Question

Derive each temperature distribution for each tip condition from the general equastion

theta(x)=C1e^mx + C2e^-mx usinf boundry condition show all work for each tip condition

Tip Condition (r = L Temperature Distribution 0/0, Convection heat transfer: ho(L) = -kdo/dx=L cosh m(L - x) + (h/mk) sinh m(L - x) cosh mL + (h/mk) sinh mL Adiabatic: do/dx = = 0 cosh m(L – x) cosh mL Prescribed temperature: 8(L) = 0 (04/06) sinh mx + sinh m(L - x) sinh mL Infinite fin (L → 0); 0(L) = 0

Answer 1

(i) Boundary condition 1: ar x=0; T=T); 1=0, (ii) Boundary condition 2: It has three possibilities (a) Insulated condition: at x=L; d = 0 (b) Convection boundary: at x=L; -k z=L = ha=L (c) Long fin: at x=L; 1=0. For constant area fin de _hp_0 = 0 (3.7) de - m² = 0 (3.8) Here m² = hop (3.9) Case 1: Insulated tip Assumption: To > T. Boundary condition 1: At x=0; 0 = 0 Boundary condition 2: At x=L; 1=0

h, T. < 3 Figure 3.2: Schematic representation of a rectangular cross section fin Let i =D Therefore equation 3.8 becomes [D– m]@ = 0 (3.10) In the above equation auxiliary term is zero in order to get roots [D2 - m²) = 0 (3.11) D= Em (3.12) Solution is complementary function and is given by 0= cosh mx + c sinh mx (3.13) Boundary condition 1: At x=0; 0 = 0 H = c +0 (3.14)

ci = 0 40 = cz sinh mx[m] + ccosh mä[m] Boundary condition 2: At x=L; 1=0 0= m[@ sinh mL + C2 cosh mL] Ca = -0, tanh mL | 0 = 0; cosh ma – Vị tanh mL sinh m2 0 = 0610 accosh mx cosh mL - sinh mL sinh mx, cosh mL 0 = 0 cosh m(L – ) cosh mL

Case 2: Long fin Fin equation is de do2-mºl = 0 General solution is 0 = Ciemx + c2e-mu Boundary condition 1: At x=0; T = T), 0 = 66

From equation 3.39 ci=0; c2=66 0 = 0,e-mx (3.40)

Case 3: Convecting tip Governing equation is d20 20 - m2 = 0 (3.46) General solution is O= ci cosh mx + C2 sinh mx (3.47) Boundary condition 1: At x=0; 0 = 06 From equation 3.47 C1 = 0 (3.48)

o= 0 cosh mx + C2 sinh mx (3.49) Boundary condition 2: -k da=L = htipc=L de 0;m sinh mc+Com cosh ma (3.50) From boundary condition 2 -km(0 sinh mL + C2 cosh mL] = htip[@ cosh mL + C2 sinh mL] (3.51) sinh mL + cosh mL] = -0 sinh mL - mk TIP As cosh mL (3.52) mk Ca = -0, Isinh mL+ tip1 (3.53) cosh mL + hrip sinh mL Temperature distribution inside the fin is given by sinh mL+m sinh 20 0 = 0, cosh mx – 06- cosh mL + huip sinh mL (3.54) A=0 cosh m(L - x) + hip sinh m(L - x) cosh mL + hatip sinh mL (3.55)