Chapter 16: Introduction to Membrane Separation Processes

Membrane separation processes¹ such as gas permeation, pervaporation, reverse osmosis (RO), and ultrafiltration (UF) are not operated as equilibrium-staged processes. Instead, these separations are based on the rate at which solutes transfer though a semipermeable membrane. The key to understanding these membrane processes is the rate of mass transfer not equilibrium. Yet, despite this difference we will see many similarities in the solution methods for different flow patterns with the solution methods developed for equilibrium-staged separations. Because the analyses of these processes are often analogous to the methods used for equilibrium processes, we can use our understanding of equilibrium processes to help understand membrane separators. These membrane processes are usually either complementary or competitive with distillation, absorption, and extraction.

¹A nomenclature list for this chapter is included in the front matter of this book.

This chapter presents an introduction to the four membrane separation methods most commonly used in industry: gas permeation, RO, UF, and pervaporation. At the level of this introduction the mathematical sophistication needed to understand the membrane processes is approximately the same as that needed for the equilibrium-staged processes. A background in mass transfer will be helpful but is not essential. Detailed descriptions of these membrane separation processes are found in Baker et al. (1990), Eykamp (1997), Geankoplis (2003), Noble and Stern (1985), Mohr et al. (1988), Mulder (1996), Osada and Nakagawa (1992), Hagg (1998), Ho and Sirkar (1992), and Wankat (1990).

Some knowledge of the membrane separations will prove to be very helpful even if the engineer will usually design equilibrium-based processes. In gas permeation components selectively transfer through the membrane. Gas permeation competes with cryogenic distillation as a method to produce nitrogen gas. Absorption and gas permeation are competitive methods for removing carbon dioxide from gas streams. In RO a tight membrane that rejects essentially all dissolved components is used. The water dissolves in the membrane and passes through under a pressure difference up to 6000 kPa (800 psi) (Li and Kulkarni, 1997). RO has in many cases displaced distillation as a method for desalinating seawater. UF membranes are fabricated to pass low molecular weight molecules and to retain high molecular weight molecules and particulates. The pressure difference, 70 to 1,400 kPa (10-100 psi) is more modest than in RO. UF is a useful method for separating proteins and other large molecules that essentially have no vapor pressure and thus, cannot be distilled. UF competes with extraction as a separation method for biochemicals. In pervaporation the feed is a liquid while the permeate product is removed as a vapor. Pervaporation is used as a method to break azeotropes and is often coupled with distillation columns. Since the membrane separations are based on different physicochemical properties than the equilibrium-staged separations, the membrane methods can often perform separations such as separation of azeotropic mixtures or separation of nonvolatile components, which cannot be done by distillation or other equilibrium-based separations.

There are several other membrane processes that are in commercial use but are not covered in this chapter. In dialysis, small molecules in a liquid diffuse through a membrane because of a concentration driving force (Wankat, 1990). The major application is hemodialysis for treatment of people whose kidneys do not function properly. Electrodialysis (ED) uses an electrical field to force cations through cation exchange membranes and anions through anion exchange membranes (Wankat, 1990). The membranes are alternated in a stack, and every alternate region becomes concentrated or diluted. ED is used for desalination of brackish water and in the food industry. Vapor permeation is similar to gas permeation except vapors that are easily condensed are processed (Huang, 1991). This process has not met its potential partly because of difficulties with condensation of liquid. Liquid membranes use a layer of liquid instead of a solid polymer to achieve the separation (Wankat, 1990). Liquid membrane systems can be operated as countercurrent processes, and, to some extent, compete with extraction. Microfiltration is similar to UF but is used for particles between the sizes processed by UF and normal filtration (Noble and Terry, 2004). Nanofiltration removes particles between those removed by RO and UF and is essentially a loose RO membrane (Wankat, 1990). The design procedures developed for RO and UF can be applied to microfiltration and nanofiltration. The entire spectrum of membrane separations is summarized in Table 16-1. Nanofiltration is not listed separately but is the same as RO except that Δp is from 0.3 to 3 MPa and the approximate size retained is 8-50Å.

Table 16-1. Properties of membrane separation systems (Drioli and Romano, 2001; Noble and Terry, 2004; Wankat, 1990)

Key: CA = cellulose acetate

16.1 MEMBRANE SEPARATION EQUIPMENT

A membrane is a physical barrier between two fluids (feed side and product side) that selectively allows certain components of the feed fluid to pass. The fluid that passes through the membrane is called permeate and the fluid retained on the feed side is called retentate. The equipment needed for the separation is deceptively simple. It consists of the membrane plus the container to hold the two fluids. The simplest arrangement is to use a stirred tank that is separated into two volumes via a membrane. Stirred-tank systems are used in laboratories but not commonly in large-scale separations.

The common commercial geometries are shown in Figure 16-1 (Leeper et al., 1984). More extensive construction details are shown by Baker et al. (1990) and Eykamp (1997). The plate-and-frame system (Figure 16-1A) is similar to a parallel plate heat exchanger or a plate-and-frame filter press, except that the filter cloth is replaced by flat sheets of membranes. This design is used for food processing applications where rigorous cleaning by disassembly may be required, for electrodialysis (which involves passing a current through the membrane), and for membrane materials that are difficult to form into more complicated shapes.

Figure 16-1. Schematic diagrams of common industrial membrane modules; A) plate-and-frame, B) tube-in-shell, C) spiral-wound, D) details of hollow-fiber module. Reprinted from Leeper et al. (1984), pp. 36-37

The tube-in-shell system (Figure 16-1B) is occasionally used. This configuration is very similar to a shell-and-tube heat exchanger. The membrane would be coated on a porous support. The main advantage of these systems is that they can be cleaned by passing sponge balls through the separators. The surface area per unit volume is more than for plate-and-frame but less than for spiral-wound.

The spiral-wound configuration (Figure 16-1C) is more complicated but has a significantly higher surface area per unit volume. With proper design of the channels there will be significant turbulence at the membrane surface that promotes mass transfer. These systems have been used for carbon dioxide recovery, UF of relatively clean solutions and RO.

The hollow-fiber configuration (Figure 16-1D) looks schematically very similar to the tube-in-shell system, except the tubes are replaced by a very large number of hollow fibers made from the membrane. This configuration has the largest area-to-volume ratio. The hollow fibers can be optimized for a particular separation. For RO the inner diameter is about 42 microns while the outer diameter is about 85 microns. The separation is done by a 0.1 to 1.0 micron skin on the outer surface. The remainder of the membrane is a structural support. For gas permeation the requirement of a small pressure drop inside the tubes dictates a larger inner fiber diameter. For UF, where the feed can be dirty, the membranes are 500 to 1,100 microns inside diameter. Typically the feed is inside the tubes, and the thin membrane skin is on the inside of the fibers. Care must be taken that particulates do not clog the fibers. Hollow-fiber membranes are technologically the most difficult to make, and the typical user buys complete modules from the manufacturer.

The most common systems are spiral-wound and hollow-fiber. Systems are usually purchased as off-the shelf modules in a limited number of sizes. Since a standard size is unlikely to provide the required separation and flow rate required for a given problem, a series of modules is cascaded to obtain the desired separation (see Figure 16-2). The parallel configuration (Figure 16-2A) allows one to increase the feed flow rate and is used with all of the membrane separations. Parallel operation is roughly analogous to increasing the diameter of a distillation or absorption column.

Figure 16-2. Membrane cascades; A) parallel, B) retentate-in-series, C) parallel and series, D) permeate-in-series, E) retentate recycle

The retentate-in-series system (Figure 16-2B) will increase the purity of the more strongly retained components and simultaneously increase the recovery of the more permeable species. Unfortunately, the purity of permeate and the recovery of retained components both decrease. This configuration is used for production of less-permeable nitrogen from air. This process produces a product nitrogen gas at relatively high pressure since the nitrogen has not permeated through the membrane. The system is roughly analogous to a cross-flow extractor or stripper. Parallel and retentate-in-series systems are often combined (Figure 16-2C).

The permeate-in-series system (Figure 16-2D) is used when the permeate product is not of high enough purity. One or more additional stages of separation are required. Designers try to avoid this configuration if possible. The major cost of operating most membrane separators is the pressure difference required to force permeate through the membrane. This configuration will require an additional compressor or pump to repressurize permeate for the next membrane separator. One of the major disadvantages of the membrane separators is that additional stages for permeate do not reuse the energy (pressure) separating agent. The equilibrium-staged separations have the advantage that the energy (heat and cooling) separating agents can easily be reused. Membrane separators are often used because a membrane can be developed which produces permeate of desired purity in one stage.

A final common membrane cascade is the retentate-recycle mode (Figure 16-2E). The recycle allows for a very high flow rate in the membrane module to increase mass transfer rates and minimize fouling. The same high flow rates can be obtained without the recycle but a very long membrane system would be required to have a sufficiently long residence time. Recycle is also used with batch UF systems. Note that the recycle in Figure 16-2E is similar to the recycle in Figure 16-2D since the high-pressure retentate is recycled in both cases. Thus, the compressor or pump only needs to boost the pressure (e.g., because of friction losses for flow inside a hollow-fiber membrane), not overcome the large pressure difference that results when fluid permeates through the membrane.

The flow patterns on the retentate (feed) and permeate sides of the membrane have major effects on the mass balances and the separation. These flow patterns can be perfectly mixed on both sides, plug flow on one side and perfectly mixed on the other side, plug flow in the same direction on both sides (co-current), plug flow in opposite directions on the two sides (countercurrent), mixed on one side and cross-flow on the other, and somewhere in-between these ideal regimes. For example, if we consider the hollow-fiber module shown in Figure 16-1D, the flow inside the hollow fibers is very close to plug flow. Depending on the shell side design, the flow on the shell side could be approximately well-mixed, co-current plug flow or countercurrent plug flow. Countercurrent plug flow usually gives the most separation.

In most of this chapter we will make the assumption that both sides of the membrane are perfectly mixed. This assumption greatly simplifies the mathematics. The resulting design will be conservative in that the actual apparatus will result in the same or better separation than predicted. The effect of other flow patterns will be explored in section 16.7.

16.2 MEMBRANE CONCEPTS

Clearly the key to the membrane separators is the membrane. The membrane needs to have a high permeability for permeate and a low permeability for retentate. It helps if the membrane has high temperature and chemical resistances, is mechanically strong, resists fouling, can be formed into the desired module shapes, and is relatively inexpensive. Commercial membranes are made from polymers, ceramics, and metals (e.g., palladium for helium purification), but polymer membranes are by far the most common. Membrane development is done by a few large chemical companies and by several specialty firms. Most chemical engineers will use membrane separators, but they will never be involved in membrane development. However, some understanding of the polymer membrane will be useful when specifying and operating membrane separators.

A large number of polymers have been used to make membranes for membrane separators. Cellulose, ethyl cellulose, and cellulose acetate [actually a polymer blend of cellulose, cellulose acetate, and cellulose triacetate (Kesting and Fritzsche, 1993)] were the first commercially successful membranes and are still used. These materials have relatively poor chemical resistance, but their low cost makes them attractive when they can be used. The most common commercial membrane is polysulfone, which has excellent chemical and thermal resistance. Various polyamide polymers are also commonly used in RO. Silicone rubber, which has very high fluxes but low selectivity, is often used in pervaporation to preferentially permeate organics and as a coating on composite membranes for gas permeation. A large variety of grafted polymers, specialty polymers, and composite membranes has been developed to optimize flux and selectivity particularly for pervaporation (Huang, 1991) but also other membrane separations. Kesting and Fritzsche (1993) is an excellent source for information on the relationship between polymer structure and membrane properties for the polymers used for gas separation. The repeating units for several polymers used to form membranes are shown in Figure 16-3.

Figure 16-3. Monomers for polymers used to form membranes; A) cellulose (GP, RO, UF), B) polysulfone (GP, RO, UF), C) silcone rubber from dimethyl siloxane (GP, pervaporation), D) example of polyamide (RO, UF), E) polystyrene (matrix for composite resins, pervaporation) (Kesting and Fritzsche, 1993; Wankat, 1990).

The basic equation for flux of permeate through the membrane is

(16.1a)

Although this equation is derived from experiments and is not a fundamental law, it is so basic and powerful that you need to memorize it and explore its implications in as many ways as possible. The exact form of the equation depends upon the type of flux and the type of separation.

The flux can be written as a volumetric flux,

(16-1b)

or as a mass flux,

(16-1c)

or as a molar flux,

(16-1d)

These fluxes are obviously related to each other,

(16-1e,f)

where and ρ are the molar and mass densities, respectively. Once the flux is known, it is used in conjunction with the feed rate and the feed concentration to determine the membrane area required.

Equation (16-1a) can be used to determine the flux once the appropriate terms on the right hand side of the equation are known. The permeability P is a transport coefficient that can be determined directly from experiment. In some cases the permeability can be estimated from more fundamental variables such as solubility and diffusivity. This is briefly discussed in the next section. The membrane separation thickness t_ms is the thickness of the portion of the membrane that is actually doing the separation. This is often a thin skin with a thickness less than one micron. Since t_ms can be difficult to measure, the ratio (P/ t_ms), the permeance, is often reported from experimental data. The units of P and (P/ t_ms) depend upon the flux and driving force employed.

The driving force depends upon the type of membrane separation. For gas permeation the driving force is the difference in partial pressure of the transferring species across the membrane. For RO the driving force is the pressure difference minus the osmotic pressure difference across the membrane. For UF the driving force is the same as for RO, but this usually simplifies to the pressure difference across the membrane since osmotic pressures are small. These driving force effects are specific to each membrane separation and are discussed in more detail in the later sections.

Since flux is flow rate per membrane area, increasing the flux (while maintaining the desired separation) will decrease the membrane area. The result will be a more compact, less expensive device. Equation (16-1a) shows that flux can be increased by increasing permeability or driving force, or by decreasing the separation thickness. Permeability of the membrane depends upon interactions between the molecular structure of the polymers and the solutes. These effects will be briefly discussed in the sections on each membrane separation. More details on molecular structure-permeability effects are in the books by Kesting and Fritzsche (1993), Ho and Sirkar (1992), Mulder (1996), Noble and Stern (1995), and Osada and Nakagawa (1992).

For many years attempts were made to make very thin membranes, but there was always difficulty in making the membranes mechanically strong enough. Sidney Loeb and Srinivasa Sourirajan (Loeb and Sourirajan, 1960, 1963) made the breakthrough discovery. They found that anisotropic (asymmetric) membranes with a very thin skin (0.1 to 1.0 µm) on a much thicker porous support had the necessary mechanical strength while having a very small separation thickness. Schematics of isotropic and anisotropic membranes are shown in Figure 16-4. Another method for producing an asymmetric membrane is to cast a thin layer of one polymer onto a porous supporting layer of another polymer to form a composite membrane. Although not all membrane separators use asymmetric membranes, they are the most common.

Figure 16-4. Schematics of A) isotropic and B) anisotropic membranes, reprinted from Leeper et al. (1984)

16.3 GAS PERMEATION

We will consider gas permeation first, since, in addition to being commercially important, in many ways these systems are closest to being ideal. Once the ideal operation is understood, we can look at deviations from ideality for other membrane separations. Gas permeation membrane systems are used commercially for separation of permanent gases. Some common applications are purification of helium, hydrogen, and carbon dioxide, production of high purity nitrogen from air, and production of air enriched in oxygen (low purity oxygen).

Gas permeation systems typically use hollow-fiber or spiral-wound membranes. Cellulose acetate membranes are used for carbon dioxide recovery, polysulfone coated with silicone rubber is used for hydrogen purification, and composite membranes are used for air separation. The feed gas is forced into the membrane module under pressure. Retentate, which does not go through the membrane, will become concentrated in the less permeable gas. Retentate exits at a pressure that will be close to the input pressure. The more permeable species will be concentrated in permeate. Permeate, which has passed through the membrane, exits at low pressure. The operating cost for a gas permeator is the cost of compression of the feed gas and the irreversible pressure difference that occurs for the gas that permeates the membrane.

For gas permeation the steady-state general flux Eq. (16-1a) for gas A becomes

(16-2a)

where J_A is the volumetric flux of gas A, is the molar transfer rate of gas which permeates through the membrane, y_t,A is the mole frac A in the gas that transfers (or permeates) through the membrane, is the molar density of the solute in the gas transferring through the membrane, A is the membrane area available for mass transfer, P_A is the permeability of species A in the membrane, the driving force for the separation Δp_A is the difference in the partial pressure of A across the portion of the membrane which does the separation, and t_ms is the thickness of the membrane skin which actually does the separation. The ratio P_A/ t_ms is known as the permanence, and is often the variable that is measured experimentally. Since partial pressure = (mole frac)(total pressure) or p_A = y_A p, Eq. (16-2a) can be expanded to

(16-2b,c)

where p_r is the total pressure on the retentate (high pressure) side, y_r,w,A is the mole frac A on the retentate side at the membrane wall, and p_p and y_p,w,A are the pressure and mole frac at the wall on the permeate (low pressure) side. The flux equation for other components will look exactly the same as for component A except the subscript A will be changed to B, C, … as appropriate. The relationship between the mole frac transferring through the membrane y_t,A and the permeate mole frac at the wall y_p,w,A depends upon the flow patterns in the permeate.

