Reference ID: MET-6FBD | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
Material balance for single-stage solid-liquid leaching is a fundamental calculation in process engineering used to determine the distribution of a solute between a solid carrier and a solvent phase. This operation is critical in industries such as oilseed processing, hydrometallurgy, and pharmaceutical extraction. By establishing the mass balance across a mixer-settler unit, engineers can predict the yield of the extract stream and the residual solute concentration in the raffinate, which is essential for optimizing solvent usage and downstream recovery processes.
Methodology & Formulas
The calculation relies on the conservation of mass for the total system and the individual solute species. The system is defined by the feed of solids (F) and solvent (S), resulting in an extract (E) and a raffinate (R). The inert solid mass (B) remains constant throughout the process. It is assumed that the leaching stage achieves equilibrium relationship in leaching, meaning the solute concentration in the solution retained by the solid is the same as in the bulk extract phase.
The inert solid mass is determined by the feed rate and the initial solute fraction:
\[ B = F \cdot (1 - x_{F}) \]
The raffinate mass is the sum of the inert solids and the solution retained within the solid matrix, where N represents the retention ratio:
\[ R = B + (B \cdot N) = B(1 + N) \]
The extract mass is derived from the overall mass balance of the system:
\[ E = F + S - R \]
The solute mass fraction in the extract (yE) is calculated by distributing the total solute from the feed and solvent across the extract and the solution retained in the raffinate:
\[ y_{E} = \frac{F \cdot x_{F} + S \cdot y_{S}}{E + B \cdot N} \]
Finally, the solute mass fraction in the raffinate (xR) is determined by the concentration of the retained solution:
\[ x_{R} = \frac{y_{E} \cdot B \cdot N}{R} \]
Parameter
Empirical Range
Description
xF
0.05 - 0.5
Feed solute mass fraction
yS
0.0 (pure solvent)
Solvent solute mass fraction
S / F
0.5 - 10.0
Solvent-to-feed mass ratio
N
0.2 - 2.0
Solution retention (kg solution / kg inert solid)
To accurately define your flow rates, you must establish a steady-state mass balance. Follow these steps:
Identify the mass flow rate of the feed solid (F) and the solvent (S).
Determine the initial concentration of the solute in the feed (xF) and in the solvent (yS).
Apply the total mass balance equation: F + S = E + R, where E is the extract phase and R is the raffinate phase.
The distribution coefficient (Kd) represents the equilibrium relationship between the extract and raffinate phases. It is critical because:
It dictates the theoretical maximum solute recovery for a given solvent-to-feed ratio.
It allows you to relate the concentrations of the solute in both phases at equilibrium (yE = Kd * xR).
It helps determine if the chosen solvent is efficient enough to achieve the desired separation targets.
In a single-stage extraction, the operating line is defined by the mass balance constraints, while the equilibrium curve is defined by the thermodynamic properties of the system.
The operating line represents the actual mass balance relationship between the phases.
The equilibrium curve represents the limit of separation.
The intersection of these two lines provides the specific concentrations (yE, xR) achieved in the single stage.
If the liquids are partially miscible, you cannot assume constant flow rates for the carrier and solvent. You must:
Use a ternary phase diagram to locate the compositions of the extract and raffinate phases.
Account for the mutual solubility of the components, which changes the total mass of each phase.
Perform the mass balance using the lever rule on the ternary diagram to find the exact phase compositions.
Worked Example: Single-Stage Leaching of Oil from Soybeans
A batch leaching process is used to extract oil from crushed soybeans using pure hexane as a solvent. The goal is to determine the masses and compositions of the extract and raffinate streams.
Calculate the mass of inert solid (oil-free beans), \( B \):
\[ B = F \cdot (1 - x_{F}) = 1000.0 \, \text{kg} \cdot (1 - 0.180) = 820.0 \, \text{kg} \]
Calculate the mass of solution retained in the raffinate:
\[ \text{Retained solution} = B \cdot N = 820.0 \, \text{kg} \cdot 0.500 = 410.0 \, \text{kg} \]
Calculate the raffinate mass, \( R \). The raffinate consists of inert solids and retained solution:
\[ R = B + \text{Retained solution} = 820.0 \, \text{kg} + 410.0 \, \text{kg} = 1230.0 \, \text{kg} \]
Calculate the extract mass, \( E \), from the overall mass balance:
\[ F + S = E + R \]
\[ E = F + S - R = 1000.0 \, \text{kg} + 500.0 \, \text{kg} - 1230.0 \, \text{kg} = 270.0 \, \text{kg} \]
Solve for the oil mass fraction in the extract, \( y_{E} \), using the solute balance. All oil from the feed distributes between the extract and the retained solution (solvent is pure):
\[ F \cdot x_{F} + S \cdot y_{S} = y_{E} \cdot E + y_{E} \cdot (B \cdot N) \]
With \( y_{S} = 0 \):
\[ y_{E} = \frac{F \cdot x_{F}}{E + B \cdot N} = \frac{1000.0 \, \text{kg} \cdot 0.180}{270.0 \, \text{kg} + 410.0 \, \text{kg}} = \frac{180.0 \, \text{kg}}{680.0 \, \text{kg}} = 0.2647 \]
(Rounded to four decimal places for precision in subsequent steps.)
Calculate the oil mass fraction in the raffinate, \( x_{R} \). The oil in the raffinate is only in the retained solution:
\[ x_{R} = \frac{y_{E} \cdot B \cdot N}{R} = \frac{0.2647 \cdot 410.0 \, \text{kg}}{1230.0 \, \text{kg}} \approx 0.0882 \]
Final Answer:
Extract stream: \( E = 270.0 \, \text{kg} \) with oil fraction \( y_{E} = 0.265 \, \text{kg oil/kg extract} \).
Raffinate stream: \( R = 1230.0 \, \text{kg} \) with oil fraction \( x_{R} = 0.088 \, \text{kg oil/kg raffinate} \). Note: Final values are rounded to three significant figures for reporting.
"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
