Question

Problem 1 (30 points): The feedwater heater for a boiler supplies 10,000 kg/hr of water at 65°C. The feedwater has an inlet temperature of 20°C and is to be heated in a single shell, two tube pass heat exchanger by condensing steam at 90°C in the shell side. The overall heat transfer coefficient is 2,000 W/m²K. (a) Find the steam condensation rate (b) Determine the required surface area using LMTD method. (c) Determine the required surface area using NTU method

Answer 1

feed water heater (Heat Exchanger). water mass flow rate = 10,000 kglar inlet temperature 20°c exit temperature -65°C. steam = 90°c temperature mas flow rate = ? for steam (saturated steam) at 90°, enthalpy hfg = 2283.2 kJ/kg. (from steam tables). a Energy from Conservation of in an (Text-Timl) water - in stem hafa 10000* 4.187 (65-20) = in (2283.2) 825.22 kylny steam m Steam Condensate rate rate of steam = 825.22 kg/mr. то

b LMTD Logarthmic Mean temperature difference) method :- T 900 90°c 1 мт, ΔΤ 65°C 20 (whether it is Parallel flow or Counter flow, same for this case. both LMIDA (Phase change) : LMTD = AT, – ATZ In (). (90-20) (90-65) In 90-20 90-65 70 - 25 In LMTD 43.70 K con c M Y2

heat transfer rate 8 = UAS (LMTD). Q = steam fg 825.22 2283.2 Kg/ X KJ kg, 19884150 kJ/my KJ hr, 1884150 kW 3600 523.375 kw. Q UA, (LMTO) (2000) ASX (4370) 3 523.375 Xlo As = 5.988 m2 Required Surface area wing LMTD method As 5.988 m2 NTU (Number of Transfer whits) method: UAS targer temp drop NTU DCO OT ON POCO X2 (mC/min LMTD.

C D in this case, for steam, (rc) steam is maximum. (mww) is minimum. NTU = UA, (c) water NTU = larger temperature drop LMTD. OTI from LMTO diagram, Larger temperature drop is = 90-20 = 70 c To NTU = = 1.602 43.70 UAS NTU= (in C) water 1-602 - (2000) As WE 10000 3600 4.187X103 X102) WK m2 As ON POCO X2 6.1758 m2.

e 6.1758 m². As = 6.1758 m². required heat trausfer how surface area wring NTU method 6. 1758 m². 825.22 kg/mr Condemate rate of steam surface area uring 5.988 m2 'b LMTD method ② surface area NTU wring method
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