Aroma Retention in Extraction

Reference ID: MET-726F | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The aroma-extraction calculation predicts the fraction of volatile compounds that can be recovered from a solid feed during a batch extraction with a supercritical solvent. It is a core tool in process engineering for designing and optimizing flavor, fragrance, and nutraceutical extraction operations, where product quality and yield are directly linked to the efficiency of mass transfer between the solid particles and the solvent phase.

This methodology is typically applied to:

Batch extraction vessels used in spice, coffee, and essential-oil processing.
Scale-up studies where solvent flow, particle size, and operating conditions must be correlated to expected aroma recovery.
Feasibility assessments that compare different solvents, temperatures, or pressures.

Methodology & Formulas

1. Material balance for a batch extractor

The bulk aroma concentration in the solvent, \(C_{b}\), is related to the extracted mass, \(M_{\text{extract}}\), by the constant solvent volume, \(V_{s}\):

\[ C_{b} \;=\; \frac{M_{\text{extract}}}{V_{s}} \]

The instantaneous mass-transfer rate from the solid surface to the solvent is expressed as:

\[ \dot{m}_{A} \;=\; k_{c}\,A\,\bigl(C^{*} - C_{b}\bigr) \]

where:

\(k_{c}\) – overall mass-transfer coefficient (m·s^-1).
\(A\) – total interfacial area between solid particles and solvent (m²).
\(C^{*}\) – equilibrium (saturation) concentration of the aroma in the solvent at the solid–solvent interface (kg·m^-3).
\(C_{b}\) – bulk aroma concentration in the solvent (kg·m^-3).

2. Correlation for the mass-transfer coefficient

The coefficient \(k_{c}\) is obtained from a Sherwood-number correlation that depends on the Reynolds and Schmidt numbers for the particle-solvent system:

Reynolds number (inertial vs. viscous forces):
\[ Re \;=\; \frac{\rho\,u\,d_{p}}{\mu} \]
Schmidt number (momentum vs. mass diffusion):
\[ Sc \;=\; \frac{\mu}{\rho\,D_{AB}} \]
Sherwood number (dimensionless mass-transfer coefficient):
\[ Sh \;=\; 2 \;+\; 0.6\,Re^{1/2}\,Sc^{1/3} \]
Conversion to \(k_{c}\):
\[ k_{c} \;=\; \frac{Sh\,D_{AB}}{d_{p}} \]

Note: All properties (\(\rho, \mu, D_{AB}\)) must be in consistent SI units (kg, m, s, Pa·s).

Symbols:

\(\rho\) – solvent density (kg·m^-3).
\(u\) – superficial solvent velocity (m·s^-1).
\(d_{p}\) – characteristic particle diameter (m).
\(\mu\) – solvent dynamic viscosity (Pa·s).
\(D_{AB}\) – binary diffusion coefficient of aroma in the solvent (m²·s^-1).

3. First-order solution for batch extraction

Substituting the expression for \(C_{b}\) into the rate equation and rearranging yields a first-order ordinary differential equation:

\[ \frac{dM_{\text{extract}}}{dt} \;=\; k_{c}\,A\left(C^{*} - \frac{M_{\text{extract}}}{V_{s}}\right) \]

Defining the overall extraction constant:

\[ k_{\text{term}} \;=\; \frac{k_{c}\,A}{V_{s}} \]

The analytical solution for a constant \(k_{c}\) and constant \(C^{*}\) is:

\[ M_{\text{extract}}(t) \;=\; C^{*}\,V_{s}\,\Bigl(1 - e^{-k_{\text{term}}\,t}\Bigr) \]

4. Aroma-extraction percentage

The fraction of the original aroma mass in the feed that has been transferred to the solvent after a time \(t\) is the extraction efficiency or recovery:

\[ \text{Extraction}\,\% \;=\; \frac{M_{\text{extract}}(t)}{M_{\text{feed}}}\times 100 \]

where \(M_{\text{feed}}\) is the total aroma mass initially present in the solid feed (kg).

