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ME 344-002 Heat Transfer (Spring 2022) MATLAB Project Assignment. Due by Monday, May 9, 2022 We consider a hot surface of heat generating system as in Fig. 1(a), and the surface dimension of interest is w 160 mm? (w x f), and the width, w is much larger than t (w >>t). A constant and uniform heat flux (9,") of 2 W/cm² flows out from the surface. The surface is exposed to convection heat transfer where the air temperature T.. is 30°C and heat transfer coefficient h=100 W/m²-K. Here, we add a metal block covering the hot surface to lower the surface temperature. The dimension of the metal block is 200 x w x 160 mm(a x w x f). Pure copper and commercial bronze are considered as a block material, and the material properties are evaluated at 300 K from Table A.1 in our textbook. We assume a steady-state condition and constant thermal properties and neglect a contact resistance between the hot surface and the block, and radiation heat transfer. Use MATLAB for analyses and plotting in the following parts. Surface M Ax 23 (x = 120 00--0-0-0 m 6-102m Th Th 90-C(y -- 80 mm) 1xm Ay nodal network - Olim Ar = Ay Tip nodes (x = 200 mm) Th BE Metal Block 1) Pure Copper 2) Commercial Bronze Base nodes (x=0) (b) Ar w W w >> 1 10 1 12 9." - 2 W/cm 1 1 1 16 17 18 T. = 30°C h = 100 W/m -K L-200 mm - 160 mm (a) 19 2011 24 1 20 T TE 12 Ar - Ay - 40 mm (c) Figure 1. (a) Rectangular straight metal block dissipating the supplied heat (q;") through convection heat ") transfer. (b) Using / x m mesh, we will analyze the thermal system of the block using the finite different method. (c) Mesh configuration when Ar= Ay= 40 mm. Part A (12 pt, 1D Analysis) (a) What is the hot surface temperature without the metal block? (b) Using the corrected length approximation for the fin analysis, find the temperature at the metal block base T(x = 0) and the effectiveness (&f). (c) Find the temperature distribution from the base (r = 0) to the tip (x = 200 mm) using the analytical solution in Table 3.4 in Chap. 3 and plot the temperature distribution using the MATLAB. Using the base temperature, T(r = 0), calculate the effectiveness (&). Part A(b) and Part A(C) should be done for both pure copper and bronze metal block. (Since w >>t, you can approximate perimeter P to 2w.)
Now, we will analyze this metal block numerically using the finite difference method. Part B (8 pt) For numerical analysis, we define a l * m nodal network as in Fig. 1(b). I and m are the numbers of rows and columns, respectively; thus, both are integers and depend on mesh size, Ar and Ay. In this project, Ar=Ay. Each nodal point has a single index n (i.e., In is the temperature at node n), and the nodal point index of the i-th row and j-th column can be expressed as n=mx (1 - 1) +j (i = 1, 2, ..., 1; j = 1, 2, ..., m). To use the finite difference method, we need to identify the finite-difference equations for all nodal points. For efficient coding, we can classify all nodal points into a total of nine cases (= 3 × 3) as three cases for each i and j can be considered (for i, i-1) i=1, 1-2) i=1, and i-3) 1<i<l; forj, j-1);= 1,j-2) j=m, and j-3) 1<j<m]. For these nine cases Case 1: 1-1) +j-1), Case 2: 1-1) +j-2), Case 3: 1-1) +,-3), Case 4: 1-2) + j-1), ..., and Case 9: 1-3) + ;-3)], we will derive the finite-difference equations. For example, in Case 9. i-3) 1<i<l + j-3) 1<j<m, the finite difference equation for nodal point n is Tn-1 + Tn-1 + In-m + Tn-m- 4Tn=0. Here, derive the finite-difference equations for the other eight cases, and express them using index n, number of column m, and properties (q”, h, k, Tso, Ar, and k, not using values). In Parts C and D, we will use the Gauss-Seidel (GS) approach to find a numerical solution for the nodal temperatures. Part C (10 pt) First, use a mesh size of Ar= Ay = 40 mm as described in Fig. 1(C). (a) Using the GS algorithm [described in class and in our textbook (Appendix D)], write a MATLAB code (copy the code to your report). Use the equations from Part B to identify matrices [A] and [C] in [A][T] = [C] [Eq. (4.48)]. (Code efficiency & scalability will also be assessed) (b) Find an appropriate convergence criterion (ə) in GS algorithm. (Check and compare nodal temperatures by reducing ε). (C) Plot the 2D temperature distribution (color heat map) for both pure copper and bronze block. Part D (20 pt) Now, we will use a finer mesh network, where the mesh size Ar = Ay = 20 mm. (a) Plot the 2D temperature distribution (color heat map) for both pure copper and bronze block. (b) Now, we compare the results from the ID analytical solution, the coarse mesh analysis, and the fine mesh calculation for both pure copper and bronze block as below i) Comparison Plot 1: Temperature distributions (analytical, coarse & fine) along with the center plane C in the r direction (0 SX S200 mm &y=80 mm).
= = ii) Comparison Plot 2: Temperature distributions (analytical, coarse & fine) along with the base surface in the y direction (r = 0 mm & 0 Sy s 160 mm). iii) Comparison Plot 3: Temperature distributions (analytical, coarse & fine) along with the surface M in the y direction (r = 120 mm & 0 Sys 160 mm). iv) Comparison Plot 4: Temperature distributions (analytical, coarse & fine) along with the tip surface in the y direction (r = 200 mm & 0 sy s 160 mm). v) Averaging the temperatures at the base nodes, calculate the average base temperatures (Tb, avg) and find the effectiveness (@j) using the calculated To, avg for both coarse and fine mesh grid. (c) From the above comparisons in Part D(b), discuss the mesh size and 2D effects. The discussion should address the following questions: Is the mesh size of 40 mm small enough? Can you justify the 1D approximation in analytical solution? When is the 2D effect more significant? (k is small or large?)