Question

A settling chamber is used for the removal of discrete particulate matter from air with the following conditions. Horizontal velocity of air = 0.2 m/s; Temperature of air stream = 77°C; Specific gravity of particle to be removed = 2.65; Chamber length = 12 m; Chamber height = 2 m; Viscosity of air at 77°C = 2.1 × 10\(^{-5}\) kg/m·s; Acceleration due to gravity (g) = 9.81 m/s²; Density of air at 77°C = 1.0 kg/m³; Assume the density of water as 1000 kg/m³ and Laminar condition exists in the chamber.
The minimum size of particle that will be removed with 100% efficiency in the settling chamber (in $\mu$m is .......... (round off to one decimal place).

Hint:

When calculating the minimum size of a particle removed with 100% efficiency in a settling chamber, use the formula involving the particle's velocity, chamber dimensions, and the physical properties of air and water. Ensure that all units are consistent and in SI units.

Answer 1

Correct Answer: 21

Given: \begin{align*} \mu &= 2.1 \times 10^{-5} { kg/m}\cdot{s}
\rho_{{air}} &= 1 { kg/m}^3
\rho_{{water}} &= 1000 { kg/m}^3
V &= 0.2 { m/s}
G &= 2.65
L &= 12 { m}
H &= 2 { m}
g &= 9.81 { m/s}^2 \end{align*}
The formula for calculating the minimum particle size that will be removed with 100% efficiency in a settling chamber is: \[ d = C \sqrt{\frac{18 \mu V H}{g L \rho_p}} \] where: $C = 1$ (constant for laminar flow) $\mu$ is the dynamic viscosity of air $V$ is the velocity of air $H$ is the height of the chamber $g$ is the acceleration due to gravity $L$ is the length of the chamber $\rho_p$ is the density of the particle Substituting the known values: \[ d = 1 \times \sqrt{\frac{18 \times 2.1 \times 10^{-5} \times 0.2 \times 2}{9.81 \times 12 \times 2.65 \times 10^3}} \] \[ d = \sqrt{\frac{2.52 \times 10^{-4}}{3.12 \times 10^3}} = 2.201 \times 10^{-5} { m} = 22 \times 10^{-6} { m} = 22 \mu{m} \] Thus, the minimum particle size that will be removed with 100% efficiency is $22 \, \mu{m}$.