Empirical Validity Checks

Parameter	Valid Range
Reynolds number, \(Re\)	1 ≤ \(Re\) ≤ 1000
Schmidt number, \(Sc\)	0.6 ≤ \(Sc\) ≤ 3000
Extracted mass, \(M_{\text{extract}}\)	\(M_{\text{extract}}\) ≤ \(M_{\text{feed}}\)

Step-by-Step Calculation Procedure

Determine the total aroma mass in the feed:
\[ M_{\text{feed}} \;=\; m_{\text{feed}} \cdot w_{A} \]

where \(m_{\text{feed}}\) is the feed mass (kg) and \(w_{A}\) is the weight fraction of aroma (kg·kg^-1).
Obtain the equilibrium concentration \(C^{*}\) from solubility or partition-coefficient data at the operating temperature and pressure.
Calculate the dimensionless groups:
- \(Re = \dfrac{\rho\,u\,d_{p}}{\mu}\) (Ensure \(\mu\) is in Pa·s)
- \(Sc = \dfrac{\mu}{\rho\,D_{AB}}\)
Apply the Sherwood correlation to find \(Sh\) and then the mass-transfer coefficient:
\[ Sh = 2 + 0.6\,Re^{1/2}\,Sc^{1/3} \qquad k_{c} = \dfrac{Sh\,D_{AB}}{d_{p}} \]
Form the overall extraction constant:
\[ k_{\text{term}} = \frac{k_{c}\,A}{V_{s}} \]
Compute the extracted aroma mass at the desired extraction time \(t\):
\[ M_{\text{extract}}(t) = C^{*}\,V_{s}\,\bigl(1 - e^{-k_{\text{term}}\,t}\bigr) \]
Calculate the extraction percentage:
\[ \text{Extraction}\,\% = \frac{M_{\text{extract}}(t)}{M_{\text{feed}}}\times 100 \]
Verify that the computed \(Re\), \(Sc\), and \(M_{\text{extract}}\) satisfy the empirical ranges listed in the table above. If any check fails, revisit the assumptions (e.g., particle size, velocity, or correlation selection).

To maximize recovery of delicate aromatics, process engineers should focus on enhancing mass transfer and managing process conditions. Key strategies include:

Operating at the optimal temperature and pressure to maximize the solubility (\(C^{*}\)) of the target compounds without causing thermal degradation.
Reducing particle size (\(d_{p}\)) to increase interfacial area (\(A\)), which directly increases the mass-transfer rate.
Optimizing solvent flow (superficial velocity, \(u\)) to increase the Reynolds number and thus the mass-transfer coefficient (\(k_{c}\)), within the valid range of the correlation.
Using a solvent with favorable transport properties (high diffusivity \(D_{AB}\), low viscosity \(\mu\)) to increase the Schmidt number and enhance \(k_{c}\).
Ensuring sufficient extraction time (\(t\)) to approach the maximum theoretical recovery (\(C^{*}V_{s}\)).

Solvent polarity is critical for targeting specific aromatic molecules while excluding unwanted matrix components like waxes or pigments. By adjusting the dielectric constant of the solvent mixture, you can:

Increase the solubility (\(C^{*}\)) of target polar esters and terpenes.
Minimize the co-extraction of heavy non-polar lipids that can mask or dilute the final aroma profile.
Fine-tune the selectivity of the extraction phase to ensure the final extract represents the authentic profile of the raw material.

Once the extract is obtained, stabilization is necessary to prevent oxidation and evaporation of the volatile components. Recommended techniques include:

Encapsulation using spray drying or freeze drying to create a protective barrier around volatile molecules.
Addition of natural antioxidants to inhibit the oxidative degradation of unsaturated aromatic compounds.
Storage under inert gas blanketing, such as nitrogen or argon, to displace oxygen and prevent chemical spoilage.
Temperature-controlled storage to maintain the integrity of the aromatic profile during the shelf life of the product.

