/*! 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 99 A fuel oil is fed to a furnace a... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 99

A fuel oil is fed to a furnace and burned with \(25 \%\) excess air. The oil contains \(87.0 \mathrm{wt} \% \mathrm{C}, 10.0 \% \mathrm{H},\) and 3.0\% S. Analysis of the furnace exhaust gas shows only \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2}, \mathrm{SO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\). The sulfur dioxide emission rate is to be controlled by passing the exhaust gas through a scrubber, in which most of the \(\mathrm{SO}_{2}\) is absorbed in an alkaline solution. The gases leaving the scrubber (all of the \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{CO}_{2}\), and some of the \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{SO}_{2}\) entering the unit) pass out to a stack. The scrubber has a limited capacity, however, so that a fraction of the furnace exhaust gas must be bypassed directly to the stack. At one point during the operation of the process, the scrubber removes \(90 \%\) of the \(\mathrm{SO}_{2}\) in the gas fed to it, and the combined stack gas contains 612.5 ppm (parts per million) \(\mathrm{SO}_{2}\) on a dry basis; that is, every million moles of dry stack gas contains 612.5 moles of \(\mathrm{SO}_{2}\). Calculate the fraction of the exhaust bypassing the scrubber at this moment.

Short Answer

Expert verified
After solving the final equation derived from the SO鈧� balance in the stack gas, the value of \(\alpha\), which represents the fraction of the exhaust bypassing the scrubber, is obtained.

Step by step solution

01

- Calculate the moles of SO鈧� produced

The first step is to determine how much SO鈧� is produced by burning the fuel oil. Given that the fuel oil is composed of 87% Carbon (C), 10% Hydrogen (H) and 3% Sulphur (S) by weight, the moles of S in 1 kg of fuel oil can be found using the formula \(\frac{weight \% of S}{molar mass of S}\), which equals \(\frac{3 \%}{32 g/mol} = 0.09375 mol\). Each mole of S produces a mole of SO鈧� when burned, so the moles of SO鈧� produced equals 0.09375 mol/kg.
02

- Calculate the SO鈧� released from scrubber

In this step, we determine the amount of SO鈧� released by the scrubber. The scrubber absorbs 90% of the incoming SO鈧�, which means 10% of the SO鈧� is released. The amount of the released SO鈧� equals to \(10 \%\) of the incoming SO鈧�, which is \(0.09375 mol/kg * 10 \% = 0.009375 mol/kg\).
03

- Determine the moles of dry stack gases

The dry stack gases were described as having 612.5 ppm of SO鈧�. This means that for every million (1e6) moles of dry stack gas, there are 612.5 moles of SO鈧�. Hence, 1 mol of dry stack gas is associated with \( \frac{612.5 mol of \, SO_{2}}{1e6 mol of \, stack \, gas} = 6.125e-4 mol of \, SO_{2}\).
04

- Calculate the fraction that bypasses the scrubber

In the final step, we can determine the fraction of exhaust that bypasses the scrubber. The total SO鈧� in stack gases is a sum of the SO鈧� from the bypassed gas and the SO鈧� released by the scrubber. If \(\alpha\) is the fraction of exhaust gas that bypasses the scrubber, the moles of SO鈧� in bypassed gas will be \(0.09375 \alpha mol/kg\). The complete balance of SO鈧� in the stack gas gives us the equation \(0.09375 \alpha + 0.009375 = 6.125e-4 \times (1 + \alpha)\). Solving this equation gives the value of \(\alpha\).

