传热学第10讲非稳态导热2.ppt

1、4.5 Transient Heat Conduction in Multidimensional Systems,Method:Product Solution,Long Solid Bar,Suppose: the Result of the solution is the products of two Planes solutions,Suppose:,Conclusion:,The Equations are the same,Also the boundary conditions are the same,Chapter 5 Numerical Methods in Heat T

2、ransfer,The limitations of analytical Methods,The Geometrical Shape and boundary Conditions Must be Simple!,5.1 Finite Difference,Difference of Differential Equation and Finite Differential Equation,Substitute:,Methods:,1)equation changing,Commutative error:,Therefore, the smaller the x, the smaller

3、 the error, and thus the more accurate the approximation.,Meaning: Tangent Line replacing,Now consider steady one-dimensional heat transfer in a plane wall of thickness L with heat generation.,With the constancy of,For each of points.,(M+1 points 0, 1, 2, . . . , m1, m, m1, . . . , M),The error will

4、 be,Note: the second derivatives,the first derivative of temperature dT/dx at the midpoints m-1/2 and m+1/2 of the sections surrounding the node m can be expressed as,Noting that the second derivative is simply the derivative of the first derivative, the second derivative of temperature at node m ca

5、n be expressed as,which is the finite difference representation of the second derivative at a general internal node m .,For each points you have designed here are called as nodes or nodal points.,The lines we used to divide the whole solving space will be called as mesh lines latter.,Steps To Solvin

6、g:,1) Making Meshes in a solving territory,2) Making Differential Equation to Finite Equations,3) Making Condition Equations to the form of Finite,4) Solving all the Linear Equations with Computer,5) Arranging the Data in a Proper Way,Problems well having:,1) The meshes are same,2) The conductivity

7、should be constant,In the most circumstances, the nodes dispatches and than the mesh lines dispatches will be designed differently with each other. Thus we use the following method.,Inner nodes:,One dimension problem：,Making integration for each of them.,In time from t to t +t, the temperature of no

8、de P will change form T t to T t+t.,The changing Enthalpy of the control volume represented by Node P,Assumption: that the temperature on the border is the average of the neighbors.,While integrating for the time then we need some assumptions.,Assumption 1：when time reaches to t+t for each points P,

9、 T t changes to T t+t suddenly,The form is:,Assumption 2：when time at t for each points P, T t changes to T t+t suddenly,The form is:,Assumption 3：the temperature changes with time in some relation like:,Boundary conditions,Methods : energy balance,有一房间的砖墙厚0.3m，导热系数0.85W/(mK)，热容量 1.05106 J/(m3K)，室内温度20保持不变，对流换热系数6 W/(m2K)。起初该墙的温度处于稳态，内墙表面温度为15。后寒潮入侵，室外温度下降到零下10，外墙表面换热系数为35 W/(m2K)。问寒潮入侵后多少时间内墙表面方可感到外界气温的变化？,习题：,问题：如何建立恰当的模型？有没有理论分析解？数值解与理论解会有什么相同点和不同点？解的结果如何表示，表示内容有哪些为好？,直角坐标系 圆柱轴对称坐标系 极坐标系,多维导热问题处理,直角坐标系下的为例，根据二维非稳态导热方程：,二维非稳态导热全隐格式的通用离散方程,条件：全隐格式 假设节点之间温度线