Chemistry - Ideal Gas Question with Solution | TestHub
Suppose you are a manager of a fish drying unit. You have to dry 200 kg of fish containing 36% water on wet basis. The drier specification is mentioned in the figure.
Saturation vapour pressure at 227°C = 38 mm of Hg
Saturation vapour pressure at 127°C = 19 mm of Hg
What is the process time of the drier (in min)?
Answer:
Solution:
Water to remove = 0.36 × 200 = 72 kg. Inlet air (227°C = 500 K): p1 = 0.20 × 38 = 7.6 mm = 0.01 atm, so moles of H2O per litre n1 = p1/RT1 = 0.01/(0.08 × 500) = 2.5 × 10-4 mol/L. Outlet air (127°C = 400 K): p2 = 0.80 × 19 = 15.2 mm = 0.02 atm, so n2 = 0.02/(0.08 × 400) = 6.25 × 10-4 mol/L. Water picked up per litre of air = 3.75 × 10-4 mol = 6.75 × 10-3 g, i.e. 6.75 g per m3. With air flow 10 m3/min, drying rate = 67.5 g/min. Process time = 72000/67.5 = 1066.7 min, which rounds to 1067 min.
