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Chapter 6: Problem 97

A stream of 5.00 wt\% oleic acid in cottonseed oil enters an extraction unit at a rate of \(100.0 \mathrm{kg} / \mathrm{h}\). The unit operates as a single equilibrium stage (the streams leaving the unit are in equilibrium) at \(85^{\circ} \mathrm{C} . \mathrm{At}\) this temperature, propane and cottonseed oil are essentially immiscible, the vapor pressure of propane is 34 atm, and the distribution coefficient (oleic acid mass fraction in propane/oleic acid mass fraction in cottonseed oil) is \(0.15 .\)(a) Calculate the rate at which liquid propane must be fed to the unit to extract \(90 \%\) of the oleic acid. (b) Estimate the minimum operating pressure of the extraction unit. Explain your answer.(c) High-pressure operation is costly and introduces potential safety hazards. Suggest two possible reasons for using propane as the solvent when other less volatile hydrocarbons are equally good solvents for oleic acid.

Short Answer

Expert verified
The extraction unit must be fed liquid propane at a rate of 30 kg/hr to extract 90% of the oleic acid. The minimum operating pressure of the unit would be 34 atm. Propane could be used for its high extraction efficiency and its cost-effectiveness.

Step by step solution

01

Calculating Required Propane Feed Rate

First, calculate the amount of oleic acid coming into the extraction unit. This is done by multiplying the rate the mixture is fed into the unit (100.0 kg/h) by the weight percent of oleic acid in the feed (0.05). This gives 5 kg/hr of oleic acid in the feed. Now, it is stated that we want to extract 90% of that, so the amount of oleic acid to extract is 0.9 * 5 kg/hr = 4.5 kg/hr. Knowing that the oleic acid is distributed between the cottonseed oil and propane according to the distribution coefficient (0.15), we can determine the amount of propane needed to achieve this extraction using the mass balance equation over the extraction unit. This yields a required propane feed rate of 4.5 kg/hr / 0.15 = 30 kg/hr.
02

Estimating Minimum Operating Pressure

The minimum operating pressure of the extraction unit would be the vapor pressure of propane at the given operating temperature, 85C. This is because at pressures below this value, propane would start to vaporize and would no longer be available for the extraction process. So, the minimum operating pressure would be 34 atm.
03

Discussing Reasons for Using Propane

Despite the high pressure required for operation, propane could be preferred as the solvent for a couple of reasons. First, it could provide a higher extraction efficiency due to its distribution coefficient with respect to oleic acid. Second, propane is relatively cheap and widely available, which can lower the costs of the extraction process substantially.

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Key Concepts

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

Equilibrium Stage Extraction
Understanding the equilibrium stage extraction process provides insight into the methodology behind separating components based on their differing solubilities in two immiscible liquids. In the context of the given problem, a mixture of oleic acid in cottonseed oil is being treated with propane in a single equilibrium stage. At this stage, two phases are formed, with oleic acid preferentially distributing itself between the cottonseed oil phase and the propane phase.

At equilibrium, the ratio of the concentrations of oleic acid in these phases is described by the distribution coefficient. To maximize the separation effectiveness, the process aims to transfer a significant portion of oleic acid into the propane, and this is achieved by leveraging the equilibrium condition at the given temperature of 85掳C. The calculation of the propane feed rate required to extract 90% of the oleic acid is a quintessential representation of applying mass balance tied to the equilibrium concept.

This exemplifies how equilibrium stage theory is imperative in optimizing the extraction processes, ensuring that the separation objectives are met efficiently.
Vapor Pressure in Extraction Processes
The role of vapor pressure in extraction processes is paramount in understanding the behaviour of solvents under different conditions of temperature and pressure. Vapor pressure, the pressure exerted by the vapor in equilibrium with its liquid at a given temperature, dictates at what point a solvent will remain in the liquid phase or start to vaporize.

