Conservation of a Quantity in Fluid Motion

2023 EESC W3400
Lec 23: FD method illustrated by channel flow with eddy
Computational Earth Science
Bill Menke, Instructor
Emily Glazer, Teaching Assistant
TR 2:40 – 3:55
Today
the Navier Stokes Equation
trick of solving for pressure
trick of repeating boundaries
trick of particle tracking
eddy in a channel
Part 1
the Navier Stokes Equation
(equation for conservation of momentum in a moving fluid)
small if the fluid is
incompressible
could use the
finite difference
method
source like waste
dumping
forces like
contact forces
and gravity
extremely difficult to solve
forces in a moving fluid
gravity
forces in a moving fluid
pressure
forces in a moving fluid
contact forces
shearing
fluid
shearing fluid
forces in a moving fluid
contact forces
stationary
stationary
viscosity
forces in a moving fluid
contact forces
net force
forces in a moving fluid
contact forces
net force
so the x-component only equation is
so the x-component only equation is
and the full Navier Stokes equation is
change in momentum with time
and the full Navier Stokes equation is
advection in momentum
gravity
pressure contact force
viscous contact forces
Navier Stokes equation for an incompressible viscous fluid
4 equations
3 equations
1 equation
4 unknowns
same number of equations as unknowns, so
solving equation may be possible
Part 2
trick to find pressure
mathematical trick to find the pressure field
Navier Stokes equation for an incompressible viscous fluid
at each time step, solve Poisson equation for pressure
Step 1: use velocities at current time to calculate pressure
 
solve Poisson Equation
with
Step 3: go to step 1
Part 3
trick of repeating boundary conditions
Finite Difference grid
Finite Difference grid
require to be the same
Finite Difference grid
require to be the same
like you turned it
into a cylinder
Part 4
trick of tracer particle tracking
velocity field
table of tracer particles
advance along velocity vector
advance along velocity vector
repeat for each time step
table of tracer particles
Part 5
eddy in a shearing channel
“background” solution
is shear flow in channel
equation satisfied because
every term is zero
satisfies equation
initial “eddy”
satisfies
because no
flow into box
The eddy rotation is in the same
sense of the background shear flow
so will the eddy grow with time?
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Today's lecture in Computational Earth Science delved into the Navier-Stokes Equation and the conservation of momentum in moving fluids. The discussion focused on the tricks involved in solving for pressure, dealing with repeating boundaries, and tracking eddies in channel flow. Through detailed illustrations and equations, the concept of conservation of a quantity in the presence of fluid motion was explained, emphasizing the increase in the quantity over time due to net flux into a volume and production within that volume.

  • Fluid Motion
  • Conservation
  • Navier-Stokes Equation
  • Channel Flow

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  1. 2023 EESC W3400 Lec 23: FD method illustrated by channel flow with eddy Computational Earth Science Bill Menke, Instructor Emily Glazer, Teaching Assistant TR 2:40 3:55

  2. Today the Navier Stokes Equation trick of solving for pressure trick of repeating boundaries trick of particle tracking eddy in a channel

  3. Part 1 the Navier Stokes Equation (equation for conservation of momentum in a moving fluid)

  4. Conservation of a quantity ? in the presence of fluid motion increase in ? with time = net flux of ? into volume + production of ? in volume

  5. Conservation of a quantity ? in the presence of fluid motion ? per unit volume ? per unit volume per unit time ??? ??? + ? ? and ? flux out flux in ? per unit area per unit time ? per unit area per unit time ? increase in ? with time = net flux of ? into volume + production of ? in volume

  6. Conservation of a quantity ? in the presence of fluid motion ? per unit volume ? per unit volume per unit time ??? ??? + ? ? and ? flux out flux in ? per unit area per unit time ? per unit area per unit time ?C ?? ??= ??? ??+ ?

  7. now suppose that ? is being transported by fluid motion the flux of ? is the concentration of ? times the fluid velocity ?? ?? ??= ??? ??= ??? ??+ ?

  8. now suppose that ? is being transported by fluid motion the flux of ? is the concentration of ? times the fluid velocity ?? ?? ??= ??? ??= ??? ??+ ? ?? ??= ? ??? + ? ??

  9. now suppose that ? is being transported by fluid motion the flux of ? is the concentration of ? times the fluid velocity ?? ?? ??= ??? ??= ??? ??+ ? ?? ??= ? ??? + ? ?? ?? ??= ???? ?? ??+ ? ?? ??

  10. now suppose that ? is being transported by fluid motion the flux of ? is the concentration of ? times the fluid velocity ?? ?? ??= ??? ??= ??? ??+ ? ?? ??= ? ??? + ? ?? ?? ??= ???? ?? ??+ ? ?? ?? ?? ??+ ?? ?? ??= ???? ??+ ?

  11. now suppose that ? is being transported by fluid motion the flux of ? is the concentration of ? times the fluid velocity ?? ?? ??= ??? ??= ??? ??+ ? ?? ??= ? ??? + ? ?? ?? ??= ???? ?? ??+ ? ?? ?? small if the fluid is incompressible ?? ?? ?? ?? ??= ???? ??+ ?

