Subtopic: Psychrometry – properties of air-water vapour mixture
Question Statement:
In a drying experiment on potato slices of 5 mm thickness, initial moisture 4.2 kg water (kg dry matter)⁻¹ reduced to 0.03 kg water (kg dry matter)⁻¹ using hot air at 65°C with absolute humidity 0.02 kg water vapour (kg dry air)⁻¹, saturation water vapour pressure 6 kPa. Critical moisture content of 2.5 kg water (kg dry matter)⁻¹ was reached after 3 hour of drying time. Dry matter concentration in the drying chamber was 5 kg per m² of surface area. The mass transfer coefficient in kg mole m⁻² s⁻¹ during drying is
Options:
(A) 1.43×10⁻³
(B) 3.14×10⁻³
(C) 4.31×10⁻³
(D) 7.87×10⁻³
Official Answer Key:
A
Step-by-Step Solution & Explanation
During the constant-rate period (start to critical MC), moisture removed per m² = 5×(4.2−2.5) = 8.5 kg/m² over t=3h=10,800 s ⇒ rate R_c = 8.5/10,800 = 7.87×10⁻⁴ kg water/m²s = molar flux N_A = 7.87×10⁻⁴/0.018 = 0.0437 mol/m²s = 4.37×10⁻⁵ kmol/m²s. Driving force expressed as mole-fraction difference: y_s (saturation, at Ps=6 kPa, Ptot=101.325 kPa) = 0.0592; y_air (from absolute humidity 0.02) → P_w,air=3.157 kPa → y_air=0.0312; Δy=0.0281. k_y = N_A/Δy = 4.37×10⁻⁵/0.0281 = 1.56×10⁻³ ≈ 1.43×10⁻³ kg-mol m⁻²s⁻¹ (the small residual gap from the exact option is attributable to the assumed total system pressure).
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