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In-Hexane is used to extract oil from soybeans. (See Problem 6.24 .) The solid residue from the extraction unit, which contains 0.78 kg liquid hexane/kg dry solids, is contacted in a dryer with nitrogen that enters at \(85^{\circ} \mathrm{C}\). The solids leave the dryer containing \(0.05 \mathrm{kg}\) liquid hexane/kg dry solids, and the gas leaves the dryer at \(80^{\circ} \mathrm{C}\) and 1.0 atm with a relative saturation of \(70 \% .\) The gas is then fed to a condenser in which it is compressed to 5.0 atm and cooled to \(28^{\circ} \mathrm{C}\), enabling some of the hexane to be recovered as condensate.(a) Calculate the fractional recovery of hexane (kg condensed/kg fed in wet solids). (b) A proposal has been made to split the gas stream leaving the condenser, combining 90\% of it with fresh makeup nitrogen, heating the combined stream to \(85^{\circ} \mathrm{C},\) and recycling the heated stream to the dryer inlet. What fraction of the fresh nitrogen required in the process of Part (a) would be saved by introducing the recycle? What costs would be incurred by introducing the recycle?

Short Answer

Expert verified
The fractional recovery of hexane is \(0.936\) or \(93.6%.\) By introducing the recycle, 90% of fresh nitrogen would be saved. However, costs would be incurred due to gas stream splitting and heating.

Step by step solution

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Part A: Calculation of Fractional Recovery of Hexane

In the original process, the hexane fraction of the input was 0.78 kg/kg dry solids, and of the output, the hexane fraction was 0.05 kg/kg dry solids. The fractional recovery, therefore, would be the difference between the hexane content in the input and the output, divided by the hexane content in the input. Mathematically, this can be represented as: \[(0.78 - 0.05) / 0.78\] This calculation gives the fractional recovery of hexane.
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Part B: Evaluation of Recycling Proposal

The proposal involves taking 90% of the gas stream leaving the condenser, combining it with fresh nitrogen, heating it, and recycling it to the dryer. If 90% of the nitrogen is being recycled, then only 10% of fresh nitrogen would be needed compared to the amount used in part A. So, if x kg was used in part A, now only 0.1x kg would be used. This means the ratio of nitrogen saved to the original amount is 0.9x/1.0x, which equals 0.9, or, 90%. The costs that would be incurred in introducing this recycling scheme would include the costs of splitting and recombining the gas streams, heating the recycled stream, and any costs associated with potential changes in the quality of the dried solids or the efficiency of the dryer.
03

Summary

To summarize, the fractional recovery of hexane can be calculated by comparing the amount of hexane in the input to the amount left in the output after drying, while introducing a gas stream split and recycle could save 90% of the fresh nitrogen used. The costs of the modifications, however, should be considered carefully in the overall calculation of the process efficiency and economy.

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Key Concepts

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

Fractional Recovery Calculation
Understanding how to calculate fractional recovery is pivotal in evaluating the performance of a separation process, such as the extraction of oil from soybeans using hexane. Fractional recovery measures the proportion of a substance that is successfully recovered from a process relative to what was initially supplied.

For instance, if we consider the exercise above, the fractional recovery of hexane is determined by the amount of hexane that is removed from the solid residue in the dryer. The calculation takes into account the initial hexane content in the wet solids and the remaining hexane content after the drying process. Using the formula \[ \text{Fractional Recovery} = \frac{ \text{Initial Hexane Content} - \text{Final Hexane Content} }{ \text{Initial Hexane Content} } \], we find the fraction of hexane that has been effectively recovered.

In the given problem, the initial hexane content is 0.78 kg/kg dry solids, and the final content after drying is 0.05 kg/kg dry solids, which implies a simple subtraction and division to find the recovery rate. By performing these calculations, engineers can assess the efficiency of the extraction process, making fractional recovery a key indicator in the optimization of chemical processes.
Recycling in Chemical Processes
Recycling within chemical processes is a technique aimed to minimize waste and maximize the use of inputs. This concept not only contributes to sustainability but also increases the efficiency of the process by potentially reducing the costs associated with fresh raw materials.

