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PROJECT #1 (2.5 Marks): The heat that is conducted through a body must frequently be removed by other heat transter processes. For example, the heat generated in an electronic device must be dissipated to the surroundings through convection by means of fins. Consider the one-dimensional aluminum fin (thickness t 3.0 mm, width Z 20 cm, length L) shown in Figure 1, that is exposed to a surrounding fluid at a temperature 1. The conductivity of the aluminum fin (k) and coefficient of heat convection of surrounding fluid (h) are given in Figure 1. The defining equation for the conduction and convection heat-transfer coefficients are recalled as qcnd -k A dT/dx and geany h A' (T-T) where the area in these equations (A and A) are the surface area for conduction and convection, respectively. ayhe temperature of the base of the fin is To. Make an energy balance on the element of the fin 6f thickness dx as shown in the figure 1 and develop a linear ODE as the mathematical model for this heat transfer problem. de=h Pdx (T-T rads dr Figure 1: Sketch illustrating one-dimensional conduction and convection through a rectangular fin. /Consider three following cases and determine the boundary conditions for each case. Find the solution of the linear ODE for each case in the form of temperature distribution along the fin if To 300C and T- 50 C CASE 1: The fin is very long, and the temperature at the end of the fin is essentially that of the surrounding fluid. What is the temperature at x L/2? CASE2: The fin is of finite length (7.5 cm long) and loses heat by convection from its end. What L) and at x L/2? are the temperatures at the tip of the fin (x CASE 3: The end of the fin is insulated so that dT/dx 0 at x L. What is the temperature at this point and at x= L/2? Compare the results and make your conclusion. 4 aseg ONbat will be the boundary conditions for the same mathematical model when the base plate contains a uniform heat source rather than just being at a constant temperature To Repeat the calculations of the above three cases if the heat source is an electrical element of 100 W and the only possible heat transfer mechanism is through the fin. PROJECT #2 (2.5 Marks): Use the electrical analogy for the heat transfer problem of the fin schematically presented in Fig. 1, and find the equivalent resistances in your electrical analogy Assume that the fin is also exposed to a space such that the radiation loss is given by qrad=E A' a (T-T) where E is a surface emissivity constant, a is the Stefan-Boltzmann constant, and the temperatures are expressed in degrees Kelvin. Derive a differential equation for the temperature in the pin fin as a function of x, the distance from the base. Let To be the base temperature, and write the appropriate boundary conditions for the differential equation. C Repeat the calculations for the three cases (Part b in Project # 1 ) by using the following parameters: a 5.670374419 x 108 W-m2.K4 E = 0.5 What mechanism of heat transfer is more efficient in this temperature range? What's about the case when the temperature at the base of the fin is To 1000 C

Answer 1

Project 1: k and h values are not given
Qztdx onv=hda T-T = hPdz{r-To) To = Qt 3)d1 thpdx(T-To) Qout Q Cenv Q+dx QinQot0 dr a(d-pd (T-To)0 ADdhpdz (T-e): AOT kAcddThpdx (T-T) 0 Divido. by kAcd drhp (T-To)0 dr2 KAC Lot - T-Too TdodT do d12 dida2 Toditmegeueus Ceustaut di2 -me let D=d-m)e dz 0 Cje (eh) +C2 e ACohme Bsinhm. .

case i Very eng enininte lengh pin T=TL Too C1e tCe : а-0.,9-Өо To eo-+C2 ob .o 0 O-T-T T-To +00 OCie t C2e O GCo)Calo) Do Ct -m2. (-m) Ooe de de di. 1-0 -meo de di1-0 -kAc Fin -A(-m) T-Too To-To hekAcOo QFin kmoo

Loedod length approauh mothod fon_tinile length with Convetve heat levs Qfip = Qrveus New anea hAepfhpoL AT Canallad ina AL> Ac As V Lc= Lt A P Rectangular Fin hp AL AcTID Lc LtAL P TTD P 2W Ac AT-Wt Ces h (mlc-m) Ac hmlc LC L+D LLt Fme o fonh(mlC) PhPKAotanh(mLe)

finte kngth aith_iniulakd tip. (No hal o Casci Fin pm tie TL inulated fip ACoshmtB Ainhmx TC T To do dz de dx (Axinhmir eeshm e-T-Too L hP o RAC de 0 di de df L Ces hmL Put ARB in eq-0 COshmL Ceshmz coshml. - SinhmL sinhmL Өо Сеs hmL CEsh(mL-m) TETeoCesh (mi-mi). To-To (osh mL I Vay vany Fuste fi at lip, x=L ,T- TL. Uе oenev TL-Te0 To-To CehmL sinpcdtn+