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yuanzhut

于 2007-04-11 发布 文件大小:1KB
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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)

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