Question Bank • QB_APE_STOR_010 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Storage Systems

Subtopic: Packaging material and machines

Question Statement:

A flexible polymer film pouch has a surface area of $A = 0.08\text{ m}^2$ and film thickness of $x = 25.0\,\mu\text{m}$. The film permeability to water vapor is $P = 3.0 \times 10^{-14}\text{ kg}\cdot\text{m}/(\text{m}^2\cdot\text{s}\cdot\text{Pa})$. Dry food is stored at $25^\circ\text{C}$ (pure water saturation vapor pressure $p_{\text{sat}} = 3169\text{ Pa}$) inside the pouch at internal water activity $a_{w,1} = 0.15$. The pouch is stored in an ambient chamber maintained at $RH = 85\%$ ($a_{w,2} = 0.85$). The steady-state rate of moisture ingress into the pouch in $\text{g/day}$ is ________ (round off to two decimal places).

Answer Key:

18.40

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Permeability Equation** 1. Partial vapor pressures: - Inside: $p_{v,1} = 0.15 \times 3169\text{ Pa} = 475.35\text{ Pa}$ - Outside: $p_{v,2} = 0.85 \times 3169\text{ Pa} = 2693.65\text{ Pa}$ $\Delta p = 2693.65 - 475.35 = 2218.30\text{ Pa}$ 2. Moisture transmission rate $\dot{m}$: $\dot{m} = \frac{P \cdot A \cdot \Delta p}{x} = \frac{(3.0 \times 10^{-14}) \times 0.08 \times 2218.30}{25.0 \times 10^{-6}}$ $\dot{m} = \frac{5.32392 \times 10^{-12}}{25.0 \times 10^{-6}} = 2.129568 \times 10^{-7}\text{ kg/s}$ 3. Converting to grams per day: $\text{Rate} = 2.129568 \times 10^{-7}\text{ kg/s} \times 1000\text{ g/kg} \times 86400\text{ s/day} = 18.399\text{ g/day} \approx 18.40\text{ g/day}$
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