16.3.1 Gas Permeation of Binary Mixtures

Commercial separation of binary mixtures by gas permeation is common. In binary mixtures we can use the substitution

(16-3a)

where y without the subscript A or B is understood to refer to the mole frac of the faster permeating species A. In a binary system, the flux of the slower moving component B becomes

(16-3b)

Since pressure and mole fracs can vary along the membrane, Eqs. (16-2a), (16-2c), and (16-3b) are applied point by point along the membrane. In almost all systems there is no concentration gradient on the permeate side perpendicular to the membrane; thus, we replaced y_p,w,A with y_p,A, which can vary along the membrane. In addition, if mass transfer on the retentate side is not rapid, the mole fracs and partial pressures at the surface of the membrane will not be the same as in the bulk gas or in the feed gas (see concentration polarization in section 16.4). Usually mass transfer rates in gas systems are high enough that the mole frac at the membrane surface is equal to the mole frac in the bulk gas.

Most gas permeation membranes work by a solution-diffusion mechanism. On the high-pressure side of the membrane the gas first dissolves into the membrane. It then diffuses through the thin skin of the membrane to the low-pressure side where permeate reenters the gas phase. For this type of membrane the permeability is the product of the gas solubility H_A in the membrane times the diffusivity D_m in the membrane

(16-4a)

The solubility parameter H_A is very similar to a Henry’s law coefficient. Representative permeabilities for several polymer membranes are shown in Figure 16-5 and in Table 16-2. Small molecules such as helium have high diffusivities but low solubilities while large gas molecules such as carbon dioxide have high solubilities but low diffusivities. The resulting product, the permeability P is relatively large for both small and large molecules, but has a minimum for molecules around the size of nitrogen.

Figure 16-5. Permeation of gases in polymer membranes (Baker and Blume, 1986), reprinted with permission from Chem. Tech., 232 (1986), copyright 1986, American Chemical Society.

Table 16-2. Permeabilities of gases in various membranes; cm³(STP)cm/[cm²·s·cm Hgl × 10⁶; Reference Code: N = Nakagawa, 1992; DR = Drioli and Romano, 2001; G = Geankoplis, 2003

If the driving forces are equal, gases with higher permeabilities transfer through the membrane at higher rates. It is useful to define a selectivity α_AB

(16-4b)

as a short-hand method of comparing the ease of separation of gases. The selectivity of two solutes can be estimated from Figure 16-5 and Table 16-2. For example, from Figure 16-5, α_CO2 − N₂ = P_CO2/P_N2 = 24/0.3 = 80. The selectivity is a function of concentration, pressure and temperature since the individual permeabilities depend on concentration, pressure, and temperature. However, α_AB tends to be less dependent on these factors than the individual permeabilities. Note that α_AB is analogous to the relative volatility defined in Chapter 2. Since we prefer to operate gas permeation as a one-stage system, α_AB values of 20 and higher are preferred instead of the much more modest values commonly employed in distillation.

16.3.2 Binary Permeation in Perfectly Mixed Systems

The mass balances for a gas permeation membrane module that is completely mixed on both sides of the membrane will have the simplest form since the mole fracs and pressures on each side of the membrane are constant. Equations (16-2) and (16-3b) can then be used in algebraic mass balances with y_p = y_t and integration is not required to find the average retentate and permeate mole fracs.

We can obtain a rate-transfer (RT) equation for a perfectly mixed permeator by writing the equation for transfer through the membrane, Eq. (16-2) for both the more permeable and the less permeable species. The perfectly mixed assumption makes y_p = y_t. We will also assume that the module has a high rate of mass transfer from the bulk fluid to the membrane surface. This assumption makes y_r = y_r,w (the mole frac of A at the membrane wall). These two consequences of the perfectly mixed assumption greatly simplify the analysis. The second assumption is reasonable for many gas permeators since diffusivities and mass transfer rates are high. From Eq. (16-2c), the transfer equation for the more permeable component A is

(16-5a)

while from Eq. (16-3b) for the less permeable component B in a binary separation,

(16-5b)

Because permeate is typically at relatively low pressures where gas molecules are far apart, we will assume that transfer of A is independent of transfer of B and vice-versa. Thus, P_A and P_B are constant. We now solve these two equations for , and set them equal to each other.

(16-5c)

This equation can be solved for either y_r or for y_p. Since solving for the former is easier, we will solve for y_r. After some algebra, the result is

(16-6a)

Equation (16-6a), which we will call the RT curve, relates the mole fracs on both sides of the membrane based on the rate transfer parameters and the driving force. This equation takes the place of the equilibrium curve in flash distillation. In deriving Eq. (16-6a) we did not invoke the assumption that the membrane separator is perfectly mixed. Thus, this equation is generally applicable if it is applied point-by-point on the membrane. Equation (16-6a) is not an equilibrium expression since it was derived based on transfer rates, but as we will see, it can take the place of an equilibrium expression for binary systems.

The average outlet concentration of the permeate can be found by dividing the permeation rate of component A integrated over the entire membrane by the total permeation rate. For a perfectly mixed separator values are constant, and y_r,out = y_r,average = y_r; y_p,out = y_p,average = y_p and F_r = F_out.

We can now do external balances for the completely mixed system shown in Figure 16-6. Note that this system is the rate equivalent of flash distillation (Hoffman, 2003). A single feed goes into a well-mixed chamber and two products are withdrawn. The more permeable species concentrates in the permeate product which is analogous to the more volatile component concentrating in the vapor. However, the two products are not in equilibrium but are related by the rate transfer expression. Because of the geometric similarity, we can use an analysis procedure analogous to that used for binary flash distillation.

Figure 16-6. Completely mixed membrane module; A) general case, B) sketch for Example 16-1

The overall mole balance is

(16-7a)

and the mole balance on the more permeable component is

(16-7b)

Equations (16-7) can be combined and solved for either y_out or for y_p. Since solving for y_p keeps the analogy with flash distillation intact, we will solve for y_p. In molar units,

(16-8)

Equation (16-8) is the operating equation for the well-mixed gas permeator. The ratio , known as the “cut,” is an important operating parameter. The RT Eq. (16-6a) and the operating Eq. (16-8) are the two equations needed to solve problems with perfectly mixed, binary gas permeators. This is illustrated in Example 16-1 when y_p is known and in Example 16-2 when θ is specified.

EXAMPLE 16-1. Well-mixed gas permeation—sequential, analytical solution

A perfectly mixed gas permeation module is separating carbon dioxide from nitrogen using a poly (2,6 – dimethylphenylene oxide) membrane. The feed is 20.0 mole % carbon dioxide and is at 25°C. The module has 50.0 m² of membrane. The module is operated with a retentate pressure of 5.5 atm and a permeate pressure of 1.01 atm. We desire a permeate that is 40.0 mole % carbon dioxide. The membrane thickness is t_ms = 1.0 × 10⁻⁴ cm. Find y_r,CO2 = y_out,CO2, J_CO2, (in gmole/min), cut θ and (in gmol/min).

Solution

A. Define. This is basically a simulation problem. We want to find how much gas can be processed by a well-mixed module with a known membrane area at specified operating conditions.

B and C. Explore and plan. The permeabilities of carbon dioxide and nitrogen for this membrane are listed in Table 16-2 The selectivity, α, is the ratio of these permeabilities. The value of y_r,CO2 can be found from the RT Eq. (16-6a). The CO₂ flux and F_p can be calculated from Eq. (16-2). The cut and F_in can then be determined from mass balances, Eqs. (16-7).

D. Do it. From Table 16-2

P_CO2 = 75 and P_N2 = 4.43 both cm³ (STP)cm/[cm²s cmHg] × 10⁻¹⁰.

Then α_CO2−N2 = 75 × 10⁻¹⁰/4.43 × 10⁻¹⁰ = 16.9.

The pressure ratio p_r/p_p = 5.5/1.01 = 5.446. For ideal gases, which is a reasonable assumption, and gmole/22.4L(STP).

For an ideal gas the RT Eq. (16-6a) simplifies to

(16-6b)

Substituting in values for the parameters, we obtain

The CO₂ flux can be determined from Eq. (16-2),

The mass balances can now be solved several different ways. For example, by solving Eqs. (16-7a) and (16-7b), simultaneously we obtain,

(16-7c)

Then θ = (0.20 – 0.1044)/(0.40 – 0.1044) = 0.3234, and .

E. Check. Recalculating all the numbers gave the same result.

F. Generalize. Notes: 1. Because one of the outlet mole fracs was known, we could calculate the other from the RT equation; thus, simultaneous solution of Eqs. (16-6b) and (16-8) was not required. If neither outlet mole frac is known, then simultaneous solution is required (see Example 16-2).

2. Units are obviously very important and should be carried throughout the solution.

3. Using the conversion 1 gmole = 22.4 L (STP), is equivalent to using the ideal gas law.

A graphical solution to perfectly mixed gas permeators is also straightforward. On a graph of y_p vs. y_out, Eq. (16-8) plots as a straight line with a slope of – (1 – θ) /θ. Note that the cut θ is analogous to f = V/F in a flash distillation system. The intersection with the y_p = y_out line occurs at y_p = y_out = y_in which is analogous to y = x = z in a flash system. The intersections with the axes can also be determined (see Example 16-2 and Figure 16-7). The simultaneous solution is obtained at the point of intersection of the straight line representing Eq. (16-8) and the curve representing Eq. (16-6a). This method is illustrated in Figure 16-7 for a number of values of θ. Figure 16-7 is plotted for the separation of carbon dioxide and methane in a cellulose acetate membrane (see Example 16-2).

EXAMPLE 16-2. Well-mixed gas permeation—simultaneous analytical and graphical solutions

Chern et al. (1985) measured the permeability of carbon dioxide and methane as P_CO2 = 15.0 × 10⁻¹⁰ and P_CH4 = 0.48 × 10⁻¹⁰ [cc(STP) cm]/[cm² s cm Hg] in a cellulose acetate membrane at 35° C and p_r = 20 atm. We wish to separate a feed gas that is 50 mole % carbon dioxide and 50.0 mole % methane at 35°C. The high pressure is 20.0 atm and the permeate pressure is p_p = 1.1 atm. The completely mixed membrane module shown in Figure 16-6 will be used.

a. Determine the values of y_p and y_out for cut values of 0, 0.25, 0.5, 0.75 and 1.0 using both analytical (θ = 0.25 only) and graphical solutions.

b. If the effective membrane thickness is t_ms = 1.0 × 10⁻⁶ m, determine the fluxes for θ = 0.25.

C. For θ = 0.25 determine the membrane area needed for a feed gas flow rate of = 1 gmol/s.

Solution

A. Define. The module is sketched in Figure 16-6. We need to solve the equations analytically and plot a graph that allows us to find y_p and y_out for different "values of θ, and when θ = 0.25 find the fluxes of carbon dioxide and methane and the membrane area.

B and C. Explore and plan. The selectivity α_AB = P_A/P_B. We will assume the gases are ideal.

Analytical solution: We need to solve Eqs. (16-6a) and (16-8a) simultaneously.

Graphical solution: If we pick values of y_p, Eq. (16-8b) can be used to calculate the corresponding values of y_r for the rate transfer curve. For a well-mixed permeator this rate transfer curve and the operating line, Eq. (16-8), can then be plotted on a y_p vs. y_r graph for different values of θ. Since the system is well mixed, the points of intersection give the solutions for y_p and y_r = y_out. The fluxes can then be calculated from Eq. (16-4).

D. Do it. Preliminary calculations: For an ideal gas the molar densities are equal, and their ratio equals 1.0. The selectivity is

α_CO2−CH4 = P_CO2/P_CH4 = (15.0 × 10⁻¹⁰)/(0.48 × 10⁻¹⁰) = 31.25

p_p/p_r = 1.1/20 = 0.055 (Note that any set of consistent units can be used for this ratio.)

Analytical Solution: Solving Eq. (16-7a) and (16-7b) for y_r,

(16-9)

After substituting Eq. (16-9) in (16-6b) (in Example 16-1) and doing considerable algebra, we obtain the quadratic equation,

(16-10a)

where

(16-10b)

(16-10c)

(16-10d)

This quadratic equation can be solved numerically or analytically. An analytical solution is obtained if we use the quadratic formula

(16-10e)

Substituting in values θ = 0.25, α = 31.25, y_in = 0.5, and p_p/p_r = 0.055, we obtain a = 11.747, b = −33.247, c = 20.833, and y_p = 0.9365 or 1.894. The second value for y_p can be eliminated since the mole frac cannot be greater than 1.0. The value of y_r can be found from Eq. (16-9), y_r = 0.3545. Solutions obtained with a spreadsheet using Solver will be identical.

Graphical Solution: The RT curve, Eq. (16-6b), that relates y_r and y_p becomes

y_r = {[(31.25 − 1)(0.055) (1 − y_p) + 1] y_p}/[31.25 − 30.25 y_p]

The following table for the RT curve is generated by selecting arbitrary values of y_p.

These values were plotted in Figure 16-7 for the RT curve.

a. For operating Eq. (16-8) the intersection with y_p = y_r line occurs at y_r = y_in = 0.5 and the slope = −[1 − θ]/θ. For example, when θ = 0.25, the slope = −0.75/0.25 = −1/3.

The solutions can be obtained at the intersections of the operating lines and the RT curve. For θ = 0.25, Figure 16-7 shows that y_p = 0.93 and y_r = y_out = 0.35.

b. When θ = 0.25, the flux can be determined from Eq. (16-4), using the y_p and y_r values from part a,

Note that care must be taken with the units.

c. θ = 0.25 and , . Since or , we can find the area once the molar flux is known. The volumetric flux is, Taking care to properly calculate units, the membrane area is,

Figure 16-7. Solution for well-mixed gas permeation system for Example 16-2

E. Check. The analytical and graphical solutions for θ = 0.25 gave identical results.

F. Generalize. 1. The rate transfer curve contains the variables that affect mass transfer rates across the membrane (α_AB, p_p/p_r). If either of these variables vary (for example, if the temperature changes α_AB will change) a new RT curve must be generated. The operating equation depends on the cut and the feed mole frac. If either of these variables is changed, then a new operating line needs to be drawn. One advantage of this graphical approach is it separates the rate and operating terms; thus, making it easier to determine the effect of varying the conditions.

2. Units are important.

3. Although this is not an equilibrium process it looks very similar to a flash distillation with a large relative volatility (Hoffman, 2003). Membrane separators are useful because they are a practical way of generating favorable rate transfer curves.

4. A commercial gas permeator would probably be close to cross-flow or countercurrent flow (see section 16.7). This would result in more separation than is predicted for the perfectly mixed system.

5. Although the permeate product is fairly pure (93 % carbon dioxide) the retentate product is impure (65% methane). Unlike distillation systems, simple membrane separators cannot produce two pure products simultaneously. More complicated membrane cascades can do this, but repressurization is required (e.g., see Wankat, 1990, Chapter 12).

6. The driving force for methane transfer is significantly higher than for carbon dioxide (check the flux calculations). Yet, because of the high selectivity the carbon dioxide transfer rate is considerably greater. The carbon dioxide is being transferred “uphill” to a higher carbon dioxide mole frac but “downhill” to a lower partial pressure of carbon dioxide. The increased pressure forces this to happen.

7. At low values of y_in the methane flux can be greater than the carbon dioxide flux even though the system is concentrating carbon dioxide. At these low feed concentrations and with a relatively high cut the retentate product can be almost pure methane, but permeate will be impure. (Try doing some of these calculations. Since the transfer rate curve is not changed, only the operating line needs to be changed. Because the analytical solution is easily set up in a spreadsheet, a large number of examples can easily be run.)

8. Greater accuracy can be obtained by expanding the scales for the portion of the diagram needed for the calculation; however, this is seldom necessary since the graphs are probably at least as accurate as the experimental values of the permeabilities.

9. Once y_p and y_out have been calculated, determination of the fluxes and the membrane area are straightforward. Note that the RT curve and hence y_p and y_r depend only upon the ratio p_r/p_p. The fluxes and the area depend upon the difference (p_r y_r − p_p y_p). Higher pressures at the same pressure ratio will produce the same products but require less membrane area.

10. If the membrane area is specified instead of the cut, the problem is a simulation, not a design problem. Although the analytical solution shown in Example 16-1 is probably simpler, a graphical solution can be used. is related to the membrane area and the flux through Eq. (16-4); however, calculation of J_A requires knowing y_p and y_r = y_out, and we need to know θ to calculate the mole fracs. This is a classical trial-and-error situation. It can be solved as follows: Generate the RT curve. Guess y_p and find y_out from the RT curve. Calculate J_A and from Eq. (16-4). Calculate F_p from the mass balance Eq. (16-7c). If the values of calculated from the flux and from the mass transfer expressions are not equal, continue the calculation.

Membrane separations are often most effective at low concentrations. This is exactly where distillation is most expensive. Thus, hybrid systems where a membrane separator is combined with distillation are often used commercially.

There is one other significant difference between membrane separators and the equilibrium-staged separations we have studied. Companies are much more likely to buy “off-the-shelf” or “turnkey” membrane units. Off-the-shelf systems are modules in standard sizes that are connected together to more-or-less perform the desired separation. The engineer needs to determine the performance of these units since they will not be exactly the same as the design desired. With a turnkey unit the company buys from a manufacturer who guarantees a given performance level. This situation arose because only a limited number of companies have the technical expertise necessary to make membrane separators. A large number of companies are capable of making distillation and absorption systems. If your company decides to buy a turnkey unit, knowledge of membrane separations will enable you to include all pertinent items in the contract and to negotiate a contract that is more favorable for your company. After delivery, you will be able to perform the appropriate tests to determine if the membrane system meets the specifications in the contract. If the specifications are not met, your company can demand that the vendor fix the problems.

