/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 A stage of a separation process ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A stage of a separation process is defined as an operation in which components of one or more feed streams divide themselves between two phases, and the phases are taken off separately. In an ideal stage or equilibrium stage, the effluent (exit) streams are in equilibrium with each other.Distillation columns often consist of a series of vertically distributed stages. Vapor flows upward and liquid flows downward between adjacent stages; some of the liquid fed to each stage vaporizes,and some of the vapor fed to each stage condenses. A representation of a section of a distillation column is shown below. (See Problem 4.42 for a more realistic representation.) Consider a distillation column operating at 0.4 atm absolute in which benzene and styrene are being separated. A vapor stream containing 65 mole\% benzene and 35 mole\% styrene enters stage 1 at a rate of \(200 \mathrm{mol} / \mathrm{h}\), and liquid containing 55 mole\% benzene and 45 mole\% styrene leaves this stage at a rate of 150 mol/h. You may assume (1) the stages are ideal, (2) Raoult's law can be used to relate the compositions of the streams leaving each stage, and (3) the total vapor and liquid molar flow rates do not change by a significant amount from one stage to the next.(a) How would you expect the mole fraction of benzene in the liquid to vary from one stage to another, beginning with stage 1 and moving up the column? In light of your answer and considering that the pressure remains essentially constant from one stage to another, how would you then expect the temperature to vary at progressively higher stages? Briefly explain. (b) Estimate the temperature at stage 1 and the compositions of the vapor stream leaving this stage and the liquid stream entering it. Then repeat these calculations for stage 2 . (c) Describe how you would calculate the number of ideal stages required to reduce the styrene content of the vapor to less than 5 mole\%.

Short Answer

Expert verified
The number of stages needed and the exact temperature and composition at each stage will depend on the specific distillation conditions, such as feed rates and compositions.

Step by step solution

01

Predict Mole Fraction Variations in Stages

Since benzene is more volatile than styrene, it would be expected that the mole fraction of benzene in the liquid would tend to decrease from one stage to the next moving up the column as it vaporizes more readily. As such, the liquid stream entering a higher stage would have a lower benzene concentration than the liquid stream exiting the previous lower stage.
02

Temperature Variation

Since the pressure is essentially constant from one stage to another, and due to decrease in the mole fraction of more volatile component, i.e., benzene, from stage to stage, one would expect the temperature to decrease at progressively higher stages as per the Raoult's law.
03

Estimate Temperature and Compositions at Stage 1

To estimate the temperature and compositions at stage 1, apply the vapor-liquid equilibrium and material balances. Using Raoult's law, apply material balances separately for benzene and styrene. Solve these equations simultaneously to estimate the temperature and compositions.
04

Estimate Temperature and Compositions at Stage 2

Repeat the same process as in step 3 for stage 2. Consider the liquid leaving stage 1 as the feed to stage 2. Solve the material balances and Raoult's law equations to estimate the temperature and compositions.
05

Calculate the Number of Stages

The number of stages can be found by iteratively achieving a target vapor composition of 5 mole% styrene. Start from the first stage and repeat the vapor-liquid equilibrium calculations while moving upwards in stages until the styrene content in the vapor is less than 5 mole%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distillation Column
A distillation column is an essential tool used in the chemical separation process for separating mixtures of liquids into individual components based on their boiling points. It consists of a series of equilibrium stages arranged vertically. As you move upwards through the column, vapor moves up and liquid moves down, allowing for the separation of components.

The primary function of the distillation column is to take advantage of differences in volatilities between components. More volatile components like benzene would preferentially move into the vapor phase while less volatile ones such as styrene remain in the liquid phase.
  • Each stage in the column functions as an independent separation process.
  • The goal is to increase the purity of the desired components as the mixture travels upward through the column.
  • The efficiency of this separation is dependent on the design and operation of the column.
Overall, the design of a distillation column balances the energy required and efficiency needed to achieve separation. This balance is critical for an effective separation process.
Equilibrium Stage
An equilibrium stage in a distillation column represents an idealized step where the liquid and vapor phases come into contact and reach thermodynamic equilibrium. At each of these stages, the exiting streams from the stage—the vapor going upward and the liquid going downward—are in equilibrium with each other.

