17.10 SUMMARY—OBJECTIVES

In this chapter the basic concepts for adsorption, chromatography, and ion exchange separations were developed. At the end of this chapter, you should be able to satisfy the following objectives:

1. Determine equilibrium constants from data and use the equilibrium equations in calculations

2. Explain in your own language how the different sorption processes (e.g., elution chromatography, adsorption with thermal regeneration, PSA, SMB, monovalent-monovalent ion exchange, and water softening) work

3. Explain the meaning of each term in the development of the solute movement equations and use this theory for both linear and nonlinear isotherms to predict the outlet concentration and temperature profiles for a variety of different operations including elution chromatography, adsorption with thermal regeneration, PSA, SMB, and ion exchange

4. Explain the meaning of each term in the column mass and energy balances, and in the mass and heat transfer equations

5. Use the Lapidus and Amundson solution plus superposition to determine the outlet concentration profiles for linear adsorption and chromatography problems

6. Use the theory of linear chromatography with very small pulses to analyze chromatography systems

7. Use the LUB theory in combination with experimental data to design columns

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HOMEWORK

A . Discussion Problems

A1. Feed Step. Column is initially clean (c_init=0). Feed at concentration c_F>0.

1. If solute follows a linear isotherm, the resulting solute wave will be

a) diffuse wave, b)shock wave, c)simple wave.

2. If solute follows a favorable isotherm, the resulting solute wave will be

a) diffuse wave, b) shock wave, c) simple wave.

3. If solute follows an unfavorable isotherm, the resulting solute wave will be

a) diffuse wave, b) shock wave, c) simple wave.

A2. Elution. Column initially is saturated with solute at concentration c_init > 0. The feed to the column is c_F = 0.

1. If solute follows a linear isotherm, the resulting solute wave will be

a) diffuse wave, b) shock wave, c) simple wave.

2. If solute follows a favorable isotherm, the resulting solute wave will be

a) diffuse wave, b) shock wave, c) simple wave.

3. If solute follows an unfavorable isotherm, the resulting solute wave will be

a) diffuse wave, b) shock wave, c) simple wave.

A3. Pure thermal waves move relatively quickly in liquid systems while they are usually slow in gas systems. Explain why.

A4. Explain why an adsorption isotherm that is too steep may not work well in a PSA process.

A5. Briefly explain why the SMB system is much more efficient (i.e., uses less solvent and less adsorbent) than an elution chromatograph doing the same binary separation. Assume that both systems are operating in the migration mode using isocratic elution. Both systems are optimized. The elution chromatograph uses repeated pulses of feed.

A6. There are a number of important industrial separations such as separation of meta and para xylene that can be operated as either gas or liquid. All of the current commercial adsorption applications are operated as liquids. What are the advantages of operation as a liquid as compared to operation as a gas?

A7. In an SMB suppose the A product purity is OK, but the B product purity is too low. We can increase the B product purity while maintaining the A product purity by,

a) Increasing all M_i, b)Decreasing all M_i, c) Increasing M_4B and increasing M_3A, d) Decreasing M₁ and increasing M_3A, e) Increasing M_4B and decreasing M_2B, f) Decreasing M_4B and decreasing M_2B.

A8. In an SMB suppose the B product purity is OK, but the A product purity is too low. We can increase the A product purity while maintaining the B product purity by,

a) Increasing all M_i, b) Decreasing all M_i, c) Increasing M_4B and increasing M_3A, d) Decreasing M₁ and increasing M_3A, e) Increasing M_4B and decreasing M_2B, f) Decreasing M_4B and decreasing M_2B.

A9. We desire to separate a dilute ternary mixture consisting of A, B, and C dissolved in D (A is least strongly adsorbed component and C is most strongly adsorbed component) in a normal SMB (modify Figure 17-14b for the ternary separation). One product is A (plus D) while the other product is B + C (plus D). The four SMB design Eqs. (17-29) are,

a) unchanged, b) the same except Eq. (17-29c) becomes u_C,2 = M_2C u_port (M_2C ≤ 1.0), c) the same except Eq. (17-29d) becomes u_B,3 = M_3B u_port (M_3B ≥ 1.0), d) the same except Eq. (17-29f) becomes u_C,4 = M_4C u_port (M_4C ≥ 1.0), e) the same except Eq. (17-29c) is changed as in answer b and Eq. (17-29f) is changed as in answer d.

A10. For a differential pulse with a linear system you have calculated the standard deviation of the Gaussian peak σ_t in time units. You now want to calculate σ_l inside the column. The formula you would use is:

a) σ_l = σ_t, b) σ_l = σ_t/v_inter, c) σ_l = σ_t(v_inter), d) σ_l = σ_t/u_s, e) σ_l = σ_t(u_s).

A11. Explain how decreasing the particle diameter will increase N in linear chromatography.

B . Generation of Alternatives

B1. Brainstorm some possible alternative adsorbents made from common agricultural and/or forest products or wastes.

B2. The PSA cycle used in Example 17-4 does not produce pure hydrogen throughout the entire feed step. Brainstorm what can be done to change the cycle so that it will produce pure hydrogen.

B3. Separation in chromatography is often limited because the selectivity is too close to 1.0. For both liquid and gas systems, list as many ways as possible that the equilibrium constants can be changed.

C . Derivations

C1. A molecular sieve zeolite adsorbent consists of pellets that are agglomerates of zeolite crystals with density ρ_crystal scattered in a continuous phase of clay binder with a density ρ_clay. In this case, there is an interpellet porosity ε_e (between pellets—this is normal ε_e) an intercrystal porosity ε_p1, (which is the porosity in the binder), and an intracrystal porosity ε_p2, (inside the crystals). If the fraction of the particle volume that is crystals (including porosity within the crystals) is f_cry, derive formulas for the total porosity, V_available, and the particle and bulk densities.

C2. Derive the value for the limiting conditions in the Langmuir isotherm, Eq. (17-6a), when c_A becomes very large (K_A,c c_A >> 1.0).

C3. Derive the amounts of adsorbate in the pore liquid and sorbed on the packing (second and third terms in denominator of Eq. (17-13)). (For situations where K_d <1.0, assume that the adsorption data was taken with the same adsorbent and that the q values include the inaccessibility of some of the adsorption sites. Since K_d is already inherently included in the values of q, do not include K_d explicitly in the calculation of the amount sorbed.)

C4. Derive Eqs. (17-14a), (17-14b), and (17-14c) from Eqs. (17-13), and (17-10).

C5. Derive the equation for the solute velocity for linear isotherms for systems using a single-porosity model. The result obtained should be,

(17-15b)

C6. Derive the terms in Eq. (17-19) for the amount of energy change in the pore fluid, the solid, and the column wall.