16.3.3 Multicomponent Permeation in Perfectly Mixed Systems

In the previous section we saw that the RT curve took the place of the equilibrium curve in binary flash distillation. For multicomponent flash distillation the y-x equilibrium curve is replaced by equilibrium expressions of the form y_i = K_i x_i and the combination of the operating equations and equilibrium are summed using Σx_i = 1, Σy_i = 1, or Σy_i − Σx_i = 0. Since the operating equations for well-mixed permeators are similar to the operating equations for flash distillation (Hoffman, 2003), if we can write the rate expression in the form y_i = K_m,i x_i, then the mathematics to solve the permeation problem will be very similar to that used for flash distillation. Of course, K_m,i has a totally different meaning than K_i in flash distillation.

The operating equation is essentially Eq. (16-8a) written for each component,

(16-8c)

For ideal gases since molar ratios are equal to the volume ratios, θ is the same in molar and volume units. In addition, for ideal gases mole frac equals volume fraction; thus, Eq. (16-8c) can be used with either molar or volumetric flow rates. The rate expression Eq. (16-4) can be written for molar flows as,

(16-11a)

or in volumetric flow rates,

(16-11b)

Solving Eq. (16-11b) for y_p,i, we obtain

(16-11c)

where, after some rearrangement,

(16-11d)

In a design problem permeate volumetric flow rate F_p will be known but the area A is unknown. If Eqs. (16-8c) and (16-11d) undergo exactly the same algebraic steps used for flash distillation [from Eq. (2-36) to (2-37)], the resulting equation for y_r,i (equivalent to x_i in flash) is,

(16-11e)

and the summation equation Σ y_r,i = 1, (equivalent to Σx_i = 1 in flash) is

(16-11f)

which is essentially the same as flash distillation Eq. (2-40) (see Problem 16.C6).

To find the area required for a multicomponent permeator, we estimate the value of F_p/A (or equivalently of y_p,i for one of the components), calculate all the values of K_m,i, check to see if Eq. (16-11f) is satisfied, and if not repeat with a new value of F_p/A. Unlike the flash distillation counter part, this procedure converges rapidly using Σ y_r,i = 1 as the basic equation, and it is easy to solve on a spread sheet or in a mathematical program such as MATLAB, Mathematica or Mathcad. Once correct values for K_m,i and F_p/A are known, we can calculate area A, y_r,i from Eq. (16-11e), and y_p,i from Eq. (16-11c). This procedure is illustrated in Example 16-3. Although shown only for gas permeation, this approach can be used with other membrane separators.

EXAMPLE 16-3. Multicomponent, perfectly mixed gas permeation

A perfectly mixed gas permeation unit is separating a mixture that is 20 mol % carbon dioxide, 5 mole % oxygen and 75 mole % nitrogen using a poly(dimethylsiloxane) membrane. Feed flow rate is 20,000 cm³ (STP)/s. The membrane thickness is 1 mil (0.00254 cm). Pressure on the feed side is 3.0 atm and on the permeate side is 0.40 atm. We desire a cut fraction = 0.225. Find the membrane area needed, and the outlet mole fracs of permeate and retentate.

Solution

Permeabilities (Table 16-2 are P_CO2 = 3240 × 10⁻¹⁰, P_O2 = 605 × 10⁻¹⁰, P_N2 = 300 × 10⁻¹⁰, all in [cm³ (STP)cm/(cm² s cm Hg)]. So that we don’t forget, change the units of the pressures: p_r = 3.0 atm (76 cm Hg/atm) = 228 cm Hg, p_p = 0.40 atm = 30.4 cm Hg.

The K values can be calculated from Eq. (16-11d). For example, for carbon dioxide

and similarly for oxygen and nitrogen. To use these terms in a trial and error procedure, we need to start with a value for F_p/A. How do we find a reasonable first guess for F_p/A?

If we guess a value for y_p,i for any of the components, we can then calculate y_r,i from the rearranged component operating Eq. (16-8c),

and determine our guess for F_p/A by rearranging Eq. (16-11b).

F_p/A_guess = [P_i/(t_msy_p,i)] (p_ry_r,i − p_py_p,i)

For example, we can arbitrarily guess that y_p,CO2 = 0.50. Then the first guess for F_p/A = 0.002689. With this guess K_m,CO2 = 4.42857, K_m,O2 = 1.6827, K_m,N2 = 0.8834 and Eq. (16-11f) becomes,

. We need a higher value of F_p/A or a lower guess for y_p,i. The following vaues were generated with a spreadsheet, by guessing the next value for y_p,CO2. (If desired, Solver can be used with Excel.)

Then A = F_p/(F_p/A) = [θ(F_in)]/(F_p/A)= [(0.225)(20,000)]/(0.004189) = 1,074,360 cm². From Eq. (16-11c) we find K_m,CO2 = 3.60554, K_m,O2 = 1.1748, and K_m,N2 = 0.5922.

The retentate mole fracs from Eq. (16-11e) are: y_r,CO2 = 0.1261, y_r,O2 = 0.0481, y_r,N2 = 0.8258; and the permeate mole fracs from Eq. (16-11c) are: y_p,CO2 = 0.4546, y_p,O2 = 0.0565, y_p,N2 = 0.4890.

To be sure you know how to do multicomponent permeation problems, set up your own spreadsheet and solve either this example or Problem 16.G1. Geankoplis (2003) solves the multicomponent permeator system by a different method, but the results are identical (see Problem 16.G1).

Note that permeate is not very pure. Although nitrogen has the lowest permeability, there is still more nitrogen in the permeate than carbon dioxide. This occurs because the large amount of nitrogen in the feed produces a large driving force to push nitrogen through the membrane. Retentate is significantly purer than the feed. To obtain a still purer nitrogen stream in the retentate with this feed, we can increase the cut and the retentate pressure; however, a better approach would be to use a membrane with higher selectivity.

16.4 REVERSE OSMOSIS

Reverse osmosis (RO) is a process commonly used to purify and desalinate water. The liquid water is forced under pressure through a nonporous membrane in the opposite direction to osmosis (osmosis will be defined below). Most salts and uncharged molecules are retained by the membrane. Thus, permeate is much purer water and retentate becomes significantly more concentrated. The commonly used membranes shown in Figure 16-3 are: 1) a blend of cellulose acetate and cellulose triacetate, 2) aromatic polyamides (aramids), and 3) cross-linked aromatic polyamides (Eykamp, 1997). Both hollow-fiber and spiral-wound modules are used for RO (Figure 16-1). Because thin tubes can withstand great compression pressures, the high-pressure feed in hollow-fiber systems is usually on the shell side (outside of the fibers).

The feed to an RO system usually requires pretreatment to remove any particulates that would clog the membrane. If there are ions or solutes in solution that have limited solubility, the design must include a solubility calculation to determine if they will precipitate onto the membrane when retentate is concentrated. If precipitation is likely, these ions or solutes must either be removed or made more soluble to prevent them from precipitating. A simple schematic of a simple RO system including the most important auxiliary equipment is shown in Figure 16-8. In practice, large-scale systems may have hundreds of membrane modules arranged both in series and in parallel Figure 16-2C. More details on equipment are available in Baker et al. (1990), Eykamp (1997), Ho and Sirkar (1992), and Noble and Stern (1995).

Figure 16-8. Schematic of RO system

16.4.1 Analysis of Osmosis and Reverse Osmosis

The driving force for solvent flux in RO is the difference between the pressure drop across the membrane and the osmotic pressure difference across the membrane (Δp −Δπ). Then volumetric solvent flux is

(16-12a,b)

and the mass flux of solvent is

(16-12c,d,e)

In these equations K_solv is the permeability of the solvent (usually water) through the membrane with an effective membrane skin thickness of t_ms, and ρ_solv is the solvent mass density. The pressure drop across the membrane, Δp = p_r − p_p. Δπ is the difference in the osmotic pressure across the membrane (see description below). The mass solute flux across the membrane J′_A can be written as

(16-13)

where K′_A is the solute permeability and Δx is the difference in wt frac of solute across the membrane. For a membrane with perfect solute retention, K′_A = 0.

To study reverse osmosis, we must first study osmosis. In osmosis solvent flows through a semipermeable membrane (one which passes solvent but not solutes or ions) from the less concentrated region to the more concentrated region. For example, if we have pure water on one side of a semipermeable membrane and a sugar solution on the other side, the water will flow into the sugar solution to dilute it. Thus, in Figure 16-9 osmotic flow is from the right side (pure water) to the left side (sugar solution). This natural direction of solvent flow is the direction that will cause chemical potentials to equalize. We can stop or reverse the flow by increasing the pressure on the sugar solution using the piston shown on the left. Osmotic pressure π is the additional pressure required on the concentrated side to stop osmotic flow assuming the permeate side is pure water that contains no solute.

Figure 16-9. Osmotic pressure apparatus

Osmotic pressure is a thermodynamic property of the solution. Thus, π is a state variable that depends upon temperature, pressure, and concentration but does not depend upon the membrane as long as the membrane is semipermeable. Osmotic equilibrium requires that the chemical potentials of the solvent on the two sides of the membrane be equal. Note that the solutes are not in equilibrium since they cannot pass through the membrane. Although osmotic pressure can be measured directly, it is usually estimated from other measurements (e.g., Reid, 1966). For an incompressible liquid osmotic pressure can be estimated from vapor pressure measurements,

(16-14)

where v_solvent is the partial molar volume of the solvent. Another common method is to relate osmotic pressure to freezing point depression (Reid, 1966). For dilute systems the osmotic pressure is often a linear function of concentration,

(16-15a,b)

where or a′ are determined by plotting the data as a straight line. As the solution becomes more concentrated, the osmotic pressure increases more rapidly than predicted by the linear relationship. For some dilute systems the linear constant can be estimated from the van’t Hoff equation,

(16-15c)

Since the van’t Hoff equation assumes that Raoult’s law is valid, this result will be incorrect if the solution associates or dissociates even though empirical Eq. (16-15a) may still be accurate.

Unfortunately, natural osmotic flow is in the opposite direction to what we want to do (produce pure water). In RO we push the solvent out of the concentrated solution into the dilute solution. Thus, it takes energy to concentrate the retentate. Because RO is reversing the natural flow direction, RO is inherently a nonequilibrium process. The increase in osmotic pressure as retentate becomes more concentrated also puts a natural limit on the recovery of pure solvent by RO. If one tries to recover too much solvent, retentate becomes very concentrated, the osmotic pressure difference becomes extremely large, and the pressure drop required by Eq. (16-12) for a reasonable flux rate becomes too large for practical operation.

The analysis procedure developed in the previous section for gas permeation forms the basis for analyzing RO. However, the RO analysis is more complicated because of 1) osmotic pressure, which is included in Eq. (16-12), and 2) mass transfer rates are much lower in liquid systems. Since the mass transfer rates are relatively low, the wt frac of solute at the membrane wall x_w will be greater than the wt frac of solute in the bulk of the retentate x_r. This buildup of solute at the membrane surface occurs because the movement of solvent through the membrane carries solute with it to the membrane wall. Since the solute does not pass through the semipermeable membrane, its concentration will build up at the wall and it must back diffuse from the wall to the bulk solution. This phenomenon, concentration polarization, is illustrated in Figure 16-10. Concentration polarization has a major effect on the separations obtained in RO and UF (see next section). Since concentration polarization causes x_w > x_r, the osmotic pressure becomes higher on the retentate side and, following Eq. (16-12), the flux declines. Concentration polarization will also increase Δx in Eq. (16-13) and flux of solute may increase which is also undesirable.

Figure 16-10. Concentration polarization: buildup and back diffusion of solute

Since [where x is wt frac in RO], we can expand Eq. (16-12) using Eq. (16-3a) as,

(16-16a)

The x values are mass fractions, and the osmotic pressure on the retentate side depends on the concentration of solute at the membrane wall. To simplify the analysis, we will temporarily assume that the osmotic pressure data are linear and are satisfactorily fit by Eq. (16-15b). Then the flux equation becomes

(16-16b)

To relate the wall concentration to retentate concentration, we define the concentration polarization modulus M in terms of the wt frac,

(16-17)

Methods to measure or predict M will be developed shortly. Substituting the definition for M into Eq. (16-16b), we obtain

(16-16c)

The solute flux Eq. (16-13) can also be expanded and written in terms of the concentration polarization modulus.

(16-18)

Essentially the same procedure used to solve for the concentrations in gas permeators will be used. That is, after assuming that K_A is not zero and is independent of the solvent transfer rate, and that K_solv is independent of the solute transfer rate, we will solve Eqs. (16-16c) and (16-18) for F′_p, set the two equations equal to each other, and solve for the desired concentration. Setting the equations equal, we obtain

(16-19)

It is convenient to define the selectivity of the membrane α as,

(16-20)

We can now solve for either x_r as a function of x_p, or vice versa. Since it is easier to solve for x_r, we will do that here.

(16-21)

This rate transfer (RT) equation represents the transfer rate of solvent and solute through the membrane in wt frac units. If K_A = 0, α will be infinite, x_p will be zero, and x_r must be determined from a mass balance. If the selectivity is not constant, α will depend upon x_w = Mx_r. It will then be convenient for calculation purposes to solve for x_p as a function of x_r. This derivation is left as Problem 16.C2.

The rate transfer Eq. (16-21) needs to be solved simultaneously with an expression for the mass balance. We will again assume the membrane module is well-mixed. The module is identical to Figure 16-6A except the y terms are replaced by liquid wt frac. The external mass balances for this well-mixed module (in weight units) are

(16-22a,b)

Solving for x_p, we obtain the operating equation

(16-23)

where is the cut in mass units. Equation (16-23) is analogous to Eq. (16-8) obtained for the well-mixed gas permeator.

In the well-mixed module x_r is constant and is equal to x_out. Thus, Eq. (16-21) is valid with x_r replaced by x_out; however, since there is usually concentration polarization, x_w = Mx_out and M must be determined. Once M is known, simultaneous solution of Eqs. (16-21) and (16-23) can be obtained either analytically (see Example 16-3) or by plotting both equations on a graph of x_p vs. x_out. Equation (16-23) is a straight line that is identical to the operating line obtained for gas permeation. The curve representing Eq. (16-21) is plotted by calculating values in the same way as in Example 16-2. Note that this curve will be below the x_p = x_out line on a graph of x_p vs. x_out. This procedure will be illustrated in Example 16-5 (Figure 16-11) after the methods for determining M are explored.

Figure 16-11. Solution of simultaneous RT curve and operating line for Example 16-7

RO data may be reported as the rejection coefficient R.

(16-24)

If R is known, Eq. (16-24) is the RT equation for a perfectly mixed RO system. If values of x_in, θ′ and R are reported, x_p and x_out can be found by solving Eqs. (16-23) and (16-24) simultaneously. Since these equations are linear, the procedure is straightforward. The inherent rejection coefficient—the rejection coefficient when there is no concentration polarization (M = 1)—is given the symbol R°.

EXAMPLE 16-4. RO without concentration polarization

A new composite membrane is being tested in a perfectly mixed membrane system with a retentate pressure of 10.0 atm and a permeate pressure of 1.0 atm. Find the outlet wt fracs of the permeate and the retentate streams. There is a very high mass transfer coefficient and M = 1.0. α_water-salt = 4.157 L/(kg atm), ρ_solv = 0.997 kg/L (density of pure water).

a. θ′ = 0.30 (in mass flow rate units) and the inlet stream is 0.009 wt frac sodium chloride.

b. θ′ = 0.40 (in mass flow rate units) and the inlet stream is 0.010 wt frac sodium chloride.

Solution

A. Define. Find x_p and x_out for both cases a and b.

B and C. Explore and plan. Equations (16-21) and (16-23) can be solved simultaneously to find x_p and x_out. Since the change in operating conditions only affects the operating Eq. (16-23), the RT equation is the same for parts a and b.

D. Do it. The RT Eq. (16-21) in weight units is,

(16-21 repeated)

The term is a′ the linear coefficient for the effect of concentration on the osmotic pressure in wt frac units. Osmotic pressures are given for aqueous sodium chloride solutions in Perry and Green, 6th edition (1984, p. 16-23). At 25°C and 0.001 mole frac the osmotic pressure is 0.05 atm. Thus, in molar units = 0.05 atm/0.001 mole frac = 50 atm/mole frac.

We must convert to a′ in wt frac units.

π = a′x or a′ = m/x where m is mole fraction. For very dilute solutions,

(m, mole frac solute) = (x, weight frac solute) (MW water)/(MW solute)

The RT Eq. (16-21) can now be used to relate x_r to x_p with M = 1, α = 0.351 L/(kg atm), a′ = 15.446 atm/wt frac, ρ_solv = 0.997 kg/L, p_r = 10.0 atm, and p_p = 1.0 atm.

This equation is quadratic in x_p; however, since x_in = 0.009, concentrations are low and it can be linearized. One point is the origin, x_r=0, x_p=0. Choosing a low value of x_p which will be in the linear range (e.g., x_p=0.004), we can find an approximate linear RT equation. At x_p=0.004 we obtain,

The slope from (x_r=0, x_p=0) to (x_r=0.01495, x_p=0.0004) is slope = Δx_p/Δx_r = (0.0004 − 0)/(0.01495 − 0) = 0.0268 and the linearized equation is, x_p=0.0268 x_r.

