Example von Kármán s Laminar Boundary Layer Problem

7.6 Example von Kármán s Laminar Boundary Layer Problem CEE 3310 External Flows (Boundary Layers & Drag, Nov. 11, 2016 157 7.5 Review Non-Circular Pipes Laminar: f = 64/Re DH ± 40% Turbulent: f(re DH, ɛ/d H ) Moody chart for f ± 15% Bernoulli-Based Flow Metering