Question Bank • QB_DFE_HMT_014 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Diffusive and convective mass transfer

Question Statement:

A packaging polymer film of thickness $L = 40.0\,\mu\text{m}$ ($4.0 \times 10^{-5}\text{ m}$) separates a food headspace from ambient air. An aroma compound has a binary molecular diffusion coefficient of $D_{AB} = 3.20 \times 10^{-11}\text{ m}^2\text{/s}$ through the film. The concentration gradient maintained across the film faces is $\Delta C = 0.080\text{ kg/m}^3$. Under steady-state Fickian diffusion ($J = D_{AB} \frac{\Delta C}{L}$), the mass flux of the aroma compound in $\text{mg/(m}^2\cdot\text{h})$ is ________ (round off to one decimal place).

Answer Key:

230.4

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Fick's First Law Calculation** 1. Mass flux in SI units ($\text{kg/(m}^2\cdot\text{s)}$): $J_A = D_{AB} \frac{\Delta C}{L} = (3.20 \times 10^{-11}\text{ m}^2\text{/s}) \times \frac{0.080\text{ kg/m}^3}{4.0 \times 10^{-5}\text{ m}}$ $J_A = 3.20 \times 10^{-11} \times 2000\text{ kg/(m}^4) = 6.40 \times 10^{-8}\text{ kg/(m}^2\cdot\text{s)}$ 2. Convert flux to $\text{mg/(m}^2\cdot\text{h})$: $J_A = 6.40 \times 10^{-8}\text{ kg/(m}^2\cdot\text{s)} \times 10^6\text{ mg/kg} \times 3600\text{ s/h} = 230.4\text{ mg/(m}^2\cdot\text{h})$ **Method 2: Dimensional Unit Conversion** $J_A = \frac{(3.20 \times 10^{-11}) \times (80000\text{ mg/m}^3) \times 3600}{0.000040\text{ m}} = \frac{9.216 \times 10^{-6}}{4.0 \times 10^{-5}} = 230.4\text{ mg/(m}^2\cdot\text{h})$
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