Question

A batch dryer removes water from a solid at a rate of 30 kg/hr during the constant rate period. Critical moisture content is 0.25 kg H2O/kg dry solid. The equilibrium moisture content of the material is 0.035 kg H2O/kg dry solid. The drier is initially charged with 750 kg of total material, which contains 2 00 000 g of moisture. The final product should contain 0.08 kg H2O/kg dry solid. The falling rate period can be represented by a straight line. The total drying area is 5 m^2 .

1. Draw a schematic diagram that depicts the process.

2. Determine the total drying time to produce the product.

Answer 1

The diagram

Given data

Feed rate = 30 kg/h

Total drying area A = 5 m2

Rate constant period Rc = (30kg/h)/(5m2) = 6 kg/m2-h

Weight of total material = 750 kg

Weight of moisture = 200000 g x 1kg/1000g = 200 kg

Weight of dry solid W = 750 - 200 = 550 kg

Initial moisture content

X1 = 200/550 = 0.36 kg H2O/kg dry solid

Critical moisture content

Xc = 0.25 kg H2O/kg dry solid

equilibrium moisture content

Xe = 0.035 kg H2O/kg dry solid

Final moisture content

X2 = 0.08 kg H2O/kg dry solid

For constant rate period

t1 = ( W/ARc) (X1 - Xc)

= (550 kg)/(5 m2 * 6 kg/m2-h) * (0.36 - 0.25)

= 2.016 hr

For falling rate period

t2 = (W/ARc) (Xc - Xe) ln [(Xc - Xe) / (X2 - Xe)]

=(550 kg)/(5 m2*6 kg/m2-h)*(0.25 - 0.035) ln[(0.25-0.035)/(0.08-0.035)]

= 18.33 * 0.215 * ln (0.215/0.045)

t2 = 6.16 hr

Total drying time = t1 + t2

= 2.016 + 6.16

= 8.176 hr