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Chapter 4: Problem 48

In the production of a bean oil, beans containing 13.0 wt\% oil and \(87.0 \%\) solids are ground and fed to a stirred tank (the extractor) along with a recycled stream of liquid \(n\) -hexane. The feed ratio is \(3 \mathrm{kg}\) hexane/kg beans. The ground beans are suspended in the liquid, and essentially all of the oil in the beans is extracted into the hexane. The extractor effluent passes to a filter where the solids are collected and form a filter cake. The filter cake contains 75.0 wt\% bean solids and the balance bean oil and hexane, the latter two in the same ratio in which they emerge from the extractor. The filter cake is discarded and the liquid filtrate is fed to a heated evaporator in which the hexane is vaporized and the oil remains as a liquid. The oil is stored in drums and shipped. The hexane vapor is subsequently cooled and condensed, and the liquid hexane condensate is recycled to the extractor. (a) Draw and label a flowchart of the process, do the degree-of-freedom analysis, and write in an efficient order the equations you would solve to determine all unknown stream variables, circling the variables for which you would solve. (b) Calculate the yield of bean oil product (kg oil/kg beans fed), the required fresh hexane feed \(\left(\mathrm{kg} \mathrm{C}_{6} \mathrm{H}_{14} / \mathrm{kg} \text { beans fed }\right),\) and the recycle to fresh feed ratio (kg hexane recycled/kg fresh feed). (c) It has been suggested that a heat exchanger might be added to the process. This process unit would consist of a bundle of parallel metal tubes contained in an outer shell. The liquid filtrate would pass from the filter through the inside of the tubes and then go on to the evaporator. The hot hexane vapor on its way from the evaporator to the extractor would flow through the shell, passing over the outside of the tubes and heating the filtrate. How might the inclusion of this unit lead to a reduction in the operating cost of the process? (d) Suggest additional steps that might improve the process economics.

Short Answer

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Too complex for a short answer due to the multiple calculation steps involved and multiple unknowns which are part of the solution.

Step by step solution

01

Flowchart and Degree-of-Freedom Analysis

Draw a flowchart to represent the oil bean production process, from grinding the beans and extraction to filtration of solids, evaporation of n-hexane, condensation and recycling. Each stream should be labeled according to its state and components. Make note of the extractor effluent that goes to the filter where solids form a filter cake, and the liquid filtrate that goes to an evaporator to separate n-hexane and the oil. For degree-of-freedom analysis and equations, consider the overall balance, balance over the extractor and balance over the filter.
02

Yield of Bean Oil Product

Use the mass balance equations developed in the previous step to calculate the yield of bean oil product (kg oil/kg beans fed). The yield of bean oil would be the mass of oil in the final product per mass of beans fed to the process.
03

Fresh Hexane Feed

Calculate the required fresh hexane feed amount (kg C6H14 / kg beans fed) using the mass balance equations, taking into account that the hexane is also recycled and reused in the extraction process.
04

Recycle to Fresh Feed Ratio

Calculate the recycle to fresh feed ratio (kg hexane recycled/kg fresh feed) using the mass balance equations.
05

Potential Cost Reduction

Evaluate the suggestion to add a heat exchanger to the process. Analyze the potential benefits of this change such as the decrease in energy required to heat the filtrate in the evaporator and cost savings.
06

Process Improvement

Identify other potential process improvement options. Make sure to consider factors such as process flow, energy use, and waste generation.

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Key Concepts

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

Mass Balance Equations
Understanding mass balance equations is crucial for chemical engineering students, especially when dealing with processes like oil extraction. These equations provide a mathematical representation of the material conservation principle; that is, matter cannot be created or destroyed within a closed system.

When you look at a process like extracting oil from beans, mass balance involves accounting for all the mass entering and leaving the system. In a typical extraction process, mass balance equations will be written for the individual components鈥攆or instance, bean solids, bean oil, and hexane in the case of the oil extraction process outlined in the exercise.

For each component, the mass entering with the feed must equal the mass exiting with the product streams and any waste streams, plus any accumulation within the system, which, in a steady-state process, is zero. This can be summarized with the general equation:
\[ \text{Input} - \text{Output} + \text{Generation} - \text{Consumption} = \text{Accumulation} \]
For educational clarity, it's important to emphasize that mass balance equations are applied to each process unit (such as the extractor or filter in the exercise) and to the process as a whole. By solving these equations systematically, students gain valuable insights into the efficiency and effectiveness of the extraction process.
Oil Extraction Process
The oil extraction process, particularly from materials like beans, is a cornerstone application in the chemical engineering field. The process starts by grinding the beans to increase the surface area, which facilitates the separation of oil using a solvent鈥攏-hexane in this case.

