Chapter 3 Steady-State Conduction Multiple Dimensions HEAT TRANSFER LECTURE CHAPER 3 Steady-State Conduction Multiple Dimensions CUMT

3-1 Introduction In Chapter 2 steady-state heat transfer was calculated in systems in which the temperature gradient and area could be expressed in terms of one space coordinate. We now wish to analyze the more general case of two-dimensional heat flow. For steady state with no heat generation, the Laplace equation applies. HEAT TRANSFER LECTURE (3-1) The solution to this equation may be obtained by analytical, numerical, or graphical techniques. CUMT

3-1 Introduction HEAT TRANSFER LECTURE The objective of any heat-transfer analysis is usually to predict heat flow or the temperature that results from a certain heat flow. The solution to Equation (3-1) will give the temperature in a two-dimensional body as a function of the two independent space coordinates x and y. Then the heat flow in the x and y directions may be calculated from the Fourier equations CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE Analytical solutions of temperature distribution can be obtained for some simple geometry and boundary conditions. The separation method is an important one to apply. Consider a rectangular plate. Three sides are maintained at temperature T1, and the upper side has some temperature distribution impressed on it. The distribution can be a constant temperature or something more complex, such as a sine-wave. CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE Consider a sine-wave distribution on the upper edge, the boundary conditions are: CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE Substitute: We obtain two ordinary differential equations in terms of this constant, where λ2 is called the separation constant. CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE We write down all possible solutions and then see which one fits the problem under consideration. This function cannot fit the sine-function boundary condition, so that the solution may be excluded. CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE This function cannot fit the sine-function boundary condition, so that the solution may be excluded. CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE It is possible to satisfy the sine-function boundary condition; so we shall attempt to satisfy the other condition. CUMT

Apply the method of variable separation, let 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE Let The equation becomes: Apply the method of variable separation, let CUMT

And the boundary conditions become: 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE And the boundary conditions become: CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE Applying these conditions,we have: CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE accordingly, and from (c), This requires that CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE then We get The final boundary condition may now be applied: which requires that Cn =0 for n >1. CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE The final solution is therefore The temperature field for this problem is shown. Note that the heat-flow lines are perpendicular to the isotherms. CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE Another set of boundary conditions CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE Using the first three boundary conditions, we obtain the solution in the form of Equation: Applying the fourth boundary condition gives CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE This series is then The final solution is expressed as CUMT

3-2 Mathematical Analysis of Two-Dimensional Heat Conduction HEAT TRANSFER LECTURE Transform the boundary condition: CUMT

3-3 Graphical Analysis HEAT TRANSFER LECTURE neglect CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE Consider a general one dimensional heat conduct- ion problem, from Fourier’s Law: let then where：S is called shape factor. CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE Note that the inverse hyperbolic cosine can be calculated from For a three-dimensional wall, as in a furnace, separate shape factors are used to calculate the heat flow through the edge and corner sections, with the dimensions shown in Figure 3-4. when all the interior dimensions are greater than one fifth of the thickness, where A = area of wall, L = wall thickness, D = length of edge CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE CUMT

3-4 The Conduction Shape Factor HEAT TRANSFER LECTURE CUMT

Example 3-1 HEAT TRANSFER LECTURE CUMT

Example 3-2 HEAT TRANSFER LECTURE CUMT

Example 3-3 HEAT TRANSFER LECTURE CUMT

Example 3-4 HEAT TRANSFER LECTURE CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE The most fruitful approach to the heat conduction is one based on finite-difference techniques, the basic principles of which we shall outline in this section. CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE 1、Discretization of the solving CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE 2、Discrete equation Taylor series expansion CUMT

