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A liquid stream consisting of 12.5 mole \(\% n\) -butane and the balance a heavy nonvolatile hydrocarbon is fed to the top of a stripping column, where it is contacted with an upward-flowing stream of nitrogen. The residual liquid leaves the bottom of the column containing all of the heavy hydrocarbon, 5\% of the butane entering the column, and a negligible amount of dissolved nitrogen.(a) The highest possible butane mole fraction in the exiting gas is that in equilibrium with the butane in the entering liquid (a condition that would require an infinitely tall column to achieve). Using Raoult's law to relate the mole fractions of butane in the entering liquid and exiting gas, calculate the molar feed-stream ratio (mol gas fed/mol liquid fed) corresponding to this limiting condition.(b) Suppose the actual mole fraction of butane in the exit gas is \(80 \%\) of its theoretical maximum value and the percentage stripped (95\%) is the same as in Part (a). Calculate the ratio (mol gas fed/mol liquid fed) for this case.(c) Increasing the nitrogen feed rate for a given liquid feed rate and butane recovery decreases the cost of the process in one way and increases it in another. Explain. What would you have to know to determine the most cost-effective value of the gas/liquid feed ratio?

Short Answer

Expert verified
For limiting condition, the molar feed stream ratio can be calculated using Raoult's law. In part b, the mole fraction of butane is less than the theoretical maximum, and approximately similar simplification can be used as in the first case. Meanwhile, understanding the effects of increasing nitrogen feed rates plays significant role in the cost efficiency of the operation. Detailed cost analysis is required to precisely determine the most cost-effective ratio.

Step by step solution

01

Application of Raoult's Law

By using Raoult's law, which states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture at a constant temperature. In this case, the pressure of butane in the vapor phase equals the mole fraction (\(x_{\text{C4}}\)) of butane in the liquid phase times the saturation pressure of butane (\(p_{\text{C4,sat}}\)). Here, \(x_{\text{C4}} = 0.125\).
02

Calculation of Molar Feed-Stream Ratio Limit Condition

Next, the molar feed-stream ratio for the limiting condition can be determined. This is when the highest possible butane mole fraction in the exiting gas is in equilibrium with the butane in the entering liquid. It can be calculated using the following: \(\frac{P_{N2}}{n_{C4_{liquid}}} = \frac{(p_{total} - x_{C4} * p_{C4,sat})}{x_{C4} * F_{liquid}}\).
03

Calculation of Actual Molar Feed-Stream Ratio

The actual mole fraction of butane in the exit gas is mentioned to be \(80 %\) of its theoretical maximum value from the first part. The same equation can be used as in Step 2, but now with actual mole fraction and same stripping percent as in part a. The ratio (mol gas fed/mol liquid fed) will change.
04

Economic analysis

When the nitrogen feed rate increases for a given liquid feed rate and butane recovery, it decreases the cost of operation in terms of separation efficiency but it increases the cost of operation in terms of nitrogen consumption and possible compressor work for nitrogen recycle. To make an economic evaluation, one would need to know the costs of the nitrogen (including recycling cost), value of butane not lost in the product, and the cost of operations of the tower including labor and maintenance costs.

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Key Concepts

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

Chemical Engineering Principles
Understanding chemical engineering principles is vital when analyzing a process such as those occurring in a stripping column. These principles are the bedrock on top of which we design and operate processes to achieve the desired separation of components based on differences in chemical properties. One fundamental principle applied in such distillation processes is the conservation of mass. This principle asserts that, in an isolated system, mass can neither be created nor destroyed. In practical terms, for our stripping column scenario, this means that the mass of the incoming streams (liquid hydrocarbon and nitrogen gas) must equal the mass of the outgoing streams (stripped liquid and exiting gas with butane).

This balance helps engineers to model and scale up the process, ensuring that the design is both effective and reflects accurately the behavior of the system. Within this process, we also utilize Raoult’s law, which connects the physical chemistry concepts of vapor pressure and mole fraction to optimize the separation of n-butane from the hydrocarbon mixture.
Vapor-Liquid Equilibrium

Role of Vapor-Liquid Equilibrium

The concept of vapor-liquid equilibrium (VLE) is crucial for the design and operation of equipment such as stripping columns. VLE describes the distribution of a chemical species between a liquid phase and a vapor phase when the two phases are in thermodynamic equilibrium with each other. This distribution is determined by the species’ volatility and the operating conditions of temperature and pressure. Raoult's Law is an important relationship used in estimating the compositions at this equilibrium.

