Tutorial 2 - Drying.pdf

# 9 a wet solid is to be dried in a pan 061 m by 061 m

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9. A wet solid is to be dried in a pan 0.61 m by 0.61 m and the depth of material is 25.4 mm under constant rate period. The sides and bottom are insulated. Air flows parallel to the top surface at a velocity of 3.05 m/s and has a dry bulb temperature of 60 o C and wet bulb temperature of 29.4 o C. The pan contains 11.34 kg of dry solid having a moisture content of 40% and the material is to be dried in the constant rate period to the critical moisture content of 20%. The equilibrium moisture was found to be 4%. All moisture contents are on the dry basis. a) Calculate the drying rate and the time required in hours.

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CPB 20104 M ass Transfer 2 FG/T2/July2011 [R = 1.599 kg/h.m 2 , t = 3.82hr] b) Calculate the drying rate and the time needed in hours if the depth of material is reduced to 12.7 mm. [R = 1.599 kg/h.m 2 , t = 1.91hr] 10. A wet solid is dried with air at 60 o C and a humidity of 0.010 kg H 2 O/kg dry air. The air flows parallel to the flat, wet surface of the solid at a velocity of 5 m/s. Under these conditions the heat transfer coefficient is given by h = 14.3( v ) 0.8 , where is the density of the mist air in kg moist air per unit volume moist air. The surface area available for heat exchange with air is 0.25m 2 and the latent heat of vaporization is 2.45 x 10 6 J/kg. Calculate the total amount of water evaporating per unit time. [R = 1.71 x 10 -4 kg/s]
• Fall '19
• Atmospheric thermodynamics

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