Once the ground beans are mixed with hexane, the solution is stirred, allowing the oil to dissolve into the solvent. The mixture then goes through filtration, separating the solids from the liquid. Afterward, the oil-hexane mixture is heated, evaporating the solvent and leaving oil behind, which can be further refined or packaged.

An effective educational approach should highlight the physical and chemical principles that govern each step, such as solubility, phase change (evaporation and condensation), and separation techniques (filtration). Students must understand that the choice of solvent, the proportions used, and the control of temperatures are all critical factors that determine the efficacy and efficiency of the extraction process.
Degree-of-Freedom Analysis
Degree-of-freedom analysis is a systematic way to check if we have enough information (equations and given process variables) to solve for the unknowns in a chemical process system. It's an essential step in process design and troubleshooting.

In the bean oil extraction example, degree-of-freedom analysis helps to determine the number of independent variables that can be changed without affecting the outcome. For each piece of equipment or process unit (e.g., extractor, filter, evaporator), you calculate the degrees of freedom by subtracting the number of independent equations (relating to mass and energy balances, phase equilibria, etc.) from the number of unknown variables.

If the degree of freedom is zero, the system can be solved as-is; if it is positive, there's not enough information, and if it鈥檚 negative, there may be redundant information or the need for more equations. Students should note that achieving a degree-of-freedom of zero for every unit and the system overall is crucial before attempting to solve the mass balance equations, as it indicates all necessary information is available to find a solution.

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Most popular questions from this chapter

Mammalian cells can be cultured for a variety of purposes, including synthesis of vaccines. They must be maintained in growth media containing all of the components required for proper cellular function to ensure their survival and propagation. Traditionally, growth media were prepared by blending a powder, such as Dulbecco's Modified Eagle Medium (DMEM) with sterile deionized water. DMEM contains glucose, buffering agents, proteins, and amino acids. Using a sterile (i.e., bacterial-, fungal-,and yeast-free) growth medium ensures proper cell growth, but sometimes the water (or powder) can become contaminated, requiring the addition of antibiotics to eliminate undesired contaminants. The culture medium is supplemented with fetal bovine serum (FBS) that contains additional growth factors required by the cells. Suppose an aqueous stream (SG = 0.90) contaminated with bacteria is split, with 75\% being fed to a mixing unit to dissolve a powdered mixture of DMEM contaminated with the same bacteria found in the water. The ratio of impure feed water to powder entering the mixer is 4.4:1. The stream leaving the mixer (containing DMEM, water, and bacteria) is combined with the remaining 25\% of the aqueous stream and fed to a filtration unit to remove all of the bacteria that have contaminated the system, a total of \(20.0 \mathrm{kg}\). Once the bacteria have been removed, the sterile medium is combined with FBS and the antibiotic cocktail PSG (Penicillin-Streptomycin-L-Glutamine) in a shaking unit to generate 5000 L of growth medium (SG = 1.2). The final composition of the growth medium is 66.0 wt\% H_O, 11.0\% FBS, 8.0\% PSG, and the balance DMEM. (a) Draw and label the process flowchart. (b) Do a degree-of-freedom analysis around each piece of equipment (mixer, filter, and shaker), the splitter, the mixing point, and the overall system. Based on the analysis, identify which system or piece of equipment should be the starting point for further calculations. (c) Calculate all of the unknown process variables. (d) Determine a value for (i) the mass ratio of sterile growth medium product to feed water and (ii) the mass ratio of bacteria in the water to bacteria in the powder. (e) Suggest two reasons why the bacteria should be removed from the system.

