Disc-Bowl Centrifuge Sigma Factor Calculation

Reference ID: MET-E265 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The Sigma factor (Σ) is a fundamental parameter in process engineering used to characterize the theoretical separation capacity of a disc‑bowl centrifuge, representing the equivalent area of a gravity settling tank that would achieve the same performance. Because centrifugal force far exceeds gravitational force, the Sigma factor enables engineers to scale laboratory results to industrial scale and to compare different centrifuge geometries, while alternative technologies such as the hydrocyclone application for starch concentration illustrate how similar separation principles can be applied to other processes.

This calculation is critical in industries such as biotechnology, food processing, and wastewater treatment, where the efficient removal of suspended solids from liquid phases is required. By calculating Σ, engineers can predict the maximum throughput—see our disc‑bowl centrifuge capacity calculation—for a target particle size and ensure that the equipment operates within its design limits.

Methodology & Formulas

The calculation of the Sigma factor relies on the geometry of the disc stack and the rotational dynamics of the bowl. The process follows these physical principles:

First, the rotational speed in revolutions per minute (RPM) is converted to angular velocity (\(\omega\)) in radians per second:

\[ \omega = \text{RPM} \cdot \frac{2\pi}{60} \]

The Sigma factor (\(\Sigma\)) is then calculated based on the number of discs (\(N\)), the inner radius (\(r_{1}\)), the outer radius (\(r_{2}\)), and the half‑cone angle (\(\alpha\)); for a comparable tubular centrifuge, see the tubular centrifuge sigma factor calculation.

\[ \Sigma = \frac{2\pi \cdot \omega^{2} \cdot N \cdot (r_{2}^{3} - r_{1}^{3})}{3 \cdot g \cdot \tan(\alpha)} \]

To determine the theoretical throughput (\(Q_{\text{theory}}\)), we first calculate the terminal settling velocity of a particle under gravity (\(v_{g}\)) using Stokes Law, assuming the particle Reynolds number is sufficiently low:

\[ v_{g} = \frac{d^{2} \cdot \Delta\rho \cdot g}{18 \cdot \mu} \] \[ Q_{\text{theory}} = v_{g} \cdot \Sigma \]

In practical applications, the actual throughput (\(Q_{\text{real}}\)) is adjusted by an empirical efficiency factor (\(\eta\)) to account for non-ideal flow, turbulence, and short-circuiting:

\[ Q_{\text{real}} = Q_{\text{theory}} \cdot \eta \]

Parameter	Condition / Limit	Engineering Significance
Disc Angle (\(\alpha\))	\(30^\circ \leq \alpha \leq 60^\circ\)	Angles outside this range lead to either solids blockage or reduced settling area.
Particle Reynolds (\(Re_{p}\))	\(Re_{p} < 0.1\)	Ensures Stokes Law validity; otherwise, drag correction factors are required.
Gap Reynolds (\(Re_{\text{gap}}\))	\(Re_{\text{gap}} < 2000\) (requires disc spacing \(h_{\text{gap}}\) for calculation)	Ensures laminar, fully-developed flow between the discs.
Efficiency Factor (\(\eta\))	\(0.5 \leq \eta \leq 0.8\)	Accounts for real-world deviations from ideal theoretical performance.

The Sigma factor represents the theoretical equivalent clarification area of a gravity settling tank. For a disc-bowl centrifuge, it quantifies the separation capacity based on the geometry of the discs and the rotational speed. It is calculated using the following parameters:

The angular velocity of the bowl.
The number of discs within the stack.
The inner and outer radii of the discs.
The angle of the discs relative to the axis of rotation.

The Sigma factor allows process engineers to predict the performance of a larger centrifuge based on data from a smaller pilot-scale unit. By maintaining a constant ratio of flow rate to Sigma factor (Q/Σ), you can ensure consistent separation efficiency when moving between different equipment sizes.

While several geometric factors contribute to the calculation, the following variables have the most substantial impact:

Rotational speed: Because the Sigma factor is proportional to the square of the angular velocity, even minor changes in RPM significantly alter the separation capacity.
Disc count: Increasing the number of discs increases the total settling area.
Disc angle: The angle determines the effective projected area for particle sedimentation.

