12.3 Adsorption, Membrane Separations, and Drying Operations

12.3 Adsorption, Membrane Separations, and Drying Operations

Beyond traditional gas-liquid and liquid-liquid contacting towers, chemical engineers frequently deploy specialized rate-controlled and solid-fluid separation operations. Adsorption captures trace pollutants or purifies products via solid surface forces; membrane separations achieve molecular-scale separations without phase changes; and drying removes volatile liquids from solid matrices to produce stable particulate products. Each of these three operations possesses unique governing equations, transport mechanisms, and design pitfalls tested on the NCEES PE Chemical Exam.

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1. Adsorption Equilibria and Fixed-Bed Dynamics

Adsorption transfers solute molecules (adsorbates) from a gas or liquid phase onto the surface of a porous solid (adsorbent). Distinguish physical adsorption (physisorption), driven by weak van der Waals forces and easily reversible with heat, from chemical adsorption (chemisorption), which involves valence electron sharing, high activation energy, and formation of a surface monolayer.

Industrial Adsorbents

• Activated Carbon: Hydrophobic; non-polar; specific surface area $600-1,500\text{ m}^2\text{/g}$. Standard for VOC abatement, organic wastewater treatment, and solvent recovery.
• Molecular Sieve Zeolites: Crystalline aluminosilicates with uniform molecular pore apertures ($3\text{ Å}, 4\text{ Å}, 5\text{ Å}, 10\text{ Å}$). Highly polar; widely used for deep gas drying, natural gas dehydration, and air separation ($N_2/O_2$).
• Silica Gel & Activated Alumina: Hydrophilic metal oxides; high affinity for water vapor and polar compounds.

1. The Langmuir Adsorption Isotherm

The Langmuir model assumes reversible adsorption onto a fixed number of equivalent, localized surface sites with zero lateral interaction between adsorbed molecules (monolayer coverage):

$$q = \frac{q_{max} \cdot K \cdot C}{1 + K \cdot C} \quad (\text{Liquid Systems})$$ $$q = \frac{q_{max} \cdot b \cdot P}{1 + b \cdot P} \quad (\text{Gas Systems})$$

Where:

• $q$ = equilibrium adsorbent loading ($\text{mg solute / g adsorbent}$ or $\text{mol/kg}$).
• $q_{max}$ = maximum theoretical monolayer saturation capacity ($\text{mg/g}$ or $\text{mol/kg}$).
• $K$ or $b$ = Langmuir equilibrium adsorption constant ($\text{L/mg}$ or $\text{atm}^{-1}$), proportional to the energy of adsorption.
• $C$ or $P$ = fluid solute equilibrium concentration ($\text{mg/L}$) or partial pressure ($\text{atm}$).

Linearization for Parameter Extraction:

$$\frac{C}{q} = \frac{1}{q_{max} K} + \frac{1}{q_{max}} C$$

A plot of $C/q$ (y-axis) versus $C$ (x-axis) yields a straight line where:

$$\text{Slope} = \frac{1}{q_{max}} \implies q_{max} = \frac{1}{\text{Slope}}$$ $$\text{y-Intercept} = \frac{1}{q_{max} K} \implies K = \frac{\text{Slope}}{\text{y-Intercept}}$$

2. The Freundlich Adsorption Isotherm

The empirical Freundlich model describes multilayer adsorption on heterogeneous surfaces with non-uniform site energies:

$$q = K_F \cdot C^{1/n}$$

Where:

• $K_F$ = Freundlich capacity parameter ($\text{(mg/g)(L/mg)}^{1/n}$).
• $n$ = heterogeneity factor ($n > 1$ represents favorable adsorption; $1/n < 1$).

Linearization:

$$\ln(q) = \ln(K_F) + \frac{1}{n} \ln(C)$$

A log-log plot of $\ln(q)$ versus $\ln(C)$ yields slope $1/n$ and y-intercept $\ln(K_F)$.

   Adsorption Isotherms                     Fixed-Bed Breakthrough Curve
   q ^                                      C/C_0 ^
     |    Langmuir (Asymptotic q_max)         1.0 |                   /--- Exhaustion (t_e)
     |       /------------------                  |                  /    (C/C_0 = 0.95)
     |      /                                     |                 / 
     |     /  Freundlich (Multilayer)             |       MTZ      /  
     |    /     /                                 |    ========   /   
     |   /     /                                  |            /  
     |  /     /                                   |  Breakthrough (t_b)
     | /     /                                0.0 |  (C/C_0 = 0.05)
     +------------------------> C                 +-----------------------> Time (t)

Fixed-Bed Adsorption Dynamics & Breakthrough

In industrial fixed-bed adsorbers, fluid containing solute at concentration $C_0$ flows downward through a bed of packed solid adsorbent of total length $L$.