C7. Derive Eq. (17-23) from (17-22) using the definition of u_th, and then derive Eq. (17-24) from (17-23) and (17-6b) using the definition of u_s.

C8. Show that if the entire column is heated (or cooled) simultaneously (known as direct heating) Eq. (17-24) simplifies to,

(17-24 direct heating)

Explain why direct heating is practical for laboratory-sized columns (small diameter) but not for commercial units with large diameters. (Hint: consider the heat transfer characteristics.)

C9. Derive the appropriate mass balance equation and solutions equivalent to Eqs. (17-22) to (17-24) but for the case where u_s(T_hot)> u_s(T_cold) > u_th. Hint: Start by redrawing the differential control volume in Figure 17-7B noting that the concentration, amount adsorbed and temperature above c₂, q₂, and T₂ are c_change, q_change and T₁, where c_change is the concentration caused by the temperature change, which moves ahead of the thermal wave.

C10. Derive Eq. (17-25) for an ideal gas system when the linear isotherm is given by Eq. (17-5b).

C11. We want to operate an SMB (Figure 17-14B) doing a binary separation at (D/F)_min. Assume that the volumetric feed rate F and the cross sectional area of the columns A_c are specified. Derive the equations for u_port, v₁, v₂, v₃, v₄, v_A,product, v_B,product, v_D, and D/F = v_D / v_Feed that will satisfy Eqs. (17-29) and (17-30) using Eq. (17-15e). Show that Eq. (17-31a) is correct. After simplification you should obtain the following result for D/F,

(17-31b)

Show that this result becomes D/F = 1.0 when all the M_i = 1.0.

Hint: Start with the equations for zones 2 and 3 and v₂ = v₃ + v_Feed, and solve for u_port, v₂ and v₃.

C12. Use the solute movement theory to analyze the conditions for complete binary separation in the TMB shown in Figure 17-14A. The solids moves downwards at a volumetric flow rate S.

C13. Use the mass action expression to prove Eqs. (17-40c) and (17-41).

C14. Use the mass action expression to develop Eqs. (17-42b) and (17-42c).

C15. Derive an equation for t_MTZ for a step input for a system with a linear isotherm using the Lapidus and Amundson solution. The time for the MTZ is defined as the time required to go from 0.05 (c_feed – c_initial) to 0.95 (c_feed – c_initial).

Show that t_MTZ is approximately proportional to L^1/2.

Note: As far as I know the solution for t_MTZ for a breakthrough curve is not available in any books. The ability to derive this sort of solution (starting with a known solution and obtaining the equation for a specific case) will be very valuable in industry and will help you stand out from most engineers.

C16. To determine the bed capacity from Eq. (17-86d), the value of q_F needs to be known. If the pattern velocity u_sh is determined experimentally, show how to estimate q_F. Assume that the initial concentration is c = 0, then feed of concentration c = c_F if fed to the column, and all parameter values (ε_e, ε_p, ε_s, K_d) are known.

D . Problems

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

D1.* Find the values of K_A and q_max for methane adsorption on PCB activated carbon at 296 and 480 K (Table 17-3).

D2.* The adsorption of anthracene from cyclohexane on activated alumina follows a Langmuir isotherm,

q = 22c /(1 + 375 c) where c is in gmole/liter and q is in gmole/kg. (Thomas, 1948)

Convert this to the form of Eq. (17-6a) in terms of q_max and K_Ac in units of g/liter for c and g/kg for q. The range of validity should be from c = 0.0 to 0.012 gmole/liter.

Data: density of cyclohexane = 0.78 kg/liter, molecular weight cyclohexane = 84, molecular weight anthracene = 178.22, ρ_p = 1.47 kg/liter, ε_e = 0.4, ε_p = 0.0.

D3. A chromatographic column is packed with an ion exchange resin in the calcium form. The column is 0.75 meters long. The superficial velocity is 15.0 cm/minute. The column initially contains pure water. At t = 0 input a feed that is 75 g/liter glucose. At t = 2.0 minutes input a feed that is 100 g/liter glucose. Use solute movement theory to predict the outlet concentration profile.

Data: ε_p = 0, ε_e = 0.4, Equilibrium: q_G = 0.51 c_G, where q_G and c_G are in g/liter.

Compare the predictions for this linear system with the predictions in Example 17-7 for a nonlinear system.

Note: Although an ion exchange resin is used, this problem is not ion exchange. The calcium on the resin forms a reversible complex with the glucose. Analyze as a linear chromatographic system.

D4. We wish to use a thermal swing adsorption process to remove traces of toluene from n-heptane using silica gel as adsorbent. The adsorber operates at 1.0 atm. The feed is 0.0011 wt frac toluene and 0.9989 wt frac n-heptane at 0°C. Superficial velocity of the feed is 10.0 cm/min. The absorber is 2.0 meters long and during the feed step is at 0°C. The feed step is continued until breakthrough occurs. To regenerate, use counterflow of pure n-heptane at 80°C. Superficial velocity during purge is 10.0 cm/min. Column is cooled to 0°C before the next feed step.

Data: At low concentrations isotherms for toluene: q = 17.46x @ 0°C, q = 1.23x @ 80°C, q and x are in g solute/g adsorbent and g solute/g fluid, respectively (Matz and Knaebel, 1991). ρ_s = 2100 kg/m³, ρ_f = 684 kg/m³, C_ps = 2000 J/kg°C, C_pf = 1841 J/kg°C, ε_e = 0.43, ε_p = 0.48, K_d = 1.0. Note: Use Eq. (17-15c) for solute velocities. Assume that wall heat capacities can be ignored, heat of adsorption is negligible, no adsorption of n-heptane.

Using the solute movement theory,

a) determine the breakthrough time for toluene during the feed step.

b) determine time for thermal wave to breakthrough.

c) determine time to remove all toluene from column.

d) determine the outlet concentration profile of the regeneration fluid.

D5. We have a column packed with a resin that immobilizes a liquid stationary phase. The column is initially clean, c_A = 0. At time t = 0, we input a feed that is c_A.feed = 1.5 g/liter. The superficial velocity is 20 cm/minute. The column is 50 cm long. The packing has ε_e = 0.4, ε_p = 0.54, K_d = 1.0, ρ_s = 1.124 kg/liter, and the equilibrium for component A is an unfavorable isotherm,

q = 1.2 c_A/(1 − 0.46 c_A) where q is in g/kg and c_A is in g/liter.

Use solute movement theory to predict the outlet concentration profile of A (c_out vs. time). (You can report this as a graph or as a table or as both.)

D6. We are adsorbing p-xylene from n-heptane on silica gel. The n-heptane does not adsorb, and the p-xylene follows a linear isotherm with Arrhenius temperature dependence. The calculated values of the equilibrium are,

q = 12.1090 c at 300 K and q = 4.423 c at 350 K where q and c are both in g xylene/liter.

