MEN 310 Exam 1 with Solution, Fall 2000

wen Vetisas Natre Dame University A foe Faculty of Engineering Mechanical Engincering Dep. TEST— MEN310 (Heat Transfer) : November 10, 2000 Consider the base plate ofa 1200-W iron that has a thickness of L=0.5em, anda. base area of A=300cm?. The base plate is made of stainless steel. Its inrier face is subjected to uniform heat flux generated by the resistance ‘heaters inside, and its outer face loses heat to the surroundings at T.=20°C by convection (h=80W/m”C). Determine the temperature distribution across the base plate and evaluate the temperature at its inner and outer surfaces. Radiation effects are assumed to be negligible. 2. Steam in a heating system flows through tubes whose outer diameter is D =3em and whose walls are maintained at a temperature of 120°C. Circular copper fins of outer diameter D,= 6cm and constant thickness 2mm are attached to the tube. The space between fins is 3mm, and thus there are 200 fins per meter length of the tube. Heat is transferred to the surroundings at T.=25 °C, with a heat transfer coefficient of h=GOW/m”C. Determine the increase in heat transfer from the tube pec meter of its length as a result of adding fins. 3. Consider an Aluminum fin of triangular cross-section with length L=Scem, base thickness b=Icm, and infinite width w in the direction normal.to the plane of Paper as shown in fig.P3. The base of the fin is maintained at a temperature of To=200°C, and the fin is loosing heat to the surrounding medium at T.=25°C with a heat transfer coefficient h=15W/m™C. Using the finite difference method with six equally spaced nodes along the fin in the x-direction, determine the temperature distribution, the rate of heat transfer from the fin for w=1m and the fin efficiency. 5 D mr —_——_____ Points: 1 (30%), 2 (30%) and 3 (40%). & ¢ = {2e0 = Sxlow 3 = mi 3rax toe. 205 ‘ 4 = 4 —_ l2a0 = h A Ct, ~Teo) = ¥a x0.0Q3 Ct, - 20) Crasechion, a : — 5° “ . Ll, ive 320°C OT Renctes, 7 a0 ge ST = (eet). a 3.85 Te Ox4e, X-0 — THT, = CL X=L —» Tl. CiCo.008) 4 Szo = S3). te4 —> G= 2392.8 & 1 TL 9352. 2x 4 S20 | _ #2| Refer to Teat- Mam 28 Zoo] . Ris = 0.0] —. { iWg X occ Ix 0.007; RK ~ ~ 7 =" Keo oer Rites = 4 a =< @ ASOF4w OT: — - 1Sx0. 010 ot4za, PB mame ye mere mn Oe - L, | - 393, Sol4dy T, + (41.4 T + 36364.5393 S20 Noda ft , “tha -e 4 Ta-h 4 2. Co “T2) _o Raa Ras ’ Race 7 : Ry = c.o | _-) ft = iofeus . K 2d — An 90.00 5m MWOZ * O04 a R ; y = G Pag T - 242. 21a, + (1 T 3 +7.53935 20 | — Miao) now Sot tee Ts 1 . 60.6u t J Ee —_ ; ees2x 0.002. R 4 | (OT, -i6l. Vol Ty + 60.6%, + 4.53 73F20 | ZI .0.0dles iG | 0.6 B- Slots ty + 20.27, + 7.83720 J - eds 3; 1¢ ~Ts + ed. (25-Ts) oo Ryy Kgue Kye ~ Ss 5 Le 9.0FS3 435 1S 0.0leatGsxo.Se RK G | 20.4 Ty - 20,350797 Te + 37650 | ee ae Seolmine—= lie (98. T2204 °C Te = (YSPC a Te IIS EL 9B. Gye Tr $96. 15C seme TY os wet Toe ° 9 “od = hk (fo n a: a] RA (25 -193.42) + hA@S-!%. “ 4 hA (25- 196.145) _ 4 kA Qs 194.3) + AR (25 -193. Sz) | t . = Uren} gs Qx\Sno.cloc 495] 2S ~198.4 14 QSANIAWS 425-186 . 425 -IGN.S[ + O.Sx ZF ZOOSK (93-6923 | “4 wha oll Pu. Ls @ 200°C i G= ABA (Ho =T) = ~ 1S Koo 1ooVARx = (200-25) 42 S) qQ=- 26%. Fo4gs Lots . ‘Vs at ~- 2 .S8l > S8.ARZo rs . — checked - CL 2 bro f \. - weet fs ca af . re teyemtesiiettr f . i.” Np. .