M

where ly (m/s) is the fluidization velocity, ps (kg/m3 the density of the solid particles, p (kg/m3) the density of the fluid, g (m/s2) the acceleration due to gravity, ¡i (N • s/m2) the viscosity of the fluid, d (m) the diameter of the particles, e the voidage of the bed.

Formulas for foods of other shapes are described in Ref. 5. The minimum air velocity needed to convey particles is found using:

3 Cdp

where ve (m/s) is the minimum air velocity and Cd (= 0.44 for Re = 500-200,000) the drag coefficient.

Sample Problem 2. Peas with an average diameter of 6 mm and a density of 880 kg/m3 are dried in a fluidized-bed dryer. The minimum voidage is 0.4 and the cross-sectional area of the bed is 0.25 m2. Calculate the minimum air velocity needed to fluidize the bed if the air density is 0.96 kg/m3 and the air viscosity is 2.15 X 10"5 N • s/m2.

Solution of Sample Problem 2. From equation 15 (880 - 0.96)9.81 (0.006)2(0.4)3

Derivations of these equations are described in Refs. 6-8.

Sample Problem 3. A conveyor dryer is required to dry peas from an initial moisture content of 78% to 16% moisture (wet-weight basis) in a bed 10 cm deep with a voidage of 0.4. Air at 85°C with a relative humidity of 10% is blown perpendicularly through the bed at 0.9 m/s. The dryer belt measures 0.75 m wide and 4 m long. Assuming that drying takes place from the entire surface of the peas and there is no shrinkage, calculate the drying time and energy consumption in both the constant and the falling-rate periods. (Additional data: The equilibrium moisture content of the peas is 9%, the critical moisture content 300% (dry-weight basis), the average diameter 6 mm, the bulk density 610 kg/m3, the latent heat of evaporation 2300 kJ/kg, the saturated water vapor pressure at wet-bulb temperature 61.5 mmHg, and the mass transfer coefficient 0.015 kg/m2/s).

Solution to Sample Problem 3. In the constant-rate period, from equation 10:

From Fig. 1 for 0a = 85°C and RH = 10%, es = 42°C

To find the area of the peas,

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