Download presentation

Presentation is loading. Please wait.

Published byCody Newnham Modified over 2 years ago

1
Basic law of heat conduction --Fourier’s Law Degree Celsius

2
Conduction may be viewed as the transfer of energy from the more energetic to the less energetic particles of a substance due to interactions between the particles The mechanism in gases: The temperature at any point can be associated with the energy of gas molecules in proximity of the point. The energy is related to the random transitional motion, as well as the rotational and vibrational motions, of the molecules. Physical mechanism of heat-conduction

3
The dependence of thermal conductivities of gases on temperature For most gases at moderate pressures the thermal conductivity is a function of temperature alone. The faster the molecules move, the faster they will transport energy. Therefore the thermal conductivity of a gas should be dependent on temperature. The thermal conductivity of a gas varies with the square root of the absolute temperature.

4
The mechanism of heat conduction in liquids is similar to that in gases, but the molecules are more closely spaced and the interactions between molecules are more stronger and more frequently.

5
Two modes of heat conduction in solids: Lattice vibration: energy transfer may be attributed to atomic activities in the form of lattice( 晶格，格子 ) vibrations. Free electrons transport: large number of free electrons are moving about in the lattice structure, and carry thermal energy from higher temperature region to lower temperature region.

6
Governing equations One-dimensional heat conduction equation: Energy conducted in left face+heat generated within element=change in internal energy+energy conducted out right face Combining the above relations gives: or Energy in left face Energy generated within element Change in internal energy Energy conducted out right face

7
For constant thermal conductivity, the above equation is written is called thermal diffusivity. The larger the value of, the faster heat will diffuse through the material.

8
Cartesian coordinates: Cylindrical coordinates: Spherical coordinates: 3-dimensional heat conduction equations

9
Newton’s law

10
Stefan-Boltzmann law Grey body emissive power Emissive power of black body, RADIATION IN AN ENCLOSURE

11
2-2 THE PLANE WALL For single layer

12
For multiple layers

13
2-4 RADIAL SYSTEMS Cylinders

15
Spheres Consider a shell, the temperatures at the inside and outside walls are maintained at Ti and To, respectively, the heat flow through the shell is ToTo TiTi

16
Convection boundary condition Convection heat transfer rate is Rearranging gives:

17
2-6 CRITICAL THICKNESS OF INSULATION Consider the right tube. The inner wall is maintained at Ti; the outer surface is exposed to a convection environment. From the thermal network the heat transfer is The maximum condition is Which gives

18
2-7 HEAT SOURCE SYSTEMS Plane wall with heat source The heat source is uniformly distributed in the plane wall, calculate the temperature distribution in the plane wall. Governing equation is With boundary condition: General solution is The temperature on each side is the same C 1 =0 C 2 is the temperature at the midplane Thereforeor

19
How to get T o ? Total heat generated must equal to the heat lost at the faces. By differentiating equation gives then and Alternative form of temperature distribution

21
2-8 CYLINDER WITH HEAT SOURCES Governing equation Boundary condition Heat generated equals heat lost at surface: At the center of the cylinder (A) Rewrite (A) Note that Integration yields and From red thus Final solution

22
Dimensionless form To is the temperature of center For a hollow cylinder The general solution is Using boundary conditions yields where

23
2-11 THERMAL CONTACT RESISTANCE —thermal contact resistance —contact coefficient

24
A c —the contact area A v —the void area L g —the thickness of the void space K f —thermal conductivity of the fluid which fills the void space

25
3-2 MATHEMATICAL ANALYSIS OF TWO- DIMENSIONAL HEAT CONDUCTION (5) (4) Problem: determine temperature distribution in a rectangular plate Approach: separation of variables method. （分离变量法） T m is the amplitude of the sine function

26
(6) (7) (8) Note that each side of the equation is independent of the other Substituting T=XY into the Laplace equation gives —Separation constant, determined from the boundary conditions.

27
3 possible solutions it is impossible possible solution (7) (8)

28
Substitution Boundary conditions Appling these conditions, we have Accordingly, (a) (b) (c) (d)

29
thus This require From © Final solution form Appling the final solution gives Which requires that C n =0 for n>1, therefore the final solution is 双曲函数

30
Problem: determine the temperature distribution in a rectangular plate. Using the first 3 boundary conditions, obtain the solution Appling the fourth boundary condition gives Comparing (a) and (b) gives Expanding T 2 -T 1 into Fourier series gives Final solution is (a) (b)

31
3-3 GRAPHICAL ANALYSIS( 作图法） —Conduction shape factor Methods of constructing flux plots （通量图） 1.Trial-and error method( 试错法） 2. Experimental measurement

32
3-4 THE CONDUCTION SHAPE FACTOR The calculation of inverse hyperbolic cosine Definition of conduction shape factor Separate shape factors of 3-dimensional wall A = area of wall L = wall thickness D = length of edge

33
3-5 NUMERICAL METHOD OF ANALYSIS Finite difference form of heat equation

34
Conclusion: the net heat flow into any node is zero at steady state conditions The finite difference scheme of governing equation with heat source.

35
External corner with convection boundary Convection boundary nodal equations Equations for plane surface nodes

38
3-6 NUMERICAL FORMULATION IN TERMS OF RESISTANCE ELEMENTS

40
3-7 GAUSS-SEIDEL ITERATION （高斯－赛德尔迭代） from We obtain The solution can be obtained by Gauss_Seidel Iteration: 1.An initial set of values for the T i is assumed. 2.Using the most recent values of T j, the new values of T i are calculated from the above equation 3.The process continues until the following requirements are satisfied. or

41
Biot number By setting Bi=0, the convection boundary can be converted into insulated boundary. Heat sources and boundary radiation exchange For radiation exchange at boundary node Net radiation transferred to node i per unit area

Similar presentations

Presentation is loading. Please wait....

OK

ERT 216 HEAT & MASS TRANSFER Sem 2/

ERT 216 HEAT & MASS TRANSFER Sem 2/

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on balanced diet and nutrition Free ppt on teachers day Download ppt on turbo generator manufacturers Ppt on accounts receivable management Ppt on 9/11 terror attack impact on world economy Ppt on limitation act Ppt on family tree Ppt on indian army weapons systems Ppt on different types of houses for kids Ppt on 3-phase squirrel cage induction motor