For the n-butane stripping process, Raoult's Law allows us to estimate the highest possible butane mole fraction in the exiting gas stream. This theoretical maximum assumes a perfect equilibrium is reached, which would imply an infinitely tall column. Engineers use this ideal scenario as a starting point to determine the necessary molar feed-stream ratio under practical, finite conditions.
Molar Feed-Stream Ratio

Calculating the Molar Feed-Stream Ratio

The molar feed-stream ratio (gas fed to liquid fed) is a critical parameter that dictates the overall efficiency of the stripping process. It is a measure of how much gas is used to strip a certain amount of liquid feed and directly influences the separation process's energy and material costs. The higher the gas to liquid feed ratio, the more driving force is available to strip the n-butane from the liquid stream. However, there is also an economic trade-off since using more gas can be more expensive. The exact ratio required to achieve the desired separation can be calculated by applying Raoult’s Law and considering the column’s performance specifics. It’s a balancing act between achieving operational efficiency and maintaining cost-effectiveness within the process.
Stripping Column Operation

Key Aspects of Stripping Column Operation

Stripping columns are utilized to separate components based on their volatilities. In a typical operation, a stream containing a more volatile component, n-butane in our case, is introduced at the top of the column while a stripping agent, such as nitrogen, flows upwards. The interaction between the agent and the feed stream in the column results in the transfer of the more volatile component to the gas phase while the less volatile components exit as a residual liquid stream from the bottom.

The column must be operated under conditions that allow for the best use of the stripping agent, ideally maximizing butane recovery while minimizing nitrogen usage. The functionality of a stripping column involves complex thermodynamic and fluid dynamic principles, and it requires careful consideration of flow rates, pressure, and temperature for optimal performance.
Economic Analysis in Chemical Processes

Economic Considerations in Process Optimization

Performing an economic analysis in chemical processes like the operation of a stripping column involves assessing various costs and benefits associated with the process. Key considerations include the cost of raw materials, in our case, nitrogen, the value of the recovered product, butane, and the operating expenses such as energy, labor, and equipment maintenance.

To identify the most cost-effective gas/liquid feed ratio, one must analyze both the direct and indirect costs. Direct costs include the nitrogen and butane prices, whereas indirect costs consist of the related expenses like compressor work for nitrogen recycling. A thorough economic analysis compares these costs against the benefits associated with different operating conditions to arrive at an optimal balance that minimizes cost while ensuring the desired quality and quantity of the separated components.

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Most popular questions from this chapter

An adult inhales approximately 12 times per minute, taking in about 500 mL of air with each inhalation. Oxygen and carbon dioxide are exchanged in the lungs, but there is essentially no exchange of nitrogen. The exhaled air has a mole fraction of nitrogen of 0.75 and is saturated with water vapor at body temperature, \(37^{\circ} \mathrm{C}\). If ambient conditions are \(25^{\circ} \mathrm{C}, 1\) atm, and \(50 \%\) relative humidity, what volume of liquid water (mL) would have to be consumed over a two-hour period to replace the water loss from breathing? How much would have to be consumed if the person is on an airplane where the temperature, pressure, and relative humidity are respectively \(25^{\circ} \mathrm{C}, 1 \mathrm{atm},\) and \(10 \% ?\)

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. 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.A 400-cell 300-volt PEMFC operates at steady state with a power output of 36 kW. The air fed to the cathode side is at \(20.0^{\circ} \mathrm{C}\) and roughly 1.0 atm (absolute) with a relative humidity of \(70.0 \%\) and a volumetric flow rate of \(4.00 \times 10^{3}\) SLPM (standard liters per minute). The gas exits at \(60^{\circ} \mathrm{C}\). (a) Explain in your own words what happens in a single cell of a PEMFC. (b) The stoichiometric hydrogen requirement for a PEMFC is given by \(\left(n_{\mathrm{Hz}}\right)_{\text {conanmad }}=I N / 2 F,\) where \(I\) is the current in amperes (coulomb/s), \(N\) is the number of single cells in the fuel cell stack, and \(F\) is the Faraday constant, 96,485 coulombs of charge per mol of electrons. Derive this expression. (Hint: Recall that since the cells are stacked in series the same current flows through each one, and the same quantity of hydrogen must be consumed in each single cell to produce that current at each anode.) (c) Use the expression of Part (b) to determine the molar rates of oxygen consumed and water generated in the unit with the given specifications, both in units of mol/min. (Remember that power = voltage \(\times\) current.) Then determine the relative humidity of the cathode exit stream, \(h_{\mathrm{r} \text { rout. }}\) (d) Determine the minimum cathode inlet flow rate in SLPM to prevent the fuel cell from flooding ( \(h_{\mathrm{r}, \text { out }}=100 \%\) ) and the maximum flow rate to prevent it from drying \(\left(h_{\mathrm{r}, \text { out }}=85 \%\right)\) .