  12. ?? ?? ?? ?? ??= ? now suppose that ? is concentration ? of ???? of in ocean water with ? unknown source like waste dumping ?? ?? ?? ?? ??= ? might not be too hard to solve if we know ?? could use the finite difference method

  13. ?? ?? ?? ?? ??= ? forces like contact forces and gravity now suppose that ? is momentum of the ocean water itself ? = ??= ??? velocity ?? is the unknown constant ? since fluid is incompressible ???? ??+ ??? ??? ??= ??

  14. ?? ?? ?? ?? ??= ? now suppose that ? is momentum of the ocean water itself ? = ??= ??? velocity ?? is the unknown constant ? since fluid is incompressible ???? ?? ??? ??? ??= ?? non-linear in ?? extremely difficult to solve

  15. forces in a moving fluid gravity ??= 0 ??= ??

  16. forces in a moving fluid pressure low ? high ? ??= ?? ??

  17. forces in a moving fluid contact forces shearing fluid ? shearing fluid low ?? ? high ??

  18. forces in a moving fluid contact forces viscosity ??= ???? ?? ? ? stationary stationary low ?? drag force ??= ???? ?? ? ? high ??

  19. forces in a moving fluid contact forces ??? net force??? ?? ??? + ?

  20. forces in a moving fluid contact forces ??? ??? ?? net force ? ?? ???? ?? ??? + ? ??2?? ??2

  21. so the x-component only equation is ??+ ??2?? ???? ?? ??? ??? ??= ?? ?? ??2 when ??? ?? is small (incompressible)

  22. so the x-component only equation is ??+ ??2?? ???? ?? ??? ??? ??= ?? ?? ??2 when ??? ?? is small (incompressible) and the full Navier Stokes equation is ??? ?? ? ? ? = ?? ? + ? 2? when ? = 0 (incompressible)

  23. and the full Navier Stokes equation is change in momentum with time advection in momentum pressure contact force viscous contact forces gravity ??? ?? ? ? ? = ? ? + ? 2? when ? = 0 (incompressible)

  24. Navier Stokes equation for an incompressible viscous fluid ??? ?? ? ? ? = ?? ? + ? 2? 3 equations 4 equations 1 equation ? = 0 ??,??,??,? 4 unknowns same number of equations as unknowns, so solving equation may be possible

  25. Part 2 trick to find pressure

  26. mathematical trick to find the pressure field at time ? = 0, start with a velocity field that satisfies ? = 0 at each subsequent time step, adjust the pressure ? so that the new velocity satisfies satisfies ? = 0, too

  27. Navier Stokes equation for an incompressible viscous fluid ??? ??= ?? ? + ? 2? + ? ? ? take ?? ? ?? = 0 = ?? 2? + ? 2? + ? ? ? at each time step, solve Poisson equation for pressure 2? = ??+ ? 2? + ? ? ?

  28. Step 1: use velocities at current time to calculate pressure solve Poisson Equation 2? = ??+ ? 2? + ? ? ? Step 2: use Navier Stokes Equation to time step velocities one ? into the future ? ? + ? = ? ? +?? ?? ? with ?? ??=1 ??? ? + ? 2? + ? ? ? Step 3: go to step 1

  29. Monitor ? to check it doesn t grow

  30. Part 3 trick of repeating boundary conditions

  31. Finite Difference grid ? ?? 1 0 ? 0 ? ?? 1 ?

  32. Finite Difference grid ? ?? 1 0 ? 0 ? ?? 1 ? require to be the same

  33. Finite Difference grid ? ?? 1 0 ? 0 ? ?? 1 like you turned it into a cylinder ? require to be the same

  34. Part 4 trick of tracer particle tracking

  35. ? velocity field 4 0 ? 4 0

  36. ? tracer particles at ? = ? table of tracer particles 4 # x y 1 1 1 7 8 9 2 2 1 4 5 6 ... 1 2 3 0 ? 4 0

  37. ? 4 advance along velocity vector ?? ??? ? ?? ???+ ?= ?? ???+ 7 8 9 4 5 6 1 2 3 0 ? 4 0

  38. ? 4 advance along velocity vector 7 8 9 4 5 6 1 2 3 0 ? 4 0

  39. ? tracer particles at ? = ? table of tracer particles 4 # x y 1 1.4 1.1 7 9 9 8 2 2.5 1.0 4 5 6 ... 1 2 3 0 ? 4 0 repeat for each time step

  40. Part 5 eddy in a shearing channel

  41. background solution is shear flow in channel ??= ?0 ? ?? ?? ??= ?0?/?? ??= 0 ? ? = 0 ?? ?0 0 ? ??= 0

  42. ??= ?0 ??= ?0?/?? ??= 0 ? ? = 0 ? ??= 0 satisfies equation ??+ ??2?? 0 ???? ?? ??? 0 ??? ??= ?? ?? 0 0 ??2 0 equation satisfied because every term is zero

  43. initial eddy ? ? satisfies because no flow into box ? = 0

  44. The eddy rotation is in the same sense of the background shear flow ? ? so will the eddy grow with time?

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