10.4 Continuous Contact Methods, Membrane Separations, and Evaporative Operations

10.4 Continuous Contact Methods and Membrane Separations

While trayed columns represent stage-wise contact, many separation processes utilize continuous contact, such as packed columns. Additionally, membrane separations have become essential unit operations in water purification and gas separation. The FE Chemical exam covers packed column design equations (HTU, NTU, HETP) and membrane performance calculations (flux, rejection, and selectivity).

Packed Column Design (Continuous Contact)

In packed columns, liquid and gas contact each other continuously along the height of the column rather than on discrete trays. The total height of the packing ($Z$) required for a separation is determined by:

$$Z = \text{HTU} \times \text{NTU}$$

Where:

• HTU (Height of a Transfer Unit): Characterizes the mass transfer efficiency of the packing material. A smaller HTU value indicates a more efficient packing.
• NTU (Number of Transfer Units): Represents the difficulty of the separation, determined by the change in concentration relative to the average driving force.

Gas-Phase Transfer Units

When design is based on gas-phase resistance:

$$Z = H_{OG} \times N_{OG}$$

The Height of an Overall Gas Transfer Unit ($H_{OG}$) is defined as:

$$H_{OG} = \frac{G'}{K_y a P}$$

Where:

• $G'$ is the molar gas flux per unit area ($\text{kmol}/(\text{m}^2\cdot\text{s})$).
• $K_y a$ is the overall gas-phase volumetric mass transfer coefficient ($\text{kmol}/(\text{m}^3\cdot\text{s}\cdot\Delta y)$).
• $P$ is the total operating pressure.

The Number of Overall Gas Transfer Units ($N_{OG}$) is given by:

$$N_{OG} = \int_{y_2}^{y_1} \frac{dy}{y - y^*}$$

For dilute systems where both the operating line and the equilibrium line ($y^* = m x$) are straight, this integrates to:

$$N_{OG} = \frac{y_1 - y_2}{\Delta y_{LM}}$$

Where the log-mean concentration driving force ($\Delta y_{LM}$) is:

$$\Delta y_{LM} = \frac{(y_1 - y_1^*) - (y_2 - y_2^*)}{\ln\left(\frac{y_1 - y_1^*}{y_2 - y_2^*}\right)}$$

Where $y_1^* = m x_1$ and $y_2^* = m x_2$ are the vapor compositions in equilibrium with the liquid at the bottom and top of the column, respectively.

HETP (Height Equivalent to a Theoretical Plate)

HETP is an empirical parameter used to relate packed column height to the number of equivalent theoretical stages ($N_{\text{stages}}$):

$$Z = N_{\text{stages}} \times \text{HETP}$$

For dilute systems, HETP is related to $H_{OG}$ by the absorption factor $A = L / (m G)$:

$$\text{HETP} = H_{OG} \frac{\ln(1/A)}{(1/A) - 1}$$

---

Membrane Separations

Membrane processes use a semi-permeable barrier to separate components based on size, charge, or solubility differences.

Reverse Osmosis (RO)

RO is a pressure-driven membrane process used primarily for water purification and desalination. The membrane allows water to pass while rejecting salts.

Water Flux ($J_w$)

The volumetric or mass flux of water through the membrane depends on the net pressure driving force:

$$J_w = A (\Delta P - \Delta \pi)$$

Where:

• $J_w$ is the water flux ($\text{kg}/(\text{m}^2\cdot\text{s})$ or $\text{L}/(\text{m}^2\cdot\text{h})$).
• $A$ is the water permeability coefficient of the membrane.
• $\Delta P = P_{\text{feed}} - P_{\text{permeate}}$ is the transmembrane hydraulic pressure difference.
• $\Delta \pi = \pi_{\text{feed}} - \pi_{\text{permeate}}$ is the transmembrane osmotic pressure difference.

