瞬态

简单墙构造由混凝土隔热计算80小时初始化整堵墙有相同的温度外墙冷却并达到新温度场

*BOUNDARY_TEMPERATURE_SET *CONTROL_SOLUTION *CONTROL_TERMINATION *CONTROL_THERMAL_SOLVER *CONTROL_THERMAL_TIMESTEP *DATABASE_BINARY_D3PLOT *DATABASE_TPRINT *DEFINE_CURVE *ELEMENT_SOLID *END *INCLUDE *INITIAL_TERMPERATURE_SET *KEYWORD *MAT_THERMAL_ISOTROPIC *NODE *PART *SECTION_SOLID *SET_NODE_LIST_TITLE *TITLE

*KEYWORD $ $=============================CONTROL DEFINITIONS ============================== $ *TITLE transient heat transfer through an insulated concrete wall *CONTROL_SOLUTION $ thermal only $     soln       nlq     isnan     lcint     lcacc     ncdcf          1 *CONTROL_THERMAL_SOLVER $ transient thermal analysis $    atype     ptype    solver     cgtol       gpt    eqheat     fwork       sbc          1         0        11 $   msglvl    maxitr    abstol    reltol     omega                           tsf  *CONTROL_THERMAL_TIMESTEP $       ts       tip       its      tmin      tmax     dtemp      tscp      lcts          1       1.0        10       1.0    3600.0                  *CONTROL_TERMINATION $   endtim    endcyc     dtmin    endeng    endmas     nosol   288000.0 $ $================================ OUTPUTDATA =================================== $ *DATABASE_BINARY_D3PLOT $       dt      300.0 *DATABASE_GLSTAT $       dt      300.0 *DATABASE_MATSUM $       dt      300.0 *DATABASE_TPRINT $       dt       60.0 $ $============================== PART DEFINITIONS =============================== $ *PART $ title concrete $      pid     secid       mid     eosid      hgid      grav    adpopt      tmid          1         1         0         0         0         0         0         1 *PART $ title insulation $      pid     secid       mid     eosid      hgid      grav    adpopt      tmid          2         1         0         0         0         0         0         2 $ $============================= SECTION PROPERTIES ============================== $ *SECTION_SOLID $    secid    elform          1         1 $ $======================== THERMAL MATERIAL PROPERTIES ========================== $ *MAT_THERMAL_ISOTROPIC $ material concrete $     tmid       tro     tgrlc    tgmult      tlat      hlat          1    2.4e-9 $       hc        tc        8.80e8       2.1 $  *MAT_THERMAL_ISOTROPIC $ material insulation $     tmid       tro     tgrlc    tgmult      tlat      hlat          2   1.0e-10 $       hc        tc         9.0e8      0.04 $ $======================== NODE/ELEMENT/SET DEFINTIONS ========================== $ *INCLUDE 01_insulated_concrete_wall_parts.k *SET_NODE_LIST_TITLE inside $      sid       da1       da2       da3       da4    solver                1       0.0       0.0       0.0       0.0MECH $     nid1      nid2      nid3      nid4      nid5      nid6      nid7      nid8         22         1        64        85        43       106       127       148        169         0         0         0         0         0         0         0 *SET_NODE_LIST_TITLE outside $      sid       da1       da2       da3       da4    solver                2       0.0       0.0       0.0       0.0MECH $     nid1      nid2      nid3      nid4      nid5      nid6      nid7      nid8        394       230       271       353       312       435       517       476        558         0         0         0         0         0         0         0 $ $==================== THERMAL BOUNDARY and INITIAL CONDITIONS ================== $ *INITIAL_TEMPERATURE_SET $  nid/sid      temp       loc          0      20.0 *BOUNDARY_TEMPERATURE_SET $  nid/sid     tlcid     tmult       loc    tdeath    tbirth          1         0      20.0          2         1       1.0 *DEFINE_CURVE $     lcid      sidr       sfa       sfo      offa      offo    dattyp          1         0         1         1         0         0         0 $                 a1                  o1                  0.0                20.0                600.0                10.0                1.0e6                10.0 $ *END