In the exercise, propane's vapor pressure at 85掳C is 34 atm, which informs us of the minimum operating pressure for the extraction unit. If the operating pressure falls below the propane's vapor pressure, it would begin to evaporate, making it unsuitable for the extraction of oleic acid. Thus, maintaining an operating pressure equal to or higher than the vapor pressure of propane ensures that the solvent remains liquid and effective in the extraction process.

Furthermore, understanding vapor pressure is essential when considering safety and efficiency, as solvents with high vapor pressures may pose greater risks and require more robust equipment to contain and manage the process safely.
Solvent Selection Criteria
Choosing the appropriate solvent for an extraction process is critical for its efficiency, safety, and cost-effectiveness. The solvent selection criteria take into account several factors, such as solubility, distribution coefficient, volatility, availability, and safety. In the scenario at hand, despite propane's high vapor pressure requiring operation under high pressure, its selection as a solvent is justified by several points.

  • Propane has a favourable distribution coefficient with oleic acid, meaning that it can efficiently extract the acid from cottonseed oil.
  • It is a commonly available and relatively inexpensive solvent, which could reduce the operational costs.
  • Propane may also be chosen for its unique properties that could provide better separation performance compared to other solvents.
Considering these reasons, propane may outweigh other hydrocarbon solvents despite the need for high-pressure systems and the associated costs and potential safety hazards. It illustrates the complexity of solvent selection and the trade-offs that must be evaluated to optimize the overall extraction process.

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Most popular questions from this chapter

A storage tank for liquid \(n\) -octane has a diameter of \(30 \mathrm{ft}\) and a height of \(20 \mathrm{ft}\). During a typical \(24-\mathrm{h}\) period the level of liquid octane falls from 18 ft to 8 ft, after which fresh octane is pumped into the tank to return the level to \(18 \mathrm{ft}\). As the level in the tank falls, nitrogen is fed into the free space to maintain the pressure at 16 psia; when the tank is being refilled, the pressure is maintained at 16 psia by discharging gas from the vapor space to the environment. The nitrogen in the tank may be considered saturated with octane vapor at all times. The average tank temperature is \(90^{\circ} \mathrm{F}\). (a) What is the daily rate, in gallons and \(1 \mathrm{b}_{\mathrm{m}}\), at which octane is used? (b) What is the variation in absolute pressure at the bottom of the tank in inches of mercury? (c) How much octane is lost to the environment during a 24 -h period? (d) Why is nitrogen used in the vapor space of the tank when air would be cheaper? (e) Suggest a means by which the octane can be recovered from the gas stream discharged to the atmosphere.

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)?

Sulfur trioxide (SO \(_{3}\) ) dissolves in and reacts with water to form an aqueous solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) .\) The vapor in equilibrium with the solution contains both \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). If enough \(\mathrm{SO}_{3}\) is added, all of the water reacts and the solution becomes pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If still more \(\mathrm{SO}_{3}\) is added, it dissolves to form a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure \(\mathrm{SO}_{3}\). Twenty percent oleum by definition contains \(20 \mathrm{kg}\) of dissolved \(\mathrm{SO}_{3}\) and \(80 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as \(\% \mathrm{SO}_{3}\) by mass, with the constituents of the oleum considered to be \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Prove that a \(15.0 \%\) oleum contains \(84.4 \% \mathrm{SO}_{3}\) (b) Suppose a gas stream at \(40^{\circ} \mathrm{C}\) and 1.2 atm containing 90 mole \(\% \mathrm{SO}_{3}\) and \(10 \% \mathrm{N}_{2}\) contacts a liquid stream of 98 wt\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (aq), producing \(15 \%\) oleum. Tabulated equilibrium data indicate that the partial pressure of \(S O_{3}\) in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of \(S O_{3}\) in the outlet gas if this gas is in equilibrium with the liquid product at \(40^{\circ} \mathrm{C}\) and 1 atm, and (ii) the ratio ( \(\mathrm{m}^{3}\) gas feed) \(/\) (kg liquid feed).

