Transcript 599-5-fall2010 - Department of Applied Mathematics and Statistics

AMS 599
Special Topics in Applied
Mathematics
Lecture 5
James Glimm
Department of Applied Mathematics
and Statistics,
Stony Brook University
Brookhaven National Laboratory
A Chemical Processing Problem
• Separation of spent nuclear fuel rods
• Plan to bury them ran into political
opposition
• New plan is to greatly reduce the volume
• Separation of different fuel components
• Re-use remaining fuel and only bury a
small fraction of spent fuel rod
• Separation has many stages and steps;
we consider only one.
A Fluid Dynamics Problem
•
•
•
•
Couette flow
Inner and outer cylinder
Inner rotates, outer is stationary
As rotation speed increases, a series of
transitions occur in the flow pattern
–
–
–
–
–
Laminar
Vortex rings (Taylor vortices)
Wavy vortices
Wavy, wiggly vortices
Turbulent flow
Couette Flow
• A widely studied flow as it serves as a test
for theories of the onset of turbulence
• We are not concerned with this and are
nowhere near the transition point. High
rotation rate well past turbulent transition
point
• We study two phase Couette flow
– Oil based fluid
– Water based fluid
Contactor
• The device we study is called a contactor
• Two immiscible fluids in a high speed Couette
flow
• Produces fine scale droplets and bubbles of
heavy and light fluid
• Produces a large surface area
• Chemical processing occurs at the surface
between the two phases
• Optimal processing has a large surface area
Overall goal
• Simulate of small region of the contactor flow
• Determine the surface area and the flow
properties of the two phases in a vicinity of the
interface
• Thus estimate a possible chemical reaction rate
as limited by surface area and diffusion to the
surface
• Verify and validate
• Use results to calibrate a macroscopic model
Macroscopic Model
• Simulations of the full contactor do not attempt
to describe the surface between the fluids
• Surface area is assumed and diffusion to
surface, giving an effective reaction rate
• Effective reaction rate is a key unknown
(parameter) in the model.
• With good parameters, macroscopic model can
succeed
• We aim to supply this parameter by simulation
• The parameter will also be determined
experimentally so there will be a cross check
(validation)
The Team/Collaborators
• Stony Brook
University
–
–
–
–
–
–
–
–
James Glimm
Xaiolin Li
Xiangmin Jiao
Hyunkyung Lim
Shuqiang Wang
Navamita Ray
Yijie Zhou
Bryan Clark
• Oak Ridge National
Laboratory
– Valmor de Almeida
• Manhattan Community
College
– Brett Sims
Contactor
Photographs of 2 phase flow
droplets 10 to 100 microns
1 mm x 1 mm field of view
Simulation Region
Droplet distribution as measured
by image processing
Simulation Equations
• Two phase incompressible Navier-Stokes
equation
– Two fluids, oleac (o) and acqueous (w)
– Different densities
– Immiscible, so that at each point, time instant, there is
only one phase present
– Boundary between the fluids is called the interface
• Three phase incompressible Navier-Stokes
equation
– Extra phase is air (a)
Incompressible Navier-Stokes
Equation (3D)
 t  v    (v  v)  P  v
 v0
  dynamic viscosity
 /  kinematic viscosity
  density; P  pressure
v  velocity
Total time derivatives
x (t )  particle streamline
v (t )  dx (t ) / dt  velocity
D
 Lagrangian time derivative
Dt
= derivative along streamline


 v
t
x
Now consider Eulerian velocity v  v ( x , t ).
On streamline, v  v ( x (t ), t )
Dv v
v

v
 acceleration of fluid particle
Dt t
x

Euler’s Equation
Forces = 0
Dv

  inertial force
Dt
Pressure = force per unit area
Force due to pressure =
 Pds    Pdx
S
V
Dv

 P  other forces  0
Dt
Conservation form of equations
Conservation of mass

 v

0
t
x
Conservation of momentum
 v
v



v
t
t
t
 

  v 
   v
 v  P  
v
x 

 x 
 v
 v  v

 P  other forces
t
x
Momentum flux
U
 F (U )  0; F  flux of U
t
 v  v  P    flux of momentum
 ik   ik P   vi vk   ik   vi vk
  stress tensor
Now include viscous forces. They are added to 
 ik  P ik   'ik
 'ik  viscous stress tensor
Viscous Stress Tensor
 ' depends on velocity gradients, not velocity itself
 ' is rotation invariant; assume  ' linear as a function of velocity gradients
Theorem (group theory)
 vi vk 2 vi 
vi
 '  

  ik
   ik
xi
 xk xi 3 xi 
Corollary: Incompressible Navier-Stokes eq. constant density
 v  v  v

 P  v
t
x
Taylor Couette Vortices
Two Phase NS Equations
immiscible, Incompressible
• Derive NS equations for variable density
• Assume density is constant in each phase with a jump
across the interface
• Compute derivatives of all discontinuous functions using
the laws of distribution derivatives
– I.e. multiply by a smooth test function and integrate formally by
parts
• Leads to jump relations at the interface
– Away from the interface, use normal (constant density) NS eq.
– At interface use jump relations
• New force term at interface
– Surface tension causes a jump discontinuity in the pressure
proportional to the surface curvature. Proportionality constant is
called surface tension