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Problem 3 (60 Points). A 1-2 shell-and-tube heat exchanger must be designed to heat 2.5 kg/s of water from 15 to 85°C. The heating is to be accomplished by passing hot engine oil at a rate of 5.2 kg/s, which is available at 160°C, through the shell side of the exchanger. The oil is known to provide an average convection coefficient of 400 W/m2K on the outside of the tubes. Ten tubes passes the water through the shell. Each tube is 3 mm thick and diameter D-25 mm. The thermal conductivity is 120 W/mK, the cooling water side heat transfer coefficient is 3061 W/m? "K, a. Draw a schematic diagram of the process (5 points) b. What are the assumptions of designing a heat exchanger (5 Points) c. Draw the temperature profile. (10 points) d. Calculate the LMTD? (10 Points) e. What is the overall heat transfer co-efficient? (15 Points) f. How long must the tubes be to accomplish the desired heating? (15 Points) Hints: The Cp of engine oil at average temperature is C, -2350 J/kg K. A(Tn - Tc) 9 = = UAAT 11 T: 1.0 0.9 lis 0.8 Correction factor F 0.7 0.6 12-11 10 0.5 0.8 0.9 0.5 06 0.7 04 03 02 0 0.1 P-

Answer 1

Solutions 25.2 Kg/8ec @ moil, P=160°C =15C = 3061 W/mK ho = 400 W/m²K in water 2.5 kg/sec $=85°C K = 120 WMK h di = = 0.025m , do= 0.031m ( 3mm thick tubes) of tubes 3- # Q = miferit (& -ta) – moly Grail* (?-@)=0, A. AEN19° * Assumptions I 1- No heat loss into the surrounding. & Expansion & contraction neglected. eriform heating & cooling focess from 9 Macester Cphtuler *42- &z)= mult Grue *(4-ą) 2:5 19 fec* 4184* (85-15)°C= * (85-15)°C = 5-24*2350J (1601-T) Te = 100.082°C © * sec kgec Temperature profile T= 166c $=854LAT = 75°c T= 100.082°C AT 85.082°C 14=15°C distance

@R ① R=9-2 160-100.082) t-t = 0.86 (Abbraxa (85-15)°C S=P= t-tq (85-15) * I-te = 0.483 (Approx) (160-15% F = 0.86 (Affox) from graph (05-05) ATUMTD = F*(TLMTD). = 0.86. sale Ideal ln (AT) AT ATLMT) = 0.86* (75–85.082) = 68.74415165°C ln (758 85.08285 # 14 TLMTD = 68.7442°C Abbook Chave d. + do bi (do/di) - @ *1 1 Ve 1 he 2K di 0.031m lor (0.031m) 1' + 0.025m 1 Vo 400 W mak 2* 120W/mK 10.031m H 0.025m) 3061W m²K # lu = 340.9616W m²K Approx

© # forom @ mwater *(6) Water* -ta) = V. AXOTEMTE 340.9616W *H*0.0310* L*68-7442%= 2.58 x 41842_*(85–15) mi Set #2 = 320.76m Abbor

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