No, the theoretical Sigma factor assumes ideal laminar flow and uniform particle distribution. In practice, process engineers must apply an efficiency factor to account for:

Turbulence at the feed inlet.
Non-uniform flow distribution between individual discs.
Boundary layer effects near the disc surfaces.
Resuspension of settled solids.

Worked Example: Disc-Bowl Centrifuge Sigma Factor Calculation

A bioprocessing engineer is evaluating two disc stack configurations for a centrifugal clarifier. The centrifuge bowl operates at a reduced speed of 1000 RPM with an outer radius of 0.30 m and inner radius of 0.05 m, and discs angled at 40°. Configuration A uses 100 discs and Configuration B uses 200 discs. The feed contains particles with diameter 1 µm and density 2500 kg/m³ in a fluid of density 1000 kg/m³ and viscosity 0.001 Pa·s. The engineer applies an empirical efficiency factor of 0.65 to account for non-ideal flow. The disc gap spacing is 0.8 mm.

Knowns

Gravitational acceleration, \(g = 9.81 \, \text{m/s}^2\)
Rotational speed, \(\text{RPM} = 1000\)
Number of discs, Config A: \(N_{A} = 100\)
Number of discs, Config B: \(N_{B} = 200\)
Outer radius, \(r_{2} = 0.30 \, \text{m}\)
Inner radius, \(r_{1} = 0.05 \, \text{m}\)
Disc half-cone angle, \(\alpha = 40^\circ\)
Disc gap spacing, \(h_{\text{gap}} = 0.8 \, \text{mm} = 8.0 \times 10^{-4} \, \text{m}\)
Particle diameter, \(d = 1 \times 10^{-6} \, \text{m}\)
Particle density, \(\rho_{p} = 2500 \, \text{kg/m}^3\)
Fluid density, \(\rho_{f} = 1000 \, \text{kg/m}^3\)
Dynamic viscosity, \(\mu = 0.001 \, \text{Pa}\cdot\text{s}\)
Efficiency factor, \(\eta = 0.65\)