We now solve the linearized RT equation simultaneously with the operating Eq. (16-23) noting that x_r = x_out for a perfectly mixed separator.

Part a. Since and x_in = 0.009, Eq. (16-23) is, x_p = (7/3)x_out + 0.009/0.3

Solving this equation simultaneously with the linearized RT equation, we obtain x_out = 0.0127 and x_p = 0.00034.

Part b. Since θ′ = 0.40 and x_in = 0.010, Eq. (16-23) is,

x_p = −(6/4)x_out + 0.010/0.4

Solving this equation simultaneously with the linearized RT equation, we obtain x_out = 0.0164 and x_p = 0.000439.

E. Check. Graphical solutions, formal linearization and fitting a linear trend line with a spreadsheet give essentially the same results.

F. Generalize. This example illustrates how to predict performance for an RO system without concentration polarization. Instead of linearizing the RT equation, we could have solved the quadratic RT equation and the linear operating equation simultaneously. At the low concentrations used in this example, the result will be essentially identical. Since concentration polarization is often important, prediction when mass transfer is not extremely fast is explored in Example 16-7.

16.4.2 Determination of Membrane Properties from Experiments

Experimental values of x_p and x_out can be used to determine α_AB and M. First, an experiment is done in a very well mixed cell under conditions where there is no concentration polarization. Solving Eq. (16-19) for the selectivity, we obtain

(16-25)

where under the conditions of the experiment M = 1 and all other terms on the right hand side are known. This method is illustrated in Example 16-5.

EXAMPLE 16-5. Determination of RO membrane properties

We continue to test the new composite membrane used in Example 16-4. Both experiments were done with a retentate pressure of 15.0 atm and a permeate pressure of 1.0 atm. The temperature is 25°C. The following data were obtained: Experiment a. The pure water flux was 1029 L/(m² day).

Experiment b. In a perfectly mixed laboratory system with wt frac sodium chloride, x_in = 0.0023, the rejection coefficient R was measured as 0.983. Operation was with θ′ = 0.30. The system was highly stirred and you can assume there was no concentration polarization (M = 1.0).

Find the pure water flux parameters and the water-salt selectivity.

Solution

A. Define. Find K_solv/t_ms and α_water-salt.

B and C. Explore and plan. Experiment a allows us to calculate K_solv/t_ms with x_r = x_p = 0.0 from Eq. (16-16c). Experiment b allows us to calculate x_p and x_out from Eqs. (16-23) and (16-24). Then, Eq. (16-25) (with x_r = x_out and M = 1.0) can be used to find the selectivity, α_water-salt.

D. Do it. Experiment a. With x_r = x_p = 0.0, we can solve Eq. (16-16c) for K_solv/t_ms.

(16-26)

Experiment b. Solving Eqs. (16-23) and (16-24) simultaneously, we obtain,

(16-27)

and once x_p is known

(16-28)

Plugging in numbers, these become

Equation (16-25) with M = 1 becomes,

(16-29)

where a′ = 15.446 atm/(wt frac) was determined in Example 16-4, and the density of solvent is 0.997 kg/L. The selectivity α can be determined from Eq. (16-29),

E. Check. The numerical values are within reasonable ranges.

F. Generalize. This example illustrates how to determine parameter values from experiments. The selectivity determined from the experiments was used in Example 16-3 to calculate expected behavior of a membrane module when there is no concentration polarization.

1. The pure water flux is often measured and used to find the value of K_solv/t_ms. This is a preferred method because it is very easy and is quite reproducible. Note that the flux declines significantly when there is salt present. Thus, the pure water flux values should never be used as a direct estimate of the feed capacity of the system.

2. Rejection data are commonly used to find the value of selectivity, α_w-salt. This selectivity does not have the same meaning as the selectivity in gas permeation. Here the selectivity is given by Eq. (16-20), but the permeabilities refer to equations with different driving forces, and the permeabilities have different units. Thus, an α less than 1.0 does not mean that salt is preferentially transferred through the membrane.

16.4.3 Determination of Concentration Polarization

Once values for α are known, experiments can be done under conditions where concentration polarization is expected. M can be determined by solving Eq. (16-19) for M.

(16-30)

All terms on the right hand side are known and M can be calculated. Estimation of M will allow us to avoid doing expensive and time consuming experiments, and is illustrated in Example 16-6.

Concentration polarization was shown schematically in Figure 16-9. This figure applies to the simplest situation that is steady-state, one-dimensional back diffusion of the solute into the bulk retentate stream with a perfectly rejecting membrane (R = 1). [If rejection is not almost complete, more detailed theories are required (e.g., Ho and Sirkar, 1992; Noble and Stern, 1995; Wankat, 1990).] The differential mass balance for this simple situation is (Problem 16.C.3)

(16-31)

where D is the diffusivity in the liquid solution. This equation is subject to the boundary conditions that concentration equals the wall concentration at z = 0,

(16-32a)

and the concentration becomes the bulk concentration x_r when z is greater than the boundary layer thickness δ. The value of δ depends on the operating conditions (geometry, velocity, T).

(16-32b)

Defining the mass transfer coefficient as

(16-33)

the solution is

(16-34)

The mass transfer coefficient depends on the operating conditions and has the same units as J_solv.

This short development is useful to determine what affects concentration polarization. If the solvent flux J_solv increases, M increases. If the mass transfer coefficient k increases, concentration polarization decreases. Increasing the diffusivity will increase k. Thus, operating at a higher temperature will decrease M although there are obvious limits based on the membrane thermal stability and the thermal stability of the solutes (e.g., most proteins are not thermally stable). Decreasing the boundary layer thickness δ by promoting turbulence or operating at very high shear rates in thin channels or narrow tubes will also increase k.

The quantitative use of Eq. (16-34) requires either experimentally determined values of the mass transfer coefficient k or a correlation for k (which is ultimately based on experimental data). If experimental data are available which allow the calculation of M from Eq. (16-30), then Eq. (16-34) can be used to find k. A large number of mass transfer correlations are available for a variety of geometries and flow conditions (Wankat and Knaebel, 1997). Four that are useful for membrane separators are the correlations for turbulent flow in tubes, for laminar flow in tubes and between parallel plates, and for well-mixed tanks (Blatt et al., 1970; Wankat, 1990; Wankat and Knaebel, 1997). For turbulent flow in tubes the mass transfer coefficient can be estimated from

(16-35a)

where Sh is the Sherwood number, and the Reynolds number Re, and Schmidt number Sc, are defined as

(16-35b,c)

and u_b is the bulk velocity in the tube. Fully developed turbulent flow will certainly occur for Re > 20,000, and usually appears in UF devices for Re > 2,000.

For laminar flow in a tube of length L and radius R with a bulk velocity u_b, the average mass transfer coefficient is,

(16-36a)

For laminar flow between parallel plates with a spacing of 2h, the average mass transfer coefficient is,

(16-36b)

For flat membranes in a well-stirred turbulent tank the mass transfer coefficient can be estimated from

(16-37a)

where k is in cm/s, ω is the stirrer speed in radians/s, d_tank is the tank diameter in cm, D is the diffusivity in cm²/s, and the kinematic viscosity ν = µ/ρ is in cm²/s. Stirred tanks are a convenient laboratory configuration.

EXAMPLE 16-6. RO with concentration polarization

We continue testing the new composite membrane explored in Examples 16-4 and 16-5. An additional experiment was done at 25°C with a retentate pressure of 15.0 atm and a permeate pressure of 1.0 atm. In addition to the data reported in Example 16-5, the following new data are obtained:

Experiment c. Experiments were done in a baffled stirred tank system with a 12 cm diameter tank. We expect that Eq. (16-37a) will be valid, except the coefficient 0.04433 has to be adjusted. The stirrer was operated at 900 rpm. The inlet solution was 0.005 wt frac sodium chloride. The measured wt frac were x_p = 0.0001287, and x_out = 0.006218.

Based on these experiments determine the value for the coefficient in the correlation for mass transfer in turbulent stirred tanks (Eq. 16-37a).

Solution

A, B, and C. Define, Explore and Plan. We can calculate the polarization modulus M from Eq. (16-30), J_solv from Eq. (16-16c), the value of the mass transfer coefficient from Eq. (16-34), and a new coefficient for the correlation in Eq. (16-37a).

D. Do it. Experiment c. Since x_p and x_out were measured, and α, a′, p_r, and p_p are all known, a straightforward plug-and-chug in Eq.(16-30),

The solvent flux can then be determined from Eq. (16-16c) using K_solv/t_ms = 73.5 determined in Example 16-5, and because x_r = x_out for a well-mixed tank,

Note that this is slightly smaller than the pure water flux because the driving force is reduced by the osmotic pressure difference.

Solving Eq. (16-34) for k,

k = J_solv/Ln M = 1020.62/ln (1.212) = 5308.2 L/(m² day) = 0.006144 cm/s.

Eq. (16-37a) can be written with an unknown constant instead of 0.0443. Solving for the constant in this equation, we obtain

(16-37b)

Since the solution is quite dilute, the properties of water can be used.

The diffusivity of NaCl in water at 18°C is 1.17 × 10⁻⁵ cm²/s (Sherwood et al., 1975, p. 37). We can estimate D at 25°C by assuming a linear dependence on temperature and then can determine the experimental value for the constant from Eq. (16-37b).

E. Check. For a general mass transfer correlation this is reasonably close to the published value of 0.04433. We will use the measured value of the constant to predict performance in Example 16-7.

F. Generalization. This example illustrates how we can take experimental data, and fine tune mass transfer correlations by adjusting the constants.

Experiment c required a number of steps to eventually find the mass transfer coefficient k. This is invariably the case since k is not a directly measured variable but depends upon interpretation of the data using a model. Once k was obtained we have a single data point to compare to the correlation Eq. (16-37a). They disagreed. You may be tempted to use the correlation and ignore the data point. However, mass transfer correlations are not very accurate. They usually predict the trends well (such as the effect of Reynolds and Schmidt numbers) but the absolute value predicted can be significantly off. A single data point can be used to adjust the constant in the correlation for application to this particular system. If more data were available, we could check the entire correlation.

EXAMPLE 16-7. Prediction of RO performance with concentration polarization

Predict the values of x_p, x_out, and J_solv if a 0.01 wt frac sodium chloride in water solution is separated by the membrane studied in Example 16-4 to 16-6 in a stirred tank which is geometrically similar to the one in Example 16-6 except the tank is 20 cm in diameter and the stirrer speed is 1,500 rpm. The retentate pressure is 10 atm and the permeate pressure is 1 atm. Operate at a cut θ′ = 0.40 and at 25°C.

Solution

A, B, and C. Define, Explore and Plan. The prediction requires that we first calculate k using Eq. (16-37a) with the modified constant. At this point the solution becomes complicated. To find M from Eq. (16-34) we need to know J_solv which could be determined from Eq. (16-16c) if we knew x_r = x_out, x_p, and M. Since the cut θ′ is specified, x_out and x_p can be found by the simultaneous solution of Eqs. (16-21) and (16-23) if M were known. This loop can be cut by using a trial-and-error solution. One relatively convenient way to do the trial-and-error, that requires manipulating the equations, is illustrated in the Do it section.

D. Do it. Since the solution is still quite dilute, we will use the properties for water and the diffusivity from Experiment c. The difficulty with this calculation is that all the variables needed are required for calculation of each other. The RT curve depends on M, and if plotted as x_r vs. x_p will have to be replotted for each value of M. An alternative is to rearrange Eq. (16-21) as,

(16-38)

Now a plot of x_p vs. (Mx_r) = (Mx_out) will be a single curve. To correspond to this new graph, we rearrange the operating Eq. (16-23) as,

(16-39)

On an x_p versus (Mx_out) graph this is a straight line with a slope of −[1 − θ′]/[Mθ′] and an x_p intercept (Mx_out = 0) of x_p = x_in/θ′.

The operating equation will need to be replotted every time M changes.

The (Mx_out) intercept (x_p = 0) is (Mx_out) = Mx_in/(1 -θ′)

Eq. (16-38) can be used to generate the values for a table.

The calculation is done only in this dilute range because the feed is very dilute. The operating line, Eq. (16-39) will have,

Mx_out intercept (x_p = 0) = 0.01M/(1-0.4) = 0.0166667M

As a first guess use M = 1.20 which is slightly less than the previously measured value. The results are plotted in Figure 16-11.

Operating line Slope = −1.5/1.2= −1.25 and Mx_out intercept (x_p = 0) = 0.020.

Solution with M = 1.20 from Figure 16-10, x_p = 0.0041, Mx_out = 0.0167

Thus, x_out = (Mx_out)/M = 0.0167/1.20 = 0.0139

Eq. (16-16c) becomes

J_solv = 73.5[(10 − 1) − 15.446(0.0167 − 0.0041)] = 647.1 L/(m²day)

Eq. (16-30) will give the calculated value of M once we determine k from Eq. (16-33) using the coefficient found from experiment c. Using D, ρ and µ from Experiment c,

Now we can calculate M from Eq. (16-30) and check our guess.

Need a lower value. Try direct substitution and use the calculated value, M = 1.0734.

New Operating Line: Slope = −1.5/1.0734 = −1.3974

Mx_out intercept = 0.0166667 (1.0734) = 0.01789

Then from Figure 16-11, x_p = 0.0037, Mx_out = 0.0152, and x_out = 0.0152/1.0734 = 0.142

J_solv from Equation (17-16c) is,

J_solv = 73.5[(10 − 1) − 15.446(0.0152 − 0.0037)] = 648.3 L/m²day

Since k is unchanged, Eq. (16-30) gives

M_calc = exp(648.3/9140.8) = 1.0735

This value of M_calc is essentially identical to the guess. Thus, use M = 1.0734

The predictions are

J_solv = 648.3 L/m²day, x_p = 0.0037 wt frac. and x_out = 0.0142 wt frac.

E. Check. The results obtained are consistent with what we expect. This is a helpful and quick check but does not guarantee there are no errors. We can also check the results for the limiting case when M = 1.0 with the results obtained in Example 16-4 for the same feed concentration, pressures and cut as this problem (x_out = 0.0143 and x_p = 0.0035). In Figure 16-11 when M = 1, the intersection of the operating line (not shown for M = 1) and the RT line occurs at x_out = 0.0144 and x_p = 0.0034. The agreement is quite good.

F. Generalization. 1. The prediction of performance was trial-and-error because the unknown variables were needed to calculate other unknowns that were needed to calculate the first unknown. This circle is broken by guessing a variable, doing the calculation, and then checking the guess. Replotting the graphical solution method for a well-mixed system made the trial-and-error simpler. Convergence was rapid. In more concentrated systems or with less vigorous stirring with much larger M and larger π values (which may be nonlinear functions of wt frac), convergence can be slower.

2. This entire calculation can also be done in a spreadsheet using Solver to find a simultaneous solution of the equations. The graphical solution then becomes a very convenient check since M will be known from the spreadsheet solution.

3. There was less concentration polarization in the larger system than in experiment c because the mass transfer coefficient was significantly larger (9140.8 compared to 5932). The larger Reynolds number (larger diameter and higher ω) caused the larger value of k.

4. RO is commonly used to produce ultrapure water in the electronics and pharmaceutical industries. In these applications R° is much closer to 1.0 and x_in is smaller. These systems have little concentration polarization (M ~ 1.0) and produce very pure permeate.

5. All of these calculations assume an undamaged membrane with no holes. Even a tiny pinprick can cause a large increase in x_p. The liquid will pass through a hole as convective flow at a salt concentration of x_w. This flux will be quite large because of the large pressure drop. In addition, undesired large molecules can also pass through holes in the membrane. Performance of RO systems needs to be monitored continuously, and damaged membranes need to be plugged or replaced.

6. Membrane life will depend upon the membrane material and the operating conditions. Membrane replacement costs should be included in the operating expenses.

7. If a limiting case does not agree with an independent calculation, then there must be an error in either the original calculation, the method to produce the limiting case or the independent calculation. If the limiting case agrees with an independent calculation (as it does for this problem for M = 1), we have not proved that the calculation for M ≠ 1 is correct. As the number of limiting cases that agree with independent calculations increase, our confidence in the general solution increases.

8. The problem statement stated the tanks are geometrically similar. If they aren’t, the constant in mass transfer correlation, Eq. (16-37a) is probably different in the two tanks. Geometric similarity allows one to scale-up.

16.4.4 RO with Concentrated Solutions

A more complicated situation occurs when x_r, and hence x_w, are concentrated and osmotic pressure depends upon x_p in a nonlinear fashion. An Advanced RT equation can be developed if the osmotic pressure of permeate is linear in x_p, π_p = ax_p. Since permeate is quite dilute, this equation is often valid even if π is not a linear function of x at x_in and x_out. Start with Eq. (16-16a) and substitute in x_w = Mx_r and π_p(x_p) = a′ x_p. Solve resulting equation and Eq. (16-18) for . Then set equal and rearrange.

(16-40a)

We assume the osmotic pressure π(Mx_r) is available in tabular or equation form. Expand in terms of x_p and collect terms.

(16-40b)

If we pick a value of Mx_r we can determine π(Mx_r) and calculate x_p. Then we can generate the RT curve including nonlinear osmotic pressure and concentration polarization. If the membrane module is perfectly mixed, the operating equation is Eq. (16-39). The solution then proceeds as in Example 16-5.