The column is initially clean (c = q = 0) and is at 300 K. At t = 0 we input a feed that contains 0.010 g/liter p-xylene at 300K. At t = 20 minutes, we input a clean solvent (c = 0) at 350K.

The interstitial velocity is constant throughout at v_inter = 30 cm/min. L = 50 cm, ε_e = 0.43, ε_p = 0.50, K_d = 1.0, ρ_s = 1.80 kg/liter, ρ_f = 0.684 kg/liter, C_p,s = 920 J/kg/K, C_p,f = 2240 J/kg/K, and heat storage in the wall can be neglected. The column is adiabatic.

a) Find u_s (T = 300 K), u_s (T = 350 K), and u_th.

b) Find the breakthrough time for the thermal wave.

c) Predict the outlet concentration profile (c_out vs. time).

D7.* A 60.0 cm long column contains activated carbon. The column is initially clean. At t = 0 we start feeding the column in an upwards direction a dilute aqueous solution of acetic acid at 4°C. The feed concentration is 0.01 kg moles/m³ acetic acid. The superficial velocity of the feed is 15.0 cm/min. After a very long time (1200 min) and when the column is certainly totally saturated (c = 0.01 everywhere), the feed is stopped, the flow direction is reversed, and the column is eluted with pure water at 60°C at a superficial velocity of 15.0 cm/min. This elution continues for another 1200 minutes.

Data: Equilibrium at 4°C, q = 0.08943 c

Equilibrium at 60°C, q = 0.045305 c, c is in kg mol/m³ and q is kg mol/kg carbon, ρ_f = 1000 and ρ_s = 1820.0 kg/m³, K_d = 1.0,ε_e = 0.434, ε_p = 0.57, C_ps = 0.25 and C_pf = 1.00 cal/(g°c). Ignore wall heat capacity effects.

a) From t = 0 to 1200 minutes predict the outlet concentration profile (top of column).

b) Predict the outlet concentration profile (bottom of column) for the 1200 minutes of elution.

D8. A 60.0 cm long column contains activated carbon. The column initially contains a concentration of 0.04 kg moles/m³ acetic acid at 60°C. At t = 0 we start eluting the column in an upwards direction with pure water (c = 0.0) at 4°C at a superficial velocity of 15.0 cm/min. This elution continues for the end of the experiment.

Data: Equilibrium at 4°C, q = 0.08943 c

Equilibrium at 60°C, q = 0.045305 c, c is in kg mol/m³ and q is kg mol/kg carbon ρ_s = 1820.0 kg/m³, K_d = 1.0, ε_e = 0.434, ε_p = 0.57, C_ps = 0.25 and C_pf = 1.00 cal/(g°c), ρ_f = 1000 kg/m³. Ignore wall heat capacity effects.

a. Predict when the thermal wave will exit the column.

b. Predict the outlet concentration profile (top of column) until the concentration leaving is zero.

D9. A PSA unit with a 1.0 m. long column is used to remove methane from hydrogen using Calgon Carbon PCB activated carbon. The feed gas is 0.003 mole fraction methane. Superficial velocity is 0.020 m/s during the feed step. The high pressure is 5.0 atm while the low pressure is 1.0 atm. The operation is at 480 K. Carbon properties: ρ_s = 2.1 g/cc, K_d = 1.0, ε_p = 0.336, ε_e = 0.43. The equilibrium isotherms are q_i = K′_i,p p_i where p_i is the partial pressure in Pa and q_i is in mol i/gram adsorbent. K′_methane,p = 7.25 E -10 mol methane/(Pa g adsorbent) and = 0.0 mol hydrogen/(Pa g adsorbent). We set z = 0 at the feed end of the column. Assume isothermal operation, constant density, and constant velocity when pressure is constant. 1.0 atm = 1.01325 E 5 Pa.

A pure purge gas is used and repressurization is with pure product gas. The column is clean at the end of the purge and the repressurization steps. We operate the feed step until the methane wave reaches z = 0.85 meters. The feed is stopped, the feed end of the column is closed off, and a co-flow blowdown (gas exits at same end of column as the pure hydrogen product, z = 1.0) is done to an intermediate pressure p_m. This pressure is chosen so that the methane solute wave is at z_after = L = 1.0 m when the co-flow blowdown is done. We then do a counterflow blowdown from p_m to p_L = 1.0 atm. The cycle would be finished with a normal counterflow purge step at p_L.

a . Calculate the intermediate pressure p_m at the end of the co-flow blowdown step.

b. Calculate the mole fraction of methane y_after in the gas at z = L at the end of the co-flow blowdown step (pressure = p_m).

(If you are unable to do part a or obtain a value of p_m greater than 5 or less than 1.0, assume that p_m = 3.0 atm and then solve part b.)

c. For the counterflow blowdown step, use p_m = your calculated value or p_m = 3.0 if you assumed this value in part b. First, find the value of z_after for the methane wave when the counterflow blowdown is done. Then calculate the mole fraction of methane y_after at this location at the end of the counterflow blowdown step.

Note: Do not do any calculations for the repressurization, feed, and purge steps.

D10.* Intermediate concentrations in the outlet concentration profile for trace PSA systems can be estimated using Eqs. (17-28a) to (17-28c). For Example 17-4:

a . Show that point 10 in Figure 17-13B was determined by starting with t = 1 s (end of blowdown) at z_after = 0.40 m (0.10 m from feed end). This can be done by using Eq. (17-28c) to find p_before and Eq. (17-28a) to find y_M,after. Since solute of this mole fraction moves as a wave at the known velocity u_M during the feed step, the time it exits the top of the column can be determined.

b . Starting at the point t = 1 s, z_after = 0.48m, calculate point 11 in Figure 17-13B.

D11. Repeat Example 17-4 but using repressurization with pure product.

D12. We have a column packed with activated carbon that is used for adsorbing acetone from water at 25°C. The isotherm can be approximated as a Langmuir isotherm (Seader and Henley, 1998), q = (0.190 c)/(1 + 0.146c), q is mol/kg carbon and c is mol/m³. The bed properties are: K_d = 1.0, ρ_s = 1820 kg/m³, ρ_f = 1000 kg/m³, ε_e = 0.434, ε_p = 0.57. The bed is 0.50 meters long and has an internal diameter of 0.08 meters. The flow rate is 0.32 liters/minute for both parts a and b.

a . If the bed is initially clean (c = q = 0), determine the outlet concentration curve (c_out vs. time) if a feed containing 50 mol/m³ of acetone is fed to it starting at t = 0.>

b . If the bed is initially saturated with a fluid containing 50 mol/m³ of acetone, predict the outlet concentration profile when the bed is eluted with pure water (c = 0) starting at t = 0. Find outlet times for concentrations of c = 50, 40, 30, 15, 5 and 0 mol/m³. Plot the curve of concentration out vs. time.