Differential equation for two-dimensional steady-state heat flow 3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE 2、Discrete equation Differential equation for two-dimensional steady-state heat flow CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE 2、Discrete equation Discrete equation at nodal point (m,n) no heat generation Δx= Δy CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE 2、Discrete equation Thermal balance (1) Interior points steady-state & no heat generation CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE Thermal balance (1) Interior points Δx= Δy steady-state with heat generation CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE 2、Discrete equation Thermal balance (2) boundary points CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE Thermal balance (2) boundary points Δx= Δy Δx= Δy CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE Thermal balance (2) boundary points Δx= Δy CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE 3、Algebraic equation CUMT

3-5 Numerical Method of Analysis HEAT TRANSFER LECTURE Matrix notation Iteration Simple Iteration & Gauss-Seidel Iteration CUMT

Example 3-5 HEAT TRANSFER LECTURE Consider the square shown in the figure. The left face is maintained at 100℃ and the top face at 500℃, while the other two faces are exposed to a environment at 100℃. h=10W/m2·℃ and k=10W/m·℃. The block is 1 m square. Compute the temperature of the various nodes as indicated in the figure and heat flows at the boundaries. CUMT

Example 3-5 [Solution] The equations for nodes 1,2,4,5 are given by HEAT TRANSFER LECTURE [Solution] The equations for nodes 1,2,4,5 are given by CUMT

Example 3-5 [Solution] Equations for nodes 3,6,7,8 are HEAT TRANSFER LECTURE [Solution] Equations for nodes 3,6,7,8 are The equation for node 9 is CUMT

Example 3-5 HEAT TRANSFER LECTURE The equation for node 9 is CUMT

Example 3-5 HEAT TRANSFER LECTURE We thus have nine equations and nine unknown nodal temperatures. So the answer is For the 500℃ face, the heat flow into the face is The heat flow out of the 100℃ face is CUMT

Example 3-5 The heat flow out the right face is HEAT TRANSFER LECTURE The heat flow out the right face is The heat flow out the bottom face is The total heat flow out is CUMT

3-6 Numerical Formulation in Terms of Resistance Elements HEAT TRANSFER LECTURE Thermal balance — the net heat input to node i must be zero qi — heat generation, radiation, etc. i — solving node j — adjoining node CUMT

3-6 Numerical Formulation in Terms of Resistance Elements HEAT TRANSFER LECTURE so CUMT

3-7 Gauss-Seidel Iteration HEAT TRANSFER LECTURE Steps Assumed initial set of values for Ti； Calculated Ti according to the equation； —using the most recent values of the Ti Repeated the process until converged. CUMT

3-7 Gauss-Seidel Iteration HEAT TRANSFER LECTURE Convergence Criterion Biot number CUMT

Example 3-6 HEAT TRANSFER LECTURE Apply the Gauss-Seidel technique to obtain the nodal temperature for the four nodes in the figure. [Solution] All the connection resistance between the nodes are equal, that is Therefore, we have CUMT

Example 3-6 HEAT TRANSFER LECTURE Because each node has four resistance connected to it and k is assumed constant, so CUMT

3-8 Accuracy Consideration HEAT TRANSFER LECTURE Truncation Error — Influenced by difference scheme Discrete Error — Influenced by truncation error & △x Round-off Error — Influenced by △x CUMT

Summary Numerical Method Solving Zone Nodal equations HEAT TRANSFER LECTURE Numerical Method Solving Zone Nodal equations thermal balance method — Interior & boundary point Algebraic equations Gauss-Seidel iteration CUMT

Summary (2)Resistance Forms (3)Convergence Convergence Criterion CUMT HEAT TRANSFER LECTURE (2)Resistance Forms (3)Convergence Convergence Criterion CUMT

Summary (4)Accuracy Truncation Error Discrete Error Round-off Error HEAT TRANSFER LECTURE (4)Accuracy Truncation Error Discrete Error Round-off Error Important conceptions Nodal equations — thermal balance method Calculated temperature & heat flow Convergence criterion How to improve accuracy CUMT

Exercises HEAT TRANSFER LECTURE Exercises: 3-16, 3-24, 3-48, 3-59 CUMT