Question Bank • QB_APE_DRY_007 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Hydrothermal treatments

Question Statement:

During hydrothermal treatment (soaking) of rough rice at $60^\circ\text{C}$, the water absorption kinetics follow Peleg's equation: $M_t = M_0 + \frac{t}{k_1 + k_2 t}$, where $M$ is moisture content in dry basis ($\text{g water / g dry matter}$) and $t$ is soaking time in hours. The initial moisture content is $M_0 = 0.14\text{ g/g db}$. The kinetic parameters are $k_1 = 12.5\text{ h}\cdot\text{g db / g water}$ and $k_2 = 1.80\text{ g db / g water}$. The moisture content ($M_t$) of paddy after $4.0\text{ hours}$ of soaking is ________ $\text{g/g db}$ (round off to three decimal places).

Answer Key:

0.343

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Direct Substitution into Peleg's Equation** $M_t = M_0 + \frac{t}{k_1 + k_2 t}$ Given: $M_0 = 0.14\text{ g/g db}$, $t = 4.0\text{ h}$, $k_1 = 12.5$, $k_2 = 1.80$. $k_1 + k_2 t = 12.5 + (1.80 \times 4.0) = 12.5 + 7.20 = 19.70\text{ h}\cdot\text{g db/g water}$ $\Delta M = \frac{4.0}{19.70} \approx 0.20305\text{ g/g db}$ $M_t = 0.14 + 0.20305 = 0.34305\text{ g/g db} \approx 0.343\text{ g/g db}$ **Method 2: Inverse Rate Verification** $\frac{t}{M_t - M_0} = k_1 + k_2 t = 12.5 + 1.80(4) = 19.70$ $M_t - M_0 = \frac{4.0}{19.70} = 0.2030 \implies M_t = 0.14 + 0.2030 = 0.343\text{ g/g db}$
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