Processing math: 75%

9. Transport phenomena

9.1 Mathematical introduction

An important relation is: if X is a property of a volume element which travels from position r to r+dr in a time dt, the total differential dX is then given by:

dX=Xxdx+Xydy+Xzdz+Xtdt  dXdt=Xxvx+Xyvy+Xzvz+Xt

This in general leads to: dXdt=Xx+(v)X.

From this follows that : ddtXd3V=tXd3V+X(vn)d2A

where the volume V is surrounded by surface A. Some properties of the operator are:

div(ϕv)=ϕdivv+gradϕvrot(ϕv)=ϕrotv+(gradϕ)×vrot gradϕ=0div(u×v)=v(rotu)u(rotv)rot rotv=grad divv2vdiv rotv=0div gradϕ=2ϕ2v(2v1,2v2,2v3)

Here, v is an arbitrary vector field and ϕ an arbitrary scalar field. Some important integral theorems are:

Gauss: (vn)d2A=(divv)d3VStokes for a scalar field:(ϕet)ds=(n×gradϕ)d2AStokes for a vector field:(vet)ds=(rotvn)d2AThis results in: (rotvn)d2A=0Ostrogradsky: (n×v)d2A=(rotv)d3A (ϕn)d2A=(gradϕ)d3V

Here, the orientable surface d2A is limited by the Jordan curve ds.

9.2 Conservation laws

Two types of forces can do work on a volume:

  1. The force f0 on each volume element. For gravity: f0=ϱg.
  2. Surface forces working only on the boundaries: t. For these: t=nT, where T is the stress tensor.

can be split in a part pI representing the normal tension and a part T' representing the shear stress: T=T'+pI, where I is the unit tensor. When viscous aspects can be ignored: div=gradp.

When the flow velocity is v at position r holds at position r+dr:

v(dr)=v(r)translation+dr(gradv)rotation, deformation, dilatation

The quantity L:=gradv can be split in a symmetric part and an antisymmetric part W. L=D+W with

Dij:=12(vixj+vjxi) ,  Wij:=12(vixjvjxi)

When the rotation or vorticity ω=rotv is introduced: Wij=12εijkωk. ω represents the local rotation velocity: drW=12ω×dr.

For a Newtonian liquid :T'=2ηD. Here, η is the dynamical viscosity. This is related to the shear stress τ by:

τij=ηvixj

For compressible media it can be stated that: T'=(η div v)I+2ηD. From equating the thermodynamic and mechanical pressure it follows: 3η+2η=0. If the viscosity is constant: div(2D)=2v+grad divv.

The conservation laws for mass, momentum and energy for continuous media can be written in both integral and differential form. They are:

  1. Conservation of mass: tϱd3V+ ϱ(vn)d2A=0
  2. Conservation of momentum: tϱvd3V+ ϱv(vn)d2A=f0d3V+ nTd2A
  3. Conservation of energy: t(12v2+e)ϱd3V+ (12v2+e)ϱ(vn)d2A=
     (qn)d2A+(vf0)d3V+ (vn T)d2A

Differential notation:

  1. Conservation of mass: ϱt+div(ϱv)=0
  2. Conservation of momentum: ϱvt+(ϱv)v=f0+div T=f0gradp+ divT'
  3. Conservation of energy: ϱTdsdt=ϱdedtpϱdϱdt=divq+T': D

Here, e is the internal energy per unit of mass E/m and s is the entropy per unit of mass S/m. q=κT is the heat flow. Further: p=EV=e1/ϱ  ,   T=ES=es so CV=(eT)V   and   Cp=(hT)p with h=H/m the enthalpy per unit of mass.

From this one can derive the Navier-Stokes equations for an incompressible, viscous and heat-conducting medium:

divv=0

ϱvt+ϱ(v)v=ϱggradp+η2v

ϱCTt+ϱC(v)T=κ2T+2η D:D

with C the thermal heat capacity. The force F on an object within a flow, when viscous effects are limited to the boundary layer, can be obtained using the momentum law. If a surface A surrounds the object outside the boundary layer:

F= [pn+ϱv(vn)]d2A

9.3 Bernoulli's equations

Starting with the momentum equation one can find that for a non-viscous medium for stationary flows, with

(vgrad)v=12grad(v2)+(rotv)×v

and the potential equation g=grad(gh):

