Question

How much moisture per pound of dry air is removed by air entering moist grain at 110 degrees F dry bulb temperature and exiting the grain at 90% RH and 68 degrees F wet bulb? This question has been posted many times and all of them are incorrect!

Answer 1

The expression for the moisture per pound of dry air, o, is, C(T; -T) +m3hig2 Q= ha - h₂2 ...... (1) of air at DBT, hy2 is the specific enthalpy of air at WBT, hg2 is the (h:- h;) at WBT, and C, is the specific heat at constant pressure. Convert the temperature from Fahrenheit to celsius scale as follows, (°F– 32)=°C T2 = 68 °F T2 = (68°F – 32) = 20°C 1 Ti = 1100F =(110– 32)C = 43.3333°C The values taken are from saturated water temperature table, he = 83.915 kJ/kgK h = 2582.4 kJ/kgK = 83.915 kJ/kgK 0 = 0.01425 C, =1.005 J.kg .K- , Substitute the values in equation (1), | (1.005 J/kg- K-)(20°C-43.3°C)+(0.01425)(83.915 kJ/kgK) =- (2582.4 kJ/kgK –83.915 kJ/kgK) ( 1 lb = 4.6211504 x 10-kg- 0.45 kg = 1.02692231x102 lbs H, O/lb DA Therefore, moisture per pound of dry air 1.02692231x10-2 lbs H,O/lb DA .