Question

Apply the finite difference method (with equally spaced 4 points grid) for solving a steady state conductive heat transfer through a metal rod and convection to the surrounding air, which are described by the following ordinary differential equation: T" – TQ2 = -a?T. T(x = 0) = 0 T(x = L) = 2 T is the temperature distribution in a rod, a = hP/kA, h is the heat transfer coefficient, P, A and L are the perimeter, the cross section and the length of the rod, respectively and k is its heat conductivity. Ta is the temperature of the surrounding air and und ©2 are the temperatures at both ends of the rod. Use numerical values given in Figure 1 to compute temperatures in nodes 2, 3 and 4. 2 3 4 77777777 Figure 1: L = 0.01m, , = 0°C,, = 100°C, a = 400m-1,T, = 0°C, Ax = 0.0025m

Answer 1

Given T" ra = a ta with B-C. T(0) - 04 Ta= oc I T(L) = Oz cohere a= hoa Taaoc 0; 2=0.0025 Fe1000 We have d'I = 0 Totat Tin Te Tiri it i iti 4%A7+ dx2 discretige ot orode i, Finite difference Method) Titi - 2Ti + Tin 1 - ? D + Ti (4x) h Titi - 27; +TE = AXTi T ( ax + 2) - Tam +, | of i=2, Ta (°Ax+2) = T2+T - Tg+ , = T3 +0 - Ty -U) at i=3, Tg (2 Ax+2) = Tartz = Tat Tz Ans. (Ax+2) T, =0, = 0°c : Tz (Cata2+2) altexts] = T — iig T = 0-015 xuºc | Iz= 6-188104 oc at 1-4, Taloar+2) = Tgt T3 = Oqt Tz = 100+73 Ta= 0.248°C using @gnis y fiantz3 - 1) = 100+T; Ts=, - 100'c =) Ts [ @*mu+z3"-2] = 100 i Ta 100 - 100 (@bu +23°-2 (400300025 +2) 2-2 = 6-18 X10" .: Tq = _II__ - 6.18X104. — = 0.015*10 4 °c ".04x+2 200 10.0025 +2 - stoc T = To {m?x412 - indavina) = 618 xo59 (6 oko vas nas elastans vis] = 0 248€