Question

Solve equation (15-19) for the temperature distribution in a plane wall if the internal heat generation per unit volume varies according to y = 4 . The boundary conditions that apply are T=To atx=0 and T=T, at x=L Equation 15-19 15.2 Special Forms of the Differential Energy Equation The applicable forms of the energy equation for some commonly encountered stations follow. In every case the dissipation term is considered negligibly small I. For an incompressible fluid without energy sources and with constant DT (15.14) II. For isoburie flow without energy sources and with constant. the energy equation is DTKAT (15-15) Note that equations (15-14) and (15-15) are identical, yet apply to completely different physical situations. III. In a situation where there is no fluid motion, all heat transfor is by conduction If this situation exists, as it most certainly does in solids where the energy equation becomes AGV KVT+ (15.16) Equation (15-16) applies in general to heat conduction. No assumption has been made concerning constant . If the thermal conductivity is constant, the energy equation is (15-17) pc where the ratio ke, has been symbolized by a and is designated the thermal differ. It is casily seen that has the units, in the SI system is expressed in m/s, and as thin the English system. If the conducting medium contains to heat sources, equation (15-17) reduces to the Fourier field equation OT (15-18) which is often referred to as Fourier's second law of heat conduction For a system in which heat sources are present but there is no time variation, quation (15.17) reduces to the Poiss o n V7+?=0 (15-19) The final form of the heat-conduction equation to be presented applies to a steady-state situation without heat sources. For this case, the temperature distribution must satisfy the Laplace equation: VT=0 (15-20

Answer 1

Given, per unit volume Heat generation - ine - Bx/L Boundary at a = o x=1 conditions - T = Towar A T = Te Eq? 15-19 list v2T + 1 - 0 where, .. . dx2 282 222 heat conduction, For lo ele ele al Er 15-18 becomes d'T + q - o . dx2ř - d²T + 90 e - Ba/2 da² = 0 Integrating the above equ twice we get

dI dT -406-82/1 () + .. gol e Bx/L te = 212 BK T (2) I - 2012 -Ball, at cat de where and d are constants From BC I - x = 0 T = To 7 To = - 201² & a BPK » d = Tot do 1² B2K B C From x= L 2 – T = Te – qol2 12 T = tol t Tot B2k & B 2012 B2K $ {{ Ti-to the so ca->} Plugging in volues of c and do

T(x) = + Tot gol? 28-Ball - gol² e Pooh B²K 3 { Ten Tot qol 2.12 ß2K which is the required solution.