$-------------------------------------------------------------------------------
$            RIGID FORMAT No. 1 (APP HEAT), Heat Conduction Analysis
$             Linear Steady State Heat Conduction Through a Washer
$                         Using Solid Elements (1-12-1)
$             Linear Steady State Heat Conduction Through a Washer
$                         Using Ring Elements (1-12-2)
$ 
$ A. Description
$ 
$ This problem illustrates the capability of NASTRAN to solve heat conduction
$ problems. The washer has a radial heat conduction with the temperature
$ specified at the outside and a film heat transfer condition at the inner edge.
$ Due to symmetry about the axis and the assumption of negligible axial
$ gradients, the temperature depends only upon the radius.
$ 
$ B. Input
$ 
$ In the first NASTRAN model, the solid elements (HEXA1, HEXA2, WEDGE, and
$ TETRA) and boundary condition element (HBDY, type AREA4) are used. The
$ conductivity of the material is specified on a MAT4 card. Temperatures are
$ specified at the outer boundary with SPC cards. Punched temperature output is
$ placed on TEMP bulk data cards suitable for static analysis.
$ 
$ Another variation of the problem uses solid of revolution elements (TRIARG and
$ TRAPRG) and boundary condition element (HBDY, type REV). The conductivity of
$ the material and the convective film coefficient are specified on a MAT4 card.
$ The CHBDY card references a scalar point at which the ambient temperature is
$ specified using an SPC card. An SPC1 card is used to constrain the temperature
$ to zero degrees at gridpoInts on the outer surface.
$ 
$ C. Theory
$ 
$ The mathematical theory for the continuum is simple, and can be solved in
$ closed form. The differential equation is
$ 
$      1  a      aU
$      -  --  (rk--) = 0                                                     (1)
$      r  ar     ar
$ 
$ The boundary conditions are
$ 
$         aU
$      -k -- = H(U  - U) at r = r                                            (2)
$         ar      a              1
$ 
$ and
$ 
$     U = 0 at r = r                                                         (3)
$                    2
$ 
$ The solution is
$ 
$                    HU
$                      a
$      U(r) = --------------------    ln(r /r)
$             (k/r ) + H ln(r /r )        2
$                 1          2  1
$ 
$           = 288.516 ln(2/r)
$ 
$ D. Results
$ 
$ A comparison with the NASTRAN results is shown in Table 1.
$ 
$           Table 1. Comparison of Theoretical and NASTRAN Temperatures
$                         for Heat Conduction in a Washer
$     ------------------------------------------------------------------------
$                  Theoretical    NASTRAN Temperatures   NASTRAN Temperatures
$      r(radius)   Temperatures         (Solids)*             (Rings)*
$     ------------------------------------------------------------------------
$         1.0         199.984          202.396                  199.932
$ 
$         1.1         172.486          173.904                  172.448
$ 
$         1.2         147.381          148.833                  147.355
$ 
$         1.3         124.288          124.783                  124.269
$ 
$         1.4         102.906          102.852                  102.894
$ 
$         1.5          83.001           82.913                   82.992
$ 
$         1.6          64.380           64.306                   64.375
$ 
$         1.7          46.889           46.832                   46.886
$ 
$         1.8          30.398           30.356                   30.397
$ 
$         1.9          14.799           14.773                   14.798
$ 
$         2.0           0.000            0.000                    0.000
$     ------------------------------------------------------------------------
$       * These are the average temperatures at a radius.
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