Question
Heat transfer 1) 2.5 m/s of air at 150°C is used to heat (0.25 + /250)=0.254...
Heat transfer
1) 2.5 m/s of air at 150°C is used to heat (0.25 + /250)=0.254 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 Cecair = 1.005 kJ/(kg °C), Crewater = 4.2 kJ/(kg). 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= (+0.01) a. the NTU for the heat exchanger b. the effectiveness of the heat exchanger € (+ 0.01) C. the net heat transfer from the hot air. q= kW (+0.5 kW) d. the exit temperature of the air. Thexit = C(+ 0.5))
Answers
1.01325 x x 18 x x2.5 2.08 kals DO ain PV RT 285 x[273+150 = 2.097 kW/ki 1 Cart aud 1.005 Cain 2.08x kwle 1. 0668 A-2 0.254 x mow Cew = Cw Ca < Cain Cmin Cw 2.2497 120 x 20 N NTU UA Crnis 1.0668 x 10 1-0668 0.5087 Cw Cair c's Comin Cman Effechreness 2.097 E I - Coop Cemp [-ne. N70.22) - 1 C. NO.22 -0.22 emp (-2.249760.5087 x 2.2497 I-emp E - 1 0.22 2 -2493 0.5083 E = O. 76488 fa Taid out E = Cair [Taid in cu [ Tair in Twis 0.76488 = 2.097 [150 - Tavi out 1.0668 [150 - 20 Tain 99.415c Enih temperature of air out q. Kan [ 150 C 150 – 99.415 106.077 kW 11 qz 106. 106.07k hy Net heat transfer from hot ait-
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