Question

QUESTION 3 Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the finis essentially that of the surrounding air, le 10°C its width is 10 cm; thickness is 2 mm: thermal conductivity is 150 Wim K and base temperature is 100'C. The heat transfer coefficient is 20 W/ m K Estimate (a) the fin temperature at a distance of 5.0 cm from the base in c to the nearest degree (b) the rate of heat loss from the entire fin in watts to the nearest watt

Answer 1

Please find attached below showing calculation of answer.

Date -> The onolpois of rectangular fin for tempesutuse distribution (pohle) and heat transfer through tin. Convection of air at To, hw/mk NI INIC . Base Wall at To M e Conduction Roots Tip OR End of fin at I L => Poolile Oyed A EZ+ = length ot fin = L - width of fin= - Thickness of fin=t =) Perimeter of fin = 2x2xt) a) The actual mechanism of heat transfer in any fin is that the heat get Conchicfed into tin from its pot frost and then conduction, into its length and from there to connection to the combient air at To with convection heat fronter Coletticrent of h whic Objective: To get temperature distribution within the fin.i.e.T= f(x) ② to get heat transfer sate throughout the fin ? -> Assumption : Steadly state heat toomster Tat time) . Consider a differential element of the fin of length dx at a distance of a measured from the base of bin.

Date +3(90) dx - Let heat Conducted into the element q = -KA Heat Corchudled out of the element = 9xx x +19). & Rite of change of heat - 2 (2x) - Total Change thoonghda = }(92).do - Enesgy balance for Steady state Condition of tin, Rate of heat transfer at x = Rate of heat transfer at arda + Rute of heat confection from dx x = 9x+di+ 9 cony i 9x = x + 2(9x) dx + 9Conv i ct will on Oro (92) dx + hpdx (T-Tw) ( A = padr) - O -KAHVx) dx + hpdr (T-T. - sth PPT - hp (T-To) -o &T - HP (T-Too O da? KA Let T-Tos) = 0 = f(x)ms di do dx di d2T - d2 and put da² dia d20 m²=0 the m=hp/kA = & Hot dx2 - This is the a Standard format of second onder differential equation in a whose sondion directly given as o centce where m= [hp/KA

Page No. oc, emx + c emx where m= /hp "KA o the constant a$ ca are interqsation constants eye can be obtain by boundary conditions - - Initial boundary condition a n d at x=o Ci.e. at the base root of the tin) T=To => 0-0 - (To-Tos) final boundary condition at x: 00 (fin is very long T-00) T = Too =) = (To-Tos) I - - S & gives T= To - 0.0 Put above value into equation O = Clem(0) + C2 em (0) 0 = C + czemos 0-0 + C2 e mo GO at x = 0 = 0 0 0 = Cie + C2 e - C + C2 GEO (16220 Obtained in above ase) Putting value of constant into equation gives 0-0 MY -

Page No 011 - T ema 0. To-Too - For Steady State heat transfer sate, the heat transter sate in tin is equat to the heat transfer sate into fin at it's base L 9tr --KA doo gydx=0 = -KA dolce fit dt - do dx xo dx dB) --KA d oo. e-mx de freon = -KAS, dlēma da 2-0 O wl of wie -KAO, (m) emx who veda tro a = KAO, mēma - - song = KAOo me = KAD, m - KA 0. * hp ( 3m. JhPIKA) КР = KA (To-To) dhe КА Prin (b-Te JhPKA

Page No. * Given Too = 10c To=100c K=150 W/m.k h=200 hk, - Area A - Zt = 10 x 102 2103 = 2x104me - Perimeter, P=2+ (Z + 1) = 2* (10 + 10 + 2x103) =0.200 m = m = /hP/KA 20 x 0 200/180*2*154 = 1).66 a when 1=5.0 cm = 0.05 m - from Temperature Protile relation T-Toop-mx --10) = (100-1o) efir.tex 005) (1-19 = 90*0.5582 T=(10+50,24) c = 60.24 € - The temperature of tin at 5.0 om from base is T = 60,24 c. 6 The rate of heat loss from the entire fin combe given cas Qon = To Too * I hPKA = (100-10)*20*0.2047-150*2*104 90* 0.3498 watt = 31.487 watt Otin =31.427 watt