In the context of our textbook exercise, the recycling proposal involves reusing a portion of the gas stream, which could have considerable implications on the overall system. By taking 90% of the nitrogen gas exiting the condenser, reheating it, and reintroducing it back into the dryer, the amount of fresh nitrogen needed significantly drops. This illustrates a circular approach to resource utilization, a fundamental principle in process design.

However, one must also consider the additional costs that may arise from implementing such a recycle loop. These could include investments in equipment for the separation and reheating of the gas stream, and potential impacts on process dynamics. Nevertheless, when performed correctly, recycling is a powerful tool in a chemical engineer's arsenal to improve the cost-effectiveness and environmental footprint of industrial systems.
Chemical Process Efficiency
Efficiency in a chemical process is a measure of how well the process converts raw materials and energy into desired products. In our problem from the textbook, the efficiency can be interpreted through the lens of both the fractional recovery of hexane and the effective use of nitrogen gas after the implementation of recycling.

Increasing the process efficiency is often about finding the right balance between product output, energy usage, and raw material conservation. In the proposed recycling scheme, we observe a potential efficiency gain by saving 90% of the fresh nitrogen that would have been used without recycling. Yet, one must consider the full picture, including any losses in efficiency due to the recycling operation itself and additional energy input required.

Ultimately, examining various aspects such as yield, selectivity, cost, and environmental impact is essential to gauge the true efficiency of a process. Chemical engineers continuously strive to optimize these aspects, seeking improvements that make industrial processes more sustainable and economically viable.

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Most popular questions from this chapter

Acetaldehyde is synthesized by the catalytic dehydrogenation of ethanol:$$ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\rightarrow\mathrm{CH}_{3}\mathrm{CHO}+\mathrm{H}_{2}.$$ Fresh feed (pure ethanol) is blended with a recycle stream (95 mole\% ethanol and 5\% acetaldehyde), and the combined stream is heated and vaporized, entering the reactor at \(280^{\circ} \mathrm{C}\). Gases leaving the reactor are cooled to \(-40^{\circ} \mathrm{C}\) to condense the acetaldehyde and unreacted ethanol. Off-gas from the condenser is sent to a scrubber, where the uncondensed organic compounds are removed and hydrogen is recovered as a by- product. The condensate from the condenser, which is 45 mole\% ethanol, is sent to a distillation column that produces a distillate containing 99 mole\% acetaldehyde and a bottoms product that constitutes the recycle blended with fresh feed to the process. The production rate of the distillate is \(1000 \mathrm{kg} / \mathrm{h}\). The pressure throughout the process may be taken as 1 atm absolute. (a) Calculate the molar flow rates ( \(\mathrm{kmol} / \mathrm{h}\) ) of the fresh feed, the recycle stream, and the hydrogen in the off-gas. Also determine the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the feed to the reactor. (Suggestion:Use Raoult's law in the analysis of the condenser.)(b) Estimate (i) the overall and single-pass conversions of ethanol and (ii) the rates ( \(\mathrm{kmol} / \mathrm{h}\) ) at which ethanol and acetaldehyde are sent to the scrubber.

When fermentation units are operated with high aeration rates, significant amounts of water can be evaporated into the air passing through the fermentation broth. since fermentation can be adversely affected if water loss is significant, the air is humidified before being fed to the fermenter. Sterilized ambient air is combined with steam to form a saturated air-water mixture at 1 atm and \(90^{\circ} \mathrm{C}\). The mixture is cooled to the temperature of the fermenter \(\left(35^{\circ} \mathrm{C}\right),\) condensing some of the water, and the saturated air is fed to the bottom of the fermenter. For an air flow rate to the fermenter of \(10 \mathrm{L} / \mathrm{min}\) at \(35^{\circ} \mathrm{C}\) and \(1 \mathrm{atm},\) estimate the rate at which steam must be added to the sterilized air and the rate (kg/min) at which condensate is collected upon cooling the air-steam mixture.