12v2+gh+dpϱ=constant along a streamline

For compressible flows holds: 12v2+gh+p/ϱ=constant along a line of flow. If also holds rotv=0 and the entropy is equal on each streamline holds 12v2+gh+dp/ϱ=constant everywhere. For incompressible flows this becomes: 12v2+gh+p/ϱ=constant everywhere. For ideal gases with constant Cp and CV, with γ=Cp/CV:

12v2+γγ1pϱ=12v2+c2γ1=constant

With a velocity potential defined by v=gradϕ  for instationary flows:

ϕt+12v2+gh+dpϱ=constant everywhere

9.4 Characterising of flows with dimensionless numbers

The advantage of dimensionless numbers is that they make model experiments possible if one makes the dimensionless numbers which are important for the specific experiment equal for both model and the real situation. One can also deduce functional equalities without solving the differential equations. Some dimensionless numbers are given by:

Strouhal:Sr=ωLvFroude:Fr=v2gLMach:Ma=vcFourier:Fo=aωL2Péclet:Pe=vLaReynolds:Re=vLνPrandtl:Pr=νaNusselt:Nu=LακEckert:Ec=v2cΔT

Here, ν=η/ϱ is the kinematic viscosity, c is the speed of sound and L is a characteristic length of the system. α follows from the equation for heat transport κyT=αΔT and a=κ/ϱc is the thermal diffusion coefficient.

These numbers can be interpreted as follows:

Now, the dimensionless Navier-Stokes equation becomes, with x=x/L, v=v/V, grad=Lgrad, 2=L22 and t=tω:

Srvt+(v)v=gradp+gFr+2vRe

9.5 Tube flows

Tube flow is laminar if Re<2300 across the diameter of a tube, and turbulent if Re is larger. For an incompressible laminar flow through a straight, circular tube the velocity profile is:

v(r)=14ηdpdx(R2r2)

For the volumetric flow: ΦV=R0v(r)2πrdr=π8ηdpdxR4

The entrance length Le is given by:

  1. 500<ReD<2300: Le/2R=0.056ReD
  2. Re>2300: Le/2R50

For gas transport at low pressures (Knudsen-gas): ΦV=4R3απ3dpdx

For flows at a small Re holds: p=η2v and divv=0. For the total force on a sphere with radius R in a flow then: F=6πηRv. For large Re the force on a surface A is: F=12CWAϱv2.

9.6 Potential theory

The circulation Γ is defined as: Γ=(vet)ds=(rotv)nd2A=(ωn)d2A

For non viscous media, if p=p(ϱ) and all forces are conservative, Kelvin’s theorem can be derived:

dΓdt=0

For rotationless flows a velocity potential v=gradϕ can be introduced. In the incompressible case It follows from conservation of mass that 2ϕ=0. For a 2-dimensional flow a flow function ψ(x,y) can be defined: with ΦAB the amount of liquid flowing through a curve s between the points A and B:

ΦAB=BA(vn)ds=BA(vxdyvydx)

and the definitions vx=ψ/y, vy=ψ/x then: ΦAB=ψ(B)ψ(A). In general:

2ψx2+2ψy2=ωz

In polar coordinates:

vr=1rψθ=ϕr  ,  vθ=ψr=1rϕθ

For source flows with power Q in (x,y)=(0,0): ϕ=Q2πln(r) so that vr=Q/2πr, vθ=0.

For a dipole of strength Q in x=a and strength Q in x=a it follows from the superposition that: ϕ=Qax/2πr2 where Qa is the dipole strength. For a vortex: ϕ=Γθ/2π.

If an object is surrounded by an uniform main flow with v=vex and such a large Re that viscous effects are limited to the boundary layer: Fx=0 and Fy=ϱΓv. The statement that Fx=0 is d’Alembert’s paradox and originates from the neglect of viscous effects. The lift Fy is also created by η because Γ0 due to viscous effects. Hence rotating bodies also create a force perpendicular to their direction of motion: the Magnus effect.

9.7 Boundary layers

9.7.1 Flow boundary layers

If for the thickness of the boundary layer : δL then: δL/Re. With v the velocity of the main flow it follows that the velocity vy at the surface will be: vyLδv. Blasius’ equation for the boundary layer is with vy/v=f(y/δ): 2f with boundary conditions f(0)=f'(0)=0, f'(\infty)=1. From this follows: C_W=0.664~{\rm Re}_x^{-1/2}.