• Mass Transfer Zone (MTZ): The active band within the bed where mass transfer occurs. Upstream of the MTZ, the adsorbent is completely saturated ($q = q_0$). Downstream of the MTZ, the bed remains virgin ($q = 0$).
• Breakthrough Time ($t_b$): The operating time at which effluent solute concentration reaches an allowable regulatory or process leakage limit (typically $C/C_0 = 0.05$, or $5\%$ of inlet concentration).
• Exhaustion Time ($t_e$): The point where the bed is essentially saturated ($C/C_0 = 0.95$, or $95\%$).
• Length of the Mass Transfer Zone ($L_{MTZ}$):

$$L_{MTZ} = L \left[ \frac{t_e - t_b}{t_e - (1 - f)(t_e - t_b)} \right]$$

Where $f$ is the fractional saturation of the MTZ (typically $f \approx 0.50$ for symmetrical S-shaped breakthrough curves). The length of bed unused at breakthrough is $LUB = (1 - f) L_{MTZ}$. A narrower MTZ yields higher bed utilization before breakthrough.

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2. Membrane Separations: Gas Permeation & Reverse Osmosis

Membranes separate components by using a semipermeable synthetic barrier that permits selective transport of certain chemical species while restricting others. Transport through non-porous dense polymeric membranes follows the solution-diffusion mechanism:

1. Sorption of penetrant molecules into the high-pressure upstream membrane surface.
2. Molecular diffusion across the dense selective layer driven by chemical potential gradient.
3. Desorption from the low-pressure downstream membrane surface into the permeate.

1. Gas Permeation

For a gas component $i$, the volumetric permeation flux ($J_i$, in $\text{cm}^3\text{(STP)/(cm}^2\cdot\text{s)}$ or $\text{mol/(m}^2\cdot\text{s)}$) through a dense selective layer of thickness $l$ is:

$$J_i = \frac{P_{M,i}}{l} \left( p_{i,feed} - p_{i,perm} \right) = \left( \frac{P_{M,i}}{l} \right) \Delta p_i$$

Where:

• $P_{M,i}$ = membrane permeability coefficient ($P_{M,i} = D_i \cdot S_i$, the product of diffusivity $D_i$ and solubility $S_i$). Standard unit is the Barrer ($1\text{ Barrer} = 10^{-10}\text{ cm}^3\text{(STP)}\cdot\text{cm}/(\text{cm}^2\cdot\text{s}\cdot\text{cmHg})$).
• $P_{M,i} / l$ = permeance, expressed in Gas Permeation Units (GPU, where $1\text{ GPU} = 10^{-6}\text{ cm}^3\text{(STP)/(cm}^2\cdot\text{s}\cdot\text{cmHg})$).
• $\Delta p_i = x_{i,f} P_{feed} - y_{i,p} P_{perm}$ = partial pressure driving force across the membrane.

The ideal membrane selectivity (separation factor) for gas $A$ over gas $B$ is:

$$\alpha_{A/B} = \frac{P_{M,A}}{P_{M,B}} = \left( \frac{D_A}{D_B} \right) \left( \frac{S_A}{S_B} \right)$$

2. Reverse Osmosis (RO) & Osmotic Pressure Balances

In liquid systems containing dissolved salts or macromolecules, natural osmosis causes pure solvent water to migrate across a semipermeable membrane into the concentrated brine solution until equilibrium osmotic pressure is established.

Osmotic Pressure ($\Pi$) via the van 't Hoff Equation: For dilute solutions:

$$\Pi = i \cdot M \cdot R \cdot T = \sum c_k R T$$

Where:

• $i$ = dimensionless van 't Hoff dissociation factor (e.g., $i \approx 2$ for $\text{NaCl} \to \text{Na}^+ + \text{Cl}^-$; $i \approx 3$ for $\text{Na}_2\text{SO}_4$).
• $M$ = molar salt concentration ($\text{mol/L}$ or $\text{kmol/m}^3$).
• $R$ = ideal gas constant ($0.08314\text{ L}\cdot\text{bar/(mol}\cdot\text{K)}$ or $8.314\text{ kPa}\cdot\text{m}^3\text{/(kmol}\cdot\text{K)}$).
• $T$ = absolute temperature ($\text{K}$).

