/*! 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 11 In the production of soybean oil... [FREE SOLUTION] | 91Ó°ÊÓ

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In the production of soybean oil, dried and flaked soybeans are brought into contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil. (a) Draw a flowchart of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract). (b) The soybeans contain 18.5 wt\% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to \(2.0 \mathrm{kg}\) hexane per \(\mathrm{kg}\) beans. The residual solids leaving the extraction unit contain 35.0 wt\% hexane, all of the non-oil solids that entered with the beans, and \(1.0 \%\) of the oil that entered with the beans. For a feed rate of \(1000 \mathrm{kg} / \mathrm{h}\) of dried flaked soybeans, calculate the mass flow rates of the extract and residual solids, and the composition of the extract. (c) The product soybean oil must now be separated from the extract. Sketch a flowchart with two units, the extraction unit from Parts (a) and (b) and the unit separating soybean oil from hexane. Propose a use for the recovered hexane.

Short Answer

Expert verified
The flowchart for this exercise represents the process of extracting oil from soybeans using hexane as a solvent. The mass flow rates of the extract and residual solids are calculated using mass balance equations. The recovered hexane can be used again in the extraction process or in other applications where a solvent is needed.

Step by step solution

01

Flowchart Creation (a)

Producing a flowchart for the extraction process includes indicating the soybeans and solvent as entering streams, and solids and extract as exiting streams. This diagram serves as a visual aid for understanding the process and simplifying the mass balance equations used in the calculations.
02

Mass Balances and Calculations (b)

Assuming steady-state operation, mass balances around the extraction unit can be written. First, the total material balance is given by \( \text{{input}} = \text{{output}} \), which translates into: F = E + R, where F represents the total input mass flowrate, E is the total extract flowrate, and R is the total residuals flowrate. The mass balance equation for each component (soybean, oil and hexane) are constructed using the provided composition information. Solving the set of equations yields the flowrate for extract and residuals as well as the composition of the extract.
03

Flowchart Completion and Proposal (c)

The flowchart from part (a) is enhanced in part (c) to add the separation unit, which separates hexane (to be reused in the process or in other industrial applications) from soybean oil. The recovered hexane can be proposed to be used in the same process again or in other industrial applications where a solvent is needed.

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Key Concepts

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

Flowchart Creation
Creating a flowchart is a vital step in understanding a chemical process. This visual tool helps us to delineate the sequence of operations, inputs, and outputs of a system more clearly. In our soybean oil extraction case, the flowchart should clearly display:
  • Two feed streams, which are the soybeans and the hexane solvent.
  • Two leaving streams, which are the residual solids (a mixture of residual oil, solids, and hexane) and the desired oil-rich extract.
Visualizing these streams and their interactions allows us to better structure the subsequent calculations. By labeling each stream, you can simplify complex mass balance equations. Imagine the flowchart as a map that guides you through the chemical process, ensuring that every step is covered quantitatively and qualitatively.
Chemical Process Calculations
Chemical process calculations are the backbone of process engineering. They help determine the flow rates and compositions of streams in chemical processes. In this scenario, we use mass balances to calculate these parameters. The key equation is the total material balance: \(\text{input} = \text{output}\). This principle applies to the entire process and to individual components, like soybeans, oil, and hexane.
To perform these calculations, we need to:
  • Define variables for each stream's flow rate and composition.
  • Set up equations based on the mass balances for these streams and components.
  • Solve the equations to find unknown values, like the extract and residuals flow rate, or the composition of the extract.
This step essentially provides a mathematical confirmation of what the flowchart visually represents, ensuring that all assumptions are validated, and the process is feasible.
Extraction and Separation Techniques
Extraction and separation techniques play a crucial role in isolating desired components in chemical processes. In this exercise, the primary goal is to separate the soybean oil from the other constituents, primarily using hexane as the extraction solvent. This technique relies on the principle that different substances have different solubilities in a solvent.
Once the oil is extracted, separating the hexane and oil becomes the focus. This can be accomplished using units like distillation columns or evaporators, which take advantage of differing boiling points or volatilities. The hexane, being volatile, can be recovered and reused. This not only makes the process more cost-effective but also environmentally friendlier by reducing waste.
Proposing a use for recovered hexane also demonstrates an understanding of the importance of recycling and sustainability in industry. The hexane might be reintroduced into the extraction cycle or employed in other solvent-involved processes, showcasing the versatility and resource efficiency.

