2. Chapter 4 第四章 Unsteady-State (Transient) Conduction 非稳态导热

3. 4-1 Introduction 1. Basic conception Unsteady-state (or transient) conduction Temperature distribution in a system varies with time Steady-state conduction Temperature distribution in a system does not varies with time

4. 2. Example The shell of steam turbine Before start t f1 =t w1 =t w2 =t f2 Admission of steam t f1  Inside of the shell q 1 =h 1 (t f1 -t w1 ) At a particular time h 1A1 (t f1 -t w1 )=h 2 A 2 (t w2 -t f2 ) Steady-state conduction penetration time 穿透时间： Thermal layer 穿透深度

5. 3. Problems to be solved 1). Temperature distribution at a given time thermal stress ( 热应力 ) 2). The time to reach a given temperature distribution or steady-state quenching ( 淬火过程 ) 3). The heat transferred

6. 3. Infinite plate subjected to sudden cooling of surfaces Set

7. Graphical form for calculation

8. Set Math model Third boundary condition no heat generation, h=const. T f =const.

9. Or From the first boundary condition Separating variables gives general solution From the second boundary condition

10. There are infinite number solutions to the equation The final series form of the solution is From initial condition

11. 4-2 lumped-heat-capacity system Generally t=f(x,y,z,  ) If internal conduction resistance  0, uniform in temperature Temperature field t=f(x,y,z,  ) reduces to t=f(  ), 0-D problem In fact, impossible for internal conduction resistance to be zero, but if it is small enough, we believe t=f(  ). This method is called lumped method or heat conduction with negligible internal conduction resistance ( 集总参数法 或 忽略物体内部导热热阻的简化分析方法 )

12. 1. Physical problem k=const. Uniform T distribution Bi  0, h=const. Find T=f(  ) 2. Mathematical model No boundary condition. Convection heat transfer taken as source item

13. Then 3. Solve Then

14. Characteristic dimension of solid (V/A=s) 4. Heat transfer rate Heat transferred from 0 to 

15. 5. Biot number Bi  L/λ has decisive effect, convection waits for conduction t w  t , become first kind of boundary condition Bi  0 L/λ is very little ， temperature tends towards uniform Bi = some certain value ， L/λ and 1/h play an important role

16. 6. Applicability of lumped-capacity Analysis Characteristic dimension of solid (V/A=s)

17. Plane wall Bi V = Bi Cylinder Bi V = Bi/2 Sphere Bi V = Bi/3 M= 1 (for plane wall), ½ (for cylinder) and 1/3 (for sohere) Reason M= Bi V /Bi Characteristic dimension of solid (V/A=s) In our country

18. 7. Time constant 36.8% 5% The rate of temperature change  c  rate 

19. 4-3 Transient heat flow in a semi-infinite solid 1. Physical problem Introduction of a new variable 2. Mathematical model

20. Substitute into the eq.

21. To eliminate , set 2m +1=0, that is m = -1/2 Then

22. From initial condition Then

23. 3. Two important parameters: 4. Heat flow

24. 5. Constant heat flux on semi-infinite solid Governing eq. and initial condition are the same Boundary condition 6. Energy pulse at surface Instantaneous pulse of energy Q 0 /A

25. 4-4 Convection boundary conditions 1.Convection boundary For the semi-infinite-solid problem, the boundary condition The solution

26. 2. Important case 1-D solids suddenly subjected to convection environment at T ∞ Infinite plate Infinite cylinder and sphere the nature of series: the first term is the largest, then the 2nd, 3rd … when Fo is large enough the decrease is rapid. when Fo>0.2, the the error of the single-term approximation of the series is less than <0.1% 。

27. The ratio of  (x,  ) to  0 is It depends on the position and boundary condition, is independent of . Initial condition has no effect on temperature distribution Take infinite plate as an example This region is called regular regime ( 正规热状况 ) or fully developed regime （ 充分发展阶段 ) Fig. 4-13 Fig. 4-7 Chart solution

28. 3. Boit number and Fourier number 4. Applicability of the Heisler Charts

29. 4-5 Multidimensional systems 1. Physical problem: Infinite rectangular bar 2. Mathematical model

30. Set dimensionless temperature difference Substitute into eq.

31. If equation Has a solution T 1 (x,  ), and has a solution of T 2 (y,  ) is the solution to the equation Prove ： performing partial differentiation of  In a similar way 3. Solution

32.  (x,y,  )= t 1 (x,  ) t 2 (y,  ) satisfies the governing equation, and also satisfies the initial condition. By the same way, it can be proved that  (x,y,  )= t 1 (x,  ) t 2 (y,  ) satisfy other two boundary conditions. Substituting these relations in the governing equation

33. By the same way * Constant 1st and 3rd kinds of boundary conditions

35. 4. Heat transfer in multidimensional systems

36. 4-6 Transient numerical method Space coordinate x 1  N Space increment  x 1.Mathematical models 2.Discretization of domain (node, grid) 3.Algebraic equations for all the nodes 4.Initial variable fields 5.Solving the algebraic equation 6.Analysis of the solution Time coordinate  1  I Time increment( 时间步长 )   T (p) n is the temperature of node (n,p) Procedure is the same as steady-state

37. 1-D, λ=const. problem Implicit finite difference scheme ( 隐示格式 ) Governing equation Forward difference backward difference Explicit finite difference scheme ( 显示格式 )

38. Explicit formulation If the time and space increment are chosen so that If △ x and △  the rate that the solution proceeds  the accuracy  △  depends △ x, If M>2 the coefficient becomes negative, the condition violates the 2 nd law of thermodynamics. Stability problem, M 2 instable

39. Boundary For uniform grid  x=  y For 1-D problem Zero coefficient To insure convergence

40. Set Fourier number in the numerical format Infinite plate Convection boundary node Interior node Biot number in the numerical format 4-6 Thermal resistance and capacity formulation Self-learning