Single-Stage Batch Adsorption Material Balance

Reference ID: MET-5922 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The Single-Stage Batch Adsorption Material Balance is a fundamental calculation in process engineering used to determine the quantity of adsorbent required to reduce the concentration of a specific solute within a liquid batch. This operation is critical in industries such as food processing (e.g., decolorization or debittering), pharmaceutical purification, and wastewater treatment.

The calculation relies on the principle of mass conservation, where the mass of the adsorbate removed from the liquid phase is equal to the mass of the adsorbate accumulated on the solid adsorbent phase. By assuming the system reaches thermodynamic equilibrium, engineers can utilize adsorption isotherms to predict the final state of the system and size the required adsorbent dosage accordingly.

Methodology & Formulas

The calculation follows a systematic approach to determine the mass of adsorbent, denoted as L, required to reach a target liquid concentration, y^{*}, from an initial concentration, y_{0}, and can be extended to a multistage cross‑current adsorption calculation for more complex process designs.

First, the equilibrium solid loading x₁ is determined using the Langmuir isotherm model calculation, which relates the equilibrium concentration in the liquid to the concentration on the solid surface.

\[ x_{1} = \frac{x_{m} \cdot K \cdot y^{*}}{1 + K \cdot y^{*}} \]

Once the equilibrium loading x_{1} is established, the mass balance for the adsorbate across the system is applied. Given that the mass of adsorbate lost by the liquid must equal the mass gained by the adsorbent, the required mass of adsorbent L is calculated as follows:

\[ L = \frac{G \cdot (y_{0} - y^{*})}{x_{1} - x_{0}} \]

Where G represents the mass of the adsorbate-free liquid (for dilute systems, this is approximately equal to the total batch mass), and x_{0} represents the initial concentration of adsorbate on the fresh adsorbent (typically zero).

Condition/Parameter	Validity/Constraint
Adsorption Capacity	x_{m} > 0 and K > 0
Concentration Gradient	0 < y^{} < y_{0}*
Dilute System Assumption	y_{0} \leq 0.01 (mass fraction)
Thermodynamic Feasibility	x_{1} - x_{0} > 0

The mass balance for a single-stage batch system is based on the principle that the solute (adsorbate) removed from the liquid phase is equal to the solute gained by the solid adsorbent. The fundamental equation is expressed as:

\( G (y_{0} - y^{*}) = L (x_{1} - x_{0}) \)
\( G \) represents the mass of the adsorbate-free liquid (for dilute systems, this is approximately the total batch mass).
\( y_{0} \) and \( y^{*} \) are the initial and target (equilibrium) mass fractions of the adsorbate in the liquid.
\( L \) is the mass of the adsorbent used.
\( x_{0} \) and \( x_{1} \) are the initial and equilibrium mass fractions of the adsorbate on the adsorbent.

To simplify the design calculations for batch adsorption, process engineers generally assume the following:

The system reaches a state of thermodynamic equilibrium.
The temperature and pressure remain constant throughout the process.
The adsorbent is initially clean, meaning \( x_{0} \) is equal to zero.
The liquid volume (and mass, for incompressible liquids) remains constant during the adsorption cycle (valid for dilute systems).
The adsorption isotherm (e.g., Langmuir) accurately describes the equilibrium relationship.

The adsorption isotherm provides the functional relationship between the equilibrium solid loading \( x_{1} \) and the equilibrium liquid concentration \( y^{*} \), which is required to solve the mass balance equation. Common approaches include:

Using the Langmuir isotherm for monolayer coverage on a homogeneous surface: \( x_{1} = \frac{x_{m} K y^{*}}{1 + K y^{*}} \).
Using the Freundlich isotherm for adsorption on heterogeneous surfaces: \( x_{1} = k_{F} (y^{*})^{1/n} \).
Substituting the isotherm equation into the mass balance \( G (y_{0} - y^{*}) = L (x_{1}(y^{*}) - x_{0}) \) allows one to solve for the required adsorbent dosage \( L \) to achieve a target final concentration \( y^{*} \), or to solve for the achievable \( y^{*} \) given a specific \( L \).

