Introduction & Context
The disc-bowl centrifuge is a critical unit operation in process engineering, widely utilized for the continuous separation of liquid-liquid-solid mixtures. By employing high-speed rotation, these devices generate centrifugal forces significantly greater than gravity, forcing denser particles to migrate toward the periphery of the bowl. The Sigma Factor (Σ) represents the theoretical equivalent settling area of a gravity sedimentation tank, providing a standardized metric to compare the performance of different centrifuge designs regardless of their specific geometry. This calculation is essential for scaling up industrial processes, such as yeast separation, milk clarification, and wastewater treatment, ensuring that the volumetric flow rate (Q) remains within the limits required to achieve a target particle separation efficiency.
Methodology & Formulas
The calculation of the theoretical capacity is derived from the integration of Stokes' Law within the rotating frame of the disc stack. The process follows these fundamental steps:
1. Angular Velocity Conversion: The rotational speed is converted from revolutions per minute to radians per second:
\[ \omega = N_{\text{rpm}} \cdot \frac{2\pi}{60} \]
2. Terminal Settling Velocity: Under the assumption of Stokes' regime, the settling velocity of a spherical particle under gravity is defined as:
\[ v_{g} = \frac{d^{2} \cdot (\rho_{s} - \rho_{l}) \cdot g}{18 \cdot \mu} \]
3. Sigma Factor Calculation: The Sigma factor quantifies the separation capability based on the centrifuge geometry and rotational speed. For a disc stack with \(N\) discs, outer radius \(r_{2}\), inner radius \(r_{1}\), and disc half-angle \(\theta\):
\[ \Sigma = \frac{2\pi \cdot \omega^{2} \cdot N \cdot \cot(\theta)}{3 \cdot g} \cdot (r_{2}^{3} - r_{1}^{3}) \]
4. Theoretical Capacity: The final volumetric capacity is determined by the product of the settling velocity, the Sigma factor, and an empirical efficiency factor (\(\eta\)) to account for non-ideal flow distribution and turbulence:
\[ Q = v_{g} \cdot \Sigma \cdot \eta \]
| Parameter |
Condition / Regime |
Threshold / Limit |
| Particle Reynolds Number |
Stokes' Law Validity |
Rep < 1.0 |
| Particle Size |
Effective Separation |
0.5 µm ≤ d ≤ 50 µm |
| Density Difference |
Practical Centrifugation |
Δρ ≥ 10 kg/m3 |
| Disc Angle |
Optimal Sliding |
35° ≤ θ ≤ 50° |
| Angular Velocity |
Mechanical Limits |
2000 rpm ≤ Nrpm ≤ 15000 rpm |
Worked Example: Disc-Bowl Centrifuge Capacity for Milk Clarification
A continuous disc-bowl centrifuge is designed to clarify raw milk by removing fat globules. The goal is to determine the maximum flow rate (capacity) for effective separation under given operating conditions. This example follows the theoretical framework for sedimentation in a disc-bowl centrifuge.
Known Input Parameters:
- Particle diameter, \( d = 1 \times 10^{-5} \, \text{m} \) (10 µm)
- Solid density (fat globules), \( \rho_{s} = 1080.0 \, \text{kg/m}^3 \)
- Liquid density (milk), \( \rho_{l} = 1030.0 \, \text{kg/m}^3 \)
- Dynamic viscosity, \( \mu = 0.001 \, \text{Pa} \cdot \text{s} \) (1 cP at 20°C)
- Number of discs, \( N = 100.0 \)
- Disc half-angle, \( \theta = 45.0^\circ \)
- Inner disc radius, \( r_{1} = 0.05 \, \text{m} \)
- Outer disc radius, \( r_{2} = 0.15 \, \text{m} \)
- Rotational speed, \( N_{\text{rpm}} = 5000.0 \, \text{min}^{-1} \)
- Efficiency factor (empirical), \( \eta = 0.7 \)
- Gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \)
Step-by-Step Calculation:
- Convert rotational speed to angular velocity:
\( \omega = N_{\text{rpm}} \cdot \frac{2\pi}{60} = 5000.0 \cdot \frac{2 \cdot \pi}{60} = 523.599 \, \text{rad/s} \).
- Calculate density difference:
\( \Delta\rho = \rho_{s} - \rho_{l} = 1080.0 - 1030.0 = 50.0 \, \text{kg/m}^3 \).
- Determine gravitational settling velocity using Stokes' law:
\( v_{g} = \frac{d^{2} \cdot \Delta\rho \cdot g}{18 \cdot \mu} = \frac{(1 \times 10^{-5})^{2} \cdot 50.0 \cdot 9.81}{18 \cdot 0.001} = 2.725 \times 10^{-6} \, \text{m/s} \).
- Verify Stokes' law validity by computing particle Reynolds number:
\( Re_{p} = \frac{\rho_{l} \cdot v_{g} \cdot d}{\mu} = \frac{1030.0 \cdot 2.725 \times 10^{-6} \cdot 1 \times 10^{-5}}{0.001} = 2.807 \times 10^{-5} \).
Since \( Re_{p} \ll 1 \), Stokes' law is valid.
- Calculate the sigma factor (\( \Sigma \)) for the disc bowl:
First, compute \( \cot\theta \) for \( \theta = 45.0^\circ \): \( \cot\theta = 1.000 \).
Then, \( r_{2}^{3} - r_{1}^{3} = (0.15)^{3} - (0.05)^{3} = 0.003375 - 0.000125 = 0.003250 \, \text{m}^{3} \).
Now, \( \Sigma = \frac{2\pi \cdot \omega^{2} \cdot N \cdot \cot\theta}{3 \cdot g} \cdot (r_{2}^{3} - r_{1}^{3}) = \frac{2 \cdot \pi \cdot (523.599)^{2} \cdot 100.0 \cdot 1.000}{3 \cdot 9.81} \cdot 0.003250 \approx 1.90 \times 10^{4} \, \text{m}^{2} \).
More precisely: \( \Sigma = 1.902 \times 10^{4} \, \text{m}^{2} \).
- Compute theoretical capacity in SI units:
\( Q = v_{g} \cdot \Sigma \cdot \eta = (2.725 \times 10^{-6}) \cdot (1.902 \times 10^{4}) \cdot 0.7 = 0.0363 \, \text{m}^{3}/\text{s} \).
- Convert capacity to practical units (liters per hour):
Conversion: \( 1 \, \text{m}^{3}/\text{s} = 3.6 \times 10^{6} \, \text{L/h} \).
\( Q_{\text{L/h}} = 0.0363 \times 3.6 \times 10^{6} = 1.31 \times 10^{5} \, \text{L/h} \).
More precisely: \( 1.31 \times 10^{5} \, \text{L/h} \) (or 131,000 L/h to three significant figures).
Final Answer:
The maximum flow rate for effective milk clarification in this disc-bowl centrifuge is approximately \( 1.31 \times 10^{5} \, \text{L/h} \)) (or \( 0.0363 \, \text{m}^{3}/\text{s} \)), based on the theoretical capacity adjusted for practical efficiency.