The constituent partial pressures of a gas in equilibrium with a liquid solution at \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{atm}\) containing \(2 \mathrm{Ib}_{\mathrm{m}} \mathrm{SO}_{2} / 100 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O}\) are \(p_{\mathrm{H}_{2} \mathrm{O}}=31.6 \mathrm{mm} \mathrm{Hg}\) and \(p_{\mathrm{SO}_{2}}=176 \mathrm{mm} \mathrm{Hg} .\) The balance of the gas is air.(a) Calculate the partial pressure of air. If you make any assumptions, state what they are. (b) Suppose the only data available on this system gave \(p_{\mathrm{SO}_{2}}=176 \mathrm{mm}\) Hg, but there was no information given on the equilibrium partial pressure of water. Use Raoult's law to estimate a value for this quantity.Assuming that the value given in the problem statement is correct, what percentage error results from using Raoult's law? (c) The same system was examined in Example \(6.4-1 .\) What percentage errors in the two calculated quantities would result from using Raoult's law for the partial pressure of water?

In this problem you will use a spreadsheet to create a \(T x y\) diagram for the benzene-chloroform system at 1 atm. Once the spreadsheet has been created, it can be used as a template for vapor-liquid equilibrium calculations for other species. The calculations will be based on Raoult's law (i.e., \(y_{i} P=x_{i} p_{i}^{*}\) ), although we recognize that this relationship may not produce accurate results for benzene-chloroform mixtures.(a) Begin by establishing bounds on the system behavior. Look up the normal boiling points of chloroform and benzene and, without performing any calculations, sketch the expected shape of a Txy diagram for these two species at 1 atm. (b) Using APEx or Table B.4, estimate the normal boiling points of the two species and compare them to the results in Part (a).(c) Prepare a spreadsheet that has a title row "Txy Diagram for Ideal Binary Solution of Chloroform and Benzene." In the first cell of Row 2, place the label " \(P(\mathrm{mm} \mathrm{Hg})="\) and in the adjacent cell enter the system pressure, which for this case is 760. In Row 3 place headings for columns: xC, xB, T, p^*C, p^* B, P, yC, yB, and yC + yB. Not all of these columns are essential, but when filled they will give a complete picture of the system and a final check of the calculations. Carry out the following procedures in each subsequent row: \(\bullet\) Enter values for the mole fraction of chloroform (the first entry should be 1.000 and the last should be 0.000).\(\bullet\) Calculate the mole fraction of benzene by subtracting the value in the previous cell from 1.000 .\(\bullet\)Enter an estimate of the equilibrium temperature that is between the two pure-component boiling points.\(\bullet\) Use APEx or Table B.4 to estimate \(p^{*} \mathrm{C}\) and \(p^{*} \mathrm{B}\) from the estimated temperature. \(\bullet\) Calculate \(p \mathbf{C}\) and \(p \mathbf{B}\) from Raoult's law.\(\bullet\) Calculate \(P=p_{\mathrm{C}}+p_{\mathrm{B}}\) and apply the Goal Seek tool to adjust the value of \(T\) until \(P=760 \mathrm{mm} \mathrm{Hg}\) \(\bullet\) Calculate \(y \mathbf{C}\) and \(y \mathbf{B}\) from the partial pressures and \(P\). \(\bullet\) Sum \(y \mathbf{C}\) and \(y \mathbf{B}\) to be sure they equal 1.000.Once you have completed a row for the first value of \(x \mathrm{C},\) you should be able to copy formulas into subsequent rows. When the calculation has been completed for all rows (i.e., \(x \mathrm{C}=0.0,0.2,0.4\) \(0.5, 0.6, 0.8, 1.0)\), draw the Txy diagram.(d) Explain what you did in the bulleted sequence of steps in Part (c) giving relevant relationships among system variables. The phrase "bubble point" should appear in your explanation. (e) The following vapor-liquid equilibrium data have been obtained for mixtures of chloroform (C) and benzene (B) at 1 atm.Plot these data on the graph generated in Part (c). Estimate the percentage errors in the Raoult's law values of the bubble-point temperature and vapor mole fraction for \(x_{\mathrm{C}}=0.44,\) taking the tabulated values to be correct. Why does Raoult's law give poor estimates for this system?
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