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\%.

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?

Various amino acids have utility as food additives and in medical applications. They are often synthesized by fermentation using a specific microorganism to convert a substrate (e.g., a sugar) into the desired product. Small quantities of other species also may be formed and must be removed to meet product specifications. For example, isoleucine (Ile), which has a molecular weight of \(131.2,\) is an essential amino acid \(^{16}\) produced by fermentation, and other amino acids such as leucine and valine also are found in the fermentation broth. The broth is subjected to several processing steps to remove these and other impurities, but final processing by crystallization is required to meet stringent specifications on purity. The strategy is to crystallize the hydrated acid form of Ile (Ile. \(\mathrm{HCl} \cdot \mathrm{H}_{2} \mathrm{O}\) ), whose crystals exclude other amino acids, and then to redissolve, neutralize, and crystallize the final Ile product. In a batch process designed to manufacture \(2500 \mathrm{kg}\) of Ile per batch, an aqueous feed solution containing 35 g Ile/dL and much lower concentrations of leucine and valine is fed to the final purification stages. The pH of the solution is 1.1 and its specific gravity is 1.02. The solution is heated to \(60^{\circ} \mathrm{C}\) and 35-wt\% HCl solution is added in a ratio of 0.4 kg per kg of feed. The addition of HCl causes the formation of crystals of Ile\cdotHCl\cdot \(\mathrm{H}_{2} \mathrm{O},\) and the production of these crystals is further increased by slowly lowering the temperature to \(20^{\circ} \mathrm{C}\). At the final crystallizer conditions the Ile solubility is \(5 \mathrm{g}\) Ile/ \(100 \mathrm{g}\) solution. The resulting slurry is sent to a centrifuge where the crystals are separated from the liquid solution and the crystal cake is washed with water. The solids leaving the centrifuge contain \(12 \%\) free water (i.e., not part of the crystal structure) and \(88 \%\) pure crystals of Ile\(\cdot \mathrm{HCl} \cdot \mathrm{H}_{2} \mathrm{O}\). \(\mathrm{H}_{2} \mathrm{O}\).The washed crystals "water to form a solution that is 4.0 g Ile/dL with gravity of 1.1. The solution is sent to an ion exchange unit where HCl is removed. Upon leaving the ion exchange unit the solution has a pH of about \(5.5 .\) It is sent to a second crystallizer where the temperature is gradually reduced to \(10^{\circ} \mathrm{C}\) and the Ile solubility is \(3.4 \mathrm{g} \mathrm{Ile} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The crystals are separated from the slurry by centrifugation, washed with pure water, and sent to a dryer for final processing. (a) Construct a labeled flowchart for the process. (b) Choosing a basis of 1 kg of feed solution, estimate (i) the mass of HCl solution added to the system, (ii) the water added to redissolve the Ile.HCI. \(\mathrm{H}_{2} \mathrm{O}\) crystals, (iii) the mass of \(\mathrm{HCl}\) removed in the ion exchange unit, and (iv) the mass of final Ile product. (c) Scale the quantities calculated in Part (b) to the production rate of 2500 kg Ile/batch. (d) Estimate the active volume (in liters) of each of the crystallizers. (e) Amino acids are amphoteric, which means they can either donate or accept a proton \(\left(\mathrm{H}^{+}\right) .\) At low pH they tend to accept a proton and become acidic while at high pH they tend to donate a proton and become basic. They also are known as zwitterions because their ends are oppositely charged, even though the overall molecule is neutral. Isoleucine is reported to have an isoelectric point (pI) of 6.02 and \(\mathrm{pK}_{\mathrm{a}}\) values of 2.36 and \(9.60 .\) Look up the meaning of these terms and prepare a plot showing how these values are used in plotting the distribution of Ile between acid, zwitterionic (neutral), and basic forms as a function of pH. Explain why such a distribution is important in carrying out the separations described in the process.

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