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Key Concepts

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

Furnace Emissions
Understanding furnace emissions is crucial, especially due to their environmental impact. When fuel oil is burned in a furnace, it emits various gases. Among these emissions are nitrogen (\( \mathrm{N}_2 \)), oxygen (\( \mathrm{O}_2 \)), carbon dioxide (\( \mathrm{CO}_2 \)), sulfur dioxide (\( \mathrm{SO}_2 \)), and water vapor (\( \mathrm{H}_2\mathrm{O} \)). The composition of the fuel directly affects the emission profile.
The excessive air introduced into the system, in this case, 25% more than needed for complete combustion, ensures that complete combustion of the carbon and hydrogen occurs. However, it also amplifies the production of nitrogen oxides and, notably, \( \mathrm{SO}_2 \) from sulfur in the fuel oil. \( \mathrm{SO}_2 \) is a major pollutant, contributing to acid rain formation and respiratory problems in humans and animals. This highlights the need for effective emission control strategies, such as scrubbers, to limit the release of these hazardous components into the atmosphere.
Scrubber Systems
Scrubber systems are a vital component in reducing harmful emissions from industrial exhausts. They work by cleaning or 'scrubbing' flue gases of undesirable pollutants, absorbing substances like \( \mathrm{SO}_2 \) in an alkaline solution. The interaction with the solution neutralizes the acidic nature of \( \mathrm{SO}_2 \), typically converting it to a salt. In the described exercise, the scrubber removes up to 90% of entering \( \mathrm{SO}_2 \). This is significant because it limits the environmental damage the emissions can cause.
However, scrubbers are not 100% efficient and have operational limits. Some gases bypass the system entirely, as seen in the problem where a portion of exhaust gases are not scrubbed. This bypass ensures that the system doesn鈥檛 exceed its capacity, but emphasizes the importance of optimizing scrubber operation to balance emissions reduction with functional constraints.
Sulfur Dioxide Control
Controlling sulfur dioxide emissions is key to reducing air pollution and its adverse effects on health and the environment. \( \mathrm{SO}_2 \) is a byproduct of burning sulfur-containing fuels, such as oil or coal. Without intervention, these emissions can lead to acid rain, harm aquatic life, damage forests, and affect air quality.
One approach in sulfur dioxide control is through chemical reactions in scrubber systems, where the \( \mathrm{SO}_2 \) is captured and neutralized. By calculating the fraction of exhaust bypassing a scrubber, operators can better manage and predict emission levels, further regulating \( \mathrm{SO}_2 \) release. It is essential to continually monitor and adapt these systems to maintain compliance with environmental standards and reduce ecological impact.

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Most popular questions from this chapter

Methane reacts with chlorine to produce methyl chloride and hydrogen chloride. Once formed, the methyl chloride may undergo further chlorination to form methylene chloride ( \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) ), chloroform, and carbon tetrachloride. A methyl chloride production process consists of a reactor, a condenser, a distillation column, and an absorption column. A gas stream containing 80.0 mole \(\%\) methane and the balance chlorine is fed to the reactor. In the reactor a single-pass chlorine conversion of essentially \(100 \%\) is attained, the mole ratio of methyl chloride to methylene chloride in the product is \(5: 1,\) and negligible amounts of chloroform and carbon tetrachloride are formed. The product stream flows to the condenser. Two streams emerge from the condenser: the liquid condensate, which contains essentially all of the methyl chloride and methylene chloride in the reactor effluent, and a gas containing the methane and hydrogen chloride. The condensate goes to the distillation column in which the two component species are separated. The gas leaving the condenser flows to the absorption column where it contacts an aqueous solution. The solution absorbs essentially all of the HCl and none of the \(\mathrm{CH}_{4}\) in the feed. The liquid leaving the absorber is pumped elsewhere in the plant for further processing, and the methane is recycled to join the fresh feed to the process (a mixture of methane and chlorine). The combined stream is the feed to the reactor. (a) Choose a quantity of the reactor feed as a basis of calculation, draw and label a flowchart, and determine the degrees of freedom for the overall process and each single unit and stream mixing point. Then write in order the equations you would use to calculate the molar flow rate and molar composition of the fresh feed, the rate at which HCI must be removed in the absorber, the methyl chloride production rate, and the molar flow rate of the recycle stream. Do no calculations. (b) Calculate the quantities specified in Part (a), either manually or with an equation-solving program. (c) What molar flow rates and compositions of the fresh feed and the recycle stream are required to achieve a methyl chloride production rate of \(1000 \mathrm{kg} / \mathrm{h} ?\)

Inside a distillation column (see Problem 4.8), a downward-flowing liquid and an upward-flowing vapor maintain contact with each other. For reasons we will discuss in greater detail in Chapter \(6,\) the vapor stream becomes increasingly rich in the more volatile components of the mixture as it moves up the column, and the liquid stream is enriched in the less volatile components as it moves down. The vapor leaving the top of the column goes to a condenser. A portion of the condensate is taken off as a product (the overhead product), and the remainder (the reflux) is returned to the top of the column to begin its downward journey as the liquid stream. The condensation process can be represented as shown below: A distillation column is being used to separate a liquid mixture of ethanol (more volatile) and water (less volatile). A vapor mixture containing 89.0 mole \(\%\) ethanol and the balance water enters the overhead condenser at a rate of \(100 \mathrm{lb}\) -mole/h. The liquid