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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

01

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.
02

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

The vapor pressure of ethylene glycol at several temperatures is given below:$$\begin{array}{|l|r|r|r|r|r|r|}\hline T\left(^{\circ} \mathrm{C}\right) & 79.7 & 105.8 & 120.0 & 141.8 & 178.5 & 197.3 \\\\\hline p^{*}(\mathrm{mm} \mathrm{Hg}) & 5.0 & 20.0 & 40.0 & 100.0 & 400.0 & 760.0 \\\\\hline\end{array}$$ a semilog plot e vapor-pressure data and determine a linear expression for \(\ln p^{*}\) function of \(1 / T(\mathrm{K}) .\) Use the results to estimate the heat of vaporization \((\mathrm{kJ} / \mathrm{mol})\) of ethylene glycol, and then use that value in the Clausius-Clapeyron equation to estimate the vapor pressures at each of the temperatures given in the table.(b) Repeat Part (a) using the Slope and Intercept functions of APEx to obtain the expression for \(\ln p^{*}\) vs. \(1 / T(\mathrm{K})\).(c) Use the results from Part (b) to estimate vapor pressures of ethylene glycol at \(50^{\circ} \mathrm{C}, 80^{\circ} \mathrm{C},\) and \(110^{\circ} \mathrm{C} .\) Also estimate the boiling point of this substance at system pressures of \(760 \mathrm{mm} \mathrm{Hg}\) and 2000 mm Hg. Compare all five results with those obtained directly using APEx functions. In which of the estimates at the given temperatures and pressures would you have the least confidence? Explain your reasoning.

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.

A fuel cell is an electrochemical device in which hydrogen reacts with oxygen to produce water and DC electricity. A 1-watt proton-exchange membrane fuel cell (PEMFC) could be used for portable applications such as cellular telephones, and a \(100-\mathrm{kW}\) PEMFC could be used to power an automobile. The following reactions occur inside the PEMFC:Anode: \(\quad \mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) Cathode: \(\quad \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}\) Overall: \(\quad \overline{\mathrm{H}}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}\) A flowchart of a single cell of a PEMFC is shown below. The complete cell would consist of a stack of such cells in series, such as the one shown in Problem 9.19.The cell consists of two gas channels separated by a membrane sandwiched between two flat carbonpaper electrodes- -the anode and the cathode- -that contain imbedded platinum particles. Hydrogen flows into the anode chamber and contacts the anode, where \(\mathrm{H}_{2}\) molecules are catalyzed by the platinum to dissociate and ionize to form hydrogen ions (protons) and electrons. The electrons are conducted throughthe carbon fibers of the anode to an extemal circuit, where they pass to the cathode of the next cell in the stack. The hydrogen ions permeate from the anode through the membrane to the cathode.Humid air is fed into the cathode chamber, and at the cathode \(\mathrm{O}_{2}\) molecules are catalytically split to form oxygen atoms, which combine with the hydrogen ions coming through the membrane and electrons coming from the external circuit to form water. The water desorbs into the cathode gas and is carried out of the cell. The membrane material is a hydrophilic polymer that absorbs water molecules and facilitates the transport of the hydrogen ions from the anode to the cathode. Electrons come from the anode of the cell at one end of the stack and flow through an extemal circuit to drive the device that the fuel cell is powering, while the electrons coming from the device flow back to the cathode at the opposite end of the stack to complete the circuit. is important to keep the water content of the cathode gas between upper and lower limits. If the content reaches a value for which the relative humidity would exceed \(100 \%,\) condensation occurs at the cathode (flooding), and the entering oxygen must diffuse through a liquid water film before it can react. The rate of this diffusion is much lower than the rate of diffusion through the gas film normally adjacent to the cathode, and so the performance of the fuel cell deteriorates. On the other hand, if there is not enough water in the cathode gas (less than \(85 \%\) relative humidity), the membrane dries out and cannot transport hydrogen efficiently, which also leads to reduced performance. 400-sell 300-yolt PEMFS anerates at stady state witha nonwer outnul of 36 k W, The air fod to It is important to keep the water content of the cathode gas between upper and lower limits. If the content reaches a value for which the relative humidity would exceed \(100 \%,\) condensation occurs at the cathode (flooding), and the entering oxygen must diffuse through a liquid water film before it can react. The rate of this diffusion is much lower than the rate of diffusion through the gas film normally adjacent to the cathode, and so the performance of the fuel cell deteriorates. 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