Osmotic Pressure ($\pi$)

Osmotic pressure is estimated using the van 't Hoff equation:

$$\pi = i C R T$$

Where $i$ is the van 't Hoff factor (number of ions formed by the solute, e.g., 2 for $\text{NaCl}$, 1 for glucose), $C$ is molar concentration ($\text{mol/L}$), $R$ is the gas constant, and $T$ is absolute temperature.

Solute Flux ($J_s$) and Rejection ($R$)

The transport of solute is driven solely by concentration difference:

$$J_s = B (C_f - C_p)$$

Where $B$ is the solute permeability coefficient, and $C_f, C_p$ are the solute concentrations in the feed and permeate, respectively. The solute rejection ($R$) is defined as:

$$R = 1 - \frac{C_p}{C_f}$$

Gas Permeation

Gas permeation separates gas mixtures based on differences in their rate of permeation through a polymer membrane. The flux of component $i$ is:

$$J_i = \frac{P_{M,i}}{l} (p_{f,i} - p_{p,i})$$

Where $P_{M,i}$ is the membrane permeability of component $i$, $l$ is the membrane thickness, and $p_{f,i}, p_{p,i}$ are the partial pressures of component $i$ in the feed and permeate streams. The membrane selectivity ($\alpha_{ij}$) is the ratio of permeabilities:

$$\alpha_{ij} = \frac{P_{M,i}}{P_{M,j}}$$

Ultrafiltration (UF)

UF separates macromolecular solutes. Due to the high molecular weight of these solutes, osmotic pressure is negligible. However, at high operating pressures, a concentrated solute layer forms on the membrane surface, creating a gel layer. This phenomenon is known as concentration polarization.

Under gel-limited conditions, the flux becomes independent of pressure and is described by the gel-polarization model:

$$J_v = k_c \ln\left(\frac{C_g - C_p}{C_b - C_p}\right)$$

Where $J_v$ is the volumetric flux, $k_c$ is the mass transfer coefficient, $C_g$ is the gel layer concentration at the membrane surface, $C_b$ is the bulk solute concentration, and $C_p$ is the permeate concentration.

---

Worked Example: Reverse Osmosis Performance

Problem: A reverse osmosis system desalines water containing $0.15\text{ M}\text{ NaCl}$ at $298\text{ K}$. The feed pressure is $50.0\text{ bar}$ ($5000\text{ kPa}$), and the permeate side is at atmospheric pressure ($1.0\text{ bar}$ or $100\text{ kPa}$). The water permeability coefficient of the membrane is $A = 2.0 \times 10^{-6}\text{ kg}/(\text{m}^2\cdot\text{s}\cdot\text{kPa})$. Assume the membrane achieves $99\%$ salt rejection ($R = 0.99$). The gas constant is $R = 8.314\text{ kPa}\cdot\text{L}/(\text{mol}\cdot\text{K})$, and the van 't Hoff factor for $\text{NaCl}$ is $i = 2.0$.

1. Calculate the osmotic pressure of the feed solution ($\pi_f$).
2. Calculate the osmotic pressure of the permeate solution ($\pi_p$).
3. Calculate the water flux ($J_w$) in $\text{kg}/(\text{m}^2\cdot\text{s})$.

Solution:

1. Compute the feed osmotic pressure:

$$\pi_f = i C_f R T = (2.0) (0.15\text{ mol/L}) (8.314\text{ kPa}\cdot\text{L}/(\text{mol}\cdot\text{K})) (298\text{ K})$$ $$\pi_f = 0.30 \times 8.314 \times 298 = 743.27\text{ kPa} \approx 7.43\text{ bar}$$

1. Determine the permeate concentration using the rejection:

$$R = 1 - \frac{C_p}{C_f} \implies C_p = C_f (1 - R) = 0.15 \times (1 - 0.99) = 0.0015\text{ M}$$

Compute the permeate osmotic pressure:

$$\pi_p = i C_p R T = (2.0) (0.0015) (8.314) (298) = 7.43\text{ kPa} \approx 0.07\text{ bar}$$

