Air at 101.325 kPa pressure is used to dry a vegetable material at 52 °C. Saturation pressure of water at 52 °C is 13.51 kPa. If the mass transfer coefficient for the case of equimolar counter diffusion (k′y) is 4.79 × 10⁻⁴ kg mole m⁻² s⁻¹ mole fraction⁻¹, then the mass transfer coefficient for the case of diffusion through non-diffusing gas (ky) in kg mole m⁻² s⁻¹ mole fraction⁻¹ is
Options:
(A) 4.96 × 10⁻⁴
(B) 5.14 × 10⁻⁴
(C) 7.83 × 10⁻⁴
(D) 1.02 × 10⁻³
Official Answer Key:
B
Step-by-Step Solution & Explanation
For diffusion of A through stagnant/non-diffusing B: k_y = k′_y / y_BM, where y_BM is the log-mean of inert-gas mole fractions between the interface (saturated, y_A1 = P_sat/P = 13.51/101.325 = 0.1333) and the bulk (y_A2 ≈ 0). y_B1 = 0.8667, y_B2 = 1.0. y_BM = (1.0−0.8667)/ln(1.0/0.8667) = 0.1333/0.1430 = 0.9317. k_y = 4.79×10⁻⁴ / 0.9317 = 5.14×10⁻⁴ kg mol m⁻² s⁻¹.
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