Sodium bicarbonate is synthesized by reacting sodium carbonate with carbon dioxide and water at \(70^{\circ} \mathrm{C}\) and \(2.0 \mathrm{atm}\) gauge pressure: $$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaHCO}_{3}$$ An aqueous solution containing 7.00 wt\% sodium carbonate and a gas stream containing 70.0 mole\% \(\mathrm{CO}_{2}\) and the balance air are fed to the reactor. All of the sodium carbonate and some of the carbon dioxide in the feed react. The gas leaving the reactor, which contains the air and unreacted \(\mathrm{CO}_{2},\) is saturated with water vapor at the reactor conditions. A liquid-solid slurry of sodium bicarbonate crystals in a saturated aqueous solution containing \(2.4 \mathrm{wt} \%\) dissolved sodium bicarbonate and a negligible amount of dissolved \(\mathrm{CO}_{2}\) leaves the reactor and is pumped to a filter. The wet filter cake contains 86 wt\% sodium bicarbonate crystals and the balance saturated solution, and the filtrate also is saturated solution. The production rate of solid crystals is \(500 \mathrm{kg} / \mathrm{h}\).Suggestion: Although the problems to be given can be solved in terms of the product flow rate of \(500 \mathrm{kg} \mathrm{NaHCO}_{3}(\mathrm{s}) / \mathrm{h},\) it might be easier to assume a different basis and then scale the process to the desired production rate of crystals.(a) Calculate the composition (component mole fractions) and volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\) of the gas stream leaving the reactor. (b) Calculate the feed rate of gas to the process in standard cubic meters/min \(\left[\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{min}\right]\) (c) Calculate the flow rate \((\mathrm{kg} / \mathrm{h})\) of the liquid feed to the process. What more would you need to know to calculate the volumetric flow rate of this stream? (d) The filtrate was assumed to leave the filter as a saturated solution at \(70^{\circ} \mathrm{C}\). What would be the effect on your calculations if the temperature of the filtrate actually dropped to \(50^{\circ} \mathrm{C}\) as it passed through the filter? (e) The reactor pressure of 2 atm gauge was arrived at in an optimization study. What benefit do you suppose would result from increasing the pressure? What penalty would be associated with this increase? The term "Henry's law" should appear in your explanation. (Hint: The reaction occurs in the liquid phase and the \(\mathrm{CO}_{2}\) enters the reactor as a gas. What step must precede the reaction?)

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?

Recovery and processing of various oils are important elements of the agricultural and food industries. For example, soybean hulls are removed from the beans, which are then flaked and contacted with hexane. The hexane extracts soybean oil and leaves very little oil in the residual solids. The solids are dried at an elevated temperature, and the dried solids are used to feed livestock or further processed to extract soy protein. The gas stream leaving the dryer is at \(80^{\circ} \mathrm{C}\) 1 atm absolute, and 50\% relative saturation.(a) To recover hexane, the gas leaving the dryer is fed to a condenser, which operates at 1 atm absolute. The gas leaving the condenser contains 5.00 mole \(\%\) hexane, and the hexane condensate is recovered at a rate of \(1.50 \mathrm{kmol} / \mathrm{min}\). (b) In an altemative arrangement, the gas leaving the dryer is compressed to 10.0 atm and the temperature simultancously is increased so that the relative saturation remains at \(50 \% .\) The gas then is cooled at constant pressure to produce a stream containing 5.00 mole \(\%\) hexane. Calculate the final gas temperature and the ratio of volumetric flow rates of the gas streams leaving and entering the condenser. State any assumptions you make.(c) What would you need to know to determine which of processes (a) and (b) is more cost- effective?

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