4. Vorticity Diffusion and Boundary Layers Flashcards
start-up flows, vorticity dynamics, high reynold's number flows
Flow Near an Impulsively Moved Boundary Description
-flow above a solid wall at y=0 -initially the fluid is at rest -at time t=0, the boundary starts to move with velocity U in the x direction -in this simple flow, we can assume that: u_ = (u(y,t),0,0)
Flow Near an Impulsively Moved Boundary Navier-Stokes Equation
-since the flow is driven by motion of the boundary and not an external pressure gradient, we assume that ∂p/∂x=0 -the Navier-Stokes equation reduces to: ρ ∂u/∂t = μ ∂²u/∂y²
Flow Near an Impulsively Moved Boundary Boundary Conditions
-boundary conditions: y=0, u=U -and y->∞, U->0 -we also impose the initial condition, u=0 at t=0
Vorticity Dynamics 2D Planar Flow
-for 2D planar flow: u_ = ( u(x,y) , v(x,y) , 0 ) -giving vorticity: ω_ = (0,0,ω) -where: ω = ∂v/∂x - ∂u/∂y -in this case the vorticity equation reduces to: Dω/Dt = ν∇²ω -hence in the limit ν->0, the vorticity of a fluid particle remains constant, it is simply advected with the flow -when viscosity is considered, vorticity obeys a diffusion equation
Vorticity Dynamics The Vorticity Equation
∂ω_/∂t + (u_.∇)ω_ = Dω_/Dt = (ω_.∇)u_ + ν∇²ω_ -the LHS represents the rate of change of vorticity of a fluid particle -the term (ω_.∇)u_ corresponds to vorticity enhancements from velocity gradients parallel to the direction of ω_, vortex stretching -the term ν∇²ω_ is the diffusion of vorticity due to viscosity
Vorticity Dynamics Steps to Derive the Vorticity Equation
-start from the Navier-Stokes equation -divide by ρ -rewrite using vector identitty: u_ x (∇xu_) = 1/2 (u_.u_) - u_ . ∇u_ -take the curl of the equation -use vector identity: ∇x(u_x_) = u_∇.ω_ + ω_.∇u_ - ω_∇.u_ - u_.∇ω_ -note that ∇.ω_ = ∇.(∇xu_)=0 -and as we are considering an incompressible flow, ∇.u_=0
Line Vortex Velocity
u_ = Γ/2πr
High Reynold’s Number Flows
-when the flow is two-dimensional, u_∇.ω_=0 and the vorticity equation reduces to the diffusion equation -balancing terms in this equation gives an estimate for the thickness of the boundary layer, δ -if Ω is the approximate magnitude of the vorticity, then: |ν∇²ω| ~ ν Ω/δ² -while: |∂ω/∂t| ~ Ω/t => δ² ~ νt -the boundary layer thickness is O(√[νt])
Boundary Layer on a Solid Boundary Vorticity
-consider a steady flow of magnitude U past a streamlined body of length L -the advection term, u_∇.ω_, balances diffusion: u_∇.ω_ = ν∇²ω
Boundary Layer on a Solid Boundary Delta
-we expect vorticity to vary over the length of the body, so that: |u_∇.ω_| ~ UΩ/L |ν∇²ω| ~ ν Ω/δ² => UΩ/L ~ ν Ω/δ² => δ² = νL/U = L² (ν/UL)
Boundary Layer on a Solid Boundary Interpretation
δ² = νL/U = L² (ν/UL) -thus the vorticity is confined to a boundary layer of thickness L*Re^(-1/2) where Re=UL/ν -provided the Reynold’s number is large, the boundary layer thickness is small compared with the size of the body
Wake Behind a Streamlined Body Definition
-we have argued that there is a thin boundary layer on the surface of the body in which the vorticity is concentrated -the fluid in this boundary layer is advected around the surface of the body until the point where the flow separates behind it -this creates a region of vorticity behind the body, the wake
Wake Behind a Streamlined Body Structure
-we can estimate the structure of the wake from the form of the flow above an impulsively started plate -in the frame of the fluid particle, the thickness of the wake grows as δ=√[νt] while the fluid is being carried away from a fixed point on the boundary at speed U -hence the thickness of the wake at a distance behind the fixed point it approximately: δ = √[νx/U] -it is parabolic, however by the time the wake has grown to the size of the body it has diffused so much that it is no longer detectable
Start-up of Shear Flow Description
-two parallel plates at y=0 and y=h -begin to move the lower plate with velocity U at t=0 -boundary conditions: u(y=0,t)=U and u(y=h,t)=0
Start-up of Shear Flow Steps
-fluid velocity satisfies: ∂u/∂t = ν ∂²u/∂y² -observe that the steady solution: us = U (1 - y/h) -satisfies the equation, we can then write: u(y,t) = us + v(y,t) -seek a separable solution of the form: v(y,t) = T(t)Y(y)