Water Flux ($J_w$): To force water from the saline solution into the fresh water permeate (Reverse Osmosis), an applied hydraulic pressure difference ($\Delta P$) exceeding the osmotic pressure difference ($\Delta \Pi$) must be imposed:

$$J_w = A_w \cdot (\Delta P - \Delta \Pi) = A_w \cdot \left[ (P_f - P_p) - (\Pi_f - \Pi_p) \right]$$

Where:

• $A_w$ = water permeability coefficient (permeance) of the membrane ($\text{m}^3/(\text{m}^2\cdot\text{s}\cdot\text{bar})$ or $\text{LMH/bar}$, where $\text{LMH} = \text{L}/(\text{m}^2\cdot\text{h})$).
• $\Delta P = P_{feed} - P_{permeate}$ = applied transmembrane hydraulic pressure difference.
• $\Delta \Pi = \Pi_{feed} - \Pi_{permeate}$ = transmembrane osmotic pressure difference.

Solute (Salt) Flux ($J_s$): Salt transport across a non-porous RO membrane occurs via molecular diffusion, driven strictly by the concentration gradient, completely independent of the applied hydraulic pressure $\Delta P$:

$$J_s = B_s \cdot (C_f - C_p)$$

Where $B_s$ is the solute permeability coefficient ($\text{m/s}$ or $\text{LMH}$), and $C_f, C_p$ are feed and permeate solute concentrations ($\text{mg/L}$).

Permeate Solute Concentration ($C_p$) and Salt Rejection ($R_j$): Since solute and solvent flow simultaneously:

$$C_p = \frac{J_s}{J_w} = \frac{B_s (C_f - C_p)}{J_w} \implies C_p = C_f \left( \frac{B_s}{J_w + B_s} \right)$$

Observed Salt Rejection ($R_j$):

$$R_j \equiv 1 - \frac{C_p}{C_f} = \frac{J_w}{J_w + B_s}$$

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3. Drying Operations & Psychrometry

Drying removes volatile liquids (typically water) from a solid, wet cake, or slurry by thermal vaporization to produce a dry solid product.

1. Solid Moisture Content Definitions

• Wet Basis Moisture ($x$): Mass fraction of water relative to total wet solid: $$x = \frac{m_w}{m_w + m_s}$$
• Dry Basis Moisture ($X$): Mass ratio of water relative to invariant dry solid: $$X = \frac{m_w}{m_s}$$
• Conversion Formulas: $$X = \frac{x}{1 - x}, \quad x = \frac{X}{1 + X}$$
• Free Moisture ($X_{free}$): The moisture evaporable under the prevailing drying air conditions: $$X_{free} = X - X_{eq}$$ Where $X_{eq}$ is the equilibrium moisture content determined by the solid's sorption isotherm at the air's relative humidity and temperature. Moisture below $X_{eq}$ cannot be removed by air drying.

   Drying Rate Curve (R vs. X)
   Drying Rate (R) ^
                   |               Constant Rate Period
              R_c  |               +--------------------+
                   |              /|                    |
                   |             / |
                   |            /  |
                   |  Falling  /   |
                   |   Rate   /    |
                   |  Period /     |
                   |        /      |
                   +-------+-------+--------------------+---> Dry Basis Moisture (X)
                          X_eq    X_c                  X_1
                          (Equil) (Critical)           (Initial)

2. The Drying Rate Curve & Drying Kinetics

Batch drying under constant external air conditions (constant temperature, humidity, and air velocity) exhibits two distinct regimes separated by the Critical Moisture Content ($X_c$):

Period 1: Constant Rate Period ($X > X_c$)

• Water evaporates freely from a continuous, saturated liquid film covering the outer solid surface.
• The drying rate is entirely externally mass- and heat-transfer controlled (independent of solid thickness or porosity).
• The solid surface temperature remains constant at the wet-bulb temperature ($T_s = T_{wb}$) of the drying air.
• The constant drying rate $R_c$ ($\text{kg water/(m}^2\cdot\text{h)}$) is given by: $$R_c = \frac{h (T_{db} - T_{wb})}{\lambda_w} = k_y (Y_{sat} - Y_{air})$$ Where $h$ is the convective heat transfer coefficient, $T_{db}$ is the dry-bulb temperature, $T_{wb}$ is the wet-bulb temperature, and $\lambda_w$ is latent heat of vaporization.
• Drying time in the constant rate period from initial moisture $X_1$ to critical moisture $X_c$: $$t_c = \frac{m_s}{A \cdot R_c} (X_1 - X_c)$$ Where $m_s$ is bone-dry solid mass ($\text{kg}$) and $A$ is exposed drying surface area ($\text{m}^2$).

