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ex_heattransfer3.m File Reference

Description

EX_HEATTRANSFER3 1D Transient heat conduction.

[ FEA, OUT ] = EX_HEATTRANSFER3( VARARGIN ) NAFEMS T3 benchmark example for one-dimensional transient heat conduction [1]. A 10 cm thick steel plate is assumed to have one surface exposed to a varying temperature, T = 100*sin(pi*t/40), while the temperature at the other side is fixed, T = 0. This problem can be seen as one dimensional along the axis aligned with the thickness.

     +---T(0.02)?---- L=0.02m ------------+
T=100*sin(pi*t/40)                        T=0

The temperature at x = 0.02 m is sought at time t = 36. The material parameters of the plate are, density 7200 kg/m^3, heat capacity 440.5 J/kgK, and thermal conductivity 35 W/mK.

Reference
  [1] The Standard NAFEMS Benchmarks,
      The National Agency for Finite Element Standards, UK, 1990.

Accepts the following property/value pairs.

Input       Value/{Default}        Description
-----------------------------------------------------------------------------------
hmax        scalar {0.005}         Grid cell size
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',     0.005;
             'sfun',     'sflag1';
             'iplot',    1;
             'fid',      1 };
 [got,opt] = parseopt(cOptDef,varargin{:});


% Grid generation.
 L        = 0.1;
 nx       = round(L/opt.hmax);
 fea.grid = linegrid( nx, 0, L );


% Problem definition.
 fea.sdim  = { 'x' };                  % 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{1,end} = 7200;   % Density
 fea.phys.ht.eqn.coef{2,end} =  440.5; % Heat capacity.
 fea.phys.ht.eqn.coef{3,end} =   35;   % Thermal conductivity.

% Boundary conditions.
 fea.phys.ht.bdr.sel = [ 1 1 ];
 fea.phys.ht.bdr.coef{1,end} = { '100*sin(pi*t/40)' 0 };


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


% Compute solution.
 [fea.sol.u, tlist] = solvetime( fea, 'fid', opt.fid, 'tmax', 32, 'tstep', 0.05 );


% Postprocessing.
 if( opt.iplot>0 )
   figure
   subplot(1,2,1)
   postplot( fea, 'surfexpr', 'T', 'axequal', 'off' )
   title('Temperature distribution at time t = 32 s')
   xlabel('x')
   ylabel('T')

   subplot(1,2,2)
   for isol=1:numel(tlist)
     T(isol) = evalexpr( 'T', 0.02, fea, isol );
   end
   plot(tlist,T)
   title('Temperature at x = 0.02 m')
   xlabel('time')
   ylabel('T')
 end


% Error checking.
 T_sol = evalexpr( 'T', 0.02, fea );
 T_ref = 36.6;
 out.err  = abs(T_sol-T_ref)/T_ref;
 out.pass = out.err<1e-2;


 if( nargout==0 )
   clear fea out
 end