Chemistry - Ideal Gas Question with Solution | TestHub
Suppose you run a grain drying unit. You have to dry 250 kg of grain containing 40% water on wet basis, loaded equally on two perforated trays inside a vertical drier column as shown in the figure.
Saturation vapour pressure at 227°C = 40 mm of Hg
Saturation vapour pressure at 127°C = 20 mm of Hg
What is the process time of the drier (in min)?
Answer:
Solution:
Water to remove = 0.40 × 250 = 100 kg. Inlet air (227°C = 500 K): p₁ = 0.20 × 40 = 8.0 mm = 8/760 atm, so n₁ = p₁/RT1 = (8/760)/(0.08 × 500) = 2.632 × 10⁻⁴ mol/L. Outlet air (127°C = 400 K): p₂ = 0.80 × 20 = 16.0 mm = 16/760 atm, so n₂ = (16/760)/(0.08 × 400) = 6.579 × 10⁻⁴ mol/L. Water picked up per litre = 3.947 × 10⁻⁴ mol = 7.105 × 10⁻³ g, i.e. 7.105 g per m³. With air flow 10 m³/min, drying rate = 71.05 g/min. Process time = 100000/71.05 = 1407.4 min, which rounds to 1407 min.
