Question

5) Consider problem 4. Instead of an insulated left surface, a known heat flux is imposed on it. Everything else remains the same. • Stating all the assumptions clearly, derive expression for temperature profile with the rod in the steady state. (25 points) • Bonus problem: If possible, draw temperature profile. (10 points) 9 gen air

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Answer 1

Solution:- Consi den K> thesmal Conductivity. ha Heat transfer Coefficient Tu> Amblent temperutire. This is treated as plate: hejt Swyace însulated. -> -> → Temper ature pofile - 111 dimendonal Steady State Conduction heat equation is One wall- For d2

Integrating the abave equation ngain integrating grven:- I:- Boundery Condition. 1-0 :- The conduction region is peefectly. însulated and hence heat flow is zeno. fiom fourier's law Q: - KA () dT the temperature alterivative must and accordingly be at X=0 ze90 Q: - KA di dz. Insultated end %3D dr from & (3 at 2=0 G=0

Boundary Condition 2:L:- The heat égnducHon equals the Convective heat fiow to the SurToundings ine, Assume Te → Tempeaature at length 'L =hA [T- To) -KA lor) 7. - 12=L 'in equation @ Substitue Again at z-L ; 4:0 di • ILI 크 L and O from expression. from equation O at x-L and G=0:- Te=

Substitde ` value in equeation and find the (2 general tempexature distibutton equation T= 2K. 2K lor) The eqpeation is a tempexature profile. above the equation is parabolic And Draw tomperatue profile :- To T- Ta t gL + as (tx) 2k Temperatuore distibuion is parabolie.

Narimum temperature SParabolic at X=0 h, Too aju 11↑