The momentum theorem of von Karman for the boundary layer is: \displaystyle \frac{d}{dx}(\vartheta v^2)+\delta^* v\frac{dv}{dx}=\frac{\tau_0}{\varrho}

where the displacement thickness \delta^*v and the momentum thickness \vartheta v^2 are given by:

\vartheta v^2=\int\limits_0^\infty (v-v_x)v_xdy~~,~~~ \delta^*v=\int\limits_0^\infty (v-v_x)dy~~\mbox{and}~~ \tau_0=-\eta\left.\frac{\partial v_x}{\partial y}\right|_{y=0}

The boundary layer is released from the surface if \displaystyle\left(\frac{\partial v_x}{\partial y}\right)_{y=0}=0. This is equivalent to \displaystyle\frac{dp}{dx}=\frac{12\eta v_\infty}{\delta^2}.

9.7.2 Temperature boundary layers

If the thickness of the temperature boundary layer \delta_T\ll L then:

  1. If {\rm Pr}\leq1: \delta/\delta_T\approx\sqrt{\rm Pr}.
  2. If {\rm Pr}\gg1: \delta/\delta_T\approx\sqrt[3]{\rm Pr}.

9.8 Heat conductance

For non-stationairy heat conductance in one dimension without flow:

\frac{\partial T}{\partial t}=\frac{\kappa}{\varrho c}\frac{\partial^2 T}{\partial x^2}+\Phi

where \Phi is a source term. If \Phi=0 the solutions for harmonic oscillations at x=0 are:

\frac{T-T_\infty}{T_{\rm max}-T_\infty}=\exp\left(-\frac{x}{D}\right)\cos\left(\omega t-\frac{x}{D}\right)

with D=\sqrt{2\kappa/\omega\varrho c}. At x=\pi D the temperature variation is in anti-phase with the surface. The one-dimensional solution at \Phi=0 is

T(x,t)=\frac{1}{2\sqrt{\pi at}}\exp\left(-\frac{x^2}{4at}\right)

This is mathematically equivalent to the diffusion problem:

\frac{\partial n}{\partial t}=D\nabla^2n+P-A where P is the production of and A the discharge of particles. The flow density J=-D\nabla n.

9.9 Turbulence

The time scale of turbulent velocity variations \tau_{\rm t} is of the order of: \tau_{\rm t}=\tau\sqrt{\rm Re}/{\rm Ma^2} with \tau the molecular time scale. For the velocity of the particles: v(t)=\left\langle v \right\rangle+v'(t) with \left\langle v'(t) \right\rangle=0. The Navier-Stokes equation now becomes:

\frac{\partial \left\langle \vec{v}\, \right\rangle}{\partial t}+(\left\langle \vec{v}\, \right\rangle\cdot\nabla)\left\langle \vec{v}\, \right\rangle=-\frac{\nabla\left\langle p \right\rangle}{\varrho}+ \nu\nabla^2\left\langle \vec{v}\, \right\rangle+\frac{{\rm div}\mbox{S}_R}{\varrho}

where {\mbox{S}_R}_{ij}=-\varrho\left\langle v_i v_j \right\rangle is the turbulent stress tensor. Boussinesq’s assumption is: \tau_{ij}=-\varrho\left\langle v_i'v_j' \right\rangle. It is stated that, analogous to Newtonian media: S_R=2\varrho\nu_t\langleD\rangle. Near a boundary: \nu_t=0, far away from a boundary: \nu_t\approx\nu{\rm Re}.

9.10 Self organization

For a (semi) two-dimensional flow: \displaystyle\frac{d\omega}{dt}=\frac{\partial \omega}{\partial t}+J(\omega,\psi)=\nu\nabla^2\omega

With J(\omega,\psi) the Jacobian. So if \nu=0, \omega is conserved. Further, the kinetic energy/mA and the enstrofy V are conserved: with \vec{v}=\nabla\times(\vec{k}\psi)

E\sim(\nabla\psi)^2\sim\int\limits_0^\infty {\cal E}(k,t)dk=\mbox{constant}~~,~~ V\sim(\nabla^2\psi)^2\sim\int\limits_0^\infty k^2{\cal E}(k,t)dk=\mbox{constant}

From this follows that in a two-dimensional flow the energy flux goes towards large values of k: larger structures become larger at the expanse of smaller ones. In three-dimensional flows the situation is just the opposite.