Period 2: Falling Rate Period ($X_{eq} < X < X_c$)

• At $X_c$, the continuous surface liquid film ruptures; dry patches appear. Liquid must diffuse or wick through internal solid pores to reach the drying surface.
• Drying is internally mass-transfer controlled; the solid surface temperature rises toward the air dry-bulb temperature ($T_{db}$).
• In the standard linear falling rate model, the drying rate is directly proportional to free moisture: $$R(X) = R_c \left( \frac{X - X_{eq}}{X_c - X_{eq}} \right)$$
• Drying time in the falling rate period from $X_c$ down to target final moisture $X_2$: $$t_f = -\frac{m_s}{A} \int_{X_c}^{X_2} \frac{dX}{R(X)} = \frac{m_s (X_c - X_{eq})}{A \cdot R_c} \ln\left( \frac{X_c - X_{eq}}{X_2 - X_{eq}} \right)$$

Total Batch Drying Time ($t_{total}$)

$$t_{total} = t_c + t_f = \frac{m_s}{A \cdot R_c} \left[ (X_1 - X_c) + (X_c - X_{eq}) \ln\left( \frac{X_c - X_{eq}}{X_2 - X_{eq}} \right) \right]$$

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4. Summary Comparison Table: Adsorption, Membrane, and Drying Parameters

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5. Comprehensive Worked Numerical Examples

Example A: Industrial Reverse Osmosis Desalination System Sizing

An industrial reverse osmosis facility desalinates $T = 25.0^\circ\text{C}$ ($298.15\text{ K}$) brackish feed containing $C_f = 4,000.0\text{ mg/L}$ of sodium chloride ($\text{NaCl}$, $M_w = 58.44\text{ g/mol}$) to produce high-purity boiler feedwater.

Operating & Membrane Parameters:

• Feed hydraulic pressure: $P_f = 25.0\text{ bar}$; Permeate pressure: $P_p = 1.0\text{ bar}$
• Feed osmotic pressure: $\Pi_f = 3.20\text{ bar}$; Permeate osmotic pressure: $\Pi_p \approx 0.10\text{ bar}$
• Membrane water permeance: $A_w = 2.00 \times 10^{-7}\text{ m}^3/(\text{m}^2\cdot\text{s}\cdot\text{bar}) = 0.720\text{ LMH/bar}$
• Membrane salt permeability: $B_s = 6.00 \times 10^{-8}\text{ m/s} = 0.216\text{ LMH}$

Calculate:

1. Net transmembrane driving pressure ($\Delta P - \Delta \Pi$) in $\text{bar}$.
2. Pure water flux $J_w$ in $\text{LMH}$ and in $\text{m/s}$.
3. Permeate salt concentration $C_p$ in $\text{mg/L}$.
4. Observed salt rejection $R_j$.

Solution:

$$\Delta P = P_f - P_p = 25.0 - 1.0 = 24.0\text{ bar}$$ $$\Delta \Pi = \Pi_f - \Pi_p = 3.20 - 0.10 = 3.10\text{ bar}$$ $$\text{Net Driving Pressure} = \Delta P - \Delta \Pi = 24.0 - 3.10 = \mathbf{20.90\text{ bar}}$$

Water flux calculation:

$$J_w = A_w \cdot (\Delta P - \Delta \Pi) = (0.720\text{ LMH/bar}) \times 20.90\text{ bar} = \mathbf{15.048\text{ LMH}}$$ $$J_w = \frac{15.048\text{ L/(m}^2\cdot\text{h)}}{1,000\text{ L/m}^3 \times 3,600\text{ s/h}} = \mathbf{4.180 \times 10^{-6}\text{ m/s}}$$

Permeate salt concentration:

$$C_p = C_f \left( \frac{B_s}{J_w + B_s} \right) = 4,000.0\text{ mg/L} \times \left( \frac{0.216\text{ LMH}}{15.048 + 0.216\text{ LMH}} \right) = 4,000.0 \times \left( \frac{0.216}{15.264} \right) = \mathbf{56.60\text{ mg/L}}$$

Salt rejection:

$$R_j = 1 - \frac{C_p}{C_f} = 1 - \frac{56.60}{4,000.0} = 1 - 0.01415 = \mathbf{0.98585} \quad (\mathbf{98.59\%})$$

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Example B: Batch Tray Dryer Sizing and Drying Time

A batch tray dryer processes a porous pharmaceutical cake containing $m_s = 400.0\text{ kg}$ of bone-dry solid. The exposed drying surface area is $A = 16.0\text{ m}^2$. The wet cake enters at an initial moisture content of $x_1 = 0.200$ ($20.0\text{ wt}\%$ wet basis) and must be dried to a final moisture content of $x_2 = 0.040$ ($4.0\text{ wt}\%$ wet basis).