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Most popular questions from this chapter

A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.4 mole \(\% \mathrm{H}_{2} \mathrm{O}\). After all the water is removed from the products, the residual gas contains 69.4 mole \(\% \mathrm{CO}_{2}\) and the balance \(\mathrm{O}_{2}\) (a) What is the mole percent of propane in the fuel? (b) It now turns out that the fuel mixture may contain not only propane and butane but also other hydrocarbons. All that is certain is that there is no oxygen in the fuel. Use atomic balances to calculate the elemental molar composition of the fuel from the given combustion product analysis (i.e., what mole percent is \(C\) and what percent is \(\mathrm{H}\) ). Prove that your solution is consistent with the result of Part (a).

\- An equimolar liquid mixture of benzene and toluene is separated into two product streams by distillation. A process flowchart and a somewhat oversimplified description of what happens in the process follow: Inside the column a liquid stream flows downward and a vapor stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving the top of the column, which contains 97 mole\% benzene, is completely condensed and split into two equal fractions: one is taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains \(89.2 \%\) of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which \(45 \%\) of it is vaporized. The vapor generated in the reboiler (the boilup) is recycled to become the rising vapor stream in the column, and the residual reboiler liquid is taken off as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation $$\frac{y_{\mathrm{B}} /\left(1-y_{\mathrm{B}}\right)}{x_{\mathrm{B}} /\left(1-x_{\mathrm{B}}\right)}=2.25$$ where \(y_{\mathrm{B}}\) and \(x_{\mathrm{B}}\) are the mole fractions of benzene in the vapor and liquid streams, respectively. (a) Take a basis of 100 mol fed to the column. Draw and completely label a flowchart, and for each of four systems (overall process, column, condenser, and reboiler), do the degree-of-freedom analysis and identify a system with which the process analysis might appropriately begin (one with zero degrees of freedom). (b) Write in order the equations you would solve to determine all unknown variables on the flowchart, circling the variable for which you would solve in each equation. Do not do the calculations in this part. (c) Calculate the molar amounts of the overhead and bottoms products, the mole fraction of benzene in the bottoms product, and the percentage recovery of toluene in the bottoms product \((100 \times\) moles toluene in bottoms/mole toluene in feed).

A drug (D) is produced in a three-stage extraction from the leaves of a tropical plant. About 1000 kg of leaf is required to produce 1 kg of the drug. The extraction solvent (S) is a mixture containing 16.5 wt\% ethanol (E) and the balance water (W). The following process is carried out to extract the drug and recover the solvent. 1\. A mixing tank is charged with \(3300 \mathrm{kg}\) of \(\mathrm{S}\) and \(620 \mathrm{kg}\) of leaf. The mixer contents are stirred for several hours, during which a portion of the drug contained in the leaf goes into solution. The contents of the mixer are then discharged through a filter. The liquid filtrate, which carries over roughly \(1 \%\) of the leaf fed to the mixer, is pumped to a holding tank, and the solid cake (spent leaf and entrained liquid) is sent to a second mixer. The entrained liquid has the same composition as the filtrate and a mass equal to \(15 \%\) of the mass of liquid charged to the mixer. The extracted drug has a negligible effect on the total mass and volume of the spent leaf and the filtrate. 2\. The second mixer is charged with the spent leaf from the first mixer and with the filtrate from the previous batch in the third mixer. The leaf is extracted for several more hours, and the contents of the mixer are then discharged to a second filter. The filtrate, which contains \(1 \%\) of the leaf fed to the second mixer, is pumped to the same holding tank that received the filtrate from the first mixer, and the solid cake- -spent leaf and entrained liquid - is sent to the third mixer. The entrained liquid mass is \(15 \%\) of the mass of liquid charged to the second mixer. 3\. The third mixer is charged with the spent leaf from the second mixer and with \(2720 \mathrm{kg}\) of solvent \(\mathrm{S}\). The mixer contents are filtered; the filtrate, which contains \(1 \%\) of the leaf fed to the third mixer, is recycled to the second mixer; and the solid cake is discarded. As before, the mass of the entrained liquid in the solid cake is \(15 \%\) of the mass of liquid charged to the mixer. 4\. The contents of the filtrate holding tank are filtered to remove the carried-over spent leaf, and the wet cake is pressed to recover entrained liquid, which is combined with the filtrate. A negligible amount of liquid remains in the wet cake. The filtrate, which contains \(\mathrm{D}, \mathrm{E},\) and \(\mathrm{W},\) is pumped to an extraction unit (another mixer). 5\. In the extraction unit, the alcohol-water-drug solution is contacted with another solvent (F), which is almost but not completely immiscible with ethanol and water. Essentially all of the drug (D) is extracted into the second solvent, from which it is eventually separated by a process of no concern in this problem. Some ethanol but no water is also contained in the extract. The solution from which the drug has been extracted (the raffinate) contains \(13.0 \mathrm{wt} \% \mathrm{E}, 1.5 \% \mathrm{F},\) and \(85.5 \%\) W. It is fed to a stripping column for recovery of the ethanol. 6\. The feeds to the stripping column are the solution just described and steam. The two streams are fed in a ratio such that the overhead product stream from the column contains \(20.0 \mathrm{wt} \% \mathrm{E}\) and \(2.6 \% \mathrm{F},\) and the bottom product stream contains \(1.3 \mathrm{wt} \% \mathrm{E}\) and the balance \(\mathrm{W}\). Draw and label a flowchart of the process, taking as a basis one batch of leaf processed. Then calculate (a) the masses of the components of the filtrate holding tank. (b) the masses of the components \(D\) and \(E\) in the extract stream leaving the extraction unit. (c) the mass of steam fed to the stripping column, and the masses of the column overhead and bottoms products.