This equilibrium ensures that the concentration of components in the respective phases is maintained as determined by the system's conditions.
  • Ideal equilibrium ensures that at every stage, the vapor phase is richer in more volatile components (like benzene in our example).
  • The liquid phase, in contrast, becomes concentrated with the less volatile components.
  • Real-world conditions often require approximations but aim to get as close to ideal equilibrium as possible for efficiency.
Understanding and achieving equilibrium stages is vital to minimizing energy usage and optimizing component separation.
Vapor-Liquid Equilibrium
Vapor-Liquid Equilibrium (VLE) is a key concept in distillation processes, describing a condition where the phases of liquid and vapor coexist at equilibrium. This balance is crucial for separation processes as it dictates how components distribute between vapor and liquid phases at each stage of the distillation column.

Under equilibrium conditions, each component in the mixture has a specific distribution between the vapor and liquid phase. The VLE relationship is used to determine the composition of each phase:
  • At equilibrium, the chemical potential of each component is the same in both phases.
  • In practice, the VLE is crucial for designing distillation columns by determining component behavior during separation.
  • VLE data helps to establish operating temperatures and pressures for efficient separation.
In our example, VLE plays a significant role in predicting the temperature and composition changes across the stages.
Raoult's Law
Raoult's Law is a principle applied to predict the phase behavior of mixtures, particularly in distillation processes. It provides a method to understand how the partial vapor pressures of components relate to their concentrations in a mixture. According to Raoult's Law, the vapor pressure of a component in a mixture is proportional to the mole fraction of the component in the liquid phase and its pure component vapor pressure.
This relationship helps in understanding the changes in vapor-liquid compositions across the stages in a distillation column:
  • It assumes ideal behavior of mixtures, which means interactions between different molecular types are similar to the interactions between same molecular types.
  • Raoult’s Law forms the basis for creating material balance equations in the distillation design process.
  • In non-ideal conditions, deviations may need addressing through modifications like activity coefficients.
In the provided exercise, Raoult’s Law is used to estimate equilibrium compositions of benzene and styrene as they interact within the distillation column.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An air conditioner is designed to bring \(10,000 \mathrm{ft}^{3} / \mathrm{min}\) of outside air \(\left(90^{\circ} \mathrm{F}, 29.8 \text { inches } \mathrm{Hg} .88 \%\right.\) relative humidity) to \(40^{\circ} \mathrm{F}\), thereby condensing a portion of the water vapor, and then to reheat the air before releasing it into a room at \(65^{\circ} \mathrm{F}\). Calculate the rate of condensation (gallons \(\mathrm{H}_{2} \mathrm{O} / \mathrm{min}\) ) and the volumetric flow rate of the air delivered to the room. (Suggestion: On the flowchart, treat the coolingcondensation and the reheating as separate process steps.)

\- A solution is prepared by dissolving \(0.5150 \mathrm{g}\) of a solute \((\mathrm{MW}=110.1)\) in \(100.0 \mathrm{g}\) of an organic solvent \((\mathrm{M} \mathrm{W}=94.10) .\) The solution is observed to have a freezing point \(0.41^{\circ} \mathrm{C}\) below that of the pure solvent. A second solution is prepared by dissolving \(0.4460 \mathrm{g}\) of a solute having an unknown molecular weight in 95.60 g of the original solvent. A freezing point depression of \(0.49^{\circ} \mathrm{C}\) is observed. Determine the molecular weight of the second solute and the heat of fusion ( \(\mathrm{kJ} / \mathrm{mol}\) ) of the solvent. The melting point of the pure solvent is \(-5.000^{\circ} \mathrm{C}\).

The vapor pressure of an organic solvent is \(50 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\) and \(200 \mathrm{mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). The solvent is the only species in a closed flask at \(35^{\circ} \mathrm{C}\) and is present in both liquid and vapor states. The volume of gas above the liquid is \(150 \mathrm{mL}\). (a) Estimate the amount of the solvent \((\mathrm{mol})\)contained in the gas phase. (b) What assumptions did you make? How would your answer change if the species dimerized (one molecule results from two molecules of the species combining)?

