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Chapter 4: Problem 55

A stream containing \(\mathrm{H}_{2} \mathrm{S}\) and inert gases and a second stream of pure \(\mathrm{SO}_{2}\) are fed to a sulfur recovery reactor, where the reaction $$2 \mathrm{H}_{2} \mathrm{S}+\mathrm{SO}_{2} \rightarrow 3 \mathrm{S}+2 \mathrm{H}_{2} \mathrm{O}$$ takes place. The feed rates are adjusted so that the ratio of \(\mathrm{H}_{2} \mathrm{S}\) to \(\mathrm{SO}_{2}\) in the combined feed is always stoichiometric. In the normal operation of the reactor the flow rate and composition of the \(\mathrm{H}_{2} \mathrm{S}\) feed stream both fluctuate. In the past, each time either variable changed the required \(\mathrm{SO}_{2}\) feed rate had to be reset by adjusting a valve in the feed line. A control system has been installed to automate this process. The \(\mathrm{H}_{2} \mathrm{S}\) feed stream passes through an electronic flowmeter that transmits a signal \(R_{\mathrm{f}}\) directly proportional to the molar flow rate of the stream, \(\dot{n}_{\mathrm{f}}\). When \(\dot{n}_{\mathrm{f}}=100 \mathrm{kmol} / \mathrm{h}\), the transmitted signal \(R_{\mathrm{f}}=15 \mathrm{mV}\). The mole fraction of \(\mathrm{H}_{2} \mathrm{S}\) in this stream is measured with a thermal conductivity detector, which transmits a signal \(R_{\mathrm{a}} .\) Analyzer calibration data are as follows: $$\begin{array}{|l|c|c|c|c|c|c|}\hline R_{\mathrm{a}}(\mathrm{mV}) & 0 & 25.4 & 42.8 & 58.0 & 71.9 & 85.1 \\ \hline x\left(\mathrm{mol} \mathrm{H}_{2} \mathrm{S} / \mathrm{mol}\right) & 0.00 & 0.20 & 0.40 & 0.60 &0.80 & 1.00 \\\\\hline\end{array}$$ The controller takes as input the transmitted values of \(R_{\mathrm{f}}\) and \(R_{\mathrm{a}}\) and calculates and transmits a voltage signal \(R_{\mathrm{c}}\) to a flow control valve in the \(\mathrm{SO}_{2}\) line, which opens and closes to an extent dependent on the value of \(R_{c} .\) A plot of the \(S O_{2}\) flow rate, \(\dot{n}_{c},\) versus \(R_{c}\) on rectangular coordinates is a straight line through the points \(\left(R_{c}=10.0 \mathrm{mV}, \dot{n}_{c}=25.0 \mathrm{kmol} / \mathrm{h}\right)\) and \(\left(R_{c}=25.0 \mathrm{mV}, \dot{n}_{c}=60.0 \mathrm{kmol} / \mathrm{h}\right)\) (a) Why would it be important to feed the reactants in stoichiometric proportion? (Hint: \(\mathrm{SO}_{2}\) and especially \(\mathrm{H}_{2} \mathrm{S}\) are serious pollutants.) What are several likely reasons for wanting to automate the \(\mathrm{SO}_{2}\) feed rate adjustment? (b) If the first stream contains 85.0 mole \(\% \mathrm{H}_{2} \mathrm{S}\) and enters the unit at a rate of \(\dot{n}_{\mathrm{f}}=3.00 \times 10^{2} \mathrm{kmol} / \mathrm{h}\) what must the value of \(\dot{n}_{c}\left(\mathrm{kmol} \mathrm{SO}_{2} / \mathrm{h}\right)\) be? (c) Fit a function to the \(\mathrm{H}_{2} \mathrm{S}\) analyzer calibration data to derive an expression for \(x\) as a function of \(R_{\mathrm{a}}\) Check the fit by plotting both the function and the calibration data on the same graph. (d) Derive a formula for \(R_{\mathrm{c}}\) from specified values of \(R_{\mathrm{f}}\) and \(R_{\mathrm{a}},\) using the result of Part (c) in the derivation. (This formula would be built into the controller.) Test the formula using the flow rate and composition data of Part (a). (e) The system has been installed and made operational, and at some point the concentration of \(\mathrm{H}_{2} \mathrm{S}\) in the feed stream suddenly changes. A sample of the blended gas is collected and analyzed a short time later and the mole ratio of \(\mathrm{H}_{2} \mathrm{S}\) to \(\mathrm{SO}_{2}\) is not the required 2: 1 . List as many possible reasons as you can think of for this apparent failure of the control system.

