Question

Problem 3: Ordinary Differential Equations A straight fin of uniform rectangular cross section (0.5 mm x 100 mm) with a length (L) of 5 cm is attached to a base surface of temperature 110°C (T). The surface of the fin is exposed to a cooling fluid at 20°C (T) with a convection heat transfer coefficient (h) of 15 W/(m²K). The conductivity (k) of the fin material is 400 W/(m.K). (a) Plot the temperature profile along the length of the fin, i.e., T(x), and (b) Calculate the temperature at x = 2.5 cm. (c) Determine the heat transferred by the fin. From energy balance, the following differential equation is obtained: d27 hP dx2 - kA. (T – Tj) = 0 where P is the fin perimeter, Ac is the fin cross-sectional area and x is the distance measured from the base. The equation has the following boundary conditions: • The temperature at the base of fin (x = 0): T(0) = T) The rate at which energy is transferred to the fluid by convection from the tip (x=L) must equal the rate at which energy reaches the tip by conduction through the fin: dT NA (T(L) – Tx) = -kacam The heat transferred by the fin can be calculated from the following q= -kacam

Answer 1

Ver Given dajo according to Question ГТ \\ \, идеа hard Base C coolela bruid) affin cross-sectionaf fin - A=0.5mm x 100mm length of Fin = L = 5cm Base Temperature=Ty = 110°c cooling fluid Temeperature If = 20° heat transfer Coefficient (h)= 150/m24 Conduc fuity of Fin(K) = 400w/mk liven Temperature pocabile d2 hP IT-TJ = 0 doc2 KA (T-TJ =0 Where P= Perimeter Ac = Crass-section Area R = fue distance measured - from base af fin. Boundary conditions - At x=0, 7(0) = To htc (T() - Tf) = -K A co lach Heat transfer by the fin q = -K A c d e

Now athly Boundary condit D(V) = TCM)-TA At x=0, TO0) - to :: 0(0) = T-TA oo (0) = ciemos temo) > P (4) = 2 + C - 6) → (-1)=,462 +(910-90) = 4 +Cg 90=64 :6+62 +90_ o in C+C2 = 90 = 0 -o Now olx GenxtGena na px Differentiating wir.t"X" doo - e, menxamene ma El_qema) = deres ma come Gent) - dob Since Boundary Condition: - -k Ac dt bine - ha (TUI) :-keretezh Dikey che reach ock -ma 99/K K

ma em-mcenil - holt) mc, em _mceml = h cc, emu se Gene Geml = tham (achetla Equation & salwing e = 0o (1-hemlo cemete musth Ceme emeg GE JE en m a G = 0 - 0o ( 1-4 cm leme an cemetemuythm lene meg Polith Jeme Cemetem th me emly - O (M) = Gema Gema Tool x-2

(b) temperature al 3 -- 215 cm - La sem = 0.05m m= home kne DE 21246) p =? (0'5-1100) =-201mm AG,0.5X100mm2 - somm? me the !! KAC 157201 400 X 50X103 T 15X201x183 7 20X10 400 450 15x201 X1 ST 20x10s m ="115X201 = 12.28 J20 mla! 12-28 X 0.05 ML =0.60g (ss units reme les s! h = 1 x 12:28 -0.003 ; é me = 0.55 9=: Ooltah meme Teme emeth lene eme) Go go (1 -0.003) €0.55) _ (1685 +0.55) + 0.0031 @ 183-0:55) G = 20.70°C

ang = 2007°c 12 = 90-2017 0 2 ą 69.3°C » 0(4) gemx & Gema 012- 2-5) = 2007 il? •28%0.025 = 28.138 + 50.9805 0 (4=2-5) = 79.1185 T(X) - TF = 79.12°C - 12.288 0.025 + 69:3 Xe T(R=2.5) = Tf +79.12€ = 20479.120 T(*=25) = 99.12°C (C) Heat transfer through fin 9 = K A c d ulazo = -800) (0*5*100 XVõ6) de desco wart =-400 5X165) debice 5 - 400 X 5xrās (me e* 10%-me, encos) . --400x5X105 ym cc, -C2) 9 = tyrox sx1Ő5x12.28 1 2017-69034) W ax 20 9 - Hellt 11.986 wato transfer through bin q = 11:3936 wate