Question

2.- Hot oil (Cp = 2200 J/kg °C) is to be cooled by water (Cp = 4180 J/kg °C) in a 2-shell-passes and 12-tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of 1.8 cm. The length of each tube pass in the heat exchanger is 3 m. Water flows through the tubes at a total rate of 0.1 kg/s, and the oil through the shell at a rate of 0.2 kg/s. The water and the oil enter at temperatures 18 °C and 160 °C, respectively. If water leave at 100 °C determine the rate of heat transfer in the heat exchanger and the overall heat transfer coefficient. Comprehension (State Definition, Free Body Diagram and/or Diagram Sketch) [10 points] Oil 160°C 0.2 kg/s 100°C Water 18°C 0.1 kg/s (12-tube passes)

Answer 1

= son - Hot oil 0.1 Kyls . Tai Given :- cold water (p = 4180 Jlky C (p= 2200 Jlky c ma m= 0.2 kg/s Tin= 18°C Tin = 160 °C = Thi Tout = 100 °C = TC2 D=1.8 cm L 3 m DD=0.018 m To find o Rute of heat transfer in heat exchanger 1 overall heut transfer coefficient. - sol":- 160°C 0.2 kg is looc < Water 18°c 0112915 ( 12 tube passes) As 2 Heat transfer surface area is As = n(TDL) = 12 (T) (0.018 m) (3 m) 2.04 m2 Rate of heat transfer in heat exechanger is @= {m cp (Tout - Tin)] water = (0.1) (4180) (100-18)] 34276 Watt [m cq (Tom-Tout)) = {molo (Tout-Tin] water >(0.2) (2200) (T160 - Tout) = (0 : 1) (4180) (100-18) 440 (160 - Tout) = 34276

This Tout = 82.1°C for oil overall heat transfer @=UA com where Om = 0,- 02 en (01/02) (thi-teq)-(tha ta) en ( th-tca that (160 - 100) - ( 82.1 - 18) -to un 160-10o ( Temperature 82.1-18 64.1 - 0.0660 0 Hotoil om = 62-02 cold water tei ce=uAcom → 34276 - UX 2.04 x 62-02 U= 270.91 W/m²°C -- Area figi- Temperature distribution Answer = 34276 watt O U=270.91 wimu Heat transfer rate overall heat transfer coefficient