Note: the solute movement theory works just as well for mole balances as for mass balances.

D13. a . Repeat Example 17-5 but with M_1A = M_2B = 0.95, M_3A = M_4B = 1.05.

b . Determine the values of L, v_F, and D/F if F = 0.01 m³/min.

D14. We wish to use the local equilibrium model to estimate reasonable flow rates for the separation of dextran and fructose using an SMB. The isotherms are linear and both q and c are in g/liter. The linear equilibrium constants are: Dextran, 0.23 and Fructose, 0.69. The inter-particle void fraction = 0.4 and the intra-particle void fraction = 0.0. The columns are 40.0 cm in diameter. We want a feed flow rate of 1.0 liter/minute. The feed has 50.0 g/liter of each component. The desorbent is water and the adsorbent is silica gel. The columns are each 60.0 cm long. The lumped parameter mass transfer coefficients using fluid concentration differences as the driving force are 2.84 1/min for both dextran and fructose. Operation is isothermal. Use multiplier values (see notation in Figure 17-14) of M_1A = 0.97, M_2B =.99, M_3A = 1.01, and M_4B = 1.03. Determine the flow rates of desorbent, dextran product, fructose product, and recycle rate; and find the ratio D/F.

D15. A column packed with gas-phase activated carbon is initially filled with clean air. At t = 0 a feed gas containing y = 0.0005 wt frac toluene in air is started. This feed continues until t = 10.0 hours at which time a feed that is y = 0.0015 wt frac toluene is introduced and continued throughout the remainder of the operation. The superficial velocity is always 15.0 cm/s. Find the minimum column length required to have a single shock wave exit the column.

Data: Equilibrium (Basmadjian, 1997)

q = kg toluene/kg carbon and y = wt frac, which is essentially kg toluene/kg air. T = 298 K, P_tot = 50 kPa. Assume gas has density of pure air, which acts as an ideal gas.

, , Kd = 1.0, ε_e = 0.40, ε_p = 0.65, ρ_s = 1500kg/m³. Watch your units.

D16. A 25 cm long column packed with gas-phase activated carbon is initially filled with air containing y = 0.0010 wt frac toluene. At t = 0 a feed gas containing y = 0.0005 wt frac toluene in air is started. This feed continues until t = 20.0 hours at which time a feed that is pure air (y = 0.0000) is introduced and continued throughout the remainder of the operation. The superficial velocity is always 21.0 cm/s.

Data: Equilibrium (Basmadjian, 1997)

q = kg toluene/kg carbon and y = wt frac, which is essentially kg toluene/kg air. T = 298 K, P_tot = 50 kPa. Assume gas has density of pure air, which acts as an ideal gas.

, , Kd = 1.0, ε_e = 0.40, ε_p = 0.65, ρ_s = 1500kg/m³.

Predict the outlet concentration profile. Specifically, find when the following concentrations exit: y = 0.0010, 0.00075, 0.0005, 0.00025 and 0.00000. Sketch the outlet concentration profile (y vs. t) and label the times when these concentrations exit the column. Watch your units.

D17. Plot ion exchange curves.

a . Monovalent-monovalent exchange. Plot y silver (resin phase) vs. x silver (solution) for the exchange of Ag⁺ and K⁺ on a strong acid resin with 8%DVB. The total resin capacity is c_RT = 2.0 equivalents/liter and the ionic concentration in solution is c_T= 2.2 equivalents/liter.

b . Divalent-monovalent exchange. Plot y copper (resin phase) vs. x copper (solution) for the exchange of Cu⁺⁺ and H⁺ on a strong acid resin with 8%DVB. The total resin capacity is c_RT = 2.0 equivalents/liter and the ionic concentration in solution is c_T= 0.6 equivalents/liter.

c . Repeat part b, but with total resin capacity c_RT = 2.0 equivalents/liter and the ionic concentration in solution c_T = 5.0 equivalents/liter.

D18. We have adsorbed phenylalanine on an Amberlite XAD-2 resin (polystyrene cross linked with divinylbenzene). The phenylalanine is dissolved in water. The column is saturated with an aqueous solution that is c = 0.0350 moles/liter. At t = 0 we start to elute the column with a feed that has c = 0.0050 moles/liter. We are interested in the concentrations of the outlet stream.

Data: Equilibrium data for phenylalanine fits Langmuir form, q = 0.00636c/(1 + 7.167c) where q is in moles/g adsorbent and c is in moles/liter fluid. Column length L = 50 cm. Superficial velocity v_super = 5.0 cm/min. ε_e = 0.4, ε_p = 0.0, solid density ρ_s = 1280 g/liter, ρ_f = 1000 g/liter, K_d = 1.0

Determine the time (minutes) that the outlet concentration profile has the following concentrations:

c = 0.03499 moles/liter

c = 0.0200 moles/liter

c = 0.00501 moles/liter

D19. We are exchanging Ag⁺ and K⁺ on a strong acid resin with 8%DVB. The total resin capacity is c_RT = 2.0 equivalents/liter and the ionic concentration of the feed solution is c_T= 1.2 equivalents/liter all of which is Ag⁺. The column is initially at a solution concentration of c_T= 0.2 equivalents/liter all of which is K⁺.

a. How long does it take for the total ion wave to breakthrough?

b . At what time does the Ag⁺ shock wave breakthrough?

c . After the Ag⁺ shock wave breaks through, the column is regenerated with pure K⁺ solution with c_T= 1.2 equivalents/liter. Predict the shape of the ensuing diffuse wave. (Reset t = 0 when start the K⁺ regeneration solution.)

Data: ε_e = 0.4, ε_p = 0.0, K_E = 1.0, v_super = 3.0 cm/min, L = 50 cm.

D20. A strong acid exchanger is exchanging Ni⁺² and H⁺. Initially the column has a total fluid concentration of c_T = 5.1 eq/L and x_Ni = 0.2. At t = 0 we feed the column with c_T = 0.1 eq/liter and x_Ni = 0.2. Superficial velocity = 10.0 cm/minute. L = 75 cm, ε_e = 0.4, ε_p = 0.0, K_E = 1.0, c_RT = 2.1 equivalents/liter, and selectivity constants can be determined from Table 17-5. Predict the outlet concentration profile using ion movement theory.

D21. We have an adsorption column that is 50.0 cm long. It is packed with a material that adsorbs glucose. At t = 0, the column is saturated with glucose at 50 g/liter. The equilibrium is linear, (ρ_p K_Glucose) = 0.23. ε_e = 0.45, ε_p = 0. Interstitial velocity v = 1.0 cm/s. E_eff = 0.77 for the Lapidus and Amundson solution.

