Question

(1) Two long fins Two infinitely long, metallic rods are joined together as shown in the figure. At the point that they are joined is a thin heat source with a known generation rate of in Watts. The two rods are surrounded by air with known h and T.. (a) Determine expressions for the temperature distribution in the rods in terms of symbols shown in the figure. (b) Determine expressions for the heat transfer rate into each of the two rods in terms of symbols shown in the figure. What fraction of goes into rod I versus rod 2? (c) Sketch the temperature distribution for the case of k>k>. Show T(x) for both rods on the same sketch thin heat generating layer of known Watts)

Answer 1

Heat generation - Igen Igen 0 Then, the Jahta 2, ki heat generated will go h Too to the two rods Let To be temperature at x=0 Let q, heat goes to rod with thermal conductivity K, and a heat goes to rod with thermal conductivity Kz Then As no heat accumulation In case of infinite long fin, we know 4 = Phka (T-Too) where heat flow rate p = perimeter of fin K= Thonmal conductivity A = cross-section area T,= Temperature at origin T Too Temperature at infinite - long distance And for temperature distribution T-Too = e-mx T.- Too where m ph VKA

Then, - 2 = √PhK, A (T.-Too) here, T, is temperature at heat is being generated x=0 uehere And 2 = √ PhKA (T.-Too) Putting in eq 4 - You = / PhK, A (Ty Too) + JP hk, A. (T,- Too) agen VP HERA, X(VK, TVK,) (T,- Too) Then - Ti= Igen VPRAVK+SK) - here, P = Perimeter = ad A = Area = Td² VPha = V Tidxhx nat - Tid jah & Ti agen + Too Tid V dh (VK, +2) -

- a) Temperature distribution- we know, T- Too - T-Too m= Ph VKA here, m fph - Ich bin Tidh kadu sehr 46 JKd Then T= Toot (T-Too-Juh da Toot (Ti-Too) é for rod 1 k=k TA)= Too + (Ti-To) erhada @ from To go on a) * To Ti-Tio na hatia Mat VI Putting in eg. TX) = Ta) = To * web Too + 2 2gen Tidrah (VR stran Vasava

for rod 2 In case of rod a heat travels in negative a direction .so, modifying equation, Temperature distribution - T- Too -emy - é mex) T.-Too Similarly for roda Too + T(x) = - 2 qgen Tdſah (JK, tk.) 5 Heat rate 9 and 22 As already derived - q = √Phk, A. (T-Too) Hous, - P = па A = Tid² derived earlier Ti-Too = 2 qgen Malah (sk, tuk2) transfer rate for Heat rod 1 - q = Tdxh K, yno 4 x 2 Igen Teddah (URAVE

Page q, td Janki x2qgen ad veih (rrituko) = Ski Igen ſkituk, Like wise for rod 2 12= K2 Igen rkt JK2 we know, Oy Puttin/ vatues of a So, fraction of agen that goes in rod 1 - Igen ſk+uk, And fraction of agen that gees in roaz = 2 Igen k Ketuk,

Temperature K, SK, distribution slope for rod 2 As K is less difference per will be more of rod 1. will be steeper. than K temperature unit length in rod 2 than that in case