16.5 ULTRAFILTRATION

Ultrafiltration (UF) is another membrane separation method used to purify liquids. UF is commonly used for recovery of proteins and in food and pharmaceutical applications. It is useful for separating “permanent” emulsions since the oil droplets will not pass through the membrane. UF is used for the removal of fine colloidal particles, and for recovery of dyes from wastewater. In many applications such as whey processing UF and RO are used in series. The valuable proteins are recovered by UF, and permeate from the UF system is sent to the RO system. The remaining sugars and salts are concentrated in the RO system by removing water. The concentrated permeate can then be fermented to produce ethanol, lactic acid or other products.

The equipment for UF systems often looks very similar to RO systems although they operate at lower pressures. However, this similarity does not extend to the molecular level. Remember that RO membranes are nonporous and separate based on a solution-diffusion mechanism. UF membranes are porous and separate based on size exclusion. Large molecules are excluded from pores in the thin membrane skin and thus, the large molecules are retained in the retentate. Small molecules fit into the pores and pass through to the permeate. Since there is usually a distribution of pore sizes, molecules within the range of pore sizes partially permeate and are partially retained. In a somewhat oversimplified picture, UF is cross-flow filtration at the molecular level.

Because of the different separation mechanism, UF membranes have significantly higher fluxes than RO membranes. Thus, concentration polarization is usually worse in UF than in RO because there is a much greater solvent flow from the bulk fluid through the wall. This concentration polarization can cause membrane fouling which not only decreases the flux but also can drastically decrease the membrane life. Hydrophilic membranes tend to foul less rapidly but have shorter lives than the more stable hydrophobic membranes. The choice of the best membrane thus depends upon the operating conditions. Cellulose acetate (Figure 16-3) membranes were the first commercial membranes and are still used where their low level of interaction with proteins is more important than their relatively short life. Polymeric membranes are used where more basic conditions are encountered. The most common polymeric membrane is polysulfone (Figure 16-3). Membranes are tailor-made to sieve molecules in different size ranges depending on the purpose of the separation (Figure 16-12). The nominal molecular weight exclusion (shown by x in Figure 16-12) is often reported by manufacturers, but it is not nearly as useful as the complete retention curve.

Figure 16-12. Solute retention on Amicon Diaflo membranes (Porter, 1997), reprinted with permission from P.A. Schweitzer (Ed.), Handbook of Separation Techniques for Chemical Engineers, 3rd ed., (1997), copyright 1997, McGraw-Hill.

UF can be initially analyzed by the same procedures used for RO; thus, Eqs. (16-12) and (16-16) are valid. However, in UF the molecules being retained are often very large and the resulting osmotic pressure is very low. For most UF applications the osmotic pressure difference can be ignored and the solvent flux equation can be simplified. The volumetric flux is,

(16-41)

In mass units the solvent flux is

(16-42)

where is the mass transfer rate of permeate (e.g., kg/s), x_p is the wt frac of solute in the permeate, and ρ_solv is the density of pure solvent. As we will see, ignoring the osmotic pressure difference is a more important simplification than it looks at first.

For sieve type membranes if a pore does not exclude solute, it will carry solute at the wall wt frac, x_w = Mx_r, through the pore. In sieve type membranes the inherent rejection R° (M = 1) can be interpreted as the fraction of flux carried by pores which exclude solute. Then 1 − R° is the fraction of flux carried by pores that do not exclude solute.

The permeate wt frac, x_p, is then

(16-43)

This equation can also be solved for the retentate wt frac,

(16-44)

Equation (16-43) or (16-44) are the RT equations for UF. They are particularly simple because the inherent rejection R° is based on experimental data. (If R° is known experimentally for RO for the operating conditions of interest, the same procedure can be used for RO. This RO result is simpler than the procedure outlined earlier but requires data that is often unavailable.) Note that the RT equations for UF depend only on the solute rejection and M. For a perfectly mixed membrane module we assume that x_r = x_out. Then either Eq. (16-43) or (16-44) written in terms of x_out can be solved simultaneously with the mass balances, Eqs. (16-22) or operating Eq. (16-23). This simultaneous solution can again be obtained by plotting the RT curve and the operating equation on a graph of x_p vs. x_r,out (see Problem 16.C4). Equations (16-23) and (16-43) or (16-44) can also be solved analytically or numerically (see Problem 16.C5).

Experimental results with low concentration feeds or under conditions where M is close to 1.0 are in good agreement with the theoretical predictions. However, when the wall concentration becomes high, the solvent flux J_solv often cannot be controlled by adjusting the pressure difference. Thus, Eq. (16-40) no longer holds! Some other phenomenon must be controlling the solvent flux. Careful examination of the membrane surface after these experiments shows a gel-like layer covering the membrane surface. This gel layer alters the flux-pressure drop relationship and controls the solvent flow rate.

The effect of gel formation at the wall can be studied using the diffusion equation. Return to Figure 16-10. We implicitly assumed that the wall concentration was a variable that could increase without bound. In gelling systems once the wall concentration equals the gel concentration, x_g, the concentration at the wall becomes constant at x_w = x_g. As additional solute builds up at the wall, the gel concentration is unchanged, but the thickness of the gel layer increases; thus, x_w becomes a constant set by the solute gelling behavior. This is illustrated in Figure 16-13. The value of x_g can vary from less than 1 wt % for polysaccharides to 50 vol % for polymer latex suspensions (Blatt et al., 1970).

Figure 16-13. Concentration polarization with gel layer at wall

To analyze gelling systems we can again use Eq. (16-31) when R = 1. (Once a gel forms it is usually quite immobile and 100% retention is reasonable.) With the coordinate system redefined as in Figure 16-13, the boundary conditions are the same as Eqs. (16-32). The solution has the same form as in Eq.(16-34); however, since x_w and x_r are fixed, the solvent flux is the variable. If we solve for the solvent flux, the result is,

(16-45)

The consequence of this equation is that once a gel has formed the solvent flux is set by the rate of back diffusion of the solute. We no longer control the solvent flux!

What happens if we increase the pressure drop across the membrane for a system with a gel present? The solvent flux will temporarily increase, but more solute is carried to the membrane by the flow through the membrane than can be removed by back diffusion. The extra solute is deposited on the gel layer, which increases the thickness of the gel layer. This increases the resistance to flow, which reduces the flux through the membrane until the steady state solvent flux given in Eq. (16-45) is again obtained. Note that this effect is usually not reversible. Reduction of the pressure will result in a flow rate less than that predicted by Eq. (16-45) because the thicker gel layer remains on the membrane. The gel layer can often be removed by shutting down the system and backflushing (running pure solvent at a higher pressure on the permeate side) or by mechanical scrubbing of the membrane surface. Unfortunately, the membrane may become fouled after a gel forms and it may be very difficult to return the membrane to its original flux behavior. The use of fouling-resistant membranes is highly recommended under gelling conditions.

The engineer does have some control since decreasing the boundary layer thickness, δ, with thin channels or turbulence will increase the mass transfer coefficient k and hence the flux. To a limited extent D can be increased by raising the temperature. The effect of these variables on the mass transfer coefficient can be explored using correlation Eqs. (16-35) to (16-37). If fouling is severe after gel formation, it may be necessary to operate so that a gel will never form: x_w = Mx_r < x_g at all times. This can be done by operating under conditions that reduce M (high values of k) and keeping x_r low (low feed concentrations and low cuts). When a gel does not form the solvent flux in UF is controlled by Eqs. (16-41), the concentration polarization modulus can be calculated from Eq. (16-34), and solution is straightforward.

Most feeds processed by UF contain a number of different solutes. How should we analyze these separations? The almost overwhelming temptation is to study each solute individually and then assume that superposition is valid. That is, we assume that each solute in the mixture will behave in the same way as it does alone. After all, this is what we did for gas permeation and RO. Thus, we would predict that large molecules will be retained and small molecules will pass through the membrane. If no gel forms, this behavior is often observed; however, if a gel forms, the gel is usually much tighter (less porous) than the membrane, and the gel layer will usually capture the small molecules. Thus, the separation behaviors with and without the gel are quite different. If the purpose of the UF operation is to separate the large and small molecules, then we must operate under conditions where a gel will not form. Gel formation and fouling have prevented UF from achieving its full potential because they often limit both the separation and the flux.

RO membranes will also cause gelling and will foul if they are operated with feeds that can gel. To prevent this it is common to use an UF system before an RO system. Then the UF system will remove particulates and large molecules that could foul the RO membrane. This procedure is followed in the processing of whey where the UF system retains proteins and the RO system retains sugars and salts.

Example 16-8. UF with gel formation

We are ultrafiltering latex particles which are known to form a gel when x_w = x_g = 0.5.

The well mixed system is operated with permeate pressure of 1.0 bar and retentate pressure of 4.5 bar. We do a series of experiments with different values of the inlet concentration with a cut of 1/3. The permeate wt frac is zero for all of the experiments. We find that gelling occurs when x_in = 0.1466, and the measured flux is 4,500 L/(m² day). Predict the solvent flux for inlet wt frac x_in = 0.20.

Solution

A. Define. Find J_solv when x_in = 0.20, θ′ = 1/3, p_p = 1.0 bar, and p_r = 4.5 bar.

B and C. Explore and plan. Use a mass balance to determine x_r = x_out for x_in = 0.1466, and Eq. (16-45) to determine k using the measured flux at this inlet wt frac. Then for x_in = 0.20 find x_r = x_out (since x_p is reliably zero, we can assume it remains zero for x_in = 0.20). Use Eq. (16-45) to find J_solv at this higher feed wt frac.

D. Do it. Experimental conditions:

With

When x_in = 0.1466, x_out = 1.5 (0.1466) = 0.220.

Rearranging Eq. (16-45),

Now solve Eq. (16-45) with x_in = 0.20 and x_out = 1.5 x_in = 0.30.

Final result is:

E. Check: Qualitatively we would expect a lower flux since there are more latex particles carried toward the wall per liter of fluid which permeates through the membrane. Other than checking the equations and calculations, a check is difficult.

F. Generalize. 1. We have assumed k is not concentration dependent. This is reasonable for particles.

2. If a correlation for k was available, we could use Eq. (16-45) and J_solv to estimate x_g.

3. The general procedure was to:

a. Use rearranged design equation to find a design parameter (k) using experimental conditions. Then

b. Use design equation to predict flux under design conditions.

This general procedure is very common in all types of separation problems.

4. It is critically important to determine if a gel layer forms.

Warning! Experimental results and equations for both UF and RO are reported in many different units. It is easy to make a unit mistake if you do not carefully carry units in the equations. It is especially easy to make a subtle mistake in the appropriate solution density to use when converting from concentration c in g/L or mole/L to wt frac x. The correct conversion from g/L to wt frac is,

(16-46)

Unfortunately, the solution density ρ_solution is a function of concentration. For a relatively pure permeate (R close to 1.0) the permeate solution density is approximately the solvent density. For more concentrated streams such as the feed or the retentate the density needs to be known or estimated as a function of concentration.

When concentrations are given in units of g/L, approximations are often made to calculate fluxes. For example, the correct equation for the solute mass flux is

(16-47a)

This is often approximated as

(16-47b)

For relatively high rejection coefficients permeate is quite pure, J_solv >> J_A and the approximation is quite good. In other cases the approximation may not be as accurate.

The rejection and concentration polarization modulus are often defined in concentration units,

(16-48)

With these definitions the solute mass flux is

(16-49)

where is the inherent rejection measured when there is no concentration polarization (M_c = 1). Equations (16-48) and (16-49) are valid ways to formulate the problem. Unfortunately, it may be assumed that and M = M_c when the exact equations are

(16-50a)

(16-50b)

If ρ_solution,out and ρ_solution,p are very different, confusing R with R_c and M with M_c can result in significant error. To be exact, check how all terms are defined, use the appropriate solution densities of permeate and retentate to convert to wt frac units, and then solve in wt frac units.

16.6 PERVAPORATION

Pervaporation is a rapidly expanding membrane separation technique because very high selectivities with reasonable fluxes are often obtained. In pervaporation a high-pressure liquid is fed to one side of the membrane, one component preferentially permeates the membrane, then evaporates on the downstream side, and a vapor product, the permeate, is withdrawn (Figure 16-14). The word “pervaporation” is a contraction of permeation and evaporation. The retentate, which does not permeate through the membrane, is a high-pressure liquid product. Either permeate or retentate may be the desired product.

Figure 16-14. Simplified schematic of single-pass pervaporation

In pervaporation the driving force for separation in Eq. (16-1) is the difference in activities across the membrane. The flux equation is (Eykamp, 1997),

(16-51)

where a_i is the activity. Unfortunately, neither this equation nor Eq. (16-1) is of great practical use. Since a detailed analysis is usually not solvable in practical situations (e.g., see Neel, 1992) we will use a greatly simplified theory, which relies heavily on experimental selectivity data. The modeling procedures used earlier for gas permeation do not work for pervaporation because the fluxes of the two components interact significantly, the membrane swells, and permeabilities are very concentration dependent. Detailed models are available in the books by Dutta et al. (1996-97), Ho and Sirkar (1992), Huang (1991), Mulder (1996), and Noble and Stern (1995).

Since both selective membrane permeation and evaporation occur, pervaporation both separates and concentrates. Evaporating the liquid at the downstream side of the membrane also increases the driving force. Assume that the driving force is adequately represented by the partial pressure difference across the membrane. The local partial pressure on the upstream side, labeled 1 in Figure 16-14, is

(16-52a)

while the local partial pressure on the downstream side, labeled 2, is

(16-52b)

and the local driving force is

(16-52c)

Since x_i,1, y_i,2 and p_tot,2 vary, the local values of the partial pressures and the driving force depend upon the flow patterns and pressure drop in the membrane module. The driving force can be increased by lowering p_tot,2 either by drawing a vacuum as shown in Figure 16-14, or, less frequently, reducing y_i,2 by using a sweep gas on the permeate side. The driving force will also be increased if the upstream partial pressure of component i is increased. This occurs for more concentrated feeds (higher x_i,1) and at higher upstream temperatures (larger VP_i,1). Use of higher upstream temperatures may require a higher upstream pressure to prevent vaporization of liquid on the upstream side. The higher upstream pressure does not increase the permeation rate significantly (Neel, 1991). Although the feed concentration is not usually a variable under the control of the designer in a single pass system (see Figure 16-14), it is a design variable in hybrid systems (see Figure 16-15).

Figure 16-15. Hybrid pervaporation system coupled with distillation; A) general two-column system, B) simplified one-column system shown for dehydration of ethanol

As usual with membrane separations, the membrane is critical for success. Currently, two different classes of membranes are used commercially for pervaporation. To remove traces of organics from water a hydrophobic membrane, most commonly silicone rubber is used. To remove traces of water from organic solvents a hydrophilic membrane such as cellulose acetate, ion exchange membrane, polyacrylic acid, polysulfone, polyvinyl alcohol, composite membrane, and ceramic zeolite is used. Both types of membranes are nonporous and operate by a solution-diffusion mechanism. Selecting a membrane that will preferentially permeate the more dilute component will usually reduce the membrane area required.

The use of evaporation to increase the driving force allows one to use a highly selective membrane and still retain a reasonable flux. However, evaporation complicates both the equipment and the analysis. Typical permeate pressures are quite low (0.1-100 Pa) (Leeper, 1992). Because of this low pressure, permeate needs a large cross-sectional area for flow or the pressure drop caused by the flow of the permeate vapor will be large. Thus, plate-and-frame, spiral-wound, and hollow-fibers with feed inside the fibers are used commercially. Figure 16-14 shows that a vacuum pump and a condenser are required to recover the dilute low-pressure vapor. Unless the cut is small, an additional energy source is required to supply the heat of evaporation. If the cut is small, the energy in the hot liquid can supply this energy. For larger cuts a portion of the retentate can be heated and then recycled. More details on equipment and equipment design are given in Huang’s (1991) book.

Although selectivity in pervaporation units can be quite high, (e.g., Leeper (1992) reports values from 1.0 to 28,000) the values are not infinite. Thus, there will always be some of the less permeable species in the permeate and some of the more permeable species in the retentate. If the feed concentration and the selectivity are high enough, the single pass system shown in Figure 16-14 can produce either a permeate or a retentate that meets product specifications. However, the other stream will contain a significant amount of valuable product. The single pass system can produce high purity or high recovery but not both simultaneously.

If both high purity and high recovery are desirable or the single pass system cannot meet the product specifications, a hybrid system with recycle is often used (Huang, 1991; Suk and Matsura, 2006; Wankat, 1990). Hybrid systems use two different types of separation to achieve the desired total separation. The most common pervaporation hybrid is to combine it with distillation with either two columns (Figure 16-15A) or a single column (Figure 16-15B). In Figure 16-15A the feed to distillation column 1 forms a minimum-boiling azeotrope. This distillation produces essentially pure component A as the bottoms product. The distillate product from column 1, which approaches the azeotrope concentration, is sent to the pervaporation unit. If component A preferentially permeates through the membrane, permeate will be more concentrated in A than the distillate. Permeate is then recycled to column 1 to recover the A product. Component B is retained in the pervaporation unit and is concentrated in retentate, which is the feed to distillation column 2. The distillate from column 2 also approaches the azeotrope concentration, and is part of the feed to the pervaporation unit. The bottoms product from column 2 is essentially pure B. The two distillation columns in Figure 16-15A will be similar to the columns in Figure 8-3A. Since the pervaporation unit replaces the liquid-liquid separator in Figure 8-3A, the azeotropic system does not have to have a heterogeneous azeotrope; thus, Figure 16-15A is more generally applicable.