If at t = 0 we start feeding a solution which is 20 g/liter glucose, predict the outlet concentration profile (concentration vs. time). Calculate the center point and two other points so that you can sketch the curve.

D22. Use the Lapidus and Amundson solution to predict the behavior of fructose in a column packed with silica gel. The column is initially clean (contains no fructose). The feed is 50 g/liter, the feed pulse lasts for 8 minutes, and then it is eluted with water. The flow rate is 20 ml/min. The other values are:

Lumped parameter with concentration driving force., k_m,ca_p = 5.52 1/min

Isotherm is linear, . Isotherm parameter, q and c in g/liter.

Calculate E_effective, and then calculate enough points on the curve to plot it. Note that this is a step up followed 8 minutes later by a step down.

D23. A column is packed with a strong cation exchanger. The column is initially in the K⁺ form and c_T = 0.02 eq/liter. The column is 75.0 cm long, the superficial velocity is 20.0 cm/min and the flow is in the same direction for all steps, ε_p = 0, ε_e = 0.4, K_E = 1.0 (for the cations), and c_RT = 2.0 equivalents/liter. Equilibrium data are listed in Table 17-5.

a . At t = 0, feed the column with a solution with c_T = 0.02 eq/liter, x_Ca = 0.80, x_K = 0.20. Plot the outlet value of x_Ca vs. time.

b . At t = 500 minutes, the column is regenerated with a pure aqueous solution of K⁺ with x_K = 1.0, x_Ca = 0.0 and c_T = 1.0 eq/liter. Plot the outlet value of c_T vs. time and the outlet value of x_Ca vs. time.

Note: if either of these steps require you to calculate a diffuse wave, calculate velocities and breakthrough times at three values of x_Ca: at the highest mole fraction, at the lowest mole fraction and at x_Ca = 0.5.

D24. For a linear system the breakthrough solution is c_out/c_feed = X (z,t). At t = 0 we have an adsorption column that is initially at a uniform concentration c_initial, and the feed concentration is also c_initial. At t = 17.5 minutes, the feed concentration is reduced to c_F1(c_F1 < c_initial). At t = 28 minutes the feed concentration is reduced to 0.0. Write the solution for c_out in terms of c_initial, c_F1, and the breakthrough solution X.

D25. We wish to use chromatography with a differential pulse to separate fructose and dextran on silica gel. The interstitial velocity is 25.0 cm/min. The external porosity = 0.4 and the particle porosity = 0.0. The solid density is ρ_s = 1.023 kg/liter. The equilibrium for both components is q_i = K_i c_i, where both q and c are in grams/ml. K_fructose = 0.69 and K_dextran = 0.23. We desire a resolution of R = 0.92. The value of HETP = 0.14 cm.

a . Find the value of N required.

b . Find the column length L required.

D26. A chromatograph is separating acetonaphthalene (A) from dinitronaphthalene (Dinitro) on 20-micron silica gel. For a single-porosity model, , and in a solvent with 23% methylene chloride, and 77% n-pentane. When the interstitial velocity v = 1.0 cm/s, HETP is 0.05 cm.

a . If we desire a resolution of R = 1.5 for an infinitesimal pulse of feed, what column length is required?

b . Plot the outlet A curve as c/c_max, vs. time using the Gaussian solution for R = 1.5.

Note: in the chromatography literature the parameter is known as the relative retention. The relative retention is easily determined from experiments.

D27. For a linear system the breakthrough solution is c_out = X (z,t). We have an adsorption column that is initially at a uniform concentration c_initial = 0, and the feed concentration is also c_initial = 0. At t = 0 minutes, the feed concentration is increased to 0.33c_F (c_F > 0). At t = 0.4 t_F minutes the feed concentration is increased to c_F. At t = 0.8 t_F the feed concentration is decreased to 0.55 c_F. Finally, at t = t_F the feed concentration is reduced to 0.0. Write the solution for c_out in terms of c_F, t_F, and the breakthrough solution X.

D28.* A 25.0 cm long laboratory column is packed with particles that have an average diameter of 0.12 mm. At a superficial velocity of 9.0 cm/min we measure the breakthrough curve for a step input of solute. The column was initially clean. The center of the symmetrical breakthrough curve exits at 35.4 minutes while the width (measured from 0.05 c_F to 0.95 c_F) is 2.8 minutes. Pore diffusion controls and the isotherm has a Langmuir type shape.

a . What is L_MTZ in the lab unit?

b . We want to design a large-scale unit with 0.80 frac bed use. The average particle diameter will be 1.0 mm. The superficial velocity will be 12 cm/minute. How long should this unit be? What is t_br (c_out = 0.05 c_F)? Assume ε_e is same in both units.

D29. For the larger scale system in Example 17-11 part b, use 2.0 mm diameter particles in a system that is pore diffusion controlled. The other conditions are the same as in part b of Example 17-11. Find the column length for 0.90 frac bed use.

F . Problems Requiring Other Resources

F1. Read the article by Agosto et al. (1989). Apply the solute movement theory to this system to determine how the moving withdrawal chromatography system works.

G . Simulator Problems

G1. a . If you have access to a simulator such as Aspen Chromatography, repeat Problem 17.D.22 on the simulator.

b . Rerun Problem 17.D.22 on the simulator, but with k_m,ca_p = 100,000 1/min and E_D = E_effective.

c . Compare your two simulator runs.

d . Compare the Lapidus and Amundson solution to the Aspen solutions. Note: you do need to consider convergence and accuracy of the Aspen runs.

G2. Set up a chromatographic column with one feed, a column and a product. The components that adsorb are acetonaphthalene (A) and dinitronaphthalene (DN). Use a model with convection with constant axial dispersion coefficients (0.25 cm²/min for both A and DN), constant pressure, and velocity. Use a linear lumped parameter model with driving force of (c – c*) and constant mass transfer coefficients (k_m,ca_p = 50.0 1/min for A and 45.0 1/min for DN). The isotherms are both linear, q = 0.003056 c (for A) and q = 0.003222 c (for DN) where q is in g adsorbed / kg adsorbent and c is in g solute/m³ of solution. Operation is isothermal.

The column will be 50.0 cm long with a 2.0 cm diameter. The adsorbent has the following properties: ε_e = 0.40, ε_p = 0.46, K_D = 1.0, ρ_s = 2222 kg/m³. The feed flow rate is 0.10 liters/minute, feed pressure = 3.0 bar, and the feed concentration for A is 2.0 g/liter while the feed concentration of D is 1.0 g/liter.

a . Run a breakthrough curve for 10 minutes. Print, label, and turn in your plot. Accurately determine the t_MTZ for component A where t_MTZ is measured from 0.05 times the A feed concentration to 0.95 times the A feed concentration. Show this calculation.

b . Input a 0.010-minute feed pulse, and develop with pure solvent for a total time of 10 minutes. Print your plot.

