/*! 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

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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%.

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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.

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Most popular questions from this chapter

An ore containing \(90 \mathrm{wt} \% \mathrm{MgSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) and the balance insoluble minerals is fed to a dissolution tank at a rate of \(60,000 \mathrm{lb}_{\mathrm{m}} / \mathrm{h}\) along with fresh water and a recycle stream. The tank contents are heated to \(120^{\circ} \mathrm{F}\), causing all of the magnesium sulfate monohydrate in the ore to dissolve, forming a solution 10^0 F above saturation. The resulting slurry of the insoluble minerals in MgSO_solution is pumped to a heated filter, where a wet filter cake is separated from a solids-free filtrate. The filter cake retains \(5 \mathrm{lb}_{\mathrm{m}}\) of solution per \(100 \mathrm{lb}_{\mathrm{m}}\) of solids. The filtrate is sent to a crystallizer in which the temperature is reduced to \(50^{\circ} \mathrm{F},\) producing a slurry of \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) crystals in a saturated solution that is sent to another filler. The product filter cake contains all of the prea entrained solution in a ratio of \(5 \mathrm{Ib}_{\mathrm{m}}\) solution per \(100 \mathrm{lb}_{\mathrm{m}}\) crystals. The filtrate from this filter is returned to the dissolution tank as the recycle stream.Solubility data: Saturated magnesium sulfate solutions at \(110^{\circ} \mathrm{F}\) and \(50^{\circ} \mathrm{F}\) contain \(32 \mathrm{wt} \%\) \(\mathrm{MgSO}_{4}\) and \(23 \mathrm{wt} \% \mathrm{MgSO}_{4},\) respectively.(a) Explain why the solution is first heated (in the dissolution tank) and filtered and then cooled (in the crystallizer) and filtered. (b) Calculate the production rate of crystals and the required feed rate of fresh water to the dissolution tank. (Note: Don't forget to include water of hydration when you write a mass balance on water.)(c) Calculate the ratio \(\mathrm{lb}_{\mathrm{m}}\) recycle/lb \(_{\mathrm{m}}\) makeup water.

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at \(58.3^{\circ} \mathrm{C}\), the height of the mercury in the arm of the manometer connected to the flask is \(747 \mathrm{mm}\) and that in the arm open to the atmosphere is \(52 \mathrm{mm} . \mathrm{At} 110^{\circ} \mathrm{C},\) the mercury level is \(577 \mathrm{mm}\) in the arm connected to the flask and \(222 \mathrm{mm}\) in the other arm. Atmospheric pressure is \(755 \mathrm{mm} \mathrm{Hg}\). (a) Extrapolate the data using the Clausius-Clapeyron equation to estimate the vapor pressure of chlorobenzene at \(130^{\circ} \mathrm{C}\). (b) Air saturated with chlorobenzene at \(130^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) is cooled to \(58.3^{\circ} \mathrm{C}\) at constant pressure. Estimate the percentage of the chlorobenzene originally in the vapor that condenses. (See Example 6.3-2.)(c) Summarize the assumptions you made in doing the calculation of Part (b).

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Recovery and processing of various oils are important elements of the agricultural and food industries. For example, soybean hulls are removed from the beans, which are then flaked and contacted with hexane. The hexane extracts soybean oil and leaves very little oil in the residual solids. The solids are dried at an elevated temperature, and the dried solids are used to feed livestock or further processed to extract soy protein. The gas stream leaving the dryer is at \(80^{\circ} \mathrm{C}\) 1 atm absolute, and 50\% relative saturation.(a) To recover hexane, the gas leaving the dryer is fed to a condenser, which operates at 1 atm absolute. The gas leaving the condenser contains 5.00 mole \(\%\) hexane, and the hexane condensate is recovered at a rate of \(1.50 \mathrm{kmol} / \mathrm{min}\). (b) In an altemative arrangement, the gas leaving the dryer is compressed to 10.0 atm and the temperature simultancously is increased so that the relative saturation remains at \(50 \% .\) The gas then is cooled at constant pressure to produce a stream containing 5.00 mole \(\%\) hexane. Calculate the final gas temperature and the ratio of volumetric flow rates of the gas streams leaving and entering the condenser. State any assumptions you make.(c) What would you need to know to determine which of processes (a) and (b) is more cost- effective?

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