If the selectivity is high enough, retentate may meet the purity specifications. Then distillation column 2 is not needed which results in obvious savings in capital and operating costs. This is illustrated with a simplified one-column flowchart used for breaking the ethanol-water azeotrope (Figure 16-15B). Open steam heating may be used instead of a reboiler (Leeper, 1992). A hydrophilic membrane that selectively permeates water is used. This figure is essentially the same as Figure 8-1B with the pervaporation system replacing the unidentified separator.

Hybrid systems are usually designed with a low cut per pass to prevent large temperature drops in the system. The total cut for the entire unit can be any desired value. Typically the vacuum pump is the highest operating expense. The load on the vacuum pump can be decreased by refrigerating the final stage of the permeate condenser. Appropriate design of the heat exchanger system can significantly reduce the energy costs.

If selectivity and flux data are available, pervaporation units can be designed without knowing the details of the concentration polarization, diffusion, and evaporation steps. For a binary separation the selectivity, α_A,B is defined as,

(16-53a,b)

Where x and y are the wt frac in the liquid and vapor, respectively. Molar units can also be used, but data are often reported as wt frac. This definition is superficially similar to the definition of relative volatility in distillation. Of course, here the selectivity is for a rate process and represents the Rate Transfer (RT) curve, not equilibrium. Selectivity data are usually obtained under conditions where concentration polarization is negligible. Experimentally determined selectivities are complex functions of temperature and liquid mole frac (e.g., see Figure 16-16).

Figure 16-16. Effects of temperature and liquid composition for cross-linked PVA membrane with 12 wt % cross-linking agent. Plot of permeation rate and separation factor (Huang and Rhim, 1991), copyright 1991. Reprinted with permission from Elsevier

If data are available as selectivities we can convert it to a y vs. x format by solving for y_A in Eq. (16-53a)

(16-53c)

Since the selectivity depends upon the liquid concentration, Eqs. (16-53a,b,c) are valid locally.

We will again consider the simplest case—membrane modules that are well-mixed on both the retentate and permeate sides. In this situation Eq. (16-53c) is valid with y replaced by y_p and x replaced by x_out,

(16-54)

where y_p and x_out refer to the wt frac of the more permeable component in the permeate and retentate, respectively. The selectivity in Eq. (16-54) must be determined at the operating temperature T and the liquid wt frac x_out. If data are available in the form of Figure 16-17, it must be at the operating temperature T of the pervaporation system.

Figure 16-17 Vacuum-pervaporation of water-ethanol mixtures through homogeneous films made from hydrophilic polymer. c, c′: Water concentration (by weight in liquid (c) and in permeate (c′)); J′: Permeation flux. Polyacrylonitrile film (20 µm thick), T = 25°C. (Neel, 1991), copyright 1991. Reprinted with permission from Elsevier

The overall mass balance for the single pass system shown in Figure 16-13 is

(16-55a)

and the mass balance for the more permeable species is

(16-55b)

These equations are identical to the balances for gas permeation, Eqs. (16-7a) and (16-7b), except that x has replaced y for the feed and retentate. These two equations can be solved simultaneously for y_p. The resulting operating equation is

(16-56)

where the cut . This result is essentially the same as Eq. (16-8). If we want to use a graphical procedure, the operating equation will plot as a straight line on a graph of y_p vs. x_out.

If x_out and T are specified (which means α_AB is known), Eqs. (16-54) and (16-55a,b) can be solved simultaneously. The resulting equation is quadratic in y_p and linear in θ′ (Wankat, 1990, pp. 707-709). We can easily solve for y_p. Unfortunately, this procedure is more complicated and less useful in real situations than it appears at first glance. Since neither x_out nor T is usually known, the selectivity is not known and the calculation becomes complicated. We will use a simultaneous graphical solution to develop a somewhat simpler procedure that will be illustrated in Examples 16-9 and 16-10.

An energy balance is needed to estimate the temperature of the pervaporation system. We will assume that the pervaporation system is operating at steady state and that it is adiabatic. Then the energy balance for the system shown in Figure 16-14 is

(16-57a)

We choose the reference point as pure liquid component A at temperature T_ref. Assuming that heat of mixing is negligible, the enthalpies of the liquid streams are

(16-57b)

and the vapor enthalpy is

(16-57c)

where λ_p is the mass latent heat of vaporization of the permeate determined at T_ref and the heat capacities are also in mass units. Combining Eqs. (16-57) we obtain

(16-58)

Since the membrane module is well mixed, it is reasonable to assume that the system is in thermal equilibrium, T_out = T_p. If we arbitrarily set the reference temperature equal to the outlet temperatures, T_ref = T_out = T_p, we obtain the simplified forms of the energy balance.

(16-59a)

(16-59b)

(16-59c)

From Eq. (16-59b) we can determine the cut necessary to obtain a specified outlet temperature, while Eq. (16-59c) allows us to determine the outlet temperature for any specified cut. Note that the relationship between cut and the temperature drop of the liquid stream is linear. Equation (16-59c) is instructive. Since the latent heat is significantly greater than the heat capacity, the outlet temperature drops rapidly as the cut is increased. To prevent this drop in temperature a recycle system with a low cut per pass is often used.

We are now ready to develop the solution procedure for completely mixed pervaporation systems. This procedure is straightforward if either the outlet temperature or the cut are specified. If the cut is specified, calculate T_out from Eq. (16-58c). To plot the RT curve, pick arbitrary values for x_out, calculate α_AB from appropriate data such as Figures 16-16 or 16-17, calculate y_p from the RT Eq. (16-54), and plot the point on the curve. If experimental y vs. x data at operating temperature T is available, plot it directly as y_p vs. x_out. The simultaneous solution to the selectivity equation and the mass balances is at the point of intersection of the RT curve and the straight operating line, Eq. (16-56) (see Figure 16-18 in Example 16-9).

Figure 16-18. Solution for Example 16-9; RT curve for water-ethanol through 20 µm polyvinyl alcohol at 60°C. (Neel, 1991). Flux curve from Neel (1991), copyright 1991. Reprinted with permission from Elsevier

Example 16-9. Pervaporation: feasibility calculation

We wish to separate water from ethanol using a single-pass pervaporation system that uses a 20-micron thick polyvinyl alcohol film. The perfectly mixed pervaporation system will operate at 60°C with 1,000 kg/hr of hot feed. The feed is 60 wt % water. If the retentate product is 20 wt % water, is this process feasible?

Solution

A. Define. In this case, feasibility means can the process be conducted with the feed at a reasonable temperature? Thus, to determine feasibility we need to determine the required inlet temperature.

B and C. Explore and Plan. The rate transfer information and the flux data for this film are given by Neel (1991), and are shown in Figure 16-18. The data required for the energy balances is available in Perry and Green (1997) on pages 2-235 and 2-306:

Ethanol: C_P,L,E = 2.78 kJ/(kg K) and λ_E = 985 kJ/kg. Estimated at 60°C.

Water: C_P,L,W = 4.185 kJ/(kg K) and λ_W = 2359 kJ/kg. Estimated at 60°C.

We will solve for the cut and the feed temperature required and then discuss feasibility.

D. Do it. The data from Neel (1991) is plotted in Figure 16-18. The operating line (not shown by Neel), Eq. (16-56) goes through the point y_p = x_out = x_in which is on the y = x line. The operating line must intersect the RT curve at x_out = 0.20. The permeate wt frac can be read at this point as y_p = 0.95 wt frac water.

Since all the wt frac are now known, it is easiest to simultaneously solve Eqs. (16-55a) and (16-55b) for the cut, θ′.

(16-60)

For this problem, θ′ = (0.6 − 0.2)/(0.95 − 0.2) = 0.533.

This cut appears to be too high. To check for feasibility, we can use the energy balance to estimate the required inlet temperature. Rearranging Eq. (16-59c), we obtain

(16-61)

The latent heat of the permeate can be estimated as

(16-62)

The heat capacity of the feed at 60°C can be estimated as

(16-63)

Then from Eq. (16-61), T_in = 60 + (0.533) (2290)/(3.62) = 396.9°C.

This is obviously too hot. The process is not feasible because the membrane will be thermally degraded and the feed will not be a liquid. Too large a cut is used.

E. Check. Since typical values for the cut are approximately 0.1 or less, our conclusion that the process is not feasible is reasonable.

F. Generalization. Example 16-9 illustrates that operation of single-pass pervaporation units is controlled by the simplified energy balance, Eq. (16-59c). Essentially, there must be sufficient energy in the feed liquid to vaporize the desired fraction of the feed. High cuts require very hot feeds to supply the energy needed for vaporization.

EXAMPLE 16-10. Pervaporation: development of feasible design

Since the operation in Example 16-9 was not feasible, choose an appropriate feed temperature, and develop a feasible pervaporation process for the feed in Example 16-9. A permeate that is greater than 95 wt % water is desired. Determine the cut, permeate and retentate wt frac and the membrane area.

Solution

A. Define. An appropriate feed temperature is within the temperature limits of the membrane, and the feed must be liquid at a reasonable pressure. There is no single correct answer.

B. and C. Explore and plan. A pressurized feed will still be liquid at 120°C. With a smaller cut, permeate will be greater than 95% water. If the cut were zero, the operating line will be vertical. From Figure 16-18, y_p = 0.98 in this case. We can use y_p = 0.98 to estimate λ_p, which is needed to calculate the cut. The accuracy of this estimate can be checked when we are finished.

D. Do it. We can estimate the latent heat of vaporization of the permeate product (y_p = 0.98) as, λ_p = 0.98 λ_W + 0.02 λ_E = 2331 kJ/kg. The value of C_PL,in = 3.62 kJ/(kg°C) was determined in 16-59b),

From Eq. (16-59) the operating line has a slope = −(1 − 0.093)/0.093 = −9.75. The operating line is drawn in Figure 16-18. The permeate wt frac is about 0.98 (thus, the approximation was accurate) and the retentate is slightly above 0.56.

The membrane flux can be determined from the flux part of Figure 16-18. J′ is about 5.2 kg/(h m²). Then, .

F. Generalization. At this retentate liquid concentration the membrane has a high flux. At the original retentate value, x_out = 0.2 used in Example 16-9, the flux is much lower and significantly more membrane area would be required.

Of course, there are many different feasible designs. If a higher permeate concentration is desired, the ethanol-water feed mixture in Example 16-10 would probably be concentrated in an ordinary distillation column to a concentration much closer to the concentration of the azeotrope (see Figure 16-15B). The retentate would typically be recycled to the distillation column. Pervaporation, azeotropic distillation (Chapter 8) and adsorption (Chapter 17) are all used commercially to break the ethanol-water azeotrope.

16.7 BULK FLOW PATTERN EFFECTS

Completely mixed membrane systems are relatively common in laboratory units since the effects of concentration polarization and gel formation can be minimized or eliminated. Commercial units usually use one of the modules shown in Figure 16-1 since a large membrane area can be packaged in a small volume. The flow patterns of the bulk fluid in these modules are unlikely to be perfectly mixed but are more likely to approximate cross-flow (Figure 16-19a), co-current (Figure 16-19b) or countercurrent (Figure 16-19c) flow. Since these bulk flow patterns all result in better separation than a perfectly mixed module, our previous results are conservative estimates. In this section we will analyze ideal examples (that is with no axial mixing and no concentration polarization) for gas permeation for these other flow patterns to determine the improvement in separation. Because real systems usually have some axial mixing, the separation is often between that predicted for the ideal case and the perfectly mixed case.

Figure 16-19. Flow patterns in gas permeation systems; A) cross-flow with unmixed permeate, B) co-current flow, C) countercurrent flow

EXAMPLE 16-11. Flow pattern effects in gas permeation

The increase in separation for the different flow patterns can be compared for separation of oxygen and nitrogen. Geankoplis (2003) presents a problem with a feed that is 20.9 % nitrogen, p_r = 190 and p_p = 19 cm Hg, P_O2/t_ms = 1.9685 × 10⁻⁵, α_O2−N2 = 10.0, F_in = 1,000,000 cm³/s, and cut = 0.2. The models discussed later in this section were implemented using the spreadsheets in the chapter appendix. The results shown in Table 16-3 illustrate that a countercurrent flow pattern provides better separation with less membrane area.

Table 16-3. Results for oxygen mole fracs for different flow patterns in gas permeation systems.

Geankoplis (2003) presents a complicated analytical solution for the completely mixed and cross-flow cases. He obtained identical results for the completely mixed case and almost identical results for cross-flow (see Table 16-3.

It is convenient to use staged models (Figure 16-20) to analyze the different flow patterns in permeation systems. Essentially, this represents a numerical integration of the differential equations representing the three configurations shown in Figure 16-19 (Coker et al., 1998).

Figure 16-20. Staged models for different flow patterns in gas permeation; A) cross-flow system, B) co-current system, C) countercurrent system

16.7.1 Binary Cross-Flow Permeation

Spiral-wound membrane units are very close to cross-flow operation with an unmixed permeate (Figure 16-19a). We can approximate this cross-flow system as a series of N well-mixed stages as shown in Figure 16-20a. Since we know how to solve each of the well-mixed stages in Figure 16-20A, we can add all the results to find the behavior of the cross-flow system. Since a complicated analytical solution exists for binary cross-flow gas permeation (Geankoplis, 2003; Hwang and Kammermeyer, 1984), we can check the accuracy of our numerical solution.

For a design problem (cut is specified and membrane area is unknown) solution is easiest if we make the permeate flow rate F_p,j for each small well-mixed stage the same; thus, each stage has different areas. Then the flow rates from the cross-flow system are

(16-64a, b)

and the area is

(16-65)

Since the stages are well-mixed and rapid mass transfer is assumed, y_p = y_t and y_r = y_r,w. Thus, we can write Eqs. (16-6a) and (16-8a) for each small, well-mixed stage j using variables y_p,j, y_r,j (both for the more permeable component), and . For constant selectivity α, these equations can be solved simultaneously by substituting Eq. (16-8a) into (16-6a), rearranging, and solving with the quadratic equation. The resulting solutions for each stage are identical to Eqs. (16-10)(in Example 16-2) except y_in is replaced with y_r,j-1 and θ is replaced by θ_j = F_p/F_r,j-1. Retentate flow rates and retentate mole fracs for each stage can be found from the external mass balances Eq. (6-7),

(16-66a)

(16-66b)

Equations (16-10) and (16-66) are solved for each stage starting with j = 1. The area A_j for each stage can be determined from a rearrangement of Eq. (16-4b),

(16-67)

Once every stage has been calculated, the mole fracs for the cross-flow system are

(16-68a, b)

and the total area is given by Eq. (16-65). This set of equations is easy to program within a mathematical package. If N is large enough (in most cases N = 100 will be more than sufficient (Coker et al., 1998), very good agreement will be obtained with the analytical solution, and probably with less effort. The numerical solution has the advantage that variable selectivity can easily be included. A spreadsheet program and example are in the chapter appendix. Shindo et al. (1985) show an alternate numerical solution method based on solution of the differential equations and apply it to multicomponent separations.

The cross-flow system with unmixed permeate increases the separation compared to a completely mixed permeation system. The cross-flow system works better than the well-mixed system because the average driving force, [Σ(p_ry_r,j – p_py_p,j)]/N, is considerably larger in the cross-flow system than the single driving force, (p_ry_r,out – p_py_p), in the well-mixed system. A real system probably has some but not complete mixing on both permeate and retentate sides; thus, the permeate mole frac is probably between the values calculated for the ideal cases.

The analysis for cross-flow can easily be extended to multicomponent systems. Now each stage is a perfectly mixed multicomponent system with a feed from the previous stage. The results for each stage can be calculated using the procedure in section 16.3.3. Thus, a trial-and-error is needed on each stage, but the entire cascade is only calculated once. Coker et al. (1998) illustrate a different procedure for multicomponent cross-flow.

16.7.2 Binary Co-current Permeation

Tubular and hollow-fiber membrane systems can be operated in co-current flow (Figure 16-19b). Binary gas permeation with co-current flow can be analyzed with a staged model similar to that used for cross-flow except now y_p ≠ y_t, and the flow pattern must be used to relate the permeate mole frac to the mole frac transferring through the membrane. The co-current model is shown in Figure 16-20b. This model again represents a numerical integration of the differential equations. Coker et al. (1998) present a solution method for multicomponent permeation when the membrane area is known but the cut is unknown, and show that N = 100 is usually sufficient.

The solution for a design problem (cut is specified, and membrane area is unknown) is again easiest if we make the total permeate mass transfer rate through the membrane for each small well-mixed stage constant, which is the same as using the same fraction of the total cut on each stage.

(16-69a, b)

thus, each stage has a different area. The flow rates leaving each stage are

(16-69c,d)

The transfer rates for gases A and B are similar to Eqs. (16-5a) and (16-5b), respectively, except y_p ≠ y_t,

(16-70a)

(16-70b)

The mass balance for the permeate is

(16-71a)

Solving for the mole frac of the transferred gas,

(16-71b)

Dividing Eq. (16-70a) by (16-70b) and simplifying we obtain

(16-72)

After substituting in Eq. (16-71b) and simplifying, this becomes

(16-73)

with

(16-74a)

(16-74b)

(16-74c)

In the limiting case of N = 1 Eqs. (16-74a) to (16-74d) simplify to Eqs. (16-10a) to (16-10d).