G3. We want to separate component 1 from component 2 using an SMB. In the feed both compounds are dissolved in water, component 1 is 40 g/liter and component 2 is 60 g/liter. Feed rate is 141.55 ml/min. The components both have linear isotherms where both q and c are in g/liter. Component 1: q₁ = 1.5 c₁; Component 2: q₂ = 3.5 c₂. The SMB has 6 columns: two columns between the feed and the extract (B) product (zone 3 in Figure 17-14B) and 2 columns between the feed and the raffinate (A) product (zone 2). There is one column between the desorbent addition and the extract (B) product (zone 4). The SMB has closed recycle and there is one column between the raffinate (A) product and the desorbent addition (zone 1). Use constant pressure and volume (pressure = 3.0 bars), linear lumped parameter with (c – c*) driving force, and isothermal operation. Each column is 50.0 cm long and has a diameter of 10.0 cm. Data: ε_e = 0.40, ε_p = 0.45, K_D = 1.0, ρ_bulk = 800 kg/m³. The density of the fluid can be assumed to be 1000 g/liter. The values of both axial dispersion coefficients are 0.35 cm²/min, the mass transfer coefficient (k_m,ca_p) for component 1 is 150 1/min and for component 2 is 100 1/min.

a . First, design your SMB to operate at the optimum point for D/F = 1.0 (all M_i = 1.0). Calculate D, E, and R, and the switching time and the recycle rate, and report these values. Run the system for at least 10 complete cycles (a cycle is 6 switching times). Turn in a plot of raffinate concentrations (components 1 and 2) and a plot of extract concentrations (components 1 and 2). Also, report the average mass fraction over the last cycle for component 1 in the raffinate, and the average mass fraction over the last cycle for component 2 in the extract.

b . Next, increase D/F to 2.0 (there are, of course, many ways to do this). Calculate D, E, and R, and the switching time and the recycle rate, and report these values. Briefly, explain your rationale for how you did the design to increase D/F to 2. Run the system for at least 10 complete cycles. Turn in a plot of raffinate concentrations (components 1 and 2) and a plot of extract concentrations (components 1 and 2). Also, report the average mass fraction over the last cycle for component 1 in the raffinate and the average mass fraction over the last cycle for component 2 in the extract.

G4. Ion Exchange. Use a bed that is 50 cm long and 20 cm in diameter. External porosity = 0.4, internal porosity = 0.0. Use axial dispersion coefficients of 0.2 cm²/min for all components. Mass transfer coefficients (k_m,ca_p) are 10.0 1/min for all components. For a strong acid resin, use c_RT = 2.0 eq/liter. K_H-H = 1.0, K_Na-H = 1.54. Operate with a feed rate of 5.0 liter/minute. Pressure is 2.0 bar.

a . Set the feed component concentration of H⁺ equal to 0.0. Use a feed concentration of 1.0 eq/liter for Na⁺. The column is initially in the H⁺ form with c_T = 1.0 eq/liter. Do a breakthrough run.

b . After the column is converted to the Na⁺ form, feed it with pure acid (H⁺) at 1.0 eq/liter (Na⁺ is 0.0 and 1.0 eq/liter H⁺). Do a breakthrough curve.

c . Now do a wash step with pure water (feed concentrations of all components are zero). Continue the run until concentrations all become zero. Explain why this is so quick.

G5. Water Softening. Use the same column, same resin, same mass transfer and axial dispersion coefficients as in Problem 17.G4, except exchange Mg⁺⁺ with H⁺. K_Mg-H = 1.9527. The column is initially in the H⁺ form with c_T = 0.10 eq/liter. We want a feed that is 0.0 eq/liter of H⁺ and 0.10 eq/liter of Mg⁺⁺ (c_T = 0.10 eq/liter). The value of c_RT = 2.0 eq/liter is unchanged. (There is no Na⁺ involved in this operation.) To reduce run time, operate with a column length of 10.0 cm.

Unfortunately, most simulators will find this problem to be difficult since the equations are very stiff. Change the convergence settings and the discretization procedure to obtain convergence.

G6. Separation of ternary mixture with a feed consisting of dextran, fructose and a heavy impurity. A 25.0 cm long column with a diameter of 2.0 cm is used. External porosity = 0.4, internal porosity = 0.0. Pressure is constant at 2.0 bar. Use axial dispersion coefficients of 0.35 cm²/min for all components. Mass transfer coefficients (k_m,ca_p) are 2.84 1/min for dextran and fructose, and 1.5 1/min for the heavy; and the driving force for mass transfer is concentration differences. The isotherms are linear with q and c both measured in g/L. The isotherm parameters are 0.23 for dextran, 0.69 for fructose and 10.0 for the heavy impurity. The feed flow rate is 20.0 ml/min, the overall concentration of the feed is 100.0 g/l, and the feed is 48 wt % dextran, 48 wt % fructose, and 4.0 wt % heavy.

a . Input a 1.5-minute feed pulse followed by pure solvent (water). Set up a plot with the outlet concentrations of dextran, fructose and heavy and run for 100 minutes.

Where is the heavy? If your plot uses the same scale for all three components, it will be difficult to see the heavy. To find it set up the plot with a separate scale for heavy concentration. Note that the heavy has an initial peak and then a main, very broad peak (all at low concentrations). The initial peak occurs because with the low mass transfer coefficient some of the heavy exits the column without ever entering the packing material. This is called “bypassing” or “instantaneous breakthrough” and is obviously undesirable. Check to see if Eq. (17-66) is satisfied, and interpret your results.

b . Change the heavy mass transfer coefficient (k_m,ca_p) to10.0 1/min and run again. Now it should look more normal, except the concentrations of the heavy are very low. This type of problem (at least two components that are difficult to separate plus very slow component(s)) is known as the “general elution problem.” The column length needs to be set to separate the difficult separation, but then slow components take a long time to come out.

c . (All mass transfer coefficients are k_m,ca_p = 10.0 1/min for this step.) Input a 1.0-minute pulse of feed followed by 4.5 minutes of pure solvent. Then reverse the flow of solvent and elute for 15 minutes. Develop plots for both the forward flow and for the reverse flow.

G7. Thermal effects. We wish to adsorb toluene from n-heptane and then use co-current desorption with a hot liquid. Set the feed rate = 16.0 cm³/min. Use a feed concentration of toluene of 0.008 g/l. Pressure = 3.0 atm. Feed temperature is 25.0°C. Column is initially pure n-heptane and is at 25.0°C.

a . Do a breakthrough curve with a run time of 125 minutes. Create plots of toluene concentration and temperature.

b . We now want to develop this saturated column co-flow using hot pure solvent. Increase the feed temperature to 80.0°C, set the toluene concentration = 0.0 and do the regeneration run. Print the plots of outlet toluene concentration and temperature. Explain the results.

