Heat Conduction RZ coordination

[Erin Kim]

Hello all-

I have been recently looking at different i-files in HeatConductionTimeDerivative module.

I was pretty surprised while I was skimming through  'modules/combined/test/tests/heat_convection/heat_convection_rz_test.i '  file.

I initially thought there should be some kind of rz related codes under HeatConductionTimeDerivative.C file (for example) before we can actually utilize the kernel in RZ coordinate instead of XYZ.

Looking at the i file, however, it seems like we can simply choose RZ coordinate under Problem w/o writting any additional codes?

Is it true or is there any hidden magic under this i file?

Any advice would be very thankful.

Best,

Erin

FYI, the following is the i file.

Hello all-

I have been recently looking at different i-files in HeatConductionTimeDerivative module.

I was pretty surprised while I was skimming through  'modules/combined/test/tests/heat_convection/heat_convection_rz_test.i '  file.

I initially thought there should be some kind of rz related codes under HeatConductionTimeDerivative.C file (for example) before we can actually utilize the kernel in RZ coordinate instead of XYZ.

Looking at the i file, however, it seems like we can simply choose RZ coordinate under Problem w/o writting any additional codes?

Is it true or is there any hidden magic under this i file?

Any advice would be very thankful.

Best,

Erin

FYI, the following is the i file.

# Test cases for convective boundary conditions. TKLarson, 11/01/11, rev. 0.
# Input file for htc_2dtest1

# TKLarson
# 11/01/11
# Revision 0
#
# Goals of this test are:
#  1) show that expected results ensue from application of convective boundary conditions
# Convective boundary condition:
#  q = h*A*(Tw - Tf)
#  where
#    q - heat transfer rate (w)
#    h - heat transfer coefficient (w/m^2-K)
#    A - surface area (m^2)
#    Tw - surface temperature (K)
#    Tf - fluid temperature adjacent to the surface (K)
# The heat transfer coefficient (h) is input as a variable called 'rate'
# Tf is a two valued function specified by 'initial' and 'final' along with a variable
#  called 'duration,' the length of time in seconds that it takes initial to linearly ramp
#  to 'final.'

# The mesh for this test case is based on an ASTM standard for the so-called Brazillian Cylinder test
# (ASTM International, Standard Test Method for Splitting Tensile Strength of Cylindrical Concrete
# Specimens, C 496/C 496M-04, 2004) (because I already had a version of the model).  While the
# Brazillian Cylinder test is for dynamic tensile testing of concrete, the model works for the present
# purposes.  The model is 2-d RZ coordinates.
#
# Brazillian Cylinder sample dimensions:
#       L = 20.3 cm, 0.203 m, (8 in)
#       r = 5.08 cm, 0.0508 m, (2 in)

# Material properties are:
#   density = 2405.28 km/m^3
#   specific heat = 826.4 J/kg-K
#   thermal conductivity 1.937 w/m-K
#  alpha (thermal conductivity/(density*specific heat) is then 9.74e-7 m^2/s
#
# Initial cylinder temperature is room temperature 294.26 K (70 F)
# The initial fluid temperature is room temperature. We will ramp it to 477.6 K (400 F) in 10 minutes.
# We will use a natural convection h (284 w/m^2-K (50 BTU/hr-ft^2-F)) on all faces of the cylinder.
# This is akin to putting the cylinder in an oven (nonconvection type) and turning the oven on.

# What we expect for this problem:
#  1) Use of h = 284 should cause the cylinder to slowly warm up
#  2) The fluid temperature should rise from initial (294 K) to final (477 K) in 600 s.
#  3) 1) and 2) should cause the cylinder to become soaked at 477.6 K after sufficient time(i.e. ~ 1/2 hr).
# This is a simple thermal soak problem.

[Problem]
  coord_type = RZ
[]

[Mesh]    # Mesh Start
# 10cm x 20cm cylinder not so detailed mesh, 2 radial, 6 axial nodes
# Only one block (Block 1), all concrete
# Sideset 1 - top of cylinder, Sideset 2 - length of cylinder, Sideset 3 - bottom of cylinder
  file = heat_convection_rz_mesh.e
[]    # Mesh END

[Variables]  # Variables Start
  [./temp]
    order = FIRST
    family = LAGRANGE
    initial_condition = 294.26 # Initial cylinder temperature
  [../]

[]    # Variables END


[Kernels]  # Kernels Start
  [./heat]
    type = HeatConduction
    variable = temp
  [../]

  [./heat_ie]
    type = HeatConductionTimeDerivative
    variable = temp
  [../]

[]    # Kernels END


[BCs]    # Boundary Conditions Start
# Heat transfer coefficient on outer cylinder radius and ends
  [./convective_clad_surface]    # Convective Start
         type = ConvectiveFluxBC  # Convective flux, e.g. q'' = h*(Tw - Tf)
         boundary = '1 2 3'    # BC applied on top, along length, and bottom
         variable = temp
   rate = 284.      # (w/m^2-K)[50 BTU/hr/-ft^2-F]
          # the above h is a reasonable natural convection value
         initial = 294.26    # initial ambient (lab or oven) temperature (K)
         final = 477.6      # final ambient (lab or oven) temperature (K)
   duration = 600.    # length of time in seconds that it takes the ambient
          #   temperature to ramp from initial to final
  [../]          # Convective End

[]    # BCs END

[Materials]    # Materials Start
  [./thermal]
    type = HeatConductionMaterial
    block = 1
    specific_heat = 826.4
#    thermal_conductivity = 1.937  # this makes alpha 9.74e-7 m^2/s
#    thermal_conductivity = 19.37  # this makes alpha 9.74e-6 m^2/s
          # thermal conductivity arbitrarily increased by a decade to
          #    make the cylinder thermally soak faster (only for the purposes
          #    of this test problem
    thermal_conductivity = 193.7  # this makes alpha 9.74e-5 m^2/s
          # thermal conductivity arbitrarily increased by 2 decade to
          #    make the cylinder thermally soak faster (only for the purposes
          #    of this test problem

  [../]
  [./density]
    type = Density
    block = 1
    density = 2405.28
  [../]

[]      # Materials END

[Executioner]    # Executioner Start
   type = Transient
#   type = Steady

  #Preconditioned JFNK (default)
  solve_type = 'PJFNK'


   petsc_options = '-snes_ksp_ew '
   petsc_options_iname = '-ksp_gmres_restart -pc_type -pc_hypre_type'
   petsc_options_value = '70 hypre boomeramg'
   l_max_its = 60
   nl_rel_tol = 1e-8
   nl_abs_tol = 1e-10
   l_tol = 1e-5

   start_time = 0.0
  dt = 60.
  num_steps = 20  # Total run time 1200 s

[]      # Executioner END

[Outputs]    # Output Start
  # Output Start
  file_base = out_rz
  exodus = true
[]      # Output END
#      # Input file END

[Shohei Ogawa]

As far as I remember from the MOOSE workshop a couple of years ago, MOOSE is not agnostic on the coordinates at least the level of implementation of individual kernels. In other words, it is designed to handle the coordinate systems on background. Also, using the axial coordinate, you implicitly assume the z-axis is at y=0, which is set by a default parameter. If you take a look at the Executioner block using Peacock, you can see the default values, and things should be more clear.

[Shohei Ogawa]

It is not the Executioner block but the Problem block in your input file. (This is the right thread for this correction.)