Short Answer

Expert verified
The required molar flow rate of SO2 is 127.5 kmol/h. Reasons for control system failure could be sensor inaccuracy, valve control failure, sudden drastic change in feed or mechanical failure.

Step by step solution

01

Stoichiometry and Automation Importance

The question states why it is important to maintain a stoichiometric proportion when feeding the reactants. The reason is H2S and SO2 are pollutants. The reaction is designed in such a way that all the available reactants fully react without leaving any excess, which could cause environmental harm. Automation is essential to adjust the flow rates in real-time for maintaining the desired stoichiometric ratio. This prevents any unplanned release of unreacted H2S or SO2 into the air, and minimizes human error.
02

Calculate feed rate for SO2

From the chemical equation, we know that the stoichiometric ratio of H2S to SO2 is 2:1. The molar flow rate of H2S in the feed stream is 85.0% of \(3.00 \times 10^{2}\) kmol/h, which equals 255 kmol/h. Therefore, the required molar flow rate of SO2 (\(\dot{n}_{c}\)) should be half of 255 kmol/h, or approximately 127.5 kmol/h.
03

Fit a function to the H2S analyzer calibration data

To find a function for the mole fraction (\(x\)) as a function of \(R_{a}\), we need to fit the analyzer calibration data. This can be done using linear regression or least square method. Once the exact function is derived, it can be plotted alongside the given data to check the accuracy of the fit.
04

Derive a formula for \(R_{c}\)

Now, we need to derive a formula for \(R_{c}\) from specified values of \(R_{f}\) and \(R_{a}\). We know that the flow rate of SO2 is linearly related to \(R_{c}\), so we can write that as \(\dot{n}_{c} = m R_{c} + b\), where \(m\) and \(b\) are constants. We can use the given points to calculate these constants. The values of \(R_{f}\) and \(R_{a}\) can be substituted into this equation to calculate the value of \(R_{c}\).
05

Reason for control system failure

The reasons for the apparent failure in the stoichiometry could be many. Likely issues could be inaccurate readings from the electronic flowmeter, poor control of the valve in the SO2 feed line, a sudden, drastic change in the H2S feed stream beyond what the control system can respond to in a short time, or mechanical failure of a component.

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Key Concepts

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

Stoichiometry
Stoichiometry is a fundamental concept in chemical process control. It refers to the precise measurement of reactants and products in a chemical reaction. This is essential to ensure that all reactants are used efficiently without leaving any unreacted substance. In a sulfur recovery process, we deal with a specific chemical equation:

\(2\ \mathrm{H}_{2}\mathrm{S} + \mathrm{SO}_{2} \rightarrow 3\ \mathrm{S} + 2\ \mathrm{H}_{2} \mathrm{O}\)

The stoichiometric ratio here is 2 moles of hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)) for every 1 mole of sulfur dioxide (\(\mathrm{SO}_{2}\)). Maintaining this ratio is crucial to ensure the maximum conversion of reactants to desired products, sulfur (\(\mathrm{S}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)).

Failing to uphold these proportions can lead to excess pollutants, particularly leftover \(\mathrm{H}_{2}\mathrm{S}\) or \(\mathrm{SO}_{2}\), which are harmful to the environment. By controlling the feeds into a reactor based on stoichiometric calculations, the process is more cost-effective and sustainable.
Sulfur Recovery
Sulfur recovery involves the extraction and purification of sulfur from gas streams. Typically found in the gas industry, the primary goal is to reduce the emission of hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)) and sulfur dioxide (\(\mathrm{SO}_{2}\)), both hazardous pollutants. The reaction used in sulfur recovery is both a purification process and an environmental control measure.

During the reaction, \(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{SO}_{2}\) react to form elemental sulfur, a valuable by-product in various industries. The controlled reaction not only produces commercial sulfur but also reduces the environmental pollution associated with gas processing.
  • Promotes refining industries to convert potential pollutants into useful commercial products.
  • Reduces environmental liabilities by ensuring minimum pollution release.
This conversion is not only economically beneficial but also critical for compliance with regulatory standards for emission control.
Pollution Control
Pollution control in chemical processes is vital to ensure that industrial activities do not adversely impact the environment. Effective control systems are designed to address the release of harmful substances such as \(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{SO}_{2}\). These gases, if released untreated, can contribute to air pollution and cause health hazards.

