91Ó°ÊÓ

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.

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—for 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—n-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’s 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.