In a laboratory test run, the rate of drying was found to be 0.5 x 10^-3 kg/m2.s, when the moisture content reduced from 0.4 to 0.1 on dry basis. The critical moisture content of the material is 0.08 on a dry basis. A tray dryer is used to dry 100 kg (dry basis) of the same material under identical conditions. The surface area of the material is 0.04 m2/kg of dry solid. The time required (in seconds) to reduce the moisture content of the solids from 0.3 to 0.2 (dry basis) is______________________?
Correct answer: C. 5000
- A. 2000
- B. 4000
- C. 5000
- D. 6000
Explanation
The material contains 100 kg of dry solid, so its exposed area is 100 × 0.04 = 4 m². The moisture removed is (0.3 − 0.2) × 100 = 10 kg, and at 0.5 × 10⁻³ kg/m²·s the time is 10/(4 × 0.5 × 10⁻³) = 5000 s.
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