Humid air enters a dehumidifier with an enthalpy of 216 Btul

Humid air enters a dehumidifier with an enthalpy of 21.6 Btu/lbm of dry air and 1100 Btu/lbm of water vapor. There are 0.02 lbm of vapor per pound of dry air at entrance and 0.009 lbm of vapor per pound of dry air at exit. The dry air at exit has an enthalpy of 13.2 Btu/lbm, and the vapor at exit has an enthalpy of 1085 Btu/lbm. Condensate leaves with an enthalpy of 22 Btu/lbm. The rate of flow of dry air is 287 lbm/min. Determine

(a) the amount of moisture removed from the air (lbm/min)
(b) the rate of heat removal required
[Ans: 3.16 lb/min, 5860 Btu/min]

Solution

Let us assume m be the mass of dry air entering the dehumidifier.

At inlet, Mass of dry air = m

Mass of water vapor = 0.02m

Total mass of humid air = m + 0.02m = 1.02m

At outlet, Mass of dry air = m

Mass of water vapor = 0.009m

Total mass of humid air = 1.009m

Amout of moisture removed = 1.02m - 1.009m

= 0.011m

= 0.011 x 287

= 3.157 lb/min

(b) Rate of heat removal from the air = Enthalpy of air at inlet - Enthalpy of air at outlet - Enthalpy of condensate

Enthalpy of air at inlet = m x 21.6 + 0.02m x 1100

= 287 x 21.6 + 0.02 x 287 x 1100

= 12513.2 Btu/min

Enthalpy of air at outlet = m x 13.2 + 0.009m x 1085

= 287 x 13.2 + 0.009 x 287 x 1085

= 6590.955 Btu/min

Enthalpy of condensate = 22 x 3.16 = 69.52 Btu/min

Rate of heat removal = 12513.2 - 6590.955 - 69.52 = 5856.325 Btu/min