文件名称:yuanzhut
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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
-cup of boiling water cooling cylinder model, the bottom radius of 0.05m, 0.1m high. temperature around 20 degrees, initial temperature of 100 degrees equation is four-dimensional equations (a constant ^ 2 = k/(c* p), k is the thermal conductivity coefficient of 0.6006 Coke/(m** seconds degrees)) initial conditions : t = 0 to 100 degree water boundary conditions : 1. upper and lower walls are free cooling, the third boundary conditions, maintain the temperature around 20 degrees (H = k/h, h for 1) 2. Beibi insulation, the second boundary condition
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
-cup of boiling water cooling cylinder model, the bottom radius of 0.05m, 0.1m high. temperature around 20 degrees, initial temperature of 100 degrees equation is four-dimensional equations (a constant ^ 2 = k/(c* p), k is the thermal conductivity coefficient of 0.6006 Coke/(m** seconds degrees)) initial conditions : t = 0 to 100 degree water boundary conditions : 1. upper and lower walls are free cooling, the third boundary conditions, maintain the temperature around 20 degrees (H = k/h, h for 1) 2. Beibi insulation, the second boundary condition
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模拟圆柱体的热传导.m