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in solving this problem. PROBLEM 4.71 KNOWN: Hot-plate tool for micro-lithography processing of 300-mm silicon wafer consisting of an aluminum alloy equalizing block (EB) heated by ring-shaped main and trim electrical heaters (MH and TH) providing two-zone control. FIND: The assignment is to size the heaters, MH and TH, by specifying their applied heat fluxes, mh thq and q ,\u2032\u2032 \u2032\u2032 and their radial extents, mh thr and r\u2206 \u2206 , to maintain an operating temperature of 140°C with a uniformity of 0.1°C. Consider these steps in the analysis: (a) Perform an energy balance on the EB to obtain an initial estimate for the heater fluxes with mh thq q\u2032\u2032 \u2032\u2032= extending over the full radial limits; using FEHT, determine the upper surface temperature distribution and comment on whether the desired uniformity has been achieved; (b) Re-run your FEHT code with different values of the heater fluxes to obtain the best uniformity possible and plot the surface temperature distribution; (c) Re-run your FEHT code for the best arrangement found in part (b) using the representative distribution of the convection coefficient (see schematic for h(r) for downward flowing gas across the upper surface of the EB; adjust the heat flux of TH to obtain improved uniformity; and (d) Suggest changes to the design for improving temperature uniformity. SCHEMATIC: < ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction with uniform and constant properties in EB, (3) Lower surface of EB perfectly insulated, (4) Uniform convection coefficient over upper EB surface, unless otherwise specified and (5) negligible radiation exchange between the EB surfaces and the surroundings. ANALYSIS: (a) To obtain initial estimates for the MH and TH fluxes, perform an overall energy balance on the EB as illustrated in the schematic below. in outE E 0\u2212 =\ufffd \ufffd ( ) ( ) ( )2 2 2 2 2mh th o o s2 1 4 3q r r q r r h r 2 r w T T 0\u3c0 \u3c0 \u3c0 \u3c0 \u221e\uf8ee \uf8f9\u2032\u2032 \u2032\u2032\u2212 + \u2212 \u2212 + \u2212 =\uf8ef \uf8fa\uf8f0 \uf8fb Continued \u2026.. PROBLEM 4.71 (Cont.) Substituting numerical values and letting mh thq q ,\u2032\u2032 \u2032\u2032= find 2 mh thq q 2939 W / m\u2032\u2032 \u2032\u2032= = < Using FEHT, the analysis is performed on an axisymmetric section of the EB with the nodal arrangement as shown below. The Temperature Contour view command is used to create the temperature distribution shown below. The temperatures at the center (T1) and the outer edge of the wafer (r = 150 mm, T14) are read from the Tabular Output page. The Temperature Gradients view command is used to obtain the heat flow distribution when the line length is proportional to the magnitude of the heat rate. From the analysis results, for this base case design ( )mh thq q ,\u2032\u2032 \u2032\u2032= the temperature difference across the radius of the wafer is 1.7°C, much larger than the design goal of 0.1°C. The upper surface temperature distribution is shown in the graph below. Continued \u2026.. PROBLEM 4.71 (Cont.) (b) From examination of the results above, we conclude that if mhq\u2032\u2032 is reduced and thq\u2032\u2032 increased, the EB surface temperature uniformity could improve. The results of three trials compared to the base case are tabulated below. Trial ( )mh 2qW / m\u2032\u2032 ( )th 2qW / m \u2032\u2032 ( )1 T C° ( ) T14 C° ( )1 14 T T C \u2212 ° Base 2939 2939 141.1 139.3 1.8 1 2880 2997 141.1 139.4 1.7 (-2%) (+2%) 2 2880 3027 141.7 140.0 1.7 (-2%) (+3%) 3 2910 2997 141.7 139.9 1.8 (-1%) (+2%) Part (c) 2939 2939 141.7 139.1 2.6 Part (d) 2939 2939 140.4 139.5 0.9 k=150 W/m\u22c5K Part (d) 2939 2939 140.0 139.6 0.4 k=300 W/m\u22c5K The strategy of changing the heater fluxes (trials 1-3) has not resulted in significant improvements in the EB surface temperature uniformity. Continued \u2026.. EB surface temperature distribution 138.5 139 139.5 140 140.5 141 141.5 0 