Question

11.3 A type-302 stainless steel tube of inner and outer diam- eters D; = 22 mm and D. = 27 mm, respectively, is used in a cross-flow heat exchanger. The fouling factors,

Answer 1

Given that Dizazmin A type -302 stainless steel tube 00 = 27mm R98,0 F Tube, ss302 | Do water -iti-v-.-it oto; Tm, is 75°C to + Um, ;= o.sm/s - air No = zomis To = 15°C overal heat transfer coefficient expressed pn terms as د ارK4 Rio + +- - + -- +- - 2 кL - Ho او ho ho UoAo hiAi Ai 1-0 earli) Reduced + R + - = convi + foul, it cond Uo no R R + R Youro Convo - (2) Thermal Resistance Recording their convenetion resistance con= tann - cas fouling Resistance -ance Rour = t - (b)

tube call conduction Resistance Rcond = in Cooles) ankel (C) Beased on ite inner surface of the tube Reynolds number Rei = Pwlmi Di rew 1. = 974.8 x 0.58 0.22 3.746-X104 48/ mison N: Rep,i = 28624.67 Reno ô 28,695 — (3) according to Ditter - Boelter equation page turbulent flow in tube 0.8 Nupi = 0.023 Rési pri 0.4 = 0.023 Re: Procu - (4) = 0.023 x 68 625) 0.8% (2.354.0.4 Nupi= 119.082 W/m kl K from me Niesselt number Nusi = hidi hi= Nuvi XX Di = 119.082 x 0.668W/m.k 0.022m ini= 3615.760 k) (5) Κω

(3) Reo,o = VODO Based on outer surface va of the tube, 20x0.027 S ohl8xh7 17.358 100 L my sec = 31123.919 2 3 1124 from the Churchill- Bernstein correlatiis no 42 Var Nupio = 0.3 + - 0.62 Reo o pra ut It 415 5/8) It 31124 V - osa ota como "clo nos ay in Germany = 0.37 0.62 (3112430 282,000 = 0.3+85044 (1.1971) Nb;0 = 102.58 from the Nusselt number N46,0 = hoDo = hobo - Ko to applied to outer surface of the tube ho= a h- Nubio oko Do 102:58x 0.02737 0.027

outer sur face avea o (AO) = IDOL Inner 11 (AP) = TOOL from eqill - o Rfi Ao Aoln (Polo & I - +- to hiai au 2T KEL am send to to - Do Do + Rti Do Do In (Ooloi) " hipi kipi +2,otho oi akt R 0.0004miklo 80.027m 0.027m numauna toy 0.022 m 3615.76 103k X0.0227 0. +0.027 in (0.027], 0.0002m ++ - 103.985 W/mK 2 X 15.1 Wlmik -4 I-13.3942 X 10 +2. $9.6167x163) Tuo = 91.8w/m.kl from eq (a) convenction resistance (enner surface) Kconvºs Is hiA; 3615.76 W/ mKX 0.022mxlm Rconv, i = 0.004 K/W?

from eq (6) conduction Resissance Couter surface) fonts 1 . - hoao 103.985 W/o-k X TX 0.027ř lm Recondi 0 = 0, 1132 K/W] from ea (c) Resistance due to tube wall conduction in (ooloil in (0.027/0.022) 21 ktl 21X is.1 W/mKX2 m Re=0.002 158k/W/ Kcond = Rt - - 1 due to Ai fouling Resistance in inner surface 0.0004 m² klo Rrouli - z FX0.022mxim Rrouri = 0.0578 Klad fouling Resistance due to outer surferce Ppoulio = 240 = 0.0002 m2 kl UX CO.027 m tim i Rfoul, o= 0.002857 k20/

b) Reynolds number assuming water flows over the tube then Re Repo = VODO = VODO Nm = imis x 0.027M 1.028 x 100 mis 26264.59 Redio ~ 26265 rom the Churchill- Bernstein correlation 0.62 Y2 13 r re 15287915 Rein Pru it( Reo o 1518/9 Nupio = 0.3 + 19&2000 + 5/8 26265 0.5 0.62 x(26265) *(7.184) /? OLUUU 1193.881761-1777) - 0.3 + 10346 INuo, o=220.997 from the Nussect number N o hooo hooo Ko Ko ho nedox AW DO 60 - 220.x 0 60 72 /,k 0.027 m Tho= 4920.866 W/m². kl

Now find overall heat transfer coefficient based on outer surtace 1= Do Riti. Do poln (Poloi) " vo hipi Di + Rt,o 2kt + 0.0004 m HW x 0.027 0.027m 3615.76 W/mK x 0.0221 0.022 m + 10.0277 0.027 in 0.022 0.0002m 160 +- 4920.866 2x 15.1 W/mK Whik ] [ . 342 x 6", 2.1545 xe", 197 x 6" I = vo 1 + 2x10+ 2.032x104 KW Luo= 676.15 w/mk1 Resistance due to convencion on me outer surface Rconvio = - = - 1 hoxao 4920.866W/mKX TX 0.027mXIM IR conv,o= 0.023 95 kW Resistance due to fouling on me outer surface R foul o = Rto = 0.000Lin" ELO TX 0.027mx lm AD [R ful, 0 - 0 0 043 si ta Note Resistance due to fouling & convencion on the inney surface, resistance due to tube wall conduction remain unchanged. =