Polluted air with particulate matters of diameter 50 µm enter with a horizontal velocity of 1.0 m/s at a height of 0.5 m from the bottom of a dry settling chamber. The density of t... Polluted air with particulate matters of diameter 50 µm enter with a horizontal velocity of 1.0 m/s at a height of 0.5 m from the bottom of a dry settling chamber. The density of the particle is 2000 kg/m³ and dynamic viscosity of the air is 1.8 × 10⁻⁵ kg/m-s. Considering particle settling follows Stokes' law, the minimum length in m, of the chamber required for settling of the particle at its bottom, is __________. (round off up to 2 decimals)
Understand the Problem
The question is asking for the minimum length of a dry settling chamber required for particulate matters to settle according to Stokes' law, given specific parameters such as particle diameter, velocity, density, and viscosity of air.
Answer
The minimum length of the chamber required for settling of the particle at its bottom is approximately $33.11 \, m$.
Answer for screen readers
The minimum length of the chamber required for settling of the particle at its bottom is approximately $33.11 , m$.
Steps to Solve
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Identify the Parameters
From the problem, we have the following parameters:
- Diameter of the particle, $d = 50 , \mu m = 50 \times 10^{-6} , m$
- Velocity of the particle (and air), $v = 1.0 , m/s$
- Density of the particle, $\rho_p = 2000 , kg/m^3$
- Dynamic viscosity of air, $\mu = 1.8 \times 10^{-5} , kg/(m \cdot s)$
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Calculate the Settling Velocity Using Stokes' Law
Stokes' law gives the settling velocity $v_s$ as:
$$ v_s = \frac{d^2 (\rho_p - \rho) g}{18 \mu} $$
Here, $\rho$ (density of air) is negligible compared to $\rho_p$, so we can approximate:
$$ v_s = \frac{d^2 \rho_p g}{18 \mu} $$
With $g \approx 9.81 , m/s^2$, we proceed to calculate $v_s$. -
Substituting the Known Values
Substitute the values into the equation:
$$ v_s = \frac{(50 \times 10^{-6})^2 (2000)(9.81)}{18(1.8 \times 10^{-5})} $$ -
Calculate the Settling Velocity
Perform the calculations step by step:
- Calculate $(50 \times 10^{-6})^2 = 2.5 \times 10^{-9}$
- Then calculate the numerator:
$$ 2.5 \times 10^{-9} \times 2000 \times 9.81 = 4.9025 \times 10^{-6} $$ - Calculate the denominator:
$$ 18 \times (1.8 \times 10^{-5}) = 3.24 \times 10^{-4} $$ - Finally, the settling velocity becomes:
$$ v_s = \frac{4.9025 \times 10^{-6}}{3.24 \times 10^{-4}} \approx 0.0151 , m/s $$
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Calculate the Minimum Length of the Chamber
To find the minimum length ( L ), we can use the formula that relates length, velocity, and settling velocity:
$$ L = \frac{v \cdot h}{v_s} $$
Here, ( h = 0.5 , m ):
$$ L = \frac{1.0 \cdot 0.5}{0.0151} $$ -
Final Calculation of Length
Calculate ( L ):
$$ L = \frac{0.5}{0.0151} \approx 33.11 , m $$
The minimum length of the chamber required for settling of the particle at its bottom is approximately $33.11 , m$.
More Information
This answer indicates the length of the settling chamber must be significant enough to allow the particles to settle out of the air stream before the gas exits the chamber.
Tips
- Neglecting Density of Air: Ensure to recognize that the density of air is negligible compared to the density of the particles.
- Using Incorrect Units: Make sure to convert all measurements into standard SI units before calculations.
- Confusing Settling Velocity with Air Velocity: The settling velocity derived from Stokes' law should not be confused with the flow velocity of the air.
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