Implementing robust pollution control mechanisms, such as automated processes, enhances the ability to manage reactant feed rates accurately. Automation helps ensure optimal stoichiometry, reducing the chance of releasing unreacted toxic gases:
  • Prevents the release of excessive pollutants by ensuring complete reactions.
  • Supports adaptability in processes when reactant compositions vary.
  • Offers real-time adjustments to process parameters to align with environmental standards.
By automating controls, processes become more reliable, efficient, and environmentally friendly.
Automated Control Systems
Automated control systems in chemical engineering streamline operations and reduce the possibility of human error in chemical processes. They are essential in maintaining the accuracy of reaction conditions, especially in handling volatile substances like hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)) and sulfur dioxide (\(\mathrm{SO}_{2}\)).

An automated system in a sulfur recovery unit typically involves electronic sensors and control valves. For example, a flowmeter measures the molar flow rate of incoming \(\mathrm{H}_{2}\mathrm{S}\) and sends a signal to adjust the \(\mathrm{SO}_{2}\) flow, ensuring optimal stoichiometric balance.

Automation benefits include:
  • Enhancing precision in feed adjustments to sustain efficient conversion rates.
  • Reducing the response time in addressing fluctuations in feed stream composition.
  • Maintaining consistent quality and safety standards.
Through effective integration of automation, operators can focus on higher-level decisions while letting systems handle routine adjustments, leading to increased process efficiency and safety.

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Most popular questions from this chapter

A mixture of 75 mole \(\%\) methane and 25 mole \(\%\) hydrogen is burned with \(25 \%\) excess air. Fractional conversions of \(90 \%\) of the methane and \(85 \%\) of the hydrogen are achieved; of the methane that reacts, \(95 \%\) reacts to form \(\mathrm{CO}_{2}\) and the balance reacts to form CO. The hot combustion product gas passes through a boiler in which heat transferred from the gas converts boiler feedwater into steam. (a) Calculate the concentration of \(\mathrm{CO}\) (ppm) in the stack gas. (b) The CO in the stack gas is a pollutant. Its concentration can be decreased by increasing the percent excess air fed to the furnace. Think of at least two costs of doing so. (Hint: The heat released by the combustion goes into heating the combustion products; the higher the combustion product temperature, the more steam is produced.)

The respiratory process involves hemoglobin (Hgb), an iron-containing compound found in red bloodcells. In the process, carbon dioxide diffuses from tissue cells as molecular \(\mathrm{CO}_{2}\), while \(\mathrm{O}_{2}\) simultaneously enters the tissue cells. A significant fraction of the \(\mathrm{CO}_{2}\) leaving the tissue cells enters red blood cells and reacts with hemoglobin; the \(\mathrm{CO}_{2}\) that does not enter the red blood cells ( \((\mathrm{D}\) in the figure below) remains dissolved in the blood and is transported to the lungs. Some of the \(\mathrm{CO}_{2}\) entering the red blood cells reacts with hemoglobin to form a compound (Hgb. \(\mathrm{CO}_{2} ;(\) 2) in the figure). When the red blood cells reach the lungs, the Hgb.CO_ dissociates, releasing free CO_ Meanwhile, the CO_ that enters the red blood cells but does not react with hemoglobin combines with water to form carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3},\) which then dissociates into hydrogen ions and bicarbonate ions ( (3) in the figure). The bicarbonate ions diffuse out of the cells ( (4) in the figure), and the ions are transported to the lungs via the bloodstream. For adult humans, every deciliter of blood transports a total of \(1.6 \times 10^{-4}\) mol of carbon dioxide in its various forms (dissolved \(\mathrm{CO}_{2}, \mathrm{Hgb} \cdot \mathrm{CO}_{2},\) and bicarbonate ions) from tissues to the lungs under normal, resting conditions. Of the total \(\mathrm{CO}_{2}, 1.1 \times 10^{-4}\) mol are transported as bicarbonate ions. In a typical resting adult human, the heart pumps approximately 5 liters of blood per minute. You have been asked to determine how many moles of \(\mathrm{CO}_{2}\) are dissolved in blood and how many moles of \(\mathrm{Hgb} \cdot \mathrm{CO}_{2}\) are transported to the lungs during an hour's worth of breathing. (a) Draw and fully label a flowchart and do a degree-of-freedom analysis. Write the chemical reactions that occur, and generate, but do not solve, a set of independent equations relating the unknown variables on the flowchart. (b) If you have enough information to obtain a unique numerical solution, do so. If you do not have enough information, identify a specific piece/pieces of information that (if known) would allow you to solve the problem, and show that you could solve the problem if that information were known. (c) When someone loses a great deal of blood due to an injury, they "go into shock": their total blood volume is low, and carbon dioxide is not efficiently transported away from tissues. The carbon dioxide reacts with water in the tissue cells to produce very high concentrations of carbonic acid, some of which can dissociate (as shown in this problem) to produce high levels of hydrogen ions. What is the likely effect of this occurrence on the blood pH near the tissue and the tissue cells? How is this likely to affect the injured person?

