Question

a) The initial moisture content of a food product is 77% (wet basis), and the critical moisture content is 30% (wet basis). If the constant drying rate in a fluidized bed dryer is 0.1 kg water removed/m2-s, determine the time required for the product to begin the falling-rate drying period. The product has a cube shape with 5-cm sides; the initial product density is 950 kg/m3.

Answer 1

≈ 53 seconds

Step-by-step explanation:

calculate Time required for the product to begin the falling-rate drying period

Initial moisture content = 0.77 kg water /kg of product

= 3.35 kg water /kg solids

Critical moisture content = 0.3 kg water / kg product

= 0.43 kg water / kg solids

∴ amount of water to be removed = 3.35 - 0.43 = 2.95kg water /kg solids

next: calculate surface are a of product during drying

= (0.05 * 0.05 ) * 6

= 0.015 m^3

Drying rate = 0.1 kg water m^2.s^-1 * 0.015 m^3 = 1.5 * 10^-3 kg water s^-1

applying product density

initial product mass = 0.11875 * 0.23 = 0.0273kg solid

hence total amount of water to removed = 2.92 * 0.0273 = 0.07972 kg

therefore : Time required for the product to begin the falling-rate drying period

= 0.07972 / 1.5 * 10^-3

= 53 seconds