Worked Example: Batch Extraction of Aroma from Spices

A process engineer needs to estimate the aroma recovery from a batch extraction process. Ground spice is contacted with supercritical CO₂ for one hour. The goal is to calculate the percentage of the available aroma that transfers into the solvent.

Known Process Parameters

Feed mass, \( m_{\text{feed}} \): 5.0 kg
Aroma weight fraction in feed, \( w_{A} \): 0.002
Solvent temperature, \( T \): 40.0 °C
Solvent pressure, \( P \): 100.0 bar
Solvent density, \( \rho \): 800.0 kg/m³
Solvent viscosity, \( \mu \): 0.1 cP = \( 1.0 \times 10^{-4} \) Pa·s
Aroma diffusivity in solvent, \( D_{AB} \): \( 2.0 \times 10^{-8} \) m²/s
Solvent volume, \( V_{s} \): 0.05 m³
Particle diameter, \( d_{p} \): 0.002 m
Superficial velocity, \( u \): 0.001 m/s
Total interfacial area, \( A \): 10.0 m²
Equilibrium concentration, \( C^{*} \): 0.1 kg/m³
Extraction time, \( t \): 3600.0 s

Step-by-Step Calculation

Calculate the total aroma mass available in the feed.
\[ M_{\text{feed}} = m_{\text{feed}} \cdot w_{A} = 5.0 \, \text{kg} \cdot 0.002 = 0.01 \, \text{kg} \]
Identify the equilibrium concentration at the interface.
\[ C^{*} = 0.1 \, \text{kg} \cdot \text{m}^{-3} \]
Compute the dimensionless Reynolds (\( Re \)) and Schmidt (\( Sc \)) numbers.
\[ Re = \frac{\rho \cdot u \cdot d_{p}}{\mu} = \frac{800.0 \cdot 0.001 \cdot 0.002}{1.0 \times 10^{-4}} = 16.0 \]

\[ Sc = \frac{\mu}{\rho \cdot D_{AB}} = \frac{1.0 \times 10^{-4}}{800.0 \cdot 2.0 \times 10^{-8}} = 6.25 \]
Calculate the Sherwood number (\( Sh \)).
\[ Sh = 2 + 0.6 \cdot Re^{1/2} \cdot Sc^{1/3} = 2 + 0.6 \cdot \sqrt{16.0} \cdot (6.25)^{1/3} \]

\[ Sh = 2 + 0.6 \cdot 4 \cdot 1.842 = 6.421 \]
Determine the mass transfer coefficient (\( k_{c} \)).
\[ k_{c} = \frac{Sh \cdot D_{AB}}{d_{p}} = \frac{6.421 \cdot 2.0 \times 10^{-8}}{0.002} = 6.42 \times 10^{-5} \, \text{m} \cdot \text{s}^{-1} \]
Calculate the overall extraction constant and the extracted mass.
\[ k_{\text{term}} = \frac{k_{c} \cdot A}{V_{s}} = \frac{6.42 \times 10^{-5} \cdot 10.0}{0.05} = 0.01284 \, \text{s}^{-1} \]

\[ M_{\text{extract}}(t) = C^{*} \cdot V_{s} \cdot \left[ 1 - \exp\left( -k_{\text{term}} \cdot t \right) \right] \]

\[ M_{\text{extract}} = 0.1 \cdot 0.05 \cdot \left[ 1 - \exp\left( -0.01284 \cdot 3600 \right) \right] = 0.005 \, \text{kg} \]

(Note: \(\exp(-46.22) \approx 0\))
Compute the extraction percentage.
\[ \text{Extraction}\,\% = \left( \frac{M_{\text{extract}}}{M_{\text{feed}}} \right) \times 100\% = \left( \frac{0.005}{0.01} \right) \times 100\% = 50.0\% \]

Final Answer

After 1 hour of extraction, 50.0% of the available aroma has been transferred from the spice into the supercritical CO₂ solvent. This is the extraction efficiency or recovery.

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

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle