Question

Required information A shell-and-tube heat exchanger is used for cooling 47 kg/s of a process stream flowing through the tubes from 160°C to 136°C. This heat exchanger has a total of 100 identical tubes. each with an inside diameter of 2.5 cm and negligible wall thickness. The average properties of the process stream are: p = 950 kg/m”, k = 0.50 W/mK, Cp=3.5 kJ/kg.K, and -20 mPas The coolant stream is water (cp=4.18 kJ/kg K) at a flow rate of 66 kg/s and an inlet temperature of 10°C, which yields an average shell-side heat transfer coefficient of 40 KW/m2K Water 10°C Process stream 160°C Calculate the tube length if the heat exchanger has a 1-shell pass and a 1-tube pass. (Round the answer to two decimal places.) The tube length is 12.12 m

Answer 1

+ 106 que calculate the tube length if the heat exchanger has a t-shell pass a 1-tube pass. Water Ans: Given para Thin= 160°c Thout = 136°6 mn=47 kgls =g50kg/m3 Process K= 0.50 W/m.k Sleom 1600 Cp=3.5 kJ/kg.k U-2.0mpas Assump: -0.002 kylmos - cin= 10°c Fluid properties are constant Mc= 66kgls heat-exchanger is well-insulated so that heat loss to the surrounding is negligible & thus heat e transfer Mtube=100 from the hot fluid is equal to the D=0.025 m heat transfer to the cold fluid. changes in kinetic & potential energies of fluid stream are negligible. :. Rate of heat transfer 9 = m ₂ x Ch x (Thin - Thour) =47 3.5 x(160 - 136) °C 19 = 3948 kW) - 4:18 MOTORI . outlet temp of water is determined Q = mcx CCX (Tcout - Tcin) Tcout - Tcin & mcxcc Page 1/4

=10°c + 3948 66 x4.18 = 10 + 14.31 cout = 24.31°C - Logarithmic mean temp. diffeerance is AT, = Thin - Tcout ! AT2 = Thout - Tein = 160-24.31 = 136- 10°c [AT, = 135.68c [ DT2 = 126°c) AT, - DT2 – 135.68 -126 9.68 AT in= AT, 135.68) 126 0.0740 Ато. [atem = 138.78°c) The Reynolds number is v=m - m SA Neube x TD ² 47 kgls 100x950x71x(0.025)2 47 46.63 Iv=1.0078 m/s Page 2/4 Re - VxD G - 1.0078 x 0.025x950 - 23.935 0.002

Re=11967.625 Hence, which is greater than 10000 flow is Turbulent. assume fully developed flow Nusselt number from p = UxCp 0.002 X 3500 0.50 & evaluate the { (p = 3.5kJ/kg. & 2 23500J/kg. c. 0.50 Pr = 14 Nu - hxD - 0.023 Re8 x Pr.3 =0.023 *(11967.625.) *1403 Nu = 92.89 | Heat transfer coefficient on the inner surface of the tube X 92.89 0.025 - 0.50 hi = 4 x Nu hi= 1857.8 W/m2 oc) overall heat transfer is - T hi ho Page 3/4 1857.8 - 4000

7.882 x 10-4 10 = 1268.59 W/m² och The correction factor for one shell pass & one tube pass heat exchanger E= 1 : The tube length 9 - UXAXFXA Tem. 3948 = 1·26859 kW/12g x[100X7 X0·025XL)(1) x 130.78 3948 = 1303.02 x L L - 3948 1303.02 L = 3.029 m Page 414