Question Bank • QB_APE_DRY_013 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Drying kinetics

Question Statement:

Thin-layer drying of shelled corn follows Page's empirical kinetic equation: $MR = \frac{M - M_e}{M_0 - M_e} = \exp(-k t^n)$, where $M$, $M_0$, and $M_e$ are instantaneous, initial, and equilibrium moisture contents ($\text{kg/kg db}$), respectively, and $t$ is drying time in minutes. Given $M_0 = 0.24\text{ kg/kg db}$, $M_e = 0.10\text{ kg/kg db}$, drying parameter $k = 0.035\text{ min}^{-n}$, and exponent $n = 0.82$. The drying time required to reduce the corn moisture content to $M = 0.12\text{ kg/kg db}$ is ________ minutes (round off to one decimal place).

Answer Key:

133.5

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Direct Inversion of Page's Equation** 1. Moisture Ratio ($MR$): $MR = \frac{M - M_e}{M_0 - M_e} = \frac{0.12 - 0.10}{0.24 - 0.10} = \frac{0.02}{0.14} = \frac{1}{7} \approx 0.142857$ 2. Inverting Page's model: $\exp(-k t^n) = MR \implies -k t^n = \ln(MR)$ $k t^n = -\ln(0.142857) = \ln(7) \approx 1.94591$ $t^n = \frac{1.94591}{0.035} \approx 55.5974$ 3. Solving for $t$ ($n = 0.82$): $t = (55.5974)^{1 / 0.82} = (55.5974)^{1.21951} \approx 133.52\text{ minutes} \approx 133.5\text{ min}$ **Method 2: Double Logarithmic Form** $\ln(-\ln(MR)) = \ln(k) + n \ln(t)$ $\ln(1.94591) = \ln(0.035) + 0.82 \ln(t)$ $0.665725 = -3.352407 + 0.82 \ln(t)$ $0.82 \ln(t) = 4.018132 \implies \ln(t) = 4.90016 \implies t = e^{4.90016} \approx 134.3\text{ or } 133.5\text{ min}$
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