Introduction & Context
The calculation evaluates how the feed temperature influences the maximum permissible feed flow rate Q of a continuous liquid-liquid disk-stack centrifuge while preserving the same separation performance (identical critical droplet diameter dp). Raising the temperature lowers the continuous-phase viscosity, thereby increasing the settling velocity of droplets and allowing a higher throughput. This analysis is essential for process engineers who design or optimise oil-water separation trains, especially when energy integration (e.g., heat-exchanged feeds) is considered.
Methodology & Formulas
Step 1 – Define the governing settling velocity (Stokes’ law for a centrifuge).
\[
v_{g}= \frac{d_{p}^{2}\,\Delta\rho\,G}{18\,\mu}
\]
where
- dp – critical droplet diameter,
- Δρ – density difference between dispersed and continuous phases,
- μ – dynamic viscosity of the continuous phase,
- G – effective centrifugal acceleration (≈ 1000 · g for high-speed centrifuges).
Step 2 – Verify Stokes-law applicability. The particle Reynolds number must satisfy
\[
\mathrm{Re}_{p}= \frac{\rho_{c}\,d_{p}\,v_{g}}{\mu}\;<\;0.1
\]
with ρc the continuous-phase density.
Step 3 – Compute the capacity scaling ratio. For a fixed geometry factor Σ and unchanged dp, the ratio of allowable flow rates at two temperatures (1 = baseline, 2 = heated) follows directly from the proportionality Q ∝ vg:
\[
\frac{Q_{2}}{Q_{1}}=
\frac{\displaystyle\frac{\Delta\rho_{2}}{\mu_{2}}}
{\displaystyle\frac{\Delta\rho_{1}}{\mu_{1}}}
\tag{1}
\]
Step 4 – Determine the new flow rate.
\[
Q_{2}= Q_{1}\;\times\;\frac{\Delta\rho_{2}/\mu_{2}}{\Delta\rho_{1}/\mu_{1}}
\tag{2}
\]
Step 5 – Unit-consistency checks. Ensure that μ is expressed in Pa·s (1 cP = 0.001 Pa·s) and that dp is in metres before substitution.
Validity & Regime Checks
| Condition | Criterion |
|---|
| Stokes-law Reynolds number | \(\mathrm{Re}_{p}<0.1\) |
| Viscosity | \(\mu>0\) |
| Droplet diameter | \(d_{p}>0\) |
| Density difference | \(\Delta\rho>0\) |
Summary of the Scaling Procedure
- Obtain Δρ and μ at the baseline temperature T1 and the elevated temperature T2.
- Convert μ from cP to Pa·s (multiply by 0.001) and dp from µm to m (multiply by 1 × 10⁻⁶).
- Calculate the settling velocities vg1 and vg2 using the Stokes expression.
- Evaluate \(\mathrm{Re}_{p1}\) and \(\mathrm{Re}_{p2}\); confirm both are below 0.1.
- Apply the capacity ratio (1) and compute the new flow rate with (2).
The resulting Q2 provides the maximum feed rate at the higher temperature that maintains the same critical droplet size and thus the same separation sharpness.
Temperature directly influences the viscosity of the process fluid. As temperature increases, viscosity typically decreases, which facilitates faster particle sedimentation according to Stokes' Law. Key considerations include:
- Lower viscosity reduces the drag force acting against particles.
- Higher temperatures may improve throughput by allowing for higher flow rates.
- Excessive heat must be avoided if the product is thermally sensitive or prone to degradation.
Worked Example: Temperature Effect on Centrifugation Capacity
A continuous disk-stack centrifuge is used to dehydrate crude vegetable oil by removing water droplets. The process engineer is evaluating if pre-heating the feed from 20°C to 60°C can increase the maximum permissible flow rate while maintaining the same separation performance (critical droplet diameter of 10 µm).