Drying Data:

• Constant drying rate: $R_c = 2.00\text{ kg water/(m}^2\cdot\text{h)}$
• Critical moisture content: $X_c = 0.100\text{ kg water/kg dry solid}$
• Equilibrium moisture content: $X_{eq} = 0.010\text{ kg water/kg dry solid}$

Calculate:

1. Initial dry-basis moisture $X_1$ and final dry-basis moisture $X_2$.
2. Drying time in the constant rate period ($t_c$) in hours.
3. Drying time in the falling rate period ($t_f$) in hours.
4. Total batch drying time ($t_{total}$) in hours.

Solution:

$$X_1 = \frac{x_1}{1 - x_1} = \frac{0.200}{1 - 0.200} = \frac{0.200}{0.800} = \mathbf{0.2500\text{ kg/kg dry solid}}$$ $$X_2 = \frac{x_2}{1 - x_2} = \frac{0.040}{1 - 0.040} = \frac{0.040}{0.960} = \mathbf{0.04167\text{ kg/kg dry solid}}$$

Because $X_1 (0.250) > X_c (0.100) > X_2 (0.04167)$, drying spans both constant and falling rate regimes.

Constant rate drying time:

$$t_c = \frac{m_s}{A \cdot R_c} (X_1 - X_c) = \frac{400.0\text{ kg}}{(16.0\text{ m}^2) \times (2.00\text{ kg/(m}^2\cdot\text{h)})} \times (0.2500 - 0.100) = \frac{400.0}{32.0} \times 0.1500 = \mathbf{1.875\text{ hours}}$$

Falling rate drying time:

$$X_c - X_{eq} = 0.100 - 0.010 = 0.0900$$ $$X_2 - X_{eq} = 0.04167 - 0.010 = 0.03167$$ $$\frac{X_c - X_{eq}}{X_2 - X_{eq}} = \frac{0.0900}{0.03167} = 2.8418$$ $$\ln(2.8418) = 1.04447$$ $$t_f = \frac{m_s (X_c - X_{eq})}{A \cdot R_c} \ln\left( \frac{X_c - X_{eq}}{X_2 - X_{eq}} \right) = \frac{400.0 \times 0.0900}{32.0} \times 1.04447 = 1.125 \times 1.04447 = \mathbf{1.175\text{ hours}}$$

Total drying time:

$$t_{total} = t_c + t_f = 1.875\text{ h} + 1.175\text{ h} = \mathbf{3.050\text{ hours}} \quad (3\text{ h } 3\text{ min})$$

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6. Critical PE Exam Traps & Pitfalls

Trap 1: Using Wet Basis Moisture in Drying Equations
The formula $t_c = (m / (A R_c)) (x_1 - x_2)$ using wet-basis moisture $x$ is completely wrong. Total mass changes as water evaporates, so $m$ is not constant. You MUST use bone-dry solid mass $m_s$ and dry-basis moistures $X_1, X_c, X_2$.

Trap 2: Ignoring Osmotic Pressure Direction in Reverse Osmosis
In RO calculations, water flux is driven by $\Delta P - \Delta \Pi$. If the applied hydraulic pressure does not exceed the osmotic pressure of the feed ($\Delta P < \Delta \Pi$), water will flow backward (forward osmosis), diluting the feed. Always verify that $\Delta P > \Delta \Pi$.

Trap 3: Assuming Salt Flux Increases with Applied Hydraulic Pressure
In reverse osmosis, salt transport ($J_s = B_s \Delta C$) is independent of applied hydraulic pressure $\Delta P$. Increasing pump discharge pressure increases pure water flux, which dilutes the permeate salt concentration and improves salt rejection. Never calculate an increased salt flux when $\Delta P$ is raised.

Trap 4: Misinterpreting Solid Temperature During Drying
During the constant rate drying period, the solid surface is saturated with liquid and remains at the wet-bulb temperature ($T_{wb}$) of the drying air, NOT the dry-bulb temperature ($T_{db}$). Heat-sensitive materials can safely be dried using hot air during this period because evaporative cooling shields the product.