Step-by-Step Calculation

Convert RPM to angular velocity
\(\omega = \text{RPM} \cdot \frac{2\pi}{60} = 1000 \cdot \frac{2\pi}{60} = \frac{100\pi}{3} \approx 104.72 \, \text{rad/s}\)
Compute tangent of disc half-cone angle
\(\alpha = 40^\circ\), \(\quad \tan\alpha = \tan(40^\circ) = 0.8391\)
Compute cubes of radii and their difference
\(r_{2}^{3} = (0.30)^3 = 0.027 \, \text{m}^3\)
\(r_{1}^{3} = (0.05)^3 = 0.000125 \, \text{m}^3\)
\(r_{2}^{3} - r_{1}^{3} = 0.027 - 0.000125 = 0.026875 \, \text{m}^3\)
Calculate sigma factor for Config A
\(\Sigma_{A} = \frac{2\pi \cdot \omega^{2} \cdot N_{A} \cdot (r_{2}^{3} - r_{1}^{3})}{3 \cdot g \cdot \tan\alpha} = \frac{2\pi \cdot (104.72)^2 \cdot 100 \cdot 0.026875}{3 \cdot 9.81 \cdot 0.8391}\)
\(\Sigma_{A} = \frac{185\,176.4}{24.6947} = 7\,498.6 \, \text{m}^2\)
Calculate sigma factor for Config B
\(\Sigma_{B} = \frac{2\pi \cdot \omega^{2} \cdot N_{B} \cdot (r_{2}^{3} - r_{1}^{3})}{3 \cdot g \cdot \tan\alpha} = 2 \cdot \Sigma_{A} = 14\,997.2 \, \text{m}^2\)
Compute density difference
\(\Delta\rho = \rho_{p} - \rho_{f} = 2500 - 1000 = 1500 \, \text{kg/m}^3\)
Compute terminal settling velocity (Stokes law)
\(v_{g} = \frac{d^{2} \cdot \Delta\rho \cdot g}{18 \cdot \mu} = \frac{(1 \times 10^{-6})^{2} \cdot 1500 \cdot 9.81}{18 \cdot 0.001} = 8.175 \times 10^{-7} \, \text{m/s}\)
Calculate theoretical and real throughput for Config A
\(Q_{\text{theory},A} = v_{g} \cdot \Sigma_{A} = (8.175 \times 10^{-7}) \cdot 7\,498.6 = 6.130 \times 10^{-3} \, \text{m}^3/\text{s}\)
\(Q_{\text{real},A} = Q_{\text{theory},A} \cdot \eta = 6.130 \times 10^{-3} \cdot 0.65 = 3.985 \times 10^{-3} \, \text{m}^3/\text{s} \approx 3.99 \times 10^{-3} \, \text{m}^3/\text{s}\)
Calculate theoretical and real throughput for Config B
\(Q_{\text{theory},B} = v_{g} \cdot \Sigma_{B} = (8.175 \times 10^{-7}) \cdot 14\,997.2 = 1.226 \times 10^{-2} \, \text{m}^3/\text{s}\)
\(Q_{\text{real},B} = Q_{\text{theory},B} \cdot \eta = 1.226 \times 10^{-2} \cdot 0.65 = 7.969 \times 10^{-3} \, \text{m}^3/\text{s} \approx 7.97 \times 10^{-3} \, \text{m}^3/\text{s}\)
Verify Stokes regime (particle Reynolds number under centrifugal acceleration)
Mid-radius: \(r_{\text{mid}} = \frac{r_{1} + r_{2}}{2} = \frac{0.05 + 0.30}{2} = 0.175 \, \text{m}\)
Centrifugal settling velocity at mid-radius: \(v_{c} = \frac{d^{2} \cdot \Delta\rho \cdot r_{\text{mid}} \cdot \omega^{2}}{18 \cdot \mu} = \frac{(1 \times 10^{-6})^{2} \cdot 1500 \cdot 0.175 \cdot (104.72)^{2}}{18 \cdot 0.001} = 1.599 \times 10^{-4} \, \text{m/s}\)
Particle Reynolds number: \(\text{Re}_{p} = \frac{\rho_{f} \cdot v_{c} \cdot d}{\mu} = \frac{1000 \cdot 1.599 \times 10^{-4} \cdot 1 \times 10^{-6}}{0.001} = 1.60 \times 10^{-4}\)
\(\text{Re}_{p} = 1.60 \times 10^{-4} < 0.1\) → Stokes law is valid.
Verify gap Reynolds number (laminar flow between discs)
Radial velocity between discs: \(v_{\text{rad}} = \frac{Q_{\text{theory},A}}{N_{A} \cdot 2\pi \cdot r_{\text{mid}} \cdot h_{\text{gap}}} = \frac{6.130 \times 10^{-3}}{100 \cdot 2\pi \cdot 0.175 \cdot 8.0 \times 10^{-4}} = 0.0697 \, \text{m/s}\)
Gap Reynolds number: \(\text{Re}_{\text{gap}} = \frac{\rho_{f} \cdot v_{\text{rad}} \cdot h_{\text{gap}}}{\mu} = \frac{1000 \cdot 0.0697 \cdot 8.0 \times 10^{-4}}{0.001} = 55.8\)
\(\text{Re}_{\text{gap}} = 55.8 < 2000\) → Laminar flow condition is satisfied.

Final Answer

Config A yields \(\Sigma_{A} = 7\,498.6 \, \text{m}^2\) and a corrected throughput of \(Q_{\text{real},A} = 3.99 \times 10^{-3} \, \text{m}^3/\text{s}\) (approximately \(4.0 \times 10^{-3} \, \text{m}^3/\text{s}\)).
Config B yields \(\Sigma_{B} = 14\,997.2 \, \text{m}^2\) and a corrected throughput of \(Q_{\text{real},B} = 7.97 \times 10^{-3} \, \text{m}^3/\text{s}\) (approximately \(8.0 \times 10^{-3} \, \text{m}^3/\text{s}\)).
Config B has double the sigma factor and theoretical capacity of Config A, but after applying the empirical efficiency factor of 0.65, the expected real-world throughput is approximately \(3.99 \times 10^{-3} \, \text{m}^3/\text{s}\) and \(7.97 \times 10^{-3} \, \text{m}^3/\text{s}\), respectively. The particle Reynolds number of \(1.60 \times 10^{-4}\) confirms that the Stokes settling assumption is valid, and the gap Reynolds number of 55.8 confirms laminar flow between the discs.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

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