Question

Enter your assigned value for Y and C: y = 47 5 = 8 (You will lose 1 pt. if missing). 2) Hot air enters a heat exchanger at 350°C and exits at (130+y). The heat extracted is used to boil (0.2 + 4/300) kg/s of 100°C water (from saturated liquid to saturated steam). The heat exchanger is a single-shell shell-and-tube heat exchanger with two tube passes. The overall heat transfer coefficient for the hot side is 240 W/(m2). Assume a constant specific heat for air of Cp = 1.005 kJ/(kg°C). The pressure of the hot air and the boiling water is P = 1.0 atm. Find the following: a. the NTU for the heat exchanger (+ 0.01) NTU = b. the effectiveness of the heat exchanger E = (+0.01) c. the net heat transfer from the hot air. q= kW (+0.5 kW) d. the required hot-side area. An = m² (+0.1 m2)

Answer 1

» Thị - 350°C The = 130+47 = 177°C Mc = (0.2+ 47 = 0.35666 kg 19 300) Toi = Tce - 100°C U = 240 1/mpoc CPi 1.005 Kg/kge P- lami Thi Soi?; Ati They ATi = Thi-Tce = 350-100 = 250° late Tee Тci A Te = The - Toi 177-100 = 77 L.M.T.D (Tm) - Ti-de en/ATE ДТе Heat transfer b/w hot of cold fluid is given as ATm: 250-77 Q = rih Cph (Thi-The) = me hfg en(2 146.90200 Where r: 2956-5407 K3/kg from steam table at P=1 atm. 250 77 Stm = thig mihx1.005(350-177) = 0.35666 X 2256.5407 0 - 4.62 898 Kgl8| Heat transfer from hot air (9) mhxcPnx (Thi-'The) = 4162898X1.005x(350-177) (น 804-8178 km -

since, Q Cheat transfer) Can also be defined as Q- UA ATM 804.8178x1000 240X AX 146.9020 A= 22.8275 m² 2. Mequired hof sicke area hoof side area A = 22.8275 m UA since, for hot side % CPH = 4652-1249 / NTU - (mi Cp) small w for cold side mic Cpc = (phase change procesos - 240 x 22.8275 4652-1249 NTU= 117765 Effectiveness of Heat Exchanger is given by (E)= 1 - NTU е. -1.77765 е. - 0,69199 80, ay NTU: 1:17765 b) E = 0.69199 c) q: 804.8178 KW 22.8275 m2 d Ant