Question Bank • QB_DFE_HMT_017 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Diffusive and convective mass transfer

Question Statement:

In equimolar counter-diffusion (EMD) of water vapor ($A$) and air ($B$) across a stagnant boundary layer of thickness $\delta = 2.0\text{ mm}$ ($0.0020\text{ m}$) at total pressure $P = 101.325\text{ kPa}$ and $T = 298\text{ K}$ ($R = 8.314\text{ J/(mol}\cdot\text{K)}$), the molecular diffusivity is $D_{AB} = 2.60 \times 10^{-5}\text{ m}^2\text{/s}$. The partial pressure difference of water vapor is $\Delta p_A = 2.50\text{ kPa} = 2500\text{ Pa}$. Under steady state ($N_A = \frac{D_{AB} \Delta p_A}{R T \delta}$), the molar flux $N_A$ is ________ $\text{mmol/(m}^2\cdot\text{s)}$ (round off to two decimal places).

Answer Key:

13.12

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: EMD Molar Flux Formula** $N_A = \frac{D_{AB} \Delta p_A}{R T \delta}$ Numerator: $\text{Num} = (2.60 \times 10^{-5}\text{ m}^2\text{/s}) \times 2500\text{ Pa} = 0.0650\text{ Pa}\cdot\text{m}^2/\text{s} = 0.0650\text{ N/(m}\cdot\text{s)}$ Denominator: $\text{Denom} = 8.3144626\text{ J/(mol}\cdot\text{K)} \times 298.15\text{ K} \times 0.0020\text{ m} = 2478.956 \times 0.0020 = 4.9579\text{ J}\cdot\text{m/mol}$ $N_A = \frac{0.0650}{4.9579} = 0.013110\text{ mol/(m}^2\cdot\text{s)} = 13.11\text{ to } 13.12\text{ mmol/(m}^2\cdot\text{s)}$ **Method 2: Molar Density Form** $C = \frac{P}{RT} = \frac{101325}{8.314 \times 298} = 40.90\text{ mol/m}^3$ $\Delta y_A = \frac{2.50}{101.325} = 0.024673$ $N_A = \frac{C D_{AB} \Delta y_A}{\delta} = \frac{40.90 \times 2.60 \times 10^{-5} \times 0.024673}{0.0020} = 0.01312\text{ mol/(m}^2\cdot\text{s)} = 13.12\text{ mmol/(m}^2\cdot\text{s)}$
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