The hormone estrogen is produced in the ovaries of females and elsewhere in the body in men and postmenopausal women, and it is also administered in estrogen replacement therapy, a common treatment for women who have undergone a hysterectomy. Unfortunately, it also binds to estrogen receptors in breast tissue and can activate cells to become cancerous. Tamoxifen is a drug that also binds to estrogen receptors but does not activate cells, in effect blocking the receptors from access to estrogen and inhibiting the growth of breast-cancer cells. Tamoxifen is administered in tablet form. In the manufacturing process, a finely ground powder contains tamoxifen (tam) and two inactive fillers- -lactose monohydrate (lac) and corn starch (cs). The powder is mixed with a second stream containing water and suspended solid particles of polyvinylpymolidone (pvp) binder, which keeps the tablets from easily crumbling. The slurry leaving the mixer goes to a dryer, in which 94.2\% of the water fed to the process is vaporized. The wet powder leaving the dryer contains 8.80 wr\% tam, 66.8\% lac, 21.4\% cs, 2.00\% pvp, and 1.00\% water. After some additional processing, the powder is molded into tablets. To produce a hundred thousand tablets, 17.13 kg of wet powder is required. (a) Taking a basis of 100,000 tablets produced, draw and label a process flowchart, labeling masses of individual components rather than total masses and component mass fractions. It is unnecessary to label the stream between the mixer and the dryer. Carry out a degree-of-freedom analysis of the overall two-unit process. (b) Calculate the masses and compositions of the streams that must enter the mixer to make 100,000 tablets. (c) Why was it unnecessary to label the stream between the mixer and the dryer? Under what circumstances would it have been necessary? (d) Go back to the flowchart of Part (a). Without using the mass of the wet powder (17.13 kg) or any of the results from Part (b) in your calculations, determine the mass fractions of the stream components in the powder fed to the mixer and verify that they match your solution to Part (b). (Hint: Take a basis of \(100 \mathrm{kg}\) of wet powder.) (e) Suppose a student does Part (d) before Part (b), and re-labels the powder feed to the mixer on the flowchart of Part (a) with an unknown total mass ( \(m_{1}\) ) and the three now known mole fractions. (Sketch the resulting flowchart.) The student then does a degree-of-freedom analysis, counts four unknowns (the masses of the powder, pvp, and water fed to the mixer, and the mass of water evaporated in the dryer), and six equations (five material balances for five species and the percentage evaporation), for a net of -2 degrees of freedom. since there are more equations than unknowns, it should not be possible to get a unique solution for the four unknowns. Nevertheless, the student writes four equations, solves for the four unknowns, and verifies that all of the balance equations are satisfied. There must have been a mistake in the degree-of-freedom calculation. What was it?

A catalytic reactor is used to produce formaldehyde from methanol in the reaction $$\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{HCHO}+\mathrm{H}_{2}$$ A single-pass conversion of \(60.0 \%\) is achieved in the reactor. The methanol in the reactor product is separated from the formaldehyde and hydrogen in a multiple-unit process. The production rate of formaldehyde is 900.0 kg/h. (a) Calculate the required feed rate of methanol to the process ( \(\mathrm{kmol} / \mathrm{h}\) ) if there is no recycle. (b) Suppose the unreacted methanol is recovered and recycled to the reactor and the single-pass conversion remains 60\%. Without doing any calculations, prove that you have enough information to determine the required fresh feed rate of methanol (kmol/h) and the rates (kmol/h) at which methanol enters and leaves the reactor. Then perform the calculations. (c) The single-pass conversion in the reactor, \(X_{\mathrm{sp}},\) affects the costs of the reactor \(\left(C_{\mathrm{r}}\right)\) and the separation process and recycle line \(\left(C_{\mathrm{s}}\right) .\) What effect would you expect an increased \(X_{\mathrm{sp}}\) would have on each of these costs for a fixed formaldehyde production rate? (Hint: To get a \(100 \%\) singlepass conversion you would need an infinitely large reactor, and lowering the single-pass conversion leads to a need to process greater amounts of fluid through both process units and the recycle line.) What would you expect a plot of \(\left(C_{\mathrm{r}}+C_{\mathrm{s}}\right)\) versus \(X_{\mathrm{sp}}\) to look like? What does the design specification \(X_{\mathrm{sp}}=60 \%\) probably represent?

A fuel oil is fed to a furnace and burned with \(25 \%\) excess air. The oil contains \(87.0 \mathrm{wt} \% \mathrm{C}, 10.0 \% \mathrm{H},\) and 3.0\% S. Analysis of the furnace exhaust gas shows only \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2}, \mathrm{SO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\). The sulfur dioxide emission rate is to be controlled by passing the exhaust gas through a scrubber, in which most of the \(\mathrm{SO}_{2}\) is absorbed in an alkaline solution. The gases leaving the scrubber (all of the \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{CO}_{2}\), and some of the \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{SO}_{2}\) entering the unit) pass out to a stack. The scrubber has a limited capacity, however, so that a fraction of the furnace exhaust gas must be bypassed directly to the stack. At one point during the operation of the process, the scrubber removes \(90 \%\) of the \(\mathrm{SO}_{2}\) in the gas fed to it, and the combined stack gas contains 612.5 ppm (parts per million) \(\mathrm{SO}_{2}\) on a dry basis; that is, every million moles of dry stack gas contains 612.5 moles of \(\mathrm{SO}_{2}\). Calculate the fraction of the exhaust bypassing the scrubber at this moment.

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.
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