The solution, as expected, is

(16-75)

To use Eqs. (16-74) and (16-75), we must know y_{p, j-1}. Then after y_p,j has been calculated, we can obtain y_r,j from a mass balance around the permeator.

(16-76)

Now we can find y_t,j from Eq. (16-71b), and A_j from either Eq. (16-70a) or (16-70b). The value of j is increased by one and the calculation is repeated for the next stage. When stage j = N has been calculated, the outlet mole fracs are y_p,N and y_r,N. The total membrane area can then be calculated by summing all the A_j, Eq. (16-65). Since no trial-and-error is involved, the calculation is very quick. A spreadsheet program and example are in the chapter appendix.

The analysis for co-current flow can easily be extended to multicomponent systems. This extension is very similar to the extension for cross-flow systems.

16.7.3 Binary Countercurrent Flow

Tubular and hollow-fiber modules can be arranged to have approximately countercurrent flow patterns (Figure 16-19c). Since these flow patterns are an advantage in equilibrium-staged systems, we would expect that they might be an advantage in membrane separations also. Example 16-11 showed that this is true. The staged model for countercurrent flow is shown in Figure 16-20c. Coker et al. (1998) present a solution method for a staged model for countercurrent, multicomponent permeation when the membrane area is known but the cut is unknown, and show that N = 100 is usually sufficient.

Developing a design method for a binary, countercurrent permeator is challenging. Typically the values of F_in, y_r,0, p_p, p_r, P_A/t_ms, P_B/t_ms, temperature, and one additional variable, typically the cut θ, will be specified. The difficulty with countercurrent flow is deciding how to get started, since generally not everything needed to start is known at either end of the system. We will again assume that F_t is the same for every stage, which requires the area of each stage to be different. Since y_r,N is unknown, we will assume a value for y_r,N so that we can step off stages from the retentate product end of the column. This makes the calculation trial-and-error, but convergence to the correct value of y_r,N appears to be rapid.

To start stepping off stages in Figure 16-20c, we know or have assumed F_in, y_r,N, y_r,0, N (~ 100), and cut θ. Then F_p,1 = θF_in, F_r,N = F_in – F_p,1, and the transfer rate F_t is,

(16-77)

The procedure to step off stages is similar to the procedure to step off stages for the co-current system, Eqs.(16-69) to (16-76), but the mass balance equations change since the permeate flow direction has changed and we are stepping off stages backwards. The steps in the derivation starting with the equation equivalent to Eq. (16-69c) and ending with Eq. (16-76) are given in Table 16-4.

Table 16-4. Development of design equations for binary, countercurrent gas permeation

For the stage by stage procedure, start with stage N since y_r,N is known. Counter i goes from 1 to N. Then for any i, stage j = N – i + 1.

1. Calculate flow rates F_r,j-1 and F_p,j from Eq. () and y_r,j-1 from Eq. () (Table 16-4).

2. Calculate y_p,j from Eq. () using a, b, and c values from Eqs. () (Table 16-4).

3. Calculate y_t,j from Eq. () (Table 16-4), and A_j from either Eq. (16-70a) or (16-70b).

4. If i < N, let i = i + 1 and repeat starting with step 1.

5. Check if correct value of y_r,N was used.

a. The calculated value of y_in,calc = y_r,0. If y_in,calc ≠ y_in, adjust the guess for y_r,N. The following empirical approach appears to work well

(16-78)

where the damping factor is to prevent excessive oscillation. A value of 0.9 is reasonable.

b. When y_in,calc = y_in within some tolerance ε, go to step 6 and finish the calculation.

6. Calculate the area from Eq. (16-65).

This calculation is fast on a spreadsheet. A spreadsheet program and example for the binary design procedure are in the chapter appendix.

Coker et al. (1998) present a staged model for multicomponent simulation problems (membrane area known) that develops a tri-diagonal matrix similar to those developed previously for distillation and for absorption. All internal flows need to be assumed initially, and a sum-rates method was used to find new flow rates. A good initial guess of flow rates is required for convergence. This procedure is recommended for multicomponent countercurrent modules instead of using a stage-by-stage calculation. The difficulty with the stage-by-stage calculation is C – 1 mole fracs (C = number of components) need to be estimated to specify the exiting retentate, and then trial and error is needed for these C – 1 mole fracs.

16.8 SUMMARY—OBJECTIVES

In this chapter we have studied membrane separations including gas permeation, RO, UF and pervaporation. At the end of this chapter you should be able to satisfy the following objectives:

1. Explain the differences and similarities between the different membrane separation systems

2. For a perfectly mixed system with no concentration polarization, predict the performance of an existing membrane separation system and design a new membrane separation system using both analytical and graphical analysis procedures

3. Explain and analyze the effect of concentration polarization and include it in the design of perfectly mixed membrane separators

4. Explain and analyze the effect of gel formation and include it in the design of UF systems

5. Use energy balances to determine appropriate operating conditions for pervaporation systems

6. Explain and analyze the effects of flow patterns on the separation achieved in membrane systems

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Kesting, R. E. and A. K. Fritzsche, Polymeric Gas Separation Membranes, Wiley, New York, 1993.

Leeper, S. A., “Membrane Separations in the Recovery of Biofuels and Biochemicals: An Update Review,” in N. N. Li and J. M. Calo (Eds.), Separation and Purification Technology, Marcel Dekker, New York, 1992, pp. 99-194.

Leeper, S. A., D. H. Stevenson, P. Y.-C. Chiu, S. J. Priebe, H. F. Sanchez, and P. M. Wikoff, Membrane Technology and Applications: An Assessment, U. S. Department of Energy, Feb. 1984.

Li, N. N. and S. S. Kulkarni, “Membrane Separations,” in McGraw-Hill Encyclopedia of Science and Technology, eighth edition, Vol. 10, pp. 670-671 (1997).

Loeb, S. and S. Sounfiguration. Explain why it is used instead of a square cascade (same number ofrirajan, “Seawater demineralization by means of a semipermeable membrane,” University of California at Los Angeles Engineering Report No. 60-60, 1960.

Loeb, S. and S. Sourirajan, “Seawater Demineralization by Means of an Osmotic Membrane,” Advan. Chem. Ser., 38, 117 (1963).

Mohr, C. M., S. A. Leeper, D. E. Englegau, and B. L. Charboneau, Membrane Applications and Research in Food Processing: An Assessment, U.S. Department of Energy, August 1988.

Mulder, M., Basic Principles of Membrane Technology, 2nd. Edition, Kluwer Academic Publishers, Dordrecht, 1996.

Nakagawa, T., “Gas Separation and Pervaporation,” in Y. Osada and T. Nakagawa (Eds.), Membrane Science and Technology, Marcel Dekker, New York, 1992, Chapter 7.

Neel, J., “Introduction to Pervaporation” in R. Y. M. Huang (Ed.), Pervaporation Membrane Separation Processes, Elsevier, Amsterdam, 1991, pp. 1-109.

Noble, R. D. and A. Stern (Eds.), Membrane Separations Technology, Elsevier, Amsterdam, 1995.

Noble, R. D. and P. A. Terry, Principles of Chemical Separations with Environmental Applications, Cambridge University Press, Cambridge, UK, 2004.

Osada, Y. and T. Nakagawa (Eds.), Membrane Science and Technology, Marcel Dekker, New York, 1992.

Perry, R. H. and D. W. Green (Eds.), Perry’s Chemical Engineers’ Handbook, seventh edition, McGraw-Hill, New York, 1997.

Porter, M. C., “Membrane Filtration,” in P. A. Schweitzer (Ed.), Handbook of Separation Techniques for Chemical Engineers, third edition, McGraw-Hill, New York, Section 2.1, 1997.

Reid, C. E., “Principles of Reverse Osmosis,” in U. Merten (Ed.), Desalination by Reverse Osmosis, MIT Press, Cambridge, MA, Chapt. 1, 1996.

Shindo, Y., T. Hakuta, H. Yoshitome, and H. Inoue, “Calculation Methods for Multicomponent Gas Separations by Permeation,” Separ. Sci Technol., 20, 445 (1985).

Suk, D. E. and T. Matsura, “Membrane-based Hybrid Processes: A Review,” Separ. Sci Technol., 41 , 595 (2006).

Wankat, P. C., Rate-Controlled Separations, Kluwer, Amsterdam, Chapters 12 and 13, 1990.

Wankat, P.C. and K. S. Knaebel, “Mass Transfer,” in R. H. Perry and D. W. Green (Eds.), Perry’s Chemical Engineers’ Handbook, seventh edition, McGraw-Hill, New York, pp. 5-59 to 5-79, 1997.

Warashina, T., Y. Hashino and T. Kobayashi, “Hollow-fiber Ultrafiltration,” Chem. Tech., 15, 558 (1985).

HOMEWORK

A. Discussion Problems

A1. Membrane systems are rate processes and flash distillation is an equilibrium process. Explain why the solution methods are so similar for well-mixed membrane separators and flash distillation.

A2. Explain why nitrogen has a minimum permeability in Figure 16-5 and Table 16-2

A3. What are the advantages of asymmetric membranes compared to symmetric membranes?

A4. What are the advantages of hollow-fiber membranes compared to other geometries? When would one use other geometries?

A5. In large membrane systems it is common to have membrane cascades with membranes arranged both in parallel and in series. What are the advantages of this? The arrangements after the first set of membranes in series usually will have fewer membranes in parallel. Draw this co modules in every part).

A6. Two membranes may be made of the same polymer but have very different behaviors. Explain why.

A7. Two membrane modules may be made with exactly the same membranes, but one produces a higher purity product than the other. Discuss possible reasons for this.

A8. RO and UF both separate liquid mixtures. Why is osmotic pressure almost always important in the design of RO systems but often unimportant in the design of UF systems?

A9. Explain why the behavior of an UF system is so different when a gel forms than when a gel does not form.

A10. Some azeotropic mixtures can be separated by sending the vapor mixture to a gas permeation system—designated as vapor permeation if the mixture is easily condensed (Huang, 1991; Neel, 1991)—and some (probably different) azeotropic mixtures can be separated by sending a liquid mixture to an RO system. Why is pervaporation a much more popular method of separating azeotropic mixtures? Note: In all cases a hybrid membrane-distillation system will probably be used.

B. Generation of Alternatives

B1. There are a number of ways distillation and membrane separators can be combined as hybrid systems. Brainstorm as many methods as you can.

C. Derivations

C1. Derive Eqs. (16-10) for a gas permeator.

C2. Solve Eq. (16-19) for x_p as a function of x_r.

C3. Derive Eq. (16-31) from shell balances. You should obtain a second-order equation. Do the first integration and apply the boundary condition R = 1.0 at z = 0.

C4. Show how to do the graphical solution for a perfectly mixed UF system.

C5. Simultaneously solve Eqs. (16-23) and (16-43) or (16-44) analytically and simplify the expression.

C6. Derive Eq. (16-11e).

D. Problems

*Answers to problems with an asterisk are at the back of the book.

D1. A gas permeation system with a cellulose acetate membrane will be used to purify a carbon dioxide-methane stream. The permeabilities are given in Example 16-2. The effective membrane thickness t_ms = 1.0 µm (a micron = 1.0 × 10⁻⁶ meter). Operation is at 35°C. p_H = 12 atm and p_L = 0.2 atm (a vacuum). The feed is 15 mole % carbon dioxide. Assume ideal gases.

a. ^*A single-stage, well-mixed membrane separator will be used. Operate with a cut, θ = F_p/F_in = 0.32. Find y_p, y_out, and flux of carbon dioxide. If we feed 1 kg mole/hr of feed, calculate the membrane area needed, F_p and F_out.

b. Design a two-step cross-flow gas permeation system. Each stage is well mixed. Use the same feed to the first stage as in part a. Pressures and feed mole frac are the same as in part a. Set F_p1 = F_p2 = 0.5 F_p,parta (then F_out,2 = F_out,parta). Find y_p1, y_p2, y_out1 = y_in2, y_out2 and F_out1. Find the methane fluxes in both stages and the membrane areas required in both stages.

D2. We are separating a gas stream by gas permeation in a perfectly mixed system. The feed is 20 mole % carbon dioxide and 80 mole % methane. The membrane permeabilities are P_CO2 = 15.0 × 10⁻¹⁰ and P_CH4 = 0.48 × 10⁻¹⁰ [ccSTP cm]/[cm² s cm Hg]. The selectivity = 31.25. The effective thickness of the membrane skin is t_ms = 1.0 cm. We operate with p_p = 3.3 atm, p_r = 60 atm, F_in = 2 gmoles/s, and F_p/F_in = 0.3. Find:

a. y_p, y_out.

b. Membrane area needed in m².

D3. We are testing a RO membrane at 25°C. The aqueous feed contains 0.5 wt % NaCl and feed flow rate is 1,250 L/h. We measure the retentate as 1.0 wt % NaCl, and the permeate is 0.012 wt % NaCl. The pressure drop is Δp = 15 atm. With pure water we measure a permeate flow rate of 1,000 L/h when Δp = 15 atm.

From Perry’s Handbook, 4th ed., p. 3-83 the densities (concentrations in g/ml), of aqueous salt solutions at 25°C are given while p. 16-23 lists osmotic pressure:

a. Find the flow rates and in kg/h for the experiment with salt present.

b. Find M.

D4^*. A cellulose acetate membrane is being used for RO of aqueous sucrose solutions at 25°C.

Data: Density of solvent (water) is ρ = 0.997 kg/L.

Density (kg/L) of dilute aqueous sucrose solutions is ρ = 0.997 + 0.4 x where x is the wt frac sucrose.

At low sucrose wt frac the osmotic pressure (in atms) can be estimated as π = 59.895 x where x is the wt frac sucrose.

Molecular weight of water is 18.016. Molecular weight of sucrose is 342.3.

(Note: Some of this data may not be needed to solve this problem.)

Experiment A. This experiment is done in a well-mixed stirred tank. At 1,000 rpm with a 3 wt % solution of sucrose in water using p_r = 75 atm and p_p = 2 atm we obtain J_solv = 400L/m² day. The mass transfer coefficient k = 6000 L/m² day. We measure x_r = x_out = 0.054 and x_p = 3.6 × 10⁻⁴.

a. Calculate the concentration polarization modulus M.

b. Calculate selectivity α_AB, K_solv/t_ms, and K_A/t_ms. Report these in units which will give J_solv in L/m² day and in kg/m² day.

c. Estimate the mass transfer coefficient k if the rpm is increased to 2000.

D5. We are using a cellulose acetate RO membrane to concentrate a dilute sucrose mixture.

Operation is at 25°C.

a. We measure the pure water flux (no sucrose present) and find at a pressure drop of 102 atm across the membrane, J_solv = 1.5 × 10⁻³ [ml water/(cm² s)]. An experiment in a highly stirred system (M=1) with a dilute sucrose solution gives a rejection R° = 0.997 for an inlet wt frac of 0.050 and cut, θ = 0.45 (in weight units). Find (K_solv/t_ms) and the selectivity.

b. We will design a perfectly mixed RO system to operate at conditions where M = 3.0. The feed is 2.0 wt % sucrose. We want a cut, θ = 2/3. Feed rate will be 5 kg/s. Pressure drop across the membrane is 78 atm. Find x_p, x_r = x_out, J_solv, and membrane area.

Data: At low concentrations the osmotic pressure (in atm.) of an aqueous sucrose solution can be estimated as π = 59.895 x, where x is wt. fraction sucrose, and the density (in g/ml) can be estimated as ρ = 0.997 + 0.4 x.

D6. UF of a dextran solution in a well-mixed system with gel formation gives the following data:

Determine k and the wt frac at which dextran gels.

D7. An UF system is being used to concentrate dextran in an aqueous solution. The dextran forms a gel layer on the membrane at x_g = 0.314 wt frac. The pure water flux is 6000 L/m² day when p_r = 3.5 bar and p_p = 1.0 bar. We operate at conditions where M = 2.5, if a gel layer does not form.

a. If operation is at p_r = 5 bar and p_p = 1.3 bar, determine the water flux assuming x_w < x_g.

b. What is the maximum value of x_r,_max which can be used and have x_w < x_g?

c. If x_r < x_r,max (no gel layer), determine the mass transfer coefficient, k. Same pressures as in part a.

d. Assume that k is the same before and after a gel layer forms. Calculate the solvent flux if x_r = 0.20 wt frac. Same pressures as part a.

D8. An UF membrane is first tested in a stirred cell where there is no concentration polarization. The experimental values obtained are listed in the table. Then the same membrane is used in a spiral-wound module where there is concentration polarization. Values listed are obtained. Assume membrane thickness and permeabilities are the same. Estimae from the stirred cell data and assume the inherent membrane rejection is unchanged. Calculate the expected solvent flux and the concentration polarization modulus M_c in the spiral-wound membrane system. No gel layer forms. Use concentration units, g/L.

D9. The following data were obtained for UF of skim milk in a spiral-wound system (Conlee, et al., 1998).

a. Estimate the gelling wt frac, x_g, and the mass transfer coefficient, k.

b. If you were the technician’s supervisor and one more run could be done, what data would you want the technician to obtain to improve the estimate of x_g?

