Question Bank • QB_DFE_EXP_008 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Diffusive and convective mass transfer

Question Statement:

Water vapor diffuses through a stagnant air film of thickness $\delta = 2.0\text{ mm}$ over a wet food surface. The molecular diffusivity of water vapor in air is $D_{AB} = 2.5 \times 10^{-5}\text{ m}^2\text{/s}$. The partial pressure difference of water vapor across the film is $\Delta p = 2000\text{ Pa}$. Gas constant $R = 8314\text{ J/(kmol}\cdot\text{K)}$ and temperature $T = 300\text{ K}$. Using Fick's first law $N_A = \frac{D_{AB} \Delta p}{R T \delta}$, the molar flux $N_A$ in $\text{mol/(m}^2\cdot\text{s)}$ is ________ $\times 10^{-2}$ (round off to 2 decimal places).

Answer Key:

1.00

Step-by-Step Solution & Explanation

1. Molar flux equation: $N_A = \frac{D_{AB} \Delta p}{R T \delta}$ Given: • $D_{AB} = 2.5 \times 10^{-5}\text{ m}^2\text{/s}$ • $\Delta p = 2000\text{ Pa}$ • $R = 8.314\text{ J/(mol}\cdot\text{K)}$ • $T = 300\text{ K} \implies R T = 8.314 \times 300 = 2494.2\text{ J/mol}$ • $\delta = 0.002\text{ m}$ $N_A = \frac{(2.5 \times 10^{-5}) \times 2000}{2494.2 \times 0.002} = \frac{0.050}{4.9884} \approx 0.010023\text{ mol/(m}^2\cdot\text{s)} = 1.0023 \times 10^{-2}\text{ mol/(m}^2\cdot\text{s)}$ Expressed as $N_A \times 10^2 = 1.00$.
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