1. ERT 216 HEAT & MASS TRANSFER Sem 2/ 2014-2015 Dr Akmal Hadi Ma’ Radzi School of Bioprocess Engineering University Malaysia Perlis

2. Conduction 1)Steady state conduction- One dimension 2)Steady state conduction- Multiple dimension 3)Unsteady state conduction

3. Steady State Conduction- MULTIPLE Dimension

4. Point of view “Steady state conduction-1D”  Heat transfer was calculated in system in which temp. gradient and area could be expressed in term of 1 space coordinate Now, our aim to analyze more general case of 2D heat flow. For steady- state with no heat generation, the Laplace Equation applies and assuming constant thermal conductivity.  Laplace Equation The solution obtained by some analysis which are; 1)Analytical, 2)Graphical or 3)Numerical

5. the temp. in a 2D body as a function of the 2 independent space coordinates x and y. Then the heat flow in the x and y direction calculated from the Fourier equation: From Laplace Equation  General Analysis The general solution give: The objective of heat transfer analysis  predict heat flow or the temperature and

6. This heat flow quantities are directed either in the x direction or in the y direction. The total heat flow at any point in the material is the resultant of the q x and q y at that point. Thus the total heat flow vector is perpendicular to the lines of constant temp. in the materials. If the temp. distribution in the material is known, we may easily establish the heat flow

7. 1) Analytical Analysis System: For the rectangular plate as shown (3 sides of the plate are maintained at the constant temp. T 1 and upper side has some temp. distribution T=f(x)). After follow the steps and consider some method also used the boundary condition, the final solution for this system using analytical analysis is: Temp. distribution could be simple as constant temp. or more complex (sin-wave distribution)

8. Conclusion: Extensive study of “Analytical technique” used in conduction heat transfer requires a background in the theory of Orthogonal Function (Fourier series, Bessel function and etc.) which is involved a complex series. That’s why the analytical solution are not always possible to obtain  very cumbersome and difficult to use. 1) Analytical Analysis

9. 2) Graphical Analysis Consider: 2D system with inside surface is maintained at T 1 and outer surface temp. is maintained at T 2. To calculated the heat transfer  sketched the isotherms and heat –flow lanes to form groupings of curvilinear The element (curvilinear-square)  The heat flow across this curvilinear section is given by Fourier ‘s Law (assuming unit depth of material):

10. So that,, the heat flow is proportional to the across the element. 2) Graphical Analysis Note: The heat flow will be same through each section within heat flow lane and the total heat flow will be the sum of the heat flows through all the lanes. If the sketch  Because the heat flow is constant, the across each element must be same within the same heat flow lane. Thus, across an element is given by: N : The no. of temp. increment between inner & outer surfaces. Total heat transfer  M : The no. of heat-flow lanes. “Conduction Shape Factor, S” 

11. 2) Graphical Analysis Conclusion To calculate hate transfer: Need only construct the curvilinear-square plots and count the no. of temp. increments and heat-flow lanes. Make sure to construct the plot so that and the lines are perpendicular. The accuracy of graphical analysis  depend on entirely on the skill of the person sketching the curvilinear-squares. It may not be expected to be used for the solution of many practical problems.

12. 2) Graphical Analysis The Conduction Shape factor, S In 2D system where only 2 temp. limits, S define as  The values of S based on geometries (refer table). For 3D wall (like furnace), separate shape factors are used to calculate the heat flow through the edge and corner sections. When all the interior dimensions > 1/5 of the wall thickness;

13. Conduction shape factors, S based on geometries:

18. Example 1

19. Example 2

20. Example 3

21. 3) Numerical Analysis Consider, 2D body that is to be divided into equal increment in both the x and y direction  Where: m: location indicating x increment n: location indicating y increment Aim to establish the temp. at any of these nodal points within the body using Laplace equations  Finite differences are used to approximate differential increments in the temp. & space coordinates Note: The smaller we choose finite increments, the more closely the true temp. distribution will be approximated.

22. 3) Numerical Analysis  The temp. gradient may be written as follows: So that; x direction  y direction  and

23. If consider heat generation happen in system, so add the term into general equation and obtain: 3) Numerical Analysis If, then  Thus the finite difference approximation for Laplace equation becomes; If, then  For case: K constant, so the heat flow may all be expressed in term of temp. differential. The net heat flow into any node is 0 at steady state conditions. General Equation

24. 3) Numerical Analysis To utilize the numerical analysis, equation  must be written for each node within the material. Example: Equation for nodes 1, 2, 3 and 4: The solution of these equations: Once the temp. are determine, the heat flow can be calculated from: The is taken at the boundaries

25. Physical situation: 3) Numerical Analysis 1)Interior node 2)Convection boundary node 3)Convection at a corner section (to get the temp. distribution and energy balance on each node)

26. 3) Numerical Analysis So that, the heat flow may be calculated at either the 500 °C face or the three 100 °C faces. Note: The value of heat flow at 500 °C face and 100 °C face should be nearly the same. As general practice, it is usually best to take the arithmetic average of the 2 values for use in the calculations.  “Interior node”

27. 3) Numerical Analysis When the solid is exposed to some convection boundary condition, the temp. at the surface must be computed differently from the method given before.  “Convection boundary node” Consider this boundary  The energy balance on node (m, n) is; If, then 

28.  “Convection at corner section” 3) Numerical Analysis If the corner section exposed to a convection boundary condition. The energy balance for the corner section is; Consider this boundary  If, then 

29. 3) Numerical Analysis Summary of nodal formula for finite-difference calculation. (Dashed lines indicate element volume)

30. 3) Numerical Analysis

32. 1)Matrix notation 2)Software packages 3)Gauss-Seidel iteration Numerical analysis is simply means of approximating a continuous temp. distribution with finite nodal elements. The more nodes taken, the closer the approximation but more equations mean more cumbersome solutions. Fortunately, computer and even programmable calculators have the capability to obtain these solutions very quickly. In practical problem, there are several solution technique:

33. 3) Numerical Analysis Numerical Formulation in term of resistance elements Used resistance concept for writing the heat transfer between nodes. Designating our node interest with subscript i and adjoining nodes with subscript j (shown in figure- general conduction node solution). At steady-state the net heat input to node I must be 0  : the heat delivered to node i by heat generation, radiation etc : the form of convection boundaries, internal conduction etc

35. Gauss-Seidel iteration # iterative technique: more efficient solution to the nodal equations than a matrix inversion From heat flow in terms of resistances, may solve for the temp. i, and temp. of the adjoining nodes, as: For, and no heat generation  Note: Convection boundary may be converted to insulated surface by setting Bi=0

37. Assignment 3 (submit: 25 Jan 2011 before 1200) Book: J.P. Holman 1)3-3 2)3-8 3)3-11 4)3-53 5)3-59 6)3-74 Just give the temp. equation and energy balance for each node.