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1、Heat Transfer Review,2,Chapter 1 Basic Concepts of Thermodynamics and Heat Transfer,Heat Transfer Modes,the mechanisms of heat exchange & the methods to compute the rate of heat exchange,3,Chapter 1 Basic Concepts of Thermodynamics and Heat Transfer,4,Fouriers Law and Thermal Conductivity,Chapter 1
2、Basic Concepts of Thermodynamics and Heat Transfer,5,Thermal Conductivity, determined by experiments ksolids(10420) kliquds(0.070.7) kgases(0.006 0.3) k0.12, insulating material (porous structure),Chapter 1 Basic Concepts of Thermodynamics and Heat Transfer,6,Thermal Diffusivity,It measures the abil
3、ity of a material to conduct thermal energy relative to its ability to store thermal energy; Materials of large will respond quickly to changes in their thermal environment; Materials of small will respond more sluggishly, taking longer to reach a new equilibrium condition.,Chapter 1 Basic Concepts
4、of Thermodynamics and Heat Transfer,7,Newtons Law of cooling and convection heat transfer coefficient,Ts: surface temperature T: temperature of the fluid sufficiently far from the surface h: convection heat transfer coefficient W/m2.oC,Chapter 1 Basic Concepts of Thermodynamics and Heat Transfer,8,S
5、tefan-Boltzmann Law the rate of radiation emission A: surface area m2; Ts: surface temperature in Kelvin; =5.67x10-8W/m2K4: Stefan-Boltzmann constant; 0 1 : emissivity of the surface.,Blackbody radiation represents the maximum amount of radiation that can be emitted from a surface at a specified tem
6、perature.,Chapter 1 Basic Concepts of Thermodynamics and Heat Transfer,9,Chapter 1 Basic Concepts of Thermodynamics and Heat Transfer Simultaneous Heat Transfer Mechanisms,Although there are three mechanisms of heat transfer, a medium may involve only two of them simultaneously.,10,如附图所示，加热炉炉墙厚=0.3m
7、，导热系数k=0.6W/(m.K)， 墙外表面的发射率2=0.8，墙外空气温度tf=250C。对流换热的表面传热系数h=12W/(m2.K)，在稳态情况下测得墙外表面温度t2=500C ，周围辐射环境温度Tsur=298K，求墙的内表面温度t1。,例题1-1,解：,已知：炉墙的几何参数、外表面温度、物性参数，以及外部换热环境的有关数据。 求：墙的内表面温度。,11,假设： （1）一维稳态导热过程；（2）墙外的辐射环境温度与周围空气温度相同；（3）物性参数为常数。,例题1-1,代入每种传热方式对应的热流速率方程，有,分析与计算：取附图所示的控制面作为能量平衡关系的分析对象。稳态条件下，从炉墙内表
8、面传出的热量必将以对流和辐射的方式散出，根据对单位表面面积的能量守恒关系，可列出,12,其中，对流散热量和辐射散热量分别等于,例题1-1,以上两式之和为：,因此，,得，,13,例题1-1,讨 论：,（1）一旦对流项与辐射项同时出现在一个方程式中，必须把温度统一写成热力学温度的形式，否则容易酿成大错。 （2）在本问题给定的参数下，辐射散热量占总量的31%，对流部分则占69%。这个比例是随表面发射率、表面传热系数,以及墙壁表面温度与周围换热环境温度的差别大小变化的。温度水平越高，温度差距越大，辐射部分的比例将快速增长。,14,Temperature Gradient : the normal of
9、 the isotherm at point P,Chapter 2 Heat Conduction Equation,For a unsteady problem it depends also on the time!,P,(vector),In rectangular coordinates,15,Chapter 2 Heat Conduction Equation,Fouriers Law In rectangular coordinates,Chapter 2 Heat Conduction Equation,General Form of Heat Conduction Equat
10、ion,Cartesian Coordinates,Cylindrical Coordinates,Spherical Coordinates,Boundary and Initial Conditions,1. Specified Temperature Boundary Condition (a boundary condition of the first kind Dirichlet condition),2. Specified Heat Flux Boundary Condition (a boundary condition of the second kind Neumann
11、condition),3. Convection Boundary Condition (a boundary condition of the third kind Robin condition),4. Radiation Boundary Condition,5. Interface Boundary Conditions,Chapter 2 Heat Conduction Equation,18,一维无内热源、平壁稳态导热的温度场如附图所示。试说明它的导热系数k是随着温度增加而增加，还是随着温度增加而减小？,例题2-1,解：由傅里叶定律得,因此，k(x)随x增加而增加，而温度 t 随
12、x 增加而降低，所以导热系数k是随着温度增加而减小。,19,大平壁某一瞬间的温度分布形状如附图所示。如果其物性均为常数，试问：此刻该平壁处于加热还是处于冷却中？,例题2-2,已知：大平壁的瞬时温度分布曲线。,解：,求：判断该平壁处于升温还是降温状态。,假设：（1）一维稳态导热过程；（2）没有内热源；（3）物性参数为常数。,分析与计算：控制容积如附图所示，取侧表面面积为A，对所取的控制容积运用能量守恒关系，可列出,20,依照傅里叶定律和平壁热力学能增量的表达式，在图示坐标系下可以写出,例题2-2,即,按照所给的温度分布曲线不难判断，左侧面的温度梯度大于右侧面，且在该坐标系中两者均为正值。因此，该
13、控制容积温度随时间的变化率必定是负值，即平壁处于降温过程中。,Chapter 3 Steady Heat Conduction,General Form of Steady Heat Conduction Equation,Cartesian Coordinates,Cylindrical Coordinates,Spherical Coordinates,Thermal Resistance &Equivalent Thermal Circuit series arrangement: parallel arrangement: combined series-parallel: analo
14、gous to the electrical circuit Analysis,RA: the thermal resistance per unit area m2.OC/W,R: the thermal resistance OC/W,! Please note the difference between,&,Chapter 3 Steady Heat Conduction,The elementary thermal resistance relations: Conduction resistance (plane wall): Conduction resistance (cyli
15、nder): Conduction resistance (sphere): Convection resistance: Interface resistance: Radiation resistance:,m2 .C/W,C/W,C/W,m2 .C/W,m2 .C/W,m2 .C/W,Chapter 3 Steady Heat Conduction,24,Multilayer Plane Walls (1-D),where,Heat Transfer Rate for the System:,To Find:,Chapter 3 Steady Heat Conduction,Known:
16、 L1, L2, L3,k1,k2,k3,T1, T2, h1, h2,25,Chapter 3 Steady Heat Conduction,Temperature Distribution: T(x),In the First Layer:,Note that , also !,26,Chapter 3 Steady Heat Conduction,1. The Cylinder (one layer),Heat Conduction in Cylinders and Spheres,The Rate of Heat Transfer through the Cylindrical Layer:,Note that (steady), but (A=f(r) !,27,2. The Sphere (one layer),The Rate of Heat Transfer through the Sph
