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一杯沸水冷却，圆柱体模型，底面半径0.05m，高0.1m，周围温度20度，初始水温100度 方程是四维输运方程（常数a^2=k/(c*p)，k是热传导系数0.6006焦/（米*秒*度）） 初始条件：t=0时水等于100度 边界条件：1.上下壁都是自由冷却，第三类边界条件，周围温度保持在20度（H=k/h,h取1） 2.杯壁绝热，第二类边界条件 图形显示格式,取过圆柱轴的截面温度变化将其做成动画. (A cup of boiling water for cooling, the cylinder model, the bottom radius of 0.05m, the high-0.1m, an ambient temperature of 20 degrees, the initial water temperature of 100 degrees is four-dimensional transport equation equation (constant a ^ 2 = k/(c* p), k is the thermal conductivity coefficient of 0.6006 J/(m* s* degrees)) initial conditions: t = 0 when the water equivalent to 100 degrees boundary conditions: 1. up and down the wall are free cooling, the third boundary condition, the ambient temperature maintained at 20 degrees (H = k/h, h check one) 2.杯壁insulation, the second boundary condition graphical display format, check-off section of cylindrical shaft temperature change to make animation.)

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