3.19b. Cross-flow cascade of ideal stages.
The drying and liquid-liquid extraction operations described in Problems 3.17 and 3.18, respectively, are examples of a flow configuration called a cross-flow cascade. Figure 3.27 is a schematic diagram of a cross-flow cascade of ideal stages. Each stage is represented by a circle, and within each stage mass transfer occurs as if in cocurrent flow. The L phase flows from one stage to the next, being contacted in each stage by a fresh V phase. If the equilibrium-distribution curve of the cross-flow cascade is everywhere straight and of slope m, it can be shown that (Treybal, 1980)

where S is the stripping factor, mVS/LS, constant for all stages, and N is the total number of stages.

Solve Problem 3.18 using equation (3-60), and compare the results obtained by the two methods.

Solution
Initial estimate
3.20a. Cross-flow cascade of ideal stages: nicotine extraction.
Consider the nicotine extraction of Problems 3.18 and 3.19. Calculate the number of ideal stages required to achieve at least 95% extraction efficiency.

Solution
Use 8 ideal stages
3.21b. Kremser equations: absorption of hydrogen sulfide.
A scheme for the removal of H2S from a flow of 1.0 std m3/s of natural gas by scrubbing with water at 298 K and 10 atm is being considered. The initial composition of the feed gas is 2.5 mole percent H2S. A final gas stream containing only 0.1 mole percent H2S is desired. The absorbing water will enter the system free of H2S. At the given temperature and pressure, the system will follow Henry's law, according to Yi = 48.3Xi, where Xi = moles H2S/mole of water; Yi = moles H2S/mole of air.
a) For a countercurrent absorber, determine the flow rate of water that is required if 1.5 times the minimum flow rate is used.

Solution
at SC
b) Determine the composition of the exiting liquid.

Solution
c) Calculate the number of ideal stages required.

Solution
3.22b. Absorption with chemical reaction: H2S scrubbing with MEA.
As shown in Problem 3-21, scrubbing of hydrogen sulfide from natural gas using water is not practical since it requires large amounts of water due to the low solubility of H2S in water. If a 2N solution of monoethanolamine (MEA) in water is used as the absorbent, however, the required liquid flow rate is reduced dramatically because the MEA reacts with the absorbed H2S in the liquid phase, effectively increasing its solubility.
For this solution strength and a temperature of 298 K, the solubility of H2S can be approximated by (de Nevers, N., Air Pollution Control Engineering, 2nd ed., McGraw-Hill, Boston, MA, 2000):
Repeat the calculations of Problem 3.21, but using a 2N monoethanolamine solution as absorbent.


Solution
3.23b. Kremser equations: absorption of sulfur dioxide.
A flue gas flows at the rate of 10 kmol/s at 298 K and 1 atm with a SO2 content of 0.15 mole %. Ninety percent of the sulfur dioxide is to be removed by absorption with pure water at 298 K. The design water flow rate will be 50% higher than the minimum. Under these conditions, the equilibrium line is (Ben’tez, J., Process Engineering and Design for Air Pollution Control, Prentice Hall, Englewood Cliffs, NJ, 1993):
where Xi = moles SO2/mole of water; Yi = moles SO2/mole of air.
a) Calculate the water flow rate and the SO2 concentration in the water leaving the absorber.

Solution

b) Calculate the number of ideal stages required for the specified flow rates and percentage SO2 removal.
Solution

3.24b. Kremser equations: absorption of sulfur dioxide.
An absorber is available to treat the flue gas of Problem 3.23 which is equivalent to 8.5 equilibrium stages.
a) Calculate the water flow rate to be used in this absorber if 90% of the SO2 is to be removed. Calculate also the SO2 concentration in the water leaving the absorber.

Solution
Initial estimate

b) What is the percentage removal of SO2 that can be achieved with this absorber if the water flow rate used is the same that was calculated in Problem 3.23 (a)?

Solution
Initial estimate
3.25b. Kremser equations: liquid extraction.
An aqueous acetic acid solution flows at the rate of 1,000 kg/hr. The solution is 1.1% (by weight) acetic acid. It is desired to reduce the concentration of this solution to 0.037% (by weight) acetic acid by extraction with 3-heptanol at 298 K. The inlet 3-heptanol contains 0.02% (by weight) acetic acid. An extraction column is available which is equivalent to a countercurrent cascade of 15 equilibrium stages. What solvent flow rate is required? Calculate the composition of the solvent phase leaving the column. For this system, equilibrium is given by
Wt ratio acetic acid in solvent = 0.828 ´ Wt ratio acetic acid in water


Solution

Let the aqueous phase be the V-phase; the solvent phase is the L-phase.
Initial estimate:
3.26c. Countercurrent versus cross-flow extraction.
A 1-butanol acid solution is to be extracted with pure water. The butanol solution contains 4.5% (by weight) of acetic acid and flows at the rate of 400 kg/hr. A total water flow rate of 1005 kg/hr is used. Operation is at 298 K and 1 atm. For practical purposes, 1-butanol and water are inmiscible. At 298 K, the equilibrium data can be represented by YAi = 0.62 XAi, where YAi is the weight ratio of acid in the aqueous phase and XAi is the weight ratio of acid in the organic phase.
a) If the outlet butanol stream is to contain 0.10% (by weight) acid, how many equilibrium stages are required for a countercurrent cascade?

Solution
b) If the water is split up equally among the same number of stages, but in a cross-flow cascade, what is the outlet 1-butanol concentration (see Problem 3.19)?

Solution
3.27c. Glucose sorption on an ion exchange resin.
Ching and Ruthven (AIChE Symp. Ser., 81,p. 242, 1985) found that the equilibrium of glucose on an ion exchange resin in the calcium form was linear for concentrations below 50 g/L. Their equilibrium expression at 303 K is YAi = 1.961 XAi, where XAi is the glucose concentration in the resin.(g of glucose per liter of resin) and YAi is the glucose concentration in solution.(g of glucose per liter of solution).
a) We wish to sorb glucose onto this ion exchange resin at 303 K in a countercurrent cascade of ideal stages. The concentration of the feed solution is 15 g/L. We want an outlet concentration of 1.0 g/L. The inlet resin contains 0.25 g of glucose/L. The feed solution flows at the rate of 100 L/min, while the resin flows at the rate of 250 L/min. Find the number of equilibrium stages required.

Solution
b) If 5 equilibrium stages are added to the cascade of part a), calculate the resin flow required to maintain the same degree of glucose sorption.

Solution