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ex_heattransfer1.m File Reference

Description

EX_HEATTRANSFER1 2D ceramic strip with radiation and convection.

[ FEA, OUT ] = EX_HEATTRANSFER1( VARARGIN ) 2D heat transfer of a ceramic strip with both radiation and convection on the top boundary.

  _                q = h*(T_inf-T) + epsilon*sigma*(T_inf^4-T^4)
  ^            +------------------+
  |            |                  |
  |            |                  |
0.01m  T=900C  |                  |  T=900C
  |            |                  |
  |            |                  |
  v            +------------------+
                    dt/dn = 0
               |<---- 0.02m ----->|

The ceramic has a thermal conductivity of 50 W/mC and the sides are fixed at a temperature of 900C while the bottom boundary is insulated. The surrounding temperature is 50C. The top boundary is exposed to both natural convection (with a film coefficient h=50W/m^2K) and radiation (with emissivity epsilon=0.7 and the Stefan-Boltzmann 5.669e-8 W/m^2K^4). The solution is sought at three points along the vertical symmetry line.

Reference
  [1] Holman, J. P., Heat Transfer, Fifth Edition, New York: McGraw-Hill,
      1981, page 96, Example 3-8.

Accepts the following property/value pairs.

Input       Value/{Default}        Description
-----------------------------------------------------------------------------------
hmax        scalar {0.001}         Grid cell size
igrid       scalar {0}/1/2         Cell type (0=quadrilaterals, 1=triangles,
sfun        string {sflag1}        Finite element shape function
iplot       scalar {1}/0           Plot solution (=1)
                                                                                  .
Output      Value/(Size)           Description
-----------------------------------------------------------------------------------
fea         struct                 Problem definition struct
out         struct                 Output stuct

Code listing

 cOptDef = { 'hmax',     0.001;
             'igrid',    0;
             'sfun',     'sflag1';
             'iplot',    1;
             'tol',      0.01;
             'fid',      1 };
 [got,opt] = parseopt(cOptDef,varargin{:});


% Geometry definition.
 gobj = gobj_rectangle( 0, 0.02, 0, 0.01 );
 fea.geom.objects = { gobj };


% Grid generation.
 switch opt.igrid
   case 0
     fea.grid = rectgrid( round(0.02/opt.hmax), round(0.01/opt.hmax), [0 0.02;0 0.01] );
   case 1
     fea.grid = gridgen( fea, 'hmax', opt.hmax, 'fid', opt.fid );
   case 2
     fea.grid = rectgrid( round(0.02/opt.hmax), round(0.01/opt.hmax), [0 0.02;0 0.01] );
     fea.grid = quad2tri( fea.grid, 1 );
 end


% Problem definition.
 fea.sdim  = { 'x', 'y' };             % Space coordinate name.
 fea = addphys( fea, @heattransfer );  % Add heat transfer physics mode.
 fea.phys.ht.sfun = { opt.sfun };      % Set shape function.

% Equation coefficients.
 fea.phys.ht.eqn.coef{3,end} = 3;      % Thermal conductivity.

% Boundary conditions.
 fea.phys.ht.bdr.sel = [3 1 4 1];
 fea.phys.ht.bdr.coef{1,end}   = { [] 900+273 [] 900+273 };
 fea.phys.ht.bdr.coef{4,end}{3}{2} = 50;
 fea.phys.ht.bdr.coef{4,end}{3}{3} = 50+273;
 fea.phys.ht.bdr.coef{4,end}{3}{4} = 0.7*5.669e-8;
 fea.phys.ht.bdr.coef{4,end}{3}{5} = 50+273;


% Parse physics modes and problem stuct.
 fea = parsephys(fea);
 fea = parseprob(fea);


% Compute solution.
 fea.sol.u = solvestat( fea, 'fid', opt.fid, 'init', {'T0_ht'} );


% Postprocessing.
 if( opt.iplot>0 )
   postplot( fea, 'surfexpr', 'T', 'isoexpr', 'T' )
   title('Temperature, T')
 end


% Error checking.
 T2_sol = evalexpr( 'T', [0.01;0.01], fea );
 T2_ref = 984;
 T5_sol = evalexpr( 'T', [0.01;0.005], fea );
 T5_ref = 1064;
 T8_sol = evalexpr( 'T', [0.01;0], fea );
 T8_ref = 1088;
 out.err  = abs([T2_sol-T2_ref T5_sol-T5_ref T8_sol-T8_ref])./[T2_ref T5_ref T8_ref];
 out.pass = all(out.err<opt.tol);

 if( nargout==0 )
   clear fea out
 end