Question

3. (45) Operating data for an air-conditioning process is shown below. Air flows through a soaked pad and then a heating section. The device is well-insulated and the volume flow rate of the incoming moist air is 150 m3/min. Please determine (a) (6) The humidity ratio of the incoming and exiting air. (b) (10) The humidity ratio and temperature of the intermediate state (i). The enthalpy of liquid water is negligible. (c) (11) The amount of water evaporated into the air stream, in kg/min. (d) (11) The rate of heat transfer to the moist air passing through the heating section, in kJ/min. (e) (7) Sketch this process on a psychrometric chart. Label the axes, the saturation line and the states at the entrance (1), between the soaked pad and the heating section (i), and at the exit (2). Liquid water at 10 C Ti 30°C Pi - 1 atm 9 -20% (AV), = 150 m/min - T: +40°C P = I atm -20% Electrie resistor

Answer 1

Soft Air conditioning process Inlet condition (1) 1 = 30°c $,204, (Av), = 150m/min outlet condition (2) T2 = 40°C 4, = 207. Intermediate condition (3) Wg = ₂ & -rooy (because mixture is saturated) (W₂ = 0, because no Change in moisture content in heating õ section) رکن = رله l ha teestdial h ha ha Para 4 pes 3 W, 5W3 c, = 5.7 - 3000 40 legofd.a e faz is for inlet condition :- -W₂ =W3 = 5.7 gm = 5.78103 kg kg of dig kg.fd-G h = 4602 kJ ks.ofdia С орт) T- V = 0.948 ms RS.fia for outlet condition (2) Co₂ = 10.2 gm = 10:2x10 kg kg.ord-a for Intermediate condition (3) h = 66.8 kJ/is fda T = 18.5°C h3 = 40 kJ/kg of dia Humidity ratio ofor incoming air (W) = 0.0067 - 3 kg of da Humidity rati dity ratio for exiting air (602) = 0.0102 legs Humidity ratio for Intermediate state ( 63) = 0.0102 kg Temperature of infermediate state Ta legiufda and water evaporated into Air stream (mas mw = na twit W2) - mass flow rate of state air (ra) = A 2 m/min - mblis (a) Amour ma= 150 0.948 - 158.22 18 min

a. Rate (d) mw = 158.22 (-5.7x00-3+ 10. 2x10-3) = 0.77199 kgloria of heat transfer to moist air ( 0) in heating section & ante 66.8 - 40 = 26.8 ltlics 7 Q = mx 26.8= 158.22x 26.8 = 4240.3 kJ/min (e) ce Drown Drawn on ast page.