Air at a dry bulb temperature of 40 C and a wet bulb tempera

Air at a dry bulb temperature of 40 C and a wet bulb temperature of 20 C is first heated in a heater to a dry bulb temperature of 90 C. Then it is pased through a bed of apricot slices to dry them. The air exiting from the top of the apricot bed is at a dry bulb temperature of 60 C. It is then passed through a dehumidifier to reduce its relative humidity to 10%. Clearly show the various paths of the process on the psychrometric chart. The air velocity through the bed and the dehumidifier is 4 m/s and te cross-sectional diameter of the bed is 0.5 m. a. Determine the amount of moisture removed, in grams of water/second from the apricot bed. b. Determine the amount of moisture removed, in g-rams of water/second in the dehumidifier.

Solution

need to know two psychometric properties for the air before and after dehumidification process and go to psychometric chart to determine the difference in humidity content in the air and know the air flow rate then u can calculate it from this relation :

moisture removal = air mass flow rate * (W1 - W2)/100 ....kg

W1 = inlet air humidity ratio g/kg

W2 = outlet air humidity ratio g/kg