What percentage of particle is removed of settling velocity 0.18 cm/sec if particle of size 5 × 10-3 cm diameter and specific gravity is 2.65? (Kinematic viscosity of water at 20oC is 1.01×10-2 cm2/sec and Reynold number is less than 0.5)?
81.00 %
Understanding particle removal efficiency in a settling tank is crucial in water treatment processes. In an ideal settling tank, the removal of particles depends on their settling velocity compared to the tank's overflow rate.
Key Concepts: Settling Velocity and Overflow Rate
- Settling Velocity ($V_s$): This is the speed at which a particle settles down through the water due to gravity. It depends on the particle's size, shape, density, and the water's density and viscosity. For small particles under low Reynolds number conditions (like $Re < 0.5$), Stokes' Law can be used to calculate $V_s$.
- Overflow Rate ($V_o$): This is the design settling velocity of the tank. It is calculated as the flow rate divided by the surface area of the tank. Particles with a settling velocity equal to or greater than the overflow rate are expected to be completely removed in an ideal settling tank.
Particle Removal in an Ideal Settling Tank
For an ideal settling tank, the percentage of particles removed depends on the ratio of their settling velocity ($V_s$) to the tank's overflow rate ($V_o$).
- If \$ V_s \ge V_o \$, the removal efficiency is 100%.
- If \$ V_s < V_o \$, the percentage removal is given by the formula: \[ \text{Percentage Removal} = \left( \frac{V_s}{V_o} \right) \times 100\% \]
Determining the Tank's Overflow Rate
The problem provides the properties of a specific particle (size and specific gravity) and the kinematic viscosity of water. These properties can be used to calculate the settling velocity for this type of particle using Stokes' Law. In the context of settling tank design problems like this, the settling velocity calculated from these properties often represents the design settling velocity or the overflow rate ($V_o$) of the tank.
Stokes' Law for settling velocity ($V_s$) is given by:
\[ V_s = \frac{g(S_p - 1)d^2}{18\nu} \]Where:
- \$ g \$ is the acceleration due to gravity (approx. 980.665 cm/s2)
- \$ S_p \$ is the specific gravity of the particle (2.65)
- \$ d \$ is the diameter of the particle (\$ 5 \times 10^{-3} \$ cm)
- \$ \nu \$ is the kinematic viscosity of water (\$ 1.01 \times 10^{-2} \$ cm2/sec)
Let's calculate the overflow rate (\$ V_o \$) using these values:
\[ V_o = \frac{980.665 \text{ cm/s}^2 \times (2.65 - 1) \times (5 \times 10^{-3} \text{ cm})^2}{18 \times 1.01 \times 10^{-2} \text{ cm}^2/\text{sec}} \] \[ V_o = \frac{980.665 \times 1.65 \times (25 \times 10^{-6}) \text{ cm}^3/\text{s}^2}{0.1818 \text{ cm}^2/\text{sec}} \] \[ V_o = \frac{980.665 \times 1.65 \times 25 \times 10^{-6}}{0.1818} \text{ cm/sec} \] \[ V_o = \frac{40447.10625 \times 10^{-6}}{0.1818} \text{ cm/sec} \] \[ V_o \approx 0.222481 \text{ cm/sec} \]So, the overflow rate of the settling tank is approximately 0.222481 cm/sec.
Calculating the Removal Percentage for the Specified Particle
The question asks for the percentage removal of a particle with a settling velocity of \$ V_s = 0.18 \$ cm/sec in this tank where \$ V_o \approx 0.222481 \$ cm/sec.
Since \$ V_s = 0.18 \text{ cm/sec} < V_o \approx 0.222481 \text{ cm/sec} \$, we use the formula for percentage removal:
\[ \text{Percentage Removal} = \left( \frac{V_s}{V_o} \right) \times 100\% \] \[ \text{Percentage Removal} = \left( \frac{0.18 \text{ cm/sec}}{0.222481 \text{ cm/sec}} \right) \times 100\% \] \[ \text{Percentage Removal} \approx 0.80996 \times 100\% \] \[ \text{Percentage Removal} \approx 80.996\% \]Rounding this to two decimal places gives 81.00%.
Conclusion
The percentage of particles with a settling velocity of 0.18 cm/sec removed in an ideal settling tank with an overflow rate of approximately 0.222481 cm/sec (determined by the properties of the reference particle) is approximately 81.00%.
| Parameter | Value |
|---|---|
| Particle Settling Velocity (\$V_s\$) | 0.18 cm/sec |
| Reference Particle Diameter (\$d\$) | \$ 5 \times 10^{-3} \$ cm |
| Reference Particle Specific Gravity (\$S_p\$) | 2.65 |
| Water Kinematic Viscosity (\$ \nu \$) | \$ 1.01 \times 10^{-2} \$ cm2/sec |
| Calculated Overflow Rate (\$V_o\$) | \$ \approx 0.222481 \$ cm/sec |
| Percentage Removal | \$ \left( \frac{V_s}{V_o} \right) \times 100\% \$ |
| Result | \$ \approx 81.00\% \$ |
Therefore, approximately 81.00% of the particle is removed.
Important Questions from Sedimentation
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3. Density of particle intended to be removed affects the overflow rate
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