Various amino acids have utility as food additives and in medical applications. They are often synthesized by fermentation using a specific microorganism to convert a substrate (e.g., a sugar) into the desired product. Small quantities of other species also may be formed and must be removed to meet product specifications. For example, isoleucine (Ile), which has a molecular weight of \(131.2,\) is an essential amino acid \(^{16}\) produced by fermentation, and other amino acids such as leucine and valine also are found in the fermentation broth. The broth is subjected to several processing steps to remove these and other impurities, but final processing by crystallization is required to meet stringent specifications on purity. The strategy is to crystallize the hydrated acid form of Ile (Ile. \(\mathrm{HCl} \cdot \mathrm{H}_{2} \mathrm{O}\) ), whose crystals exclude other amino acids, and then to redissolve, neutralize, and crystallize the final Ile product. In a batch process designed to manufacture \(2500 \mathrm{kg}\) of Ile per batch, an aqueous feed solution containing 35 g Ile/dL and much lower concentrations of leucine and valine is fed to the final purification stages. The pH of the solution is 1.1 and its specific gravity is 1.02. The solution is heated to \(60^{\circ} \mathrm{C}\) and 35-wt\% HCl solution is added in a ratio of 0.4 kg per kg of feed. The addition of HCl causes the formation of crystals of Ile\cdotHCl\cdot \(\mathrm{H}_{2} \mathrm{O},\) and the production of these crystals is further increased by slowly lowering the temperature to \(20^{\circ} \mathrm{C}\). At the final crystallizer conditions the Ile solubility is \(5 \mathrm{g}\) Ile/ \(100 \mathrm{g}\) solution. The resulting slurry is sent to a centrifuge where the crystals are separated from the liquid solution and the crystal cake is washed with water. The solids leaving the centrifuge contain \(12 \%\) free water (i.e., not part of the crystal structure) and \(88 \%\) pure crystals of Ile\(\cdot \mathrm{HCl} \cdot \mathrm{H}_{2} \mathrm{O}\). \(\mathrm{H}_{2} \mathrm{O}\).The washed crystals "water to form a solution that is 4.0 g Ile/dL with gravity of 1.1. The solution is sent to an ion exchange unit where HCl is removed. Upon leaving the ion exchange unit the solution has a pH of about \(5.5 .\) It is sent to a second crystallizer where the temperature is gradually reduced to \(10^{\circ} \mathrm{C}\) and the Ile solubility is \(3.4 \mathrm{g} \mathrm{Ile} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The crystals are separated from the slurry by centrifugation, washed with pure water, and sent to a dryer for final processing. (a) Construct a labeled flowchart for the process. (b) Choosing a basis of 1 kg of feed solution, estimate (i) the mass of HCl solution added to the system, (ii) the water added to redissolve the Ile.HCI. \(\mathrm{H}_{2} \mathrm{O}\) crystals, (iii) the mass of \(\mathrm{HCl}\) removed in the ion exchange unit, and (iv) the mass of final Ile product. (c) Scale the quantities calculated in Part (b) to the production rate of 2500 kg Ile/batch. (d) Estimate the active volume (in liters) of each of the crystallizers. (e) Amino acids are amphoteric, which means they can either donate or accept a proton \(\left(\mathrm{H}^{+}\right) .\) At low pH they tend to accept a proton and become acidic while at high pH they tend to donate a proton and become basic. They also are known as zwitterions because their ends are oppositely charged, even though the overall molecule is neutral. Isoleucine is reported to have an isoelectric point (pI) of 6.02 and \(\mathrm{pK}_{\mathrm{a}}\) values of 2.36 and \(9.60 .\) Look up the meaning of these terms and prepare a plot showing how these values are used in plotting the distribution of Ile between acid, zwitterionic (neutral), and basic forms as a function of pH. Explain why such a distribution is important in carrying out the separations described in the process.

An adult inhales approximately 12 times per minute, taking in about 500 mL of air with each inhalation. Oxygen and carbon dioxide are exchanged in the lungs, but there is essentially no exchange of nitrogen. The exhaled air has a mole fraction of nitrogen of 0.75 and is saturated with water vapor at body temperature, \(37^{\circ} \mathrm{C}\). If ambient conditions are \(25^{\circ} \mathrm{C}, 1\) atm, and \(50 \%\) relative humidity, what volume of liquid water (mL) would have to be consumed over a two-hour period to replace the water loss from breathing? How much would have to be consumed if the person is on an airplane where the temperature, pressure, and relative humidity are respectively \(25^{\circ} \mathrm{C}, 1 \mathrm{atm},\) and \(10 \% ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.