A stream consisting of 44.6 mole \(\%\) benzene and \(55.4 \%\) toluene is fed at a constant rate to a process unit that produces two product streams, one a vapor and the other a liquid. The vapor flow rate is initially zero and asymptotically approaches half of the molar flow rate of the feed stream. Throughout this entire period, no material accumulates in the unit. When the vapor flow rate has become constant, the liquid is analyzed and found to be 28.0 mole\% benzene. (a) Sketch a plot of liquid and vapor flow rates versus time from startup to when the flow rates become constant. (b) Is this process batch or continuous? Is it transient or steady-state before the vapor flow rate reaches its asymptotic limit? What about after it becomes constant? (c) For a feed rate of 100 mol/min, draw and fully label a flowchart for the process after the vapor flow rate has reached its limiting value, and then use balances to calculate the molar flow rate of the liquid and the composition of the vapor in mole fractions.

Certain vegetables and fruits contain plant pigments called carotenoids that are metabolized in the body to produce Vitamin A. Lack of Vitamin A causes an estimated 250,000 to 500,000 children worldwide to become blind every year. An approach to reducing blindness and other childhood health problems resulting from this deficiency is to use genetic engineering of rice- -a food staple in developing countries and economically disadvantaged regions of the world \(-\) so that rice becomes a dietary source of Vitamin A. For example, a strain known as Golden Rice has been genetically engineered so that it can produce and store carotenoids such as \(\beta\) -carotene (which helps give carrots and squash their yellow-orange color). One type of Golden Rice contains approximately 30 micrograms of carotenoids (81\% \beta-carotene, 16\% \alpha- carotene, and 3\% \beta-cryptoxanthin) per gram of uncooked rice. A study has reported that when a person eats Golden Rice, their body metabolizes 1 microgram of Vitamin A for every 3.8 micrograms of \beta-carotene they consume. (a) It is recommended that children between 1 and 3 years of age should get 300 micrograms of Vitamin A per day. Considering only the metabolism of \(\beta\) -carotene given above, how many grams of Golden Rice would a child have to eat in order to obtain this much Vitamin A? Does this seem like a reasonable amount of rice to eat in one day, if one cup of cooked rice is approximately 175 g? (b) \(\alpha\) -carotene and \(\beta\) -cryptoxanthin can also be converted into Vitamin \(A\), but when compared to \beta-carotene, it takes twice as much of each of these compounds to produce one unit of Vitamin A. Considering all of the carotenoids in Golden Rice as potential sources of Vitamin A, how many grams of Golden Rice would a three-year-old child have to eat in order to obtain the recommended daily amount of Vitamin A? (c) Some individuals are not convinced that genetically modified foods are safe to grow or to eat. What kinds of risks or uncertainties are cited by these individuals? What kinds of measures are taken by farmers and suppliers of genetically modified seeds to minimize these risks? (d) Some people do not believe that Golden Rice is a practical, viable solution to Vitamin A deficiency around the world. Summarize the major arguments for and against production and distribution of Golden Rice.

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