MEN 310 Exam 1 with Solution, Spring 2001

a | | a S9\ Notre Dame University ' Faculty of Engineering - + Mechanicat Engineering Dep, TEST—- MEN310 Heat Transfer) March 28, 2001 1. Consider a pipe (8-cm ID, 10-cm OD) made of stainless steel in which energyis generated at a constant -rate of LOOMW/m’. The outer side is insulated while the | _- nner dissipates heat by convection witha heat transfer .coefficient of 1000 ' W/m”C into a fluid at 80°C. Find the temperature distribution as well as the heat tate along the thickness of the pipe watl. Determine the températures of the outer and inner sides as well as the heat lost on both sides of the pipe wall. : . 2. Steam in a heating system flows through tubes whose outer diameter is Dj= 3cm — and whose walls are mainiained at a temperature of 126°C. Circular copper ins of outer diameter D3= 6cm and constant thickness t=2mm are attached to the tube. 39 The space between fins is 3mm, and thus there are 200 fins per meter length of the 3 ' fube. Heat is transferred to the surroundings at T.=25°C, with a heat transfer coefficient of h=60W/m™C. Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins. i H \ . 3. Solve problem 1 using the finite difference technique with the following mesh. (dimensions in cm): Compare the resulting numerical solution with the corresponding analytical . solution of problem 1 and show whether such a numerical solution is good or not. What can be done in order to improve the accuracy of results (to get closer to the analytical one). : \ Points: 1 (30%), 2 (30%) and 3 (40%). FO — k. Fe) oo LZ O valde dL e { = G . + . _ : { qevcuile A | Ca wrechine ly Sth. gut l qv = AACE) * va {, T. - qv + FAT - loo xto x Ti (« 05 - 0.04") x 1+ topoxZilroeth A , loco, 2Tx0.04 | b | t= 1205°C | — Ts -9F 4G dea G _ Tle i205°C bouncer Cardi Biee — * =9 2 tf {40 dc Jez Sean. ete. 6 a: - “ae +h S — fe = ge — Ge 4352. WTC AC vo aR L Cis 12054 Ge — Car 4 G = AFZZE 1HSTF GE | T- ~ {G4OS8R . 935 07 4 1BS2 V4 Lar + 24326.14594| . 7) = {972 .4236C C2 Say 43.3333 Wot = S62442. 2.04 x (1205 ~¥o)= 2 Tie) = toon x 2 x q= fA CT -Te KA dT = = -Ka(-gr +S) oak nek. —» q= he cx Gi) ~The 21 0.04 -§F % 4 ye a = RA (Tt) = 60x1 10.03% (20-28) = 537, 2123 Weitts 4 + 4 7 = 57 “ft poesia. an onl _ ; : . on . 0270.09, fun tars fo 2 y ; _ \ fe en _ fn. = ; as _ aso : at = QA G - a) = G0xT20.03 « 0.6 (I29- Q5)= ee ea ~eS4lGe BA cen —) AY. i} %A.,/, = 2.0666F Sk — “- 337 G__ = RAC Te) 5 60x 2x7 (a.021*- 0.015") (120-30) Wray = QU. AFISFIIGR Wot P » al is -} \g? Z tr, L =. Y9GF. 09926 Welt, Ds, = Tastes * Tou = 186F 8 ° sae = Jf <=> Ag = U4 29.8851) Wath Te. may < ET / g bV= ‘) Ry = dal 3h4,1) ot ATRL AV= TI (0.042°- 0.0405") U32S. 4ST, - CSER 42ST, + 22472.ETT + BVH. V090S=0 Raa = Ju OAs) ZTAL AVETT (o.04¥s*-0.042") G [2848.G9T, — 386.4997, 4 IS83. 8091, 4 GF93C. 4113 =0 | Ned. 3; Te. Toe! (re 4-13 , T-B 5 @OVs0 Rey= £4 (5/16) Rox ZY Ro, . 2TT RL AVA VW (Co.ot8?_ 0.0445") [1583-8097 — 2864.84 Ty + 128102297, + 101203.06 22:0]