1. Calculate the net osmotic pressure difference:

$$\Delta \pi = \pi_f - \pi_p = 743.27 - 7.43 = 735.84\text{ kPa}$$

Calculate the net hydraulic pressure difference:

$$\Delta P = P_f - P_p = 5000 - 100 = 4900\text{ kPa}$$

1. Compute the water flux:

$$J_w = A (\Delta P - \Delta \pi) = (2.0 \times 10^{-6}\text{ kg}/(\text{m}^2\cdot\text{s}\cdot\text{kPa})) (4900 - 735.84\text{ kPa})$$ $$J_w = 2.0 \times 10^{-6} \times 4164.16 = 8.33 \times 10^{-3}\text{ kg}/(\text{m}^2\cdot\text{s})$$

The water flux is $8.33 \times 10^{-3}\text{ kg}/(\text{m}^2\cdot\text{s})$.

---

Humidification, Drying, and Evaporation

These operations use vaporization and gas-phase contact to remove water or solvent from a process stream. They are grouped in the NCEES FE Chemical Mass Transfer and Separation knowledge area because the same mass-transfer coefficients and equilibrium relationships drive them.

Psychrometrics and Humidification

Psychrometrics describes the thermodynamic properties of moist air. Key definitions shown on the FE Reference Handbook psychrometric chart are:

• Dry-bulb temperature ($T_{DB}$): Air temperature measured by a standard thermometer.
• Wet-bulb temperature ($T_{WB}$): The temperature reached by evaporative cooling when air flows over a water-saturated wick. It approximates the temperature of liquid water in an adiabatic saturator.
• Dew point ($T_{DP}$): The temperature at which air becomes saturated and water just begins to condense.
• Humidity ratio ($\omega$): The mass of water vapor per mass of dry air. For ideal gases: $\omega = 0.622 \, p_v / (P - p_v)$, where $p_v$ is the partial pressure of water vapor and $P$ is total pressure.
• Relative humidity ($\phi$): The ratio of actual water vapor pressure to the saturation vapor pressure at the same dry-bulb temperature: $\phi = p_v / p_{\text{sat}}(T_{DB})$.

An adiabatic humidification process (such as a cooling tower or air washer) follows the constant-enthalpy adiabatic saturation line on the psychrometric chart. In a cooling tower, warm process water is cooled by contact with ambient air; the water evaporates, removing its latent heat of vaporization from the remaining liquid. A cooling tower's approach is $T_{\text{water,out}} - T_{WB,\text{air,in}}$, and its range is $T_{\text{water,in}} - T_{\text{water,out}}$. Smaller approach indicates better tower performance but higher capital cost.

Drying Operations

Drying removes liquid (usually water) from a wet solid by vaporization. Drying proceeds in two periods:

1. Constant-rate period: The solid surface stays saturated; drying rate is controlled by external heat and mass transfer. The rate is approximately $R_c = h (T_{DB} - T_{WB}) / \lambda$, where $h$ is the heat transfer coefficient and $\lambda$ is the latent heat of vaporization.
2. Falling-rate period: Surface unsaturation develops and the rate is controlled by internal moisture diffusion. The critical moisture content marks the transition between the two periods.

The total drying time is the sum of constant-rate and falling-rate contributions; the FE exam most often tests the constant-rate calculation, which reduces to a heat/mass transfer balance.

Evaporation

Evaporation is the concentration of a solution by vaporizing the solvent. A single-effect evaporator heats the feed with steam in a shell-and-tube heat exchanger; vapor leaves the top and concentrated liquor leaves the bottom. The mass balance is:

$$F = V + L, \quad F x_F = L x_L$$

where $F$ is the feed, $V$ is the vapor, $L$ is the concentrated liquor, and $x$ are mass fractions of solute. The energy balance is:

$$F \hat{H}_F + Q_s = V \hat{H}_V + L \hat{H}_L$$

where $Q_s$ is the heat supplied by steam. Multiple-effect evaporators reuse vapor from one effect as the heating steam for the next effect at lower pressure; the steam economy (kg vapor produced per kg steam consumed) approaches the number of effects for $N$ effects operating ideally.