Column conditions and data: L= 100.0 cm, D_col = 2.0 cm, ε_e = 0.43, ε_p = 0.5, d_p = 0.0335 cm, ρ_s = 1800.0 kg/m³, E_D = 0.2 cm²/s for both toluene and n-heptane, k_m,ca_p = 5.5656 1/min. The isotherm is linear with Arrhenius temperature dependence: pre-exponential factor K_Toluene,o = 0.0061, and exponential factor (−ΔH/R) = 2175.27. C_P,s = 920.0 J/kg/K, and the effective liquid-solid heat transfer coefficient, = 0.005 J/(sm²K).

CHAPTER 17 APPENDIX INTRODUCTION TO THE ASPEN CHROMATOGRAPHY SIMULATOR

The Aspen Chromatography simulator is fairly complicated. This appendix contains material from two of the nine labs that were developed for an elective course (Wankat, 2006). The complete set of labs is available from the author by sending a request to [email protected]. The numerical method used by Aspen Chromatography is the method of lines, (e.g., see Schiesser, 1991).

Lab 1

The goal of this lab is to get you started in Aspen Chromatography. It consists of a cookbook on running Aspen Chromatography, some helpful hints, and the simulation of a real separation. The assistance of Dr. Nadia Abunasser in developing this lab was critically helpful.

1. Log in to the computer. Use your local operating system method of getting into Aspen Chromatography.

2. We will first develop a simple chromatography (or adsorption) column system. To do this, go to the menu bar and on the left hand side (LHS) select File. Go to Templates. In that window click on “Blank trace liquid batch flowsheet,” and then click on Copy. It will ask for a file name. Use something like, “column1.” This will be saved in your working file. NOTE: In all file names and names for components, columns, streams, and so forth there must be no spaces.

3. On the “Exploring simulation” box (LHS), click on “component list.” Then in the Contents box below double-click on Default. A and B are listed. Change these names to the names of the components to be separated (fructose and dextran T6). First, click the “Remove all” button. In the window below, type in the first component name (e.g., fructose), and click on the “add” button. Do the same for all the other components. Click “OK.”

4. Now to draw the column. Click on the + to the left of “Chromatography” in the Exploring Simulation box. This opens other possibilities. Click on the word “chromatography.” This should give “Contents of Chromatography” in a box below. Double-click on the model you want to use (Reversible—since it is most up-to-date). Click and drag the specific model you want—in this case “chrom_r_column”—and move to the center of the Process Flowsheet Window. This gives a column labeled B1. Left-click on B1, then right-click to open a menu. Click on “rename.” Call the block something like “column.” Click on “OK.”

5. Now for some confusion. Go to the Nonreversible model by clicking on the word “Nonreversible” in the upper box of “Exploring.” This will get you new contents. (It will be easier to read if you right-click inside the box and then left-click to obtain the menu. Choose “small icons.”) Click and drag a feed stream (“chrom_feed”), and put it near the top of the column (since the nonreversible column has flow downwards by arbitrary convention). Now click and drag a product stream (“chrom_product”), and put it near the bottom of the column. Rename both of these. (We are using the code in the reversible model since it is up to date but are using nonreversible for feed, product, and connecting lines as it is simpler and less likely to cause problems.)

6. Now to connect everything together. In Explore “All Items,” click on the word “Stream Types.” This gives contents below. Click and drag “chrom_Material_Connection” (avoid the one with “r”). Click on the blue arrow of a source (e.g., the feed), move over to the inlet blue arrow for the column, and left-click. Repeat for column outlet and product arrow. Rename these streams. (Note for Aspen Plus users: In Aspen Plus we would click on the Next button, it would tell us that the connections were complete, and would then lead us through the necessary steps. Unfortunately, Aspen Chromatography does not have this feature.)

7. We will now set up the column operating conditions. Double-click on the column. This gives a box labeled, “Configure Block/Stream Column.” For now keep the UDS1 PDE discretization scheme, but change the number of nodes to 50. Note that Aspen Chromatography has a number of integration schemes that can be used for more complex systems or longer columns, but they are not required for this lab. The important issue of accuracy of numerical integration will be explored in detail in a later lab.

a . Go to the Material balance tab. Select Convection with constant dispersion coefficient. Constant pressure and velocity (the default) is fine for this liquid system.

b . Under the Kinetic tab, the fructose-dextran T6 data we have uses (c – c*) as the driving force; thus, choose “fluid” for film model assumption. For Kinetic model assumption use linear lumped resistance. For mass transfer coefficient use “constant.”

c . In the Isotherm tab, the fructose-dextran T6 system has a linear isotherm that is based on fluid concentration (select “Volume base, g/l”) for loading base.

d . Energy balance tab—isothermal should be checked.

e . Procedures tab: should be empty. If not, an error was made.

8. Now click on the Specify button. This opens an imposing table where you specify the column dimensions, isotherms, and so forth. Note that it lists fructose and dextran T6 because in step 3 you told it to do this. Use the following values to start with: L = 25 cm, column diameter Db = 2.0 cm, external porosity Ei = 0.4, particle porosity Ep = 0.0, dispersion coefficients Ez = 0.15 for both components. Mass transfer coefficient MTC=5.52 min⁻¹ (fructose) and = 2.84 min⁻¹ (dextran T6). Isotherm coefficients: Aspen uses the formula for linear isotherm: q = (IP1) c + IP2.

Set IP2 = 0.0 for both components. The adsorbent used was silica gel, and the solvent was water. (This information is never entered into Aspen when a template is used.) For fructose IP1 = 0.69, and for dextran T6, IP1 = 0.23. (Units on q and c are in g/volume solid and g/volume fluid.) Hit Enter, and then close this window (click on X).

9. Click on the button for “Presets/Initials.” For an initially clean column (which is what we want) all values should be zero. If they aren’t, insert 0.0 values, hit enter, and close window.

10. Click on the “Initialize “button. This makes all concentrations the proper values. Close the Configure Block/Streams window.

11. To set the feed concentrations and flow rates, double-click on the arrow for the feed. This opens a window labeled Configure Block/Stream Feed. Click on the Specify button. Use a feed concentration of 50 g/liter for both components. Set pressure at 2 bar (this has little effect). Flow rate of 20.0 cm³ /min is reasonable (Note: you have to change the units in the menu to the right of the number. Change the units first, and then enter the desired value.) Hit enter. Check that your values are correct, and close the window. Close the Configure block/stream feed window.

