1. CE 8022 09 An Intro Problem—will lead to first homework Fluid is contained in along square duct, see cross section below, three faces of the duct are kept at temperature T =0. The top face is kept at T =100. If the thermal properties are constant, the duct is long and there is no fluid flow the value Of the temperature at any point in the duct can be calculated from the PDE With boundary conditions The solution is T=100 T=0 How can we solve this problem Numerically (1) (2)

2. Key Information Task 1 is to derive the governing Eq. Flux --- is a vector is the flow of a quantity per unit area per time A Diffusive Flux -- is a flux proportional to the gradient of a potential In heat transfer the Diffusive Flux (in a homogeneous medium) is the Fourier Flux where K is the thermal conductivity [Jm -1 s -1 Ke -1 ] The outward pointing normal on a plane area A is The amount of heat crossing the plane area (in the opposite direction to the outward pointing normal) due to diffusion is The Divergence Theorem sayshttp://en.wikipedia.org/wiki/Divergence_theorem Whereis the surface that encloses the volume V Also hold in 2D

3. We can now derive the governing Equation (1) T=100 T=0 Choose an arbitrary area located in our domain with surface S and OUTWARD Pointing normal n Steady State Heat Balance says All you really need to know By divergence theorem But area A is arbitrary so argument on right hand side must be zero throughout Our domain, i.e., or since K is constant As in our governing Eq (1)

4. Developing a numerical solution. The key step in a numerical solution is the discretize (break-up) the domain into Non-over lapping parts—Many Many ways to do this One approach is to use control volumes --for our test problem we discretize our domain into n x n square control volumes of size  x  and place a node point at the center of each volume—values at the node point are representative of values in the volume P N W E S Consider control volume P with faces w, s, e and n (Note upper case directions N(orth) S(outh) etc refer to the nodes That neighbor P and lower case directions w(est) e(ast) ect refer to the faces of P P N EW S w n e s Treating the volume P like our arbitary area Used in deriving the governing equation We can write the heat balance for volume P as

5. P N EW S w n e s The heat balance on P is On canceling K (constant) approximating the integrals and derivatives as follows We can write down the balance equation as Note if node P is in the a volume adjacent to the North domain boundary then With similar treatments at other boundaries At a corner volume (e.g the one in the South East corner The result is a set of linear algebraic equations n x n for each unknown nodal value

6. yNum Ana(lost of terms) 1100 0.95454590.731590.87 0.86363673.0157273.31 0.77272757.3250557.6 0.68181844.1599344.34 0.59090933.4887733.56 0.525 0.40909118.2844218.24 0.31818212.9328212.88 0.2272738.5752778.529 0.1363644.8884114.86 0.0454551.5875141.576 000