Question

4. An exothermic material is encased in a 1-m-long thin tubular sleeve. The material generates heat at a steady rate of 4800 W/m. The thermal conductivity of the exothermic material is 22 W/mK. The radius of the tube is 30 cm, and you can neglect the thickness of the sleeve. Also assume the ends of the tube are perfectly insulated. (a) If the tube is exposed to air at 20°C, and if the convection coefficient between air and the tube is 18 W/m-K, find out the surface and center temperature of the tube. (8 pt) (b) To achieve a surface temperature of 40°C, it was decided to add straight rectangular fins longitu- dinally on the tube as shown in Fig. 1. The de- signer came up with a design consisting of equis- paced 18 fins of 4-сm height, 4-mm thickness, and 1-m length made of a material of conductivity of 180 W/mK. For the next parts, assume that the air temperature and convection coefficient remain as specified before. Assume the contact between the sleeve and the fins are perfect and the tip is adiabatic. Figure 1: Fin for exothermic material i. What is the efficiency of each fin? (6 pt) ii. Is this design sufficient to achieve the desired surface temperature? (6 pt) (Hint: Does the fin array give enough surface area or efficiency required.)

Answer 1

Answer: Given that length im. Heat at a steady orate 4 - 48004 Thermal conductivitu k = 22 w imk CD Given that, Radius 7 : 30 cm = 0.3m Tube is exposed to air at 20'cie Ta-20°C convection coefficient h. 18 Wak Surface temperature! off is calculated using relation below qv = hA(Ts-Tow) where ve volume = TTY () ha convection coefficient To = Surface temperature A = wross sectional area A = ( 21YX 2) = 21YL Bubstitute values in equation 4800 (LITY") = 18 (2.11 v2] Cuts - 20) TS - 20 4800 (1997) 18(21771 4800 X Y = 4800 X 0.3 36 18X2 Ts = 40 + 20 = 60°C :. Swrface temperature is 60'c Center temperature of tube Tc = tw +

To = Twit qr BK Tc = 60 + 4800XIV 8x22 Tc = 87,27°C center temperature is 87,27c. (b)) Efficiency of fin :- mis home - KETA m' _ 18 1146 0.04. 22x10.004 x 0.04 4 mm0.004 m I 240.004+0,04) Loim - 18 22 x 1.818 x 13 - 450 m = 1950 : 21.2132, 1:1 n = tanh (m2) - mL . Hanh(212132x) _ 21. 213 2X) 0,0472 2: 4.72% total surface corca with fio = Arca due to fin + Remaining tin +- Remaining surface. oua of tubuhan sword As = [18 X (2x (0.004+0.041) X7] + [2 mY_ 18 (2x (1+0.001)x 0.047 = [18(axo.oyu)] + [2x3114x(0.3)– (18*0,08032]] 1,584 +1.884 - 1.44576 = 1.584 +0.43824 As 2.022 my

Required Surface area AS - - h (Ts-Ta) hore Surface temperature maintained 40° As - 4800x ( 77 xD . 78 (40-20) As = 4800 x1 :X3114x (0.3) 18x20 As - 1356,48 = 3,768 mm 360 Required Surface area> total surface area with fin .: due to this condition design is not scefficient to acheive qoc