Question

Heat is supplied to the exposed surface. Assume unidirectional unsteady state heat conduction according to the following equation: a?T at dx² ac with the boundary conditions ат x = 0, дх ат x = 1, 0 дх 0 < x < 0.5 t = 0, T= {. 0.5 <x<1 ,where: a and L are constants. Solve the above differential equation using separation of variable method.

Answer 1

Q71 Qatda 20 Go T Consider a iron bar as shown in figure where boue temp. To and exposed to a temperature of To. - Assumptions: unidirectional heat conduction. Consider an elemental volume of a with thickness da. Heat entering ax and leaving be Qatar In this case both conduction arod convection is taking place Q be beat transferred through convection Balance Energy Qg Qar dx + QA

Qt londuction -KA di da Fourier's law of Heal *) Qdt da = axtd dy [ au dx] Sy [ Qx dx ] = 2 2 - kA d Jaz Qua Qytdy = =KA d 27 dx +/2 dA2 Now as I hade (i ta) [ Heat Givechion] hp dx (T-To) Let o= T-Too de -> (3 let p- perimeter idi of bar. dx2- a da hp o KA a Zodra hp da o dna 2) 2²0 KA Let m²= hp a dzo KA da? dx2 m²020 (1)

Let D=d Tx (h) = So (6² m² )0=0 (DAM)0=0 or 6-mo=0 do = mda mo do 이기 dy - lino mx +A A o- oma e c, ema C is const If + m) 0 = 0 do mo dx do - - ma eno -mat B O = Cae whore G is const. ma 0=c, em+e

Oy - KA/2) = HAO 0 [LETL= To] Adiabatic tip- (do an 12 = 1 d da [ciem? t came => [c, mema ame It Came me My x 2 - 1 çen me C2 2 ml é ml сі since (+ C2 = 0 at cepend og a = 00 oo and iteme e me ozme q=c, e e me e me Doc e me ml e me

O=c, eme teon em(l-a) è mcl-x) eme a me Coshm (1-2) Cosh me oo Temperature disto bution is: cosh m(L-2) Oo Coshme