Seawater containing 3.50 wt\% salt passes through a series of 10 evaporators. Roughly equal quantities of water are vaporized in each of the 10 units and then condensed and combined to obtain a product stream of fresh water. The brine leaving each evaporator but the tenth is fed to the next evaporator. The brine leaving the tenth evaporator contains \(5.00 \mathrm{wt} \%\) salt. (a) Draw a flowchart of the process showing the first, fourth, and tenth evaporators. Label all the streams entering and leaving these three evaporators. (b) Write in order the set of equations you would solve to determine the fractional yield of fresh water from the process \(\left(\mathrm{kg} \mathrm{H}_{2} \mathrm{O} \text { recovered } / \mathrm{kg} \mathrm{H}_{2} \mathrm{O}\) in process feed) and the weight percent of salt in the \right. solution leaving the fourth evaporator. Each equation you write should contain no more than one previously undetermined variable. In each equation, circle the variable for which you would solve. Do not do the calculations. (c) Solve the equations derived in Part (b) for the two specified quantities. (d) The problem statement made no mention of the disposition of the 5 wt\% effluent from the tenth evaporator. Suggest two possibilities for its disposition and describe any environmental concerns that might need to be considered.

\- An equimolar liquid mixture of benzene and toluene is separated into two product streams by distillation. A process flowchart and a somewhat oversimplified description of what happens in the process follow: Inside the column a liquid stream flows downward and a vapor stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving the top of the column, which contains 97 mole\% benzene, is completely condensed and split into two equal fractions: one is taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains \(89.2 \%\) of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which \(45 \%\) of it is vaporized. The vapor generated in the reboiler (the boilup) is recycled to become the rising vapor stream in the column, and the residual reboiler liquid is taken off as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation $$\frac{y_{\mathrm{B}} /\left(1-y_{\mathrm{B}}\right)}{x_{\mathrm{B}} /\left(1-x_{\mathrm{B}}\right)}=2.25$$ where \(y_{\mathrm{B}}\) and \(x_{\mathrm{B}}\) are the mole fractions of benzene in the vapor and liquid streams, respectively. (a) Take a basis of 100 mol fed to the column. Draw and completely label a flowchart, and for each of four systems (overall process, column, condenser, and reboiler), do the degree-of-freedom analysis and identify a system with which the process analysis might appropriately begin (one with zero degrees of freedom). (b) Write in order the equations you would solve to determine all unknown variables on the flowchart, circling the variable for which you would solve in each equation. Do not do the calculations in this part. (c) Calculate the molar amounts of the overhead and bottoms products, the mole fraction of benzene in the bottoms product, and the percentage recovery of toluene in the bottoms product \((100 \times\) moles toluene in bottoms/mole toluene in feed).

Fermentation of sugars obtained from hydrolysis of starch or cellulosic biomass is an alternative to using petrochemicals as the feedstock in production of ethanol. One of the many commercial processes to do this \(^{16}\) uses an enzyme to hydrolyze starch in corn to maltose (a disaccharide consisting of two glucose units) and oligomers consisting of several glucose units. A yeast culture then converts the maltose to ethyl alcohol and carbon dioxide: $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O}(+\text { yeast }) \rightarrow 4 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+4 \mathrm{CO}_{2}\left(+\text { yeast }+\mathrm{H}_{2} \mathrm{O}\right)$$ As the yeast grows, \(0.0794 \mathrm{kg}\) of yeast is produced for every \(\mathrm{kg}\) ethyl alcohol formed, and \(0.291 \mathrm{kg}\) water is produced for every kg of yeast formed. For use as a fuel, the product from such a process must be around 99.5 wt\% ethyl alcohol. Corn fed to the process is 72.0 wt\% starch on a moisture-free basis and contains 15.5 wt\% moisture. It is estimated that 101.2 bushels of corn can be harvested from an acre of com, that each bushel is equivalent to \(25.4 \mathrm{lb}_{\mathrm{m}}\) of corn, and that \(6.7 \mathrm{kg}\) of ethanol can be obtained from a bushel of corn. What acreage of farmland is required to produce 100,000 kg of ethanol product? What factors (economic and environmental) must be considered in comparing production of ethanol by this route with other routes involving petrochemical feedstocks?
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