Question

1) 2.5 m/s of air at 150°C is used to heat 0.35 kg/s of 20°C water. The heat exchanger is a finned-tube cross flow heat exchanger with both sides unmixed. The overall heat transfer coefficient associated with the hot side is 120 W/(m2 °C). The hot-side area is 20 m². Assume a constant specific heat for air and water of Gezic = 1.005 kJ/(kg °C), Senator = 4.2 kJ/(kg °C). The pressure of the hot air is P = 1.0 atm, The pressure of the water is high enough so that it does not boil. Find the following: NTU = a. the NTU for the heat exchanger b. the effectiveness of the heat exchanger (+0.01) (+0.01) E = c. the net heat transfer from the hot air. q= _kW (+0.5 kW) d. the exit temperature of the air. Trans = °C (+0.5 °C))

Answer 1

solution Giveni Pa I alm. = 1.01325x105 Pa. V=2.5 m3/s. T: 150c 428k u = 120 wim2c. mio = 0.35 regis. we know ideal gar en is given by. pvz MRT. .: mair PV 1.01825x105 x 25 . = 2.08 kg/s. RT 287 x (423) To find Comin, we will find Cair = mairt epair 9.081.005 2.097 kwilk. Cwater : mwx Cpw 0.35 X 4.2 - 1.47 kolk Here, e water is minimum. .: NTU VA 1.6326 120x20 1.47X103 Cmin we know effectiveness is given by, Cr= 0.7010 Cmin Cmax 1,47 7.097 1 E 21 - Coop { escp Lane. N10:22 1 CN-0.22 1 eocp { escp ľ 0.22 -1.6 326 x 0.7010 x 1.6326 200 201 - 017010 1.6826-0.22 06.866171. = 0.5547

Orfectiveness can be given by formula { Car ( Tain Talout Cw C Tairin Twin). 0.5547 2-097 (150 - Taout ) 1249 (150-20). Tar out 99.632 Netheat transfer by hot air. e. Cair AT. = Cair (150-99.632) 22.097 (150 - 991632 ) - 105.62 KW NTU 1.6326 a. b] c] Effectiveness 0, 5547 Q 105.62 kw d] Tescit 99.632.c