Tau method in full 3D simulation

[Kyriacos Papademos]

Dear all,

I am not sure how to use the tau method with a 3D simulation, I couldn't find examples. I have:

dealias = 3/2
dtype = np.float64

# Domain and Bases
coords = d3.CartesianCoordinates('x', 'y', 'z')
dist = d3.Distributor(coords, dtype=dtype)
xbasis = d3.RealFourier(coords['x'], size=nx, bounds=(0, lx), dealias=dealias)
ybasis = d3.RealFourier(coords['y'], size=ny, bounds=(0, ly), dealias=dealias)
zbasis = d3.Chebyshev(coords['z'], size=nz, bounds=(-lz/2, lz/2), dealias=dealias)

# Fields
p = dist.Field(name='p', bases=(xbasis,ybasis,zbasis))
u = dist.VectorField(coords, name='u', bases=(xbasis,ybasis,zbasis))
tau_p = dist.Field(name='tau_p')
tau_u1 = dist.VectorField(coords, name='tau_u1', bases=(xbasis,ybasis)) #is this correct?
tau_u2 = dist.VectorField(coords, name='tau_u2', bases=(xbasis,ybasis))

# Substitutions
x, y, z = dist.local_grids(xbasis, ybasis, zbasis)
ex = dist.VectorField(coords, name='ex')
ey = dist.VectorField(coords, name='ey')
ez = dist.VectorField(coords, name='ez')
ex['g'][0] = 1
ey['g'][1] = 1
ez['g'][2] = 1
integ = lambda A: d3.Integrate(d3.Integrate(A, ('x','y')), 'z')  #is this correct?
lift_basis = zbasis.clone_with(a=1/2, b=1/2) # First derivative basis
lift = lambda A: d3.Lift(A, lift_basis, -1)
grad_u = d3.grad(u) + ez*lift(tau_u1) + ey*lift(tau_u1)# First-order reduction  #is this correct?

problem.add_equation("dt(u) + grad(p) - div(grad_u)/Reynolds + lift(tau_u2) =  -dot(u,grad(u))") #is this correct?

# Boundary conditions (just one boundary)
problem.add_equation("dot(ez, dot(ex,e_ij))(z=-lz/2) = 0")
problem.add_equation("dot(ey, dot(ex,e_ij))(z=-lz/2) = 0")
problem.add_equation("dot(ez, u)(z=-lz/2) = 0")

Thank you.

Regards,

Kyriacos

[Daniel Lecoanet]

Hi Kyriacos,

That’s almost right. For grad_u you want to only include the ez*lift(tau_u1) term. Then you need to specify 6 BCs. Also, for stress-free BCs, typically you would set the normal-tangential components of the stress to be zero (the yz and xz components), whereas you’re setting some tangential-tangential components of the stress to zero.

Daniel

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