12. Do not touch the product stream—specifying something here will over specify system.

13. Set integration step size: Go to Run in the toolbar. In the Run menu click on “solver options, which opens the Solver Properties window. In the Integration tab, pick the “Implicit Euler” method, and click on the Fixed radio button. Typically use a step size that is approximately (run time)/200. Might try 0.025 as a first try. For now, ignore the other tabs. Click OK.

14. We now need to set up the plots we want. Left-click and then right-click on the name of your product stream. In the menu that opens go to Forms, and click on “All Variables.” This opens a table for this product stream. We will use this for dragging variables. Click on the icon of a plot with “t” (in toolbar). Name the plot (e.g., “Chromatography1”). Drag the variable “Process_in.C(“fructose”)” to the y axis. Do this for dextran T6 as well. Double-click on the plot, which opens a window, “Pfs Plot Control Properties.” Click on the second tab, Axis Map, and click on “all in one.” Then click OK. This puts the two concentrations on the same scale. (The problem with two scales is Aspen has a tendency to use different ranges, which makes comparisons rather difficult.) Close the product line table.

(Note: One nice thing about Aspen Chromatography is once you set up the plot it hangs around and you can reuse it for different runs.) If you “lose” the plot, go to Window (in toolbar), and at the bottom of the menu you should find the plot name. Click on this to recover the plot.

15. We will now do a breakthrough run: Go to Run on the toolbar, and click on Run Options in the menu. Click on “Pause at.” Try 10 minutes. Click OK.

16. To initialize the separation, in the center of the toolbar, select Menu and choose Initialize. (This is not strictly necessary, but it is good practice because it allows you to use the rewind button later.) Click on the Play button (to the right of the menu). Click on OK after it runs. The run is initialized. Now, go back to the menu, and select Dynamic. Cross your fingers, and click on the Play button. After it runs, click OK. Right-click on the figure. Select “Zoom Full.” You can print the figure if you want. Right-click on the figure again. In the menu select, “Show as history.” This gives a table of both concentrations. This table can be cut and pasted into Excel if further manipulation of the numbers is desired. It is better to not close the plot—we will reuse the plot. If you minimize it, you can find it under Window. Note: If you close the plot (or close and later reopen the entire file), go to the All Items box in the “Exploring Simulation,” and click on the word “flowsheet.” The plot will be next to an icon that looks like a square donut and will be listed by name in the “contents box. Double-click on the name to recover the plot.

17. Now do a pulse input of 1.0 minute. Start with a clean column. If you initialized, then you can simply restart to the initial state (this is the rewind button on a VCR—fourth button over from the menu in toolbar). Now you will have a clean column. To check this, double-click on the column, then click on the results button. All concentrations should be zero. If it is not zero, input zero values, and click on the Initialize button. Close the windows. There are a number of ways to add a pulse. One easy way is as follows: First, go to Run in toolbar, and click on Run Options. Select “Pause at,” and insert your pulse length (1.0 minute). Initialize the simulation again (using the menu in the center of toolbar). Select “dynamic,” and click on the Start button. (Since a plot was already drawn, you will get one automatically.) The run for this pulse is very short (lasts one minute). We have now input the pulse, and we must now develop the pulse with pure solvent. Double-click on the feed, click on specify, set the concentrations to zero (the template, since it is for a dilute system, “knows” there is solvent present), hit enter, close the windows. Go back to Run (toolbar), click on Run Options, and change the “Pause at” time to the desired run time (10 minutes will work for this system). Click OK. Do not initialize this time. We want the pulse to remain in the column. The menu on the toolbar should read “Dynamic.” Click on the start button. When run is finished, click on OK, and look at the plot and the history.

18. When you look at the figure, it will probably appear to be a series of connected straight lines instead of a smooth curve. We can fix this by using more points in the plot. Click on rewind to get a clean column. Go to Run (toolbar) and then Run Options. Change “Communication” to 0.1 minute. Follow the procedure for running the pulse [input the feed for one minute (use “Pause at”) and Run. Then set the feed concentration to zero, set “Pause at” to 10 minutes, and Run.] Look at your plot. It should be smooth curves. Print this plot, and label it. Note: “Communication” just sets the number of points that are plotted but does not affect the integration. Thus, you can keep communication at 0.1 minutes for the remaining runs.

19. The two peaks are not completely separated. There are a number of ways they can be separated more completely. For example, double the value of L to L = 50 cm. Hit rewind, change L in the column dimensions table, and then rerun the one minute pulse input. When you run pure solvent, a pause time greater than 10 minutes is needed since doubling column length will double time for material to exit. Do this run and look at the result. Separation is better, but still not complete. Save your file (remember the file name), and exit Aspen.

Lab 2 The goal of this lab is to explore numerical convergence and accuracy of the simulator.

1. You need to increase accuracy of the discretization procedure. We will first do this by increasing the number of nodes with UDS1.

a . Rerun the breakthrough curve from Lab 1 (step 15), but use UDS1 with 200 nodes. To do this, double-click on the column, and in the Configure Block/ Stream Column window use the General tab. Change the number of nodes and then hit Enter. The integration method is Implicit Euler with time step at 0.025 min. Check that preset/initials are all zero, and initialize in this block. Then initialize the run. Do the run. Compare to your previous result with 50 nodes. If there is significant change, 50 nodes was not enough.

b . Make the comparison quantitative by calculating the width of the MTZ for each solute. The time of the MTZ, t_MTZ, is measured from c = c_initial + 0.05(c_feed − c_initial) to c = c_initial + 0.95(c_feed − c_initial). If the initial concentration is zero, measure from 0.05c_feed to 0.95c_feed. (The reason for using 5% and 95% of the change is that with experiments it is very difficult to tell when one is exactly at c_initial or at c_feed.) Note: The graphs are not accurate enough—use the history for the calculation of t_MTZ.

c . Try UDS1 with 800 nodes. Calculate t_MTZ values and compare to previous runs.

d . Change to either the BUDS or the QDS integration schemes with 50 nodes. Integration is still Implicit Euler with a time step of 0.025. [Change the number of nodes (no need to hit Enter), and then change to BUDS or QDS using the menu. Hit Enter (probably not necessary), and reinitialize.] Compare. Both BUDS and QDS work pretty well on this linear problem. Calculate the t_MTZ for both components and compare to previous.

e . Repeat the breakthrough runs for L = 50 cm and 100 cm, but using BUDS or QDS with 100 nodes and integration time step of 0.025. Calculate t_MTZ for both components for each run. Theory says that the widths should be proportional to L^1/2. (Runs with UDS1 with 50 nodes will not satisfy this. If you have time, try this.) Plot your widths (for each component separately) vs. L^1/2.

Note: BUDs and QDS are great for linear isotherms but often have convergence problems with nonlinear isotherms.