/*! 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 78 Dehydration of natural gas is ne... [FREE SOLUTION] | 91Ó°ÊÓ

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Dehydration of natural gas is necessary to prevent the formation of gas hydrates, which can plug valves and other components of a gas pipeline, and also to reduce potential corrosion problems. Water removal can be accomplished as shown in the following schematic diagram: Natural gas containing \(80 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O} / 10^{6} \mathrm{SCF}\) gas \(\left[\mathrm{SCF}=\mathrm{ft}^{3}(\mathrm{STP})\right]\) enters the bottom of an absorber at a rate of \(4.0 \times 10^{6}\) SCF/day. A liquid stream containing triethylene glycol (TEG, molecular weight \(=150.2\) ) and a small amount of water is fed to the top of the absorber. The absorber operates at 500 psia and \(90^{\circ} \mathrm{F}\). The dried gas leaving the absorber contains \(10 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O} / 10^{6} \mathrm{SCF}\) gas. The solvent leaving the absorber, which contains all the TEG-water mixture fed to the column plus all the water absorbed from the natural gas, goes to a distillation column. The overhead product stream from the distillation column contains only liquid water. The bottoms product stream, which contains TEG and water, is the stream recycled to the absorber.(a) Draw and completely label a flowchart of this process. Calculate the mass flow rate ( \(\left(\mathrm{b}_{\mathrm{m}} / \mathrm{day}\right)\) and volumetric flow rate (ft \(^{3}\) /day) of the overhead product from the distillation column. (b) The greatest possible amount of dehydration is achieved if the gas leaving the absorption column is in equilibrium with the solvent entering the column. If the Henry's law constant for water in TEG at \(90^{\circ} \mathrm{F}\) is \(0.398 \mathrm{psia} / \mathrm{mol}\) fraction, what is the maximum allowable mole fraction of water in the solvent fed to the absorber?(c) A column of infinite height would be required to achieve equilibrium between the gas and liquid at the top of the absorber. For the desired separation to be achieved in practice, the mole fraction of water in the entering solvent must be less than the value calculated in Part (b). Suppose it is \(80 \%\) of that value and the flow rate of TEG in the recirculating solvent is 37 Ib \(_{\mathrm{m}}\) TEG/lb \(_{\mathrm{m}}\) water absorbed in the column. Calculate the flow rate ( \(\left(\mathrm{b}_{\mathrm{m}} / \mathrm{day}\right)\) of the solvent stream entering the absorber and the mole fraction of water in the solvent stream leaving the absorber. (d) What is the purpose of the distillation column in the process? (Hint: Think about how the process would operate without it.)

Short Answer

Expert verified
The mass flow rate and volumetric flow rate of the overhead product from the distillation column are determined. The maximum allowable mole fraction of water in the solvent fed to the absorber is calculated using Henry's law. The flow rate of the solvent and the mole fraction of water in the solvent leaving the absorber are also calculated. The distillation column separates the absorbed water from the TEG and recycles the TEG back to the absorber for reuse in dehydrating the natural gas.

Step by step solution

01

Draw a flowchart

Make a flowchart of the process, labeling all incoming and outgoing streams, process units like the absorber and distillation column, and the compositions and flow rates of the streams.
02

Calculate the mass flow rate

The mass flow rate of the overhead product from the distillation column can be calculated by taking the difference between the water content in the gas entering and leaving the absorber, and multiplying it by the gas flow rate.
03

Calculate the volumetric flow rate

The volumetric flow rate of the overhead product can be calculated by dividing the mass flow rate by the density of water.
04

Determine the maximum mole fraction of water

Using Henry's law, the maximum mole fraction of water in the solvent can be calculated by dividing the pressure of water in the gas by Henry's law constant.
05

Calculate the flow rate of the solvent

The flow rate of the solvent entering the absorber can be estimated by dividing the mass flow rate of water absorbed in the column by the flow rate of TEG in the recirculating solvent, assuming the mole fraction of water in the solvent is 80% of the maximum value.
06

Calculate the mole fraction of water in the solvent leaving the absorber

The water balance around the absorber can be used to estimate the mole fraction of water in the solvent leaving the absorber.
07

Describe the purpose of the distillation column

The distillation column serves to separate the TEG-water mixture, removing the absorbed water and recycling the TEG back to the absorber.

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Key Concepts

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

Gas Hydrates Prevention
Gas hydrates, often referred to as 'clathrates', are crystalline water-based solids that resemble ice, and they are known to form under high-pressure and low-temperature conditions typically found in natural gas pipelines. The presence of hydrates in these pipelines can be problematic, leading to blockages and potential operational disruptions. To prevent the formation of gas hydrates, natural gas must undergo dehydration, a process where water vapor is removed from the gas before transportation or storage.

One of the most common dehydration methods used in the industry is through the absorption of water using triethylene glycol (TEG), as illustrated in the textbook exercise. The TEG absorbs water from the natural gas in an absorber column, becoming rich in water content. It is then cycled through a distillation column, where the water is removed, allowing the regenerated TEG to be recirculated back for further absorption. This continuous loop is crucial for effective gas hydrates prevention and maintaining the flow of natural gas through pipelines.

By understanding the risks associated with gas hydrates and incorporating systems to remove moisture effectively, significant downtime and maintenance costs can be avoided. The challenge lies in designing these dehydration systems to ensure optimized performance, considering factors such as gas composition, temperature, pressure, and flow rates. Chemical engineers must thus apply rigorous principles when sizing equipment, determining TEG circulation rates, and ensuring the quality of the dehydrated gas meets the required specifications.
Natural Gas Processing
Processing natural gas to make it suitable for commercial and industrial use is a complex series of operations that involve removing impurities and various non-methane hydrocarbons and fluids to produce pipeline-quality dry natural gas. Dehydration of natural gas, which is the main focus of the given exercise, is one of these critical processing steps. This step is essential not only for preventing gas hydrates but also to minimize corrosion in pipelines and processing equipment due to the presence of water.