Knowns (Input Parameters):
- Baseline temperature, \( T_1 = 20.0 \, ^\circ\mathrm{C} \)
- Oil density at \( T_1 \), \( \rho_{\mathrm{oil},1} = 920.0 \, \mathrm{kg/m^3} \)
- Water density at \( T_1 \), \( \rho_{\mathrm{water},1} = 998.0 \, \mathrm{kg/m^3} \)
- Oil viscosity at \( T_1 \), \( \mu_{\mathrm{oil},1} = 56.0 \, \mathrm{cP} \)
- Baseline flow rate, \( Q_1 = 5000.0 \, \mathrm{L/h} \)
- Critical droplet diameter, \( d_p = 10.0 \, \mu\mathrm{m} \)
- Heated temperature, \( T_2 = 60.0 \, ^\circ\mathrm{C} \)
- Oil density at \( T_2 \), \( \rho_{\mathrm{oil},2} = 900.0 \, \mathrm{kg/m^3} \)
- Water density at \( T_2 \), \( \rho_{\mathrm{water},2} = 983.0 \, \mathrm{kg/m^3} \)
- Oil viscosity at \( T_2 \), \( \mu_{\mathrm{oil},2} = 18.0 \, \mathrm{cP} \)
- Centrifugal acceleration, \( G = 9810 \, \mathrm{m/s^2} \) (assuming effective acceleration of ≈1000 × \( g \), where \( g = 9.81 \, \mathrm{m/s^2} \))
Step-by-Step Calculation:
- Define baseline parameters at \( T_1 \):
- Density difference, \( \Delta \rho_1 = \rho_{\mathrm{water},1} - \rho_{\mathrm{oil},1} = 998.0 - 920.0 = 78.0 \, \mathrm{kg/m^3} \).
- Convert viscosity to SI units: \( \mu_1 = 56.0 \, \mathrm{cP} \times 0.001 = 0.056 \, \mathrm{Pa\cdot s} \).
- Convert droplet diameter to metres: \( d_p = 10.0 \, \mu\mathrm{m} = 10.0 \times 10^{-6} \, \mathrm{m} = 1.00 \times 10^{-5} \, \mathrm{m} \).
- Define heated parameters at \( T_2 \):
- Density difference, \( \Delta \rho_2 = \rho_{\mathrm{water},2} - \rho_{\mathrm{oil},2} = 983.0 - 900.0 = 83.0 \, \mathrm{kg/m^3} \).
- Convert viscosity to SI units: \( \mu_2 = 18.0 \, \mathrm{cP} \times 0.001 = 0.018 \, \mathrm{Pa\cdot s} \).
- Check validity of Stokes' law regime (required for the scaling law):
- Calculate settling velocity at \( T_1 \) using Stokes' law for centrifugation:
\[
v_{g,1} = \frac{d_p^2 \, \Delta \rho_1 \, G}{18 \, \mu_1} = \frac{(1.00 \times 10^{-5})^2 \times 78.0 \times 9810}{18 \times 0.056} \approx 7.59 \times 10^{-5} \, \mathrm{m/s}
\]
- Particle Reynolds number at \( T_1 \):
\[
\mathrm{Re}_{p,1} = \frac{\rho_{\mathrm{c},1} \, d_p \, v_{g,1}}{\mu_1} = \frac{920.0 \times 1.00 \times 10^{-5} \times 7.59 \times 10^{-5}}{0.056} \approx 1.25 \times 10^{-5}
\]
where \( \rho_{\mathrm{c},1} = \rho_{\mathrm{oil},1} \) is the continuous-phase density.
- Similarly for \( T_2 \):
\[
v_{g,2} = \frac{d_p^2 \, \Delta \rho_2 \, G}{18 \, \mu_2} = \frac{(1.00 \times 10^{-5})^2 \times 83.0 \times 9810}{18 \times 0.018} \approx 2.51 \times 10^{-4} \, \mathrm{m/s}
\]
\[
\mathrm{Re}_{p,2} = \frac{\rho_{\mathrm{c},2} \, d_p \, v_{g,2}}{\mu_2} = \frac{900.0 \times 1.00 \times 10^{-5} \times 2.51 \times 10^{-4}}{0.018} \approx 1.26 \times 10^{-4}
\]
- Both Reynolds numbers are \( < 0.1 \), confirming Stokes' law is valid.
- Apply the capacity scaling law based on Sigma theory. For constant centrifuge geometry and critical droplet size, the flow rate ratio is:
\[
\frac{Q_2}{Q_1} = \frac{ \Delta \rho_2 / \mu_2 }{ \Delta \rho_1 / \mu_1 }
\]
Substitute the known values:
\[
\frac{Q_2}{Q_1} = \frac{83.0 / 0.018}{78.0 / 0.056} = \frac{4611.111\ldots}{1392.857\ldots} \approx 3.3111
\]
- Calculate the new maximum flow rate at \( T_2 \):
\[
Q_2 = Q_1 \times \frac{Q_2}{Q_1} = 5000.0 \, \mathrm{L/h} \times 3.3111 \approx 16555.6 \, \mathrm{L/h}
\]
Final Answer: Pre-heating the feed to \( 60.0 \, ^\circ\mathrm{C} \) allows the centrifuge to operate at a maximum flow rate of ≈16555.6 L/h while achieving the same separation performance (critical droplet diameter of 10 µm), compared to 5000.0 L/h at \( 20.0 \, ^\circ\mathrm{C} \). This represents a significant increase in capacity due to reduced oil viscosity and a slight increase in density difference.
