Question

A shell-and tube heat exchanger has one-shell pass and 2-tube passes. This heat exchanger is used to cool oil, flowing through the tube-side from 140°C to 50°C. The cooling is accomplished by water, flowing through the shell-side, which enters the heat exchanger at 15°C and leaves at 32°C. Each tube pass consists of 60, 2.54-cm-O.D. tubes with a wall thickness of 1.65 mm.if the inside and outside heat transfer coefficients are h1=260 W/(m^2°C) and h°=970 W/(m^2°C), respectively, and the fouling factor on the water-side is Rf,o=4*10^-4 m^2°C/W, calculate the rate of heat transfer in this heat exchanger neglecting the tube wall thermal resistance.

Answer 1

Given that shell and tube Heat exchanges with (Ishell, 2-tube passes) DST LMAO DTI - DTZ = The 18 in ATI fin The S& AT2 AT=Thi-Tcz/ot, Th= The Test STL = 5045=135c = 140-32 ATLM TO 108 – 35 = 108 in 108 Tei= 1156 35 Ex=32°C. A. AT Lugo = 64.800 But It ý not - fass shell and tube. Heat exchanges, so that We need to correct LMTD. for (1-2 pass ssuth Hx) 'corrected = LMTD Ę X OT, 'LMID I Correction facta Ę can be found from graph Farid R= capacity f =0.99 R=02 두 Ratio. Rolor Room R-3) R= the the tatti R= 0.2 0.3 Tuo - 50 32-85 R=5.3 Temparature (f) Ratio 5 teata Pa 32-15 thi-ta 17. 125 - 0.14 140-AS

- - toteoreeted = 58.32°C) At P = 0.14 are for a given curve R= 5:3, then Corseponding F1=0.9. Here al attached graph below, where R=5.3 is not there but we should ass draw that R=5.3 Cusne at P=0.14 at R=5.3 Fy cod. =FqX ATENTO = 0.9 X 64.8°C at correctes LMAD LANATO * Now 0.02 sum=0.03m im autes Diameter (do) = 2:54 cm x 10m Toys wall thickness t = 1:65 moyon =1:65X163 m 1000 myn innere diamotes Oi) = Do-2 (t) -0.03.04 – 2(1.65X163) DS 0.027 Fouling factos Rfo = 4x104 mrc W hi = 260 w mir ho =970 w mroca

Uo= DO Do o Bo + Rot Di hi ho No inside fouling factor Refi=0 Neglect wall resistance ... De on Do to Di - Vo = Di hi Root the Vos 0.03 + uxiou + 0.027 260 970 Uo = 4.27 x 103 + uxiol +1.03X103 Uo=175.4 w my C Here you did not mention (Length of Heat ene Area (A)= Ti do L. (m2) enchanges) If I assume L=1m =15m Area = Tix0.03 x19 0.28 1.413 mr.

Qe VO A DTUMTD correcte Q = 175.4 W x 1.41304 x 58.32% vog og Q = 14454.ou Walte uusu.ou walts 1 = 14454.ou I J Rate of Heatleangtes If I assume Asa (A) = 1 mm U OT = 1.75.4 xinh x 58.324 =10229 3 Ion watts X