Drying Problem

Psychrometrics and Drying Problem

1. The air in a room is at 37.8C and a total pressure of 1.00 atm abs containing water vapor with a partial pressure pA = 3.59 kPa. Calculate: humidity, wet bulb temperature, dew point temperature, specific volume and relative humidity.

2. The air in a room has a humidity of 0.021 kg water/kg dry air at 32.2C and 1 atm abs pressure, Calculate wet bulb temperature, dew point temperature, specific volume and relative humidity.

3. Cooling and dehumidifying air: Air entering an adiabatic cooling chamber has a temperature of 32.2C and a

percentage humidity of 65%. It is cooled by a cold water spray and saturated with water vapor in the chamber. After

leaving, it is heated to 23.9C. The final air has a percentage humidity of 40%. What are the initial humidity and final humidity after heating?

4. Time for drying in constant-rate period: A batch of wet solid was dried on a tray dryer using constant drying

conditions and a thickness of material on the tray is 25.4 mm. Only the top surface was exposed. The drying rate

during constant-rate period was R = 2.05 kg water/h.m . The 2

ratio Ls/A used was 24.4 kg dry solid/m exposed surface. 2

The initial free moisture was X1 = 0.55 and the critical moisture content Xc = 0.22 kg free moisture/kg dry solid.

Calculate the time to dry a batch of this material from X1 = 0.45 to X2 = 0.30 using the same drying conditions but a

thickness of 50.8 mm, with drying from the top and bottom surfaces.

5. Prediction in constant-rate drying region: A granular insoluble solid material wet with water is being dried in the

constant rate period in a pan 0.61m X 0.61m and the depth of material is 25.4 mm. The sides and bottom are insulated. Air flow parallel to the top drying surface at a velocity of

3.05 m/s and has a dry bulb temperature of 60C and wet

bulb temperature of 29.4C. The pan contains 11.34 kg of dry solid having a free moisture content of 0.35 kg water/kg dry solid. Predict the drying rate and the time in hours

needed. And predict the time needed if the depth of material is increased to 44.5 mm.

6. Graphical integration for drying in falling rate region: A wet solid is to be dried in a tray dryer under steady state

conditions from a free moisture content of X1 = 0.40 kg water/kg dry solid to X2 = 0.02 kg water/kg dry solid. The

dry solid weight is 99.8 kg dry solid and the top surface area for drying is 4.645 m . The drying rate curve is shown here. 2

(a)Calculate the time for drying using graphical integration in the falling rate period.

(b)

Repeat but use a straight line through the origin

for the drying rate in falling rate period.

7. Mass and Heat balance: A continuous countercurrent dryer is being used to dry 453.6 kg dry solid/h containing 0.04 kg total moisture/kg dry solid to a value of 0.002 kg total

moisture/kg dry solid. The granular solid enters at 15.6C and is to be discharged at 60C. The dry solid has a heat capacity of 1.465 kJ/kg.K, which is assumed constant.

Heating air enters at 87.8C, having a humidity of 0.010 kg water/kg dry air, and is to leave at 32.2C.

(a)Calculate the air flow rate and the outlet humidity, assuming no heat losses in the dryer.

(b)

Calculate the air flow rate and the outlet humidity,

heat losses from the dryer are estimated as 2931 W.