D10. A gas permeation system is being used to separate carbon dioxide from methane. The feed rate to the system is F_in = 15.0 gmoles/min, and we desire a cut, θ = F_p/F_in = 0.25. The feed is 30 mole % carbon dioxide. The permeate pressure is p_p = 0.33 atm and the retentate pressure is p_r = 6 atm. The system is perfectly mixed and has high rates of mass transfer so that the concentration polarization modulus M = 1.0. The membrane permeabilities are P_CO2 = 12.5 × 10⁻¹⁰ and P_CH4 = 0.40 × 10⁻¹⁰ [ccSTP cm]/[cm² s cm Hg]. The effective thickness of the membrane skin is 1.1 × 10⁻⁴ cm. Both gases can be assumed to be ideal.

a. Find the permeate mole frac of carbon dioxide y_p and the outlet retentate mole frac of carbon dioxide y_r,out.

b. Find the membrane area needed in m². Watch your units!

D11. We wish to use pervaporation to increase the ethanol concentration in an ethanol water mixture to 98.5 wt %. The feed from the distillation column to the pervaporation unit is 90 wt % ethanol. This feed is pressurized and heated to 85°C. It is then fed to a system similar to Figure 16-15, except the retentate can be reheated to 85°C and be sent to a second pervaporation stage. Additional stages can be added if needed. A 20-micron thick polyacrylonitrile film membrane is used (see Figure 16-17). Assume that each stage of pervaporation is perfectly mixed and operates at 25°C. The final retentate (the product) needs to be at 98.5 wt % ethanol or higher. The permeate streams are pressurized and recycled to the distillation column. If we desire to process 100 kg/hr of the 90 wt % ethanol distillate, find:

a. Number of stages needed in the pervaporation system.

b. For each stage: the cut, the permeate flow rate and the product flow rate. Also the wt frac of the combined permeate streams and the retentate wt frac.

c. The membrane area required for each stage.

D12. We are doing UF in a perfectly mixed tank with a constant stirrer speed. The flux with pure water is 6000 L/m²day with a retentate pressure of p_r = 3.5 atm, and p_p = 1 atm. We know that the solute we are ultrafiltering forms a gel at x_g = 0.65 (wt frac). An experiment containing the solute with the same pressures used above gives a retentate wt frac of x_out = 0.22 and a solvent flux of J_solv = 500 L/m²day. With a gel present x_p = 0.0. We do another experiment where we expect a gel to form. If x_in = 0.05, A = 1.0 m², F_in = 1,000 L/day, x_p = 0.0, p_p = 1 atm, and p_r = 2.2 atm what are values of x_out, F_p and F_out? Use the same stirrer speed and hence same mass transfer as in the first experiment where a gel formed.

D13. We are doing RO in a very well mixed RO system with no concentration polarization (M = 1). We measure the outlet wt frac salt and the permeate wt frac salt. These results are:

a. What is the rejection coefficient R for this membrane?

b. For the same well mixed system (M = 1), if x_in = 0.0100, and x_p = 0.0005, what are the values of x_out and the cut?

D14. A pervaporation system will be used to remove water from n-butanol using a cellulose 2.5 acetate membrane in a perfectly mixed module. Feed is 90 mole % n-butanol. Feed rate is 100 lb/hr. Flux is 0.2 lb/ft²h. The pervaporation unit operates at T_p = T_out = 30°C where the selectivity of water compared to butanol is 43.

Data: Latent heats: butanol = 141.6 cal/g; water = 9.72 kcal/gmole.

C_p,B (45°C) = 0.635 cal/g°C, C_p,w (45°C) = 1.0 cal/g°C, MW_B = 74.12, MW_w = 18.016. Assume that the heat capacities and latent heats are independent of temperature.

a. If the feed is at 60°C, find the cut, permeate mole frac, outlet liquid mole frac and the membrane area.

b. If a cut of θ = 0.08 is used, find the permeate mole frac, the outlet liquid mole frac, and the feed inlet temperature.

c. If the outlet liquid mole frac is 0.05, find the permeate mole frac, cut, and inlet feed temperature.

D15. ^* An UF system is being used to concentrate latex particles in an aqueous suspension. The membrane system is perfectly mixed on the retentate side and operates with a permeate pressure of 1.0 bar and a retentate pressure of 2.2 bar. Because of the extensive stirring the concentration polarization modulus is M = 1.2 and no gel forms (x_wall < x_gel = 0.5). The osmotic pressure of latex particles is negligible. The density of the solutions can be assumed to equal the density of pure water = 0.997 kg/L. All of the latex particles are retained by the membrane (R° = 1.0). The feed to the UF module is x_r,in = 0.10 wt frac latex. We operate with a cut θ′ = 0.2. The feed rate of the suspension, F′_r,in, is 100 kg/hr. The flux rate of the membrane with pure water is 2500 L/(m² day). Find F′_r,out, x_p and x_r,out. Note: Not all of the data presented is needed to solve this problem.

E. More Complex Problems

E1. (Do this gas permeation problem after you have studied RO and concentration polarization.) Determine the appropriate form of the equations for gas permeation with concentration polarization by repeating the derivation of the gas permeation equations but for the slower moving component, B, and including concentration polarization.

We do the experiment discussed in Example 16-2 with θ = 0.25. We find y_A,p = 0.88 and y_A,out = 0.373 where A = carbon dioxide. The cell appears to be well mixed although there is not vigorous stirring. Values of the other parameters in the equations all are the same as in Example 16-2. One hypothesis is that there is concentration polarization even though this is a gas system. Determine the concentration polarization modulus, M = y_B,wall /y_B,r, which would cause the observed changes in gas concentration for a well-mixed cell.

F. Problems Requiring Other Resources

F1. We are using pervaporation to separate a benzene–isopropyl alcohol mixture. The pervaporation unit is perfectly mixed. Figure 12-23 in Wankat (1990) shows the separation factor of benzene with respect to isopropyl alcohol vs. the wt. fraction of benzene in the liquid for pervaporation. If operation is at 50°C = T_p = T_out, x_in = 0.30 wt frac benzene and the θ = 0.10, find y_p, x_out, and T_in. (Assume heat capacities and latent heats are independent of temperature.)

Note: Use the lines on the selectivity diagram not the data points.

G. Simulator Problems

G1. ^* In section 16.3.3 the following statement appears, “Geankoplis (2003) solves the multicomponent permeator system by a different method but the results are identical.” Solve Example 13.5-1 in Geankoplis (2003) using the method in section 16.3.3 and show that the results agree with Geankoplis’ solution.

G2. Redo Example 16-2 for a cross-flow system.

G3. Redo Example 16-2 for a co-current system.

G4. Redo Example 16-2 for a countercurrent system.

CHAPTER 16 APPENDIX SPREADSHEETS FOR FLOW PATTERN CALCULATIONS FOR GAS PERMEATION

The flow pattern calculations can be done with an Excel spreadsheet using Visual Basic. In the spreadsheets shown here, data are input into the spreadsheet and all calculations are done in the Visual Basic program. All programs are done for constant values of the selectivity α, although it would not be difficult to use an α that depended on gas mole frac. These programs are meant to show how easy it is to program the flow pattern calculations in a spreadsheet using Visual Basic. The best way to learn this material is to write your own spreadsheet, not copy these spreadsheets.

16.A.1 Cross-Flow

The cross-flow calculation requires no trial-and-error and is very fast. The spreadsheet inputs the required values and records results. All calculations are done in the Visual Basic program. The spreadsheet is shown with values for the separation of oxygen and nitrogen in Example 16-11. Results are shown for the first seven stages out of N = 100 and the final results. Note that the word “Fin” is in cell (2,1).

The Visual Basic program calculates cross-flow permeation based on numbers from the spreadsheet.

Sub Main()

Dim N, i As Integer

Dim pr, pp, a, b, c, Fp, Fin, Fptotal, permA, permB, Fr, sumyp, Area, Areatot, yr, yp, ypavg, alpha, thetatot, theta As Double

Sheets(“Sheet2”).Select

Range(“B7”, “EU16”).Clear [cleans out up to 100 cells (N = 100)]

Fin = Cells(2, 2).Value [reads input values from spreadsheet]

Fr = Fin

yr = Cells(2, 4).Value

pr = Cells(3, 4).Value

pp = Cells(3, 6).Value

permA = Cells(3, 2).Value [P_A/t_ms]

permB = Cells(2, 8).Value

alpha = permA / permB

thetatot = Cells(2, 6).Value

Fptotal = Fin * thetatot

N = Cells(4, 4).Value

theta = thetatot / N

Fp = Fptotal / N

sumyp = 0 [Initialize calculation for summed values]

Areatot = 0

For i = 1 To N [Start loop for calculation for each stage]

Cells(7, i + 1).Value = i

a = (theta / (1 − theta) + pp / pr) * (alpha − 1) [terms for Eq. (16-10e) for cross-flow]

b = (1 − alpha) * (pp / pr + theta / (1 − theta) + yr / (1 − theta)) − 1 / (1 − theta)

c = alpha * yr / (1 − theta)

yp = (-b − (b * b − 4 * a * c) ^ 0.5) / (2 * a) [quadratic formula]

sumyp = sumyp + yp

Frold = Fr

Fr = Frold – Fp [Overall external mass balance]

yr = (Frold * yr − Fp * yp) / Fr [Component external mass balance]

Area = Fp * yp / (permA * (pr * yr − pp * yp))[Eq. (16-67)]

Areatot = Areatot + Area [Eq. (16-65a)]

Cells(8, i + 1).Value = Fr [Print values for stage j on spreadsheet]

Cells(9, i + 1).Value = yp

Cells(10, i + 1).Value = yr

Cells(11, i + 1).Value = Area

Next i [Increase i by 1, and return to statement “For i = 1 To N”]

ypavg = sumyp / N [Eq. (16-68a)]

Cells(5, 4).Value = Fp [Print final values on spreadsheet]

Cells(5, 2).Value = Fptotal

Cells(12, 2).Value = ypavg

Cells(13, 2).Value = Areatot

Cells(14, 2).Value = yr

End Sub

16.A.2 Co-current Flow

The co-current calculation requires no trial-and-error and is very fast. The spreadsheet inputs the required values. All calculations are done in the Visual Basic program. The spreadsheet is shown with values for the separation of oxygen and nitrogen in Example 16-11. Results are shown for the first seven stages out of N = 100 and the final results. Note that the word “Fin” is in cell (2,1).

The Visual Basic program calculates co-current permeation based on numbers from the spreadsheet.

Sub Main()

Dim N, i As Integer

Dim pr, pp, a, b, c, ConA, ConB, Cj, Cjold, Ft, Fp, Fptotal, Fpold, Frout, permA, permB, Fr, Fin, Area, Areatot, yt, yin, yr, yp, ypold, thetatot, alpha As Double

Sheets(“Sheet1”).Select

Range(“B7”, “GS16”).Clear [cleans out up to 200 cells (N = 200)]

Fin = Cells(2, 2).Value [Input values from the spread sheet]

yin = Cells(2, 4).Value

pr = Cells(3, 4).Value

pp = Cells(3, 6).Value

thetatot = Cells(2, 6).Value

permA = Cells(3, 2).Value [P_A/t_ms]

permB = Cells(2, 8).Value

alpha = permA / permB

N = Cells(4, 4).Value

Ft = Fin * thetatot / N

Fp = 0 [Initialize calculations]

Fpold = 0

yp = 0

Fptotal = Ft * N [Overall mass balance values]

Frout = Fin − Fptotal

Areatot = 0

For i = 1 To N [Start loop for calculation for each stage]

Cells(7, i + 1).Value = i

ypold = yp [Initialize calculation]

Fpold = Fp

Fp = Fpold + Ft

Fr = Fin − Fp

ConA = Fin * yin / Fr [constant to simplify Eqs. for b and c]

ConB = Fpold * ypold [constant to simplify Eqs. for b and c]

a = (1 − alpha) * (Fp * Fp / Fr + Fp * pp / pr) [terms for quadratic Eq. (16-73 to 74c)]

b = Fp * ((alpha − 1) * ConA + (1 − pp / pr)) + (alpha − 1) * ConB * (Fp / Fr + pp / pr) + alpha * Ft * (Fp / Fr + pp / pr)

c = (1 − alpha) * ConA * ConB + ConB * (pp / pr − 1) − alpha * Ft * ConA

yp = (-b + (b * b − 4 * a * c) ^ 0.5) / (2 * a) [solution of quadratic]

yr = −yp * Fp / Fr + ConA

Area = Ft * yp / (permA * (pr * yr − pp * yp)) [Area from Eq. (16-70a)]

Areatot = Areatot + Area [Eq. (16-65)]

Cells(8, i + 1).Value = Fr [Print values for stage j on spreadsheet]

Cells(9, i + 1).Value = yp

Cells(10, i + 1).Value = yr

Cells(11, i + 1).Value = Area

Cells(12, i + 1).Value = Fp

Next i [Increase i by 1, and return to statement “For i = 1 To N”]

Massbal = yin * Fin − Fptotal * yp − yr * (Fin − Fptotal)

Cells(13, 2).Value = Massbal [Print final values on spreadsheet]

Cells(14, 2).Value = yp

Cells(15, 2).Value = Areatot

Cells(16, 2).Value = yr

Cells(17, 2).Value = Frout

Cells(18, 2).Value = Fptotal

End Sub

16.A.3 Countercurrent Flow

This calculation is trial-and-error since y_r,N must be guessed. A loop using Eq. (16-78) to correct y_r,N is used with an arbitrary number of trials M (in this example M = 10). The spreadsheet inputs the required values. All calculations are done in the Visual Basic program. The spreadsheet is shown with values for the separation of oxygen and nitrogen in Example 16-11. The calculations for the first seven stages and the final results (N = 100) are shown in the spreadsheet. Note that a very poor first guess for y_r,N (0.001) converges to the correct y_in. [The spreadsheet starts with the word “Fin” in cell (2, 1).]

The visual basic program calculates terms based on input in the spreadsheet.

Sub Main()

Dim N, M, k, j, i As Integer

Dim pr, pp, a, b, c, ConA, ConB, Cj, Cjold, df, Ft, Fp, Fpold, Fpout, permA, permB, Fr, Frout, Fin, Area, Areatot, yt, yin, yr, yrout, yroutold, yrold, yp, ypold, alpha, thetatot, thetaj As Double

Sheets(“Sheet1”).Select

Range(“B7”, “GS16”).Clear [cleans out up to 200 cells (N = 200)]

Fin = Cells(2, 2).Value [Input values from the spreadsheet]

thetatot = Cells(2, 6).Value

Fpout = thetatot * Fin

Frout = Fin − Fpout

yin = Cells(2, 4).Value

yrout = Cells(4, 6).Value

pr = Cells(3, 4).Value

pp = Cells(3, 6).Value

permA = Cells(3, 2).Value [P_A/t_ms]

permB = Cells(2, 8).Value

Fptotal = Ft * N

alpha = permA / permB

M = Cells(4, 2).Value [counter for loop to find correct y_r,N]

df = Cells(5, 2).Value [damping factor for Eq. (16-78)]

N = Cells(4, 4).Value

Ft = Fpout / N

yincalc = yin [Initialize values]

yroutold = yrout

For k = 1 To M [loop to converge on y_r,N]

yrout = yroutold + df * (yin − yincalc) [Eq. (16-78), when k=1, yrout = y_r,N,guess]

yr = yrout [Initialize values]

Fr = Frout

Fp = 0

Fpold = 0

yp = 0

Areatot = 0

For i = 1 To N [Start loop for calculation for each stage]

j = N − i + 1 [allows counting backwards from stage N]

Cells(7, i + 1).Value = j

ypold = yp [Initialize values for this stage}

Fpold = Fp

Fp = Fpold + Ft

a = (1 − alpha) * Fp * pp / pr [Calculate terms for quadratic equation]

b = Fp * (1 − pp / pr) + (alpha − 1) * Fpold * ypold * pp / pr + alpha * Ft * pp / pr + Fp * (alpha − 1) * yr

c = Fpold * ypold * (−1 − (alpha − 1) * yr + pp / pr) − alpha * Ft * yr

yp = (-b + (b * b − 4 * a * c) ^ 0.5) / (2 * a) [Eq. (16-75)]

Area = Ft * yp / (permA * (pr * yr − pp * yp)) [Eq. (16-70A)]

Areatot = Areatot + Area [Eq. (16-65)]

Cells(8, i + 1).Value = Fr [Print values for stage j on spreadsheet]

Cells(9, i + 1).Value = yp

Cells(10, i + 1).Value = yr

Cells(11, i + 1).Value = Area

Cells(12, i + 1).Value = Fp

Fr = Fp + Frout

thetaj = Fp / Fr

Cells(13, i + 1).Value = thetaj

yr = Fp * yp / Fr + yrout * Frout / Fr

Next i [Increase i by 1, and return to statement “For i = 1 To N”]

yincalc = yr

yroutold = yrout

Next k [Increase k by 1, and return to statement “For k = 1 To M”]

Massbal = yr * Fr − Fp * yp − yrout * Frout

Cells(18, 2).Value = Massbal [Print final values on spreadsheet]

Cells(14, 2).Value = yp

Cells(15, 2).Value = Areatot

Cells(16, 2).Value = yincalc

Cells(17, 2).Value = Fr

Cells(19, 2).Value = yrout

End Sub