Worked Example: Single-Stage Batch Adsorption for Bitterness Reduction

A 1000 kg batch of fruit juice has an initial bitterness concentration of 40 mg/kg (ppm). To improve taste, a process engineer needs to reduce this to 10 mg/kg using fresh polyamide powder as an adsorbent. The adsorption equilibrium follows a Langmuir isotherm. Determine the mass of adsorbent required for a single-stage batch process.

Knowns (Input Parameters):

Mass of liquid batch (approximating adsorbate-free liquid mass), \( G = 1000.0 \, \text{kg} \)
Initial adsorbate concentration in liquid, \( y_{0} = 4.0 \times 10^{-5} \, \text{kg adsorbate/kg liquid} \) (from 40 ppm: \( 40 \, \text{mg/kg} = 40 \times 10^{-6} \, \text{kg/kg} \))
Target final adsorbate concentration in liquid, \( y^{*} = 1.0 \times 10^{-5} \, \text{kg adsorbate/kg liquid} \) (from 10 ppm)
Initial adsorbate loading on solid, \( x_{0} = 0.0 \, \text{kg adsorbate/kg adsorbent} \) (fresh adsorbent)
Langmuir maximum capacity, \( x_{m} = 0.120 \, \text{kg adsorbate/kg adsorbent} \)
Langmuir equilibrium constant, \( K = 8.0 \times 10^{4} \, \text{kg liquid/kg adsorbate} \) (Note: Derived from \( K = 0.080 \, \text{L/mg} \times (1\, \text{kg}/10^{6}\, \text{mg}) \times (1000\, \text{L} / \rho_{\text{liq}} \, \text{kg}) \). Assuming liquid density \( \rho_{\text{liq}} \approx 1 \, \text{kg/L} \), \( K \approx 80,000 \, \text{L/kg} \) which is dimensionally equivalent to \( \text{kg liquid/kg adsorbate} \) for dilute systems.)

Step-by-Step Calculation:

Define Inputs: The known values are listed above, all in consistent units.
Calculate Equilibrium Solid Loading \( x_{1} \): Using the Langmuir isotherm equation: \[ x_{1} = \frac{x_{m} \cdot K \cdot y^{*}}{1 + K \cdot y^{*}} \] Substitute the values: \[ x_{1} = \frac{(0.120) \cdot (8.0 \times 10^{4}) \cdot (1.0 \times 10^{-5})}{1 + (8.0 \times 10^{4}) \cdot (1.0 \times 10^{-5})} = \frac{0.09600}{1 + 0.80000} = \frac{0.09600}{1.80000} = 0.053333 \, \text{kg adsorbate/kg adsorbent} \] (Here, numerator \( = 0.09600 \) and denominator \( = 1.80000 \), leading to \( x_{1} = 0.053333 \, \text{kg/kg} \).)
Apply Material Balance: The mass of adsorbate removed from the liquid equals the mass adsorbed on the solid: \[ G (y_{0} - y^{*}) = L (x_{1} - x_{0}) \] Solve for \( L \): \[ L = \frac{G (y_{0} - y^{*})}{x_{1} - x_{0}} \] Substitute the values: \[ L = \frac{1000.0 \cdot (4.0 \times 10^{-5} - 1.0 \times 10^{-5})}{0.053333 - 0.0} = \frac{1000.0 \cdot (3.0 \times 10^{-5})}{0.053333} = \frac{0.03000}{0.053333} = 0.5625 \, \text{kg} \] (The difference \( y_{0} - y^{*} = 3.0 \times 10^{-5} \, \text{kg/kg} \), and \( x_{1} - x_{0} = 0.053333 \, \text{kg/kg} \).)

Final Answer: The required mass of fresh polyamide adsorbent is \( L = 0.5625 \, \text{kg} \). For practical purposes, this might be reported as \( 0.563 \, \text{kg} \).

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