Physical, Thermal, Aerodynamic and Mechanical Properties of Agricultural and Food Materials

Calculation of Sensible Cooling Heat Load:

• Formula: The sensible heat ($Q$) removed during product precooling above freezing is given by:
  $$Q = m \cdot c \cdot \Delta T$$
  where:
  $m$ = mass of product = $2\text{ tons} = 2 \times 1000\text{ kg} = 2000\text{ kg}$,
  $c$ = specific heat of apples = $0.1\text{ kcal/kg}^\circ\text{C}$,
  $\Delta T = T_{\text{initial}} - T_{\text{final}} = 25^\circ\text{C} - 5^\circ\text{C} = 20^\circ\text{C}$.
• Calculation:
  $$Q = 2000\text{ kg} \times 0.1\text{ kcal/kg}^\circ\text{C} \times 20^\circ\text{C} = 4000\text{ kcal}$$

Principles of Friction in Granular Agricultural Materials:
In agricultural process engineering (Mohsenin, 1986):

• Proportional to normal load: Governed by Coulomb's friction law ($F = \mu N$), frictional resistance increases linearly with applied normal load.
• Independent of apparent sliding surface area: In accordance with classical Amontons-Coulomb laws, macro-level sliding friction is independent of apparent surface area.
• Proportional to actual area of contact: Modern adhesion friction theory (Bowden-Tabor) dictates that friction occurs at micro-asperity junctions, making friction directly proportional to the true microscopic contact area.
• Dependent on nature of contact material: Frictional coefficients vary markedly based on material composition, moisture, and surface finish (e.g., grain on polished steel vs wood or galvanized iron).

Calculation of Geometric Shape Ratio:

• Formula: In agricultural material physical property analysis (Mohsenin, 1986), the two-dimensional roundness/sphericity index defined by Curray is given by the ratio of the diameter of the largest inscribing circle ($d_i$) to the diameter of the smallest circumscribing circle ($d_c$):
  $$\text{Shape Index} = \frac{d_i}{d_c}$$
• Given Data:
  $d_i = 20\text{ mm}$
  $d_c = 30\text{ mm}$
  $$\text{Ratio} = \frac{20}{30} = \frac{2}{3} \approx 0.667$$
  Among the provided options (0.6, 1.5, 1, 0.5), Option 1 (0.6) represents this value.

Container volume $V = 20\text{ cm} \times 30\text{ cm} \times 10\text{ cm} = 6000\text{ cm}^3$. Mass of wheat $M = 6\text{ kg} = 6000\text{ g}$. Bulk density $\rho_b = \frac{\text{Mass}}{\text{Volume}} = \frac{6000\text{ g}}{6000\text{ cm}^3} = 1.0\text{ g/cm}^3$ (or $1000\text{ kg/m}^3$). Option analysis: (A) $2\text{ g/cm}^3$ is incorrect. (B) $0.5\text{ g/cm}^3$ is incorrect. (C) $10\text{ g/cm}^3$ is an order of magnitude too high. (D) $1\text{ g/cm}^3$ is the exact calculated bulk density.