Question

Hot water flows thorough a parallel flow heat exchanger at a rate of 10 kg/min and is cooled by a cold water stream of flow rate 25 kg/min. The inlet temperatures of hot and cold water streams are 70 oC and 25 oC, respectively. The outlet temperature of the hot water is expected to be 50 oC. The individual convective heat transfer coefficient on both sides of the heat transfer area is 600 W/m2 .K. Take the specific heat for water in both streams c = 4.18 kJ/kg.K. a) Calculate the effectiveness, heat transfer between the two streams and the area of the heat exchanger. [15 marks] b) During operation of this heat exchanger the hot water flow rate is increased to 20 kg/min. Find the exit temperatures of the streams. Assume that the heat transfer coefficients are not affected by the changes in flow rates.

Answer 1

This question belong to Heat Exchanger. At first outlet temperatures are calculated in these type of question. Please go through the detailed solution provided in the image for clear understanding. Please dont forget to like the answer if you find it useful. Have a good day....
Given - mn= lokg/min = Hot water flow = 0.166 kg/sec Mc = 25 kg/min = cold water flow Inlet hot water Temp = Thi = 70°C Inlet cold water Temp = Tc;= 25°C Outlet temp of hot water - Tho= 50°C Heat transfer coolen- 600 W/mK Specific heat of water-C-H18 145/1gK Thu = 50°C This 70°C. mni - 10kg/min Tc; = 25 ch Teo=2 mc = 25ky/min Foom Energy balance- mn (Thi Thu) = mc (Tco - Tci). 10170-50) = 25 (Tco-25) To= 33°C Effectiveness. Effective Web = Thị - Thu - 20 - 50 04H Thi - Tci 70-25 Heat Transfer between Streams = mn c (Tni - Tho) = 0.166X 418x (70-50) = 13.93 RW

* Log mean Temp clifferce, LMTD - LMTD = (Thi - Tci) - (Tno-Tco) In 1 Thi-Tci) (Tho - Tco) LMTD - 070-25) - (50-33). 28.76 In 170-25) (50-33)

U = 1 + 12 - t = too + 600 TU=300] ⓇS UA (LMTD) 13930 = 300xAX 28.76 A = 1.614 m? - Area of Heat Exchanger

(6) gf inn = 20kg /min then- By Energy balance- mn (Thi-Tho) = mc [-Tci +Tco) 20x (70-50) = 25 (25+ Tco). Tco=41