文件名称:HeatTransport
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数值方法计算传热问题实例
边长为200×200mm的二维矩形材料内的稳态温度分布。上侧和左侧维持恒壁温；下侧为绝热；右侧为对流换热条件。具体参数为：
材料的导热系数=20w/m• k；
上侧温度t1=100℃
右侧流体温度tf=20℃ 表面传热系数 h分别为1w/m2• k和100 w/m2• k
左侧温度t2=20℃
下侧绝热-Numerical method to calculate the heat transfer problem instance side of 200 × 200mm rectangular two-dimensional steady-state temperature distribution within the material. On the left side and to maintain constant wall temperature lower side of insulation the right side of the convective heat transfer conditions. Specific parameters are: thermal conductivity of the material = 20w/m • k on the side of the temperature t1 = 100 ℃ right side of the fluid temperature tf = 20 ℃ surface heat transfer coefficient h were 1w/m2 • k and 100 w/m2 • k temperature t2 = 20 ℃ left side of the insulation under
边长为200×200mm的二维矩形材料内的稳态温度分布。上侧和左侧维持恒壁温；下侧为绝热；右侧为对流换热条件。具体参数为：
材料的导热系数=20w/m• k；
上侧温度t1=100℃
右侧流体温度tf=20℃ 表面传热系数 h分别为1w/m2• k和100 w/m2• k
左侧温度t2=20℃
下侧绝热-Numerical method to calculate the heat transfer problem instance side of 200 × 200mm rectangular two-dimensional steady-state temperature distribution within the material. On the left side and to maintain constant wall temperature lower side of insulation the right side of the convective heat transfer conditions. Specific parameters are: thermal conductivity of the material = 20w/m • k on the side of the temperature t1 = 100 ℃ right side of the fluid temperature tf = 20 ℃ surface heat transfer coefficient h were 1w/m2 • k and 100 w/m2 • k temperature t2 = 20 ℃ left side of the insulation under
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no_problem.m