Question Bank • QB_APE_SIZE_009 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Size Reduction and Material Handling

Subtopic: Centrifugal separation of solids, liquids and gases

Question Statement:

A standard cyclone separator has an inlet duct width of $B = 0.12\text{ m}$ and operates with an effective number of gas turns $N_e = 5$. Dust-laden air at $20^\circ\text{C}$ (dynamic viscosity $\mu = 1.81 \times 10^{-5}\text{ Pa}\cdot\text{s}$) enters the cyclone at an inlet velocity of $v_i = 15.0\text{ m/s}$. The grain dust particle density is $\rho_p = 1350\text{ kg/m}^3$ (air density is negligible). Using Lapple's formula for the cut diameter ($50\%$ collection efficiency), $d_{pc} = \sqrt{\frac{9 \mu B}{2 \pi N_e v_i \rho_p}}$, the cut diameter $d_{pc}$ is ________ $\mu\text{m}$ (round off to two decimal places).

Answer Key:

5.54

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Direct Substitution into Lapple's Formula** ``` Inlet (vi = 15 m/s) ---> [ Cyclone Cylinder ] [ Outer Vortex ] [ \ Inner Vortex / ] [ \ (Clean Air) / ] [ \ | / ] [ \ v / ] [ Dust Cone ] [ Dust Exit ] ``` 1. Calculate numerator: $\text{Num} = 9 \mu B = 9 \times (1.81 \times 10^{-5}\text{ Pa}\cdot\text{s}) \times 0.12\text{ m} = 1.9548 \times 10^{-5}\text{ kg/s}$ 2. Calculate denominator: $\text{Denom} = 2 \pi N_e v_i \rho_p = 2 \times 3.14159265 \times 5 \times 15.0\text{ m/s} \times 1350\text{ kg/m}^3 = 636172.5\text{ kg/(m}^2\cdot\text{s)}$ 3. Calculate $d_{pc}$: $d_{pc}^2 = \frac{1.9548 \times 10^{-5}}{636172.5} = 3.07275 \times 10^{-11}\text{ m}^2$ $d_{pc} = \sqrt{3.07275 \times 10^{-11}} = 5.5432 \times 10^{-6}\text{ m} \approx 5.54\,\mu\text{m}$ **Method 2: Centrifugal Force and Stokes Drag Balance** Equating centrifugal force $\frac{\pi}{6} d_p^3 \rho_p \frac{v_i^2}{r}$ to Stokes drag $3 \pi \mu d_p v_r$ over residence time $t = \frac{2 \pi r N_e}{v_i}$ yields the same Lapple semi-empirical formulation.
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