Question

2. (35P) Ashell-and-tube heat exchanger with 1-shell pass and 20-tube passes is used to heat glycerin (Cp = 2480 J/kg) in the shell, with hot water in the tubes. The tubes are thinwalled and have a diameter of 1.5 cm and length of 2 m per pass. The hot water (Cp=4180 J/kg Centers the tubes at 102°C at a rate of 9 kg/s and leaves at 55°C. Overall heat transfer coefficient U=13900 W/m2C. The glycerin enters the shell at 15°C and leaves at 57°C. Determine: (a) heat transfer rate Glycerin (b) the mass flow rate of the glycerin, (c) LMTD (d) Correction factor "F". (Adjusting blue colored vertical and horizontal lines to fix the F) (d) Length of the each tubes 1.0 Hot Water 0.9 0.8 Correction factor F R = 4.0 3.0 2.0 1.5 1.0 0.8 0.6 0.4 0.2 0.7 0.6 RE T, -T2 1-1 0.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 (a) One-shell pass and 2.4.6. cte. (any multiple of 2). tube passes TO P= 1,1

Answer 1

hot water glycosin Glycerin - Inside shell Ti=15c Tz=57c (CP)=2.48 4 #, = 1022 t2 = 55°c mw = 9 kg / C = 4.180 A fu lkg'c 3 tol 4 water mg = glyusin mass kgc glow sale U = 13900 wlmĆ > I shell 20 tube pass d=15cm. old ㄴ Q L=2m per pass (9) Heat pans for sale a mg (og (T2-T ) = mwllt -tz, mall tit) = 9 x4.180 ( 102-55) Q=1768.14 kW And ① mass low rate a glycerin =1768-19 = mg (Cp), (To-T, ) = ngx X 2.48 (57-15) mg = 160975 kg Ang Now & hot water :72 for glyann 2 TI (c) LM70 LMTD = 01-02 Tz enlo o = t-T2 = 102-57€ LMTD = 45 -90 0,=45°C en (4) Oz =S5-15 LMD = 42.473 O2=40 Ans ta-ti

F The o correction factor R = Ti-T2 ta-ti Now 13-J7 = 0.894 s5 -102 pa SS-102 ta-fl Titi = 0.54 18-102 Il/ From chart plotting value of P&R wo get F = 0.84 X (€) L = length of each tube 1768-14 x 103 3 ZFUA LMTD A = susjone NOW Q = aP4 = 13900X AX 4245 Eo proiems A = 3.567 m² A = * n=noo pass A =nxade' 53.567 ->20x1X00015 E=3.567 =>/L'=3.785 m Total length = 2083,785 = 75.69 m