20 40 60 80 100 120 140 160 Radial position, r (mm) T(r ,z) , (C ) PROBLEM 4.71 (Cont.) (c) Using the same FEHT code as with part (b), base case, the boundary conditions on the upper surface of the EB were specified by the function h(r) shown in the schematic. The value of h(r) ranged from 5.4 to 13.5 W/m2\u22c5K between the centerline and EB edge. The result of the analysis is tabulated above, labeled as part (c). Note that the temperature uniformity has become significantly poorer. (d) There are at least two options that should be considered in the re-design to improve temperature uniformity. Higher thermal conductivity material for the EB. Aluminum alloy is the material most widely used in practice for reasons of low cost, ease of machining, and durability of the heated surface. The results of analyses for thermal conductivity values of 150 and 300 W/m\u22c5K are tabulated above, labeled as part (d). Using pure or oxygen-free copper could improve the temperature uniformity to better than 0.5°C. Distributed heater elements. The initial option might be to determine whether temperature uniformity could be improved using two elements, but located differently. Another option is a single element heater spirally embedded in the lower portion of the EB. By appropriately positioning the element as a function of the EB radius, improved uniformity can be achieved. This practice is widely used where precise and uniform temperature control is needed. PROBLEM 4.72 KNOWN: Straight fin of uniform cross section with insulated end. FIND: (a) Temperature distribution using finite-difference method and validity of assuming one- dimensional heat transfer, (b) Fin heat transfer rate and comparison with analytical solution, Eq. 3.76, (c) Effect of convection coefficient on fin temperature distribution and heat rate. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fin, (3) Constant properties, (4) Uniform film coefficient. ANALYSIS: (a) From the analysis of Problem 4.45, the finite-difference equations for the nodal arrangement can be directly written. For the nodal spacing \u2206x = 4 mm, there will be 12 nodes. With " >> w representing the distance normal to the page, ( )22 2 2 33 c hP h 2 h 2 500 W m K 2 x x x 4 10 mm 0.0533 kA k w kw 50 W m K 6 10 m \u2212 \u2212 \u22c5 \u22c5 \u22c5 × \u22c5\u2206 \u2248 \u2206 = \u2206 = × = \u22c5 \u22c5 \u22c5 × × \ufffd \ufffd Node 1: ( )2 1100 T 0.0533 30 2 0.0533 T 0+ + × \u2212 + = or -2.053T1 + T2 = -101.6 Node n: n 1 n 1 nT T 1.60 2.0533T 0+ \u2212+ + \u2212 = or n 1 n n 1T 2.053T T 1.60\u2212 \u2212\u2212 + = \u2212 Node 12: ( ) ( )11 12T 0.0533 2 30 0.0533 2 1 T 0+ \u2212 + = or 11 12T 1.0267T 0.800\u2212 = \u2212 Using matrix notation, Eq. 4.52, where [A] [T] = [C], the A-matrix is tridiagonal and only the non-zero terms are shown below. A matrix inversion routine was used to obtain [T]. Tridiagonal Matrix A Column Matrices Nonzero Terms Values Node C T a1,1 a1,2 -2.053 1 1 -101.6 85.8 a2,1 a2,2 a2,3 1 -2.053 1 2 -1.6 74.5 a3,2 a3,3 a3,4 1 -2.053 1 3 -1.6 65.6 a4,3 a4,4 a4,5 1 -2.053 1 4 -1.6 58.6 a5,4 a5,5 a5,6 1 -2.053 1 5 -1.6 53.1 a6,5 a6,6 a6,7 1 -2.053 1 6 -1.6 48.8 a7,6 a7,7 a7,8 1 -2.053 1 7 -1.6 45.5 a8,7 a8,8 a8,9 1 -2.053 1 8 -1.6 43.0 a9,8 a9,9 a9,10 1 -2.053 1 9 -1.6 41.2 a10,9 a10,10 a10,11 1 -2.053 1 10 -1.6 39.9 a11,10 a11,11 a11,12 1 -2.053 1 11 -1.6 39.2 a12,11 a12,12 a12,13 1 -1.027 1 12 -0.8 38.9 The assumption of one-dimensional heat conduction is justified when Bi \u2261 h(w/2)/k < 0.1. Hence, with Bi = 500 W/m2\u22c5K(3 × 10-3 m)/50 W/m\u22c5K = 0.03, the assumption is reasonable. Continued... PROBLEM 4.72 (Cont.) (b) The fin heat rate can be most easily found from an energy balance on the control volume about Node 0, ( )0 1f 1 conv 0T T xq q q k w h 2 T T x 2 \u221e \u2212 \u2206 \u2032 \u2032 \u2032= + = \u22c5 + \u2212 \u2206 \uf8eb \uf8f6\uf8ec \uf8f7\uf8ed \uf8f8 ( ) ( ) ( )33 2f 3100 85.8 C 4 10 mq 50 W m K 6 10 m 500 W m K 2 100 30 C24 10 m \u2212 \u2212 \u2212 \u2212 ×