The use of desiccants such as triethylene glycol in the dehydration process, as showcased in the exercise, is a common practice. The dried gas, after being stripped of water vapor, meets pipeline transportation standards and prevents hydrate formation. High-level engineering decisions must be taken into account when designing such processing systems. These include the selection of appropriate dehydration techniques, the capacity of the processing plant, and the ensured recoverability and purity of the processed gas.

Other important steps in natural gas processing include acid gas removal, mercury removal, and nitrogen rejection, all of which are aimed at ensuring that the final product is safe, efficient to transport, and meets market or regulatory standards. Through the proper application of chemical engineering principles and technology, natural gas can be treated for safe and efficient end-use.
Chemical Engineering Principles
Chemical engineering involves the application of various fundamental principles to innovate and optimize processes within the chemical and energy industries. In the context of natural gas processing, such as the dehydration process described in the exercise, these principles include mass and energy balances, thermodynamics, fluid mechanics, and transport phenomena.

Mass balance is particularly prominent in the problem exercise, where the flow rates and compositions of water in the TEG solvent stream must be calculated based on the water absorbed from the gas. Energy balance considerations would also be crucial in a real-world scenario, especially when it comes to the distillation process, where heat integration could improve efficiency. Thermodynamics enables the understanding and application of phase equilibria (like Henry's law in the exercise) and helps predict the behavior of the gas and absorbent under varying conditions. Fluid mechanics principles aid in designing pipelines and absorbers ensuring that the pressure drops and flow patterns are conducive to an optimum separation process.

Furthermore, chemical engineering principles are not only applied to these technical aspects but are also critical in areas such as safety, environmental impact assessments, and economic feasibility studies. Modern chemical engineers must balance these considerations while adhering to industry standards and achieving the desired process outcomes. By leveraging these principles, engineers aim to design processes that are efficient, sustainable, and operate safely within the defined economic and regulatory frameworks.

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Most popular questions from this chapter

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.

In the manufacture of an active pharmaceutical ingredient (API), the API goes through a final purification step in which it is crystallized, filtered, and washed. The washed crystals contain \(47 \%\) water. They are fed to a tunnel dryer and leave the dryer at a rate of \(165 \mathrm{kg} / \mathrm{h}\) containing \(5 \%\) adhered moisture. Dry air enters the dryer at \(145^{\circ} \mathrm{F}\) and \(1 \mathrm{atm},\) and the outlet air is at \(130^{\circ} \mathrm{F}\) and 1 atm with a relative humidity of \(50 \% .\) Calculate the rate \((\mathrm{kg} / \mathrm{h})\) at which the API enters the dryer and the volumetric flow rate \(\left(\mathrm{ft}^{3} / \mathrm{h}\right)\) of inlet air.

The separation of aromatic compounds from paraffins is essential in producing many polyesters that are used in a variety of products. When aromatics and paraffins have the same number of carbon atoms, they often have similar vapor pressures, which makes them difficult to separate by distillation. Extraction is a viable alternative, as illustrated by the following simple system.Sulfolane (an industrial solvent) and octane may be considered completely immiscible. At \(25^{\circ} \mathrm{C},\) the ratio of the mass fraction of xylene in the octane-rich phase to the mass fraction of xylene in the sulfolanerich phase is 0.25. One hundred kg of pure sulfolane are added to 100 kg of a mixture containing 75 wt\% octane and \(25 \%\) xylene, and the resulting system is allowed to equilibrate. How much xylene transfers to the sulfolane phase?

Using Raoult's law or Henry's law for each substance (whichever one you think appropriate), calculate the pressure and gas-phase composition (mole fractions) in a system containing a liquid that is 0.3 mole \(\% \mathrm{N}_{2}\) and 99.7 mole \(\%\) water in equilibrium with nitrogen gas and water vapor at \(80^{\circ} \mathrm{C}\).

An aqueous solution of potassium hydroxide (KOH) is fed to an evaporative crystallizer at a rate of 875 kg/h. The crystallizer operates at 10^'C and produces crystals of KOH-2H_O. Water evaporated from the crystallizer flows to a condenser, and the resulting condensate is collected in a tank. During a 30-minute period, 73.8 kg of water is collected. Five-gram samples of the feed to the crystallizer and the liquid removed with the crystals are taken for analysis and subsequently titrated with \(0.85 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) It is found that \(22.4 \mathrm{mL}\) of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is required for the feed and \(26.6 \mathrm{mL}\) is required for the product liquid.(a) What fraction of the KOH in the feed is crystallized? (b) Later you learn that a solution in equilibrium with KOH \(\cdot 2 \mathrm{H}_{2} \mathrm{O}\) crystals at \(10^{\circ} \mathrm{C}\) has a concentration of \(103 \mathrm{kg} \mathrm{KOH} / 100 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} .\) How would this information cause you to reconsider the procedure by which a sample of the mother liquor was obtained? (Hint: Consider removing a slurry sample- -i.e., one containing both solution and KOH \(\cdot 2 \mathrm{H}_{2} \mathrm{O}\) crystals - that is maintained at \(10^{\circ} \mathrm{C},\) but that initially had a solute concentration of \(121 \mathrm{kg} \mathrm{KOH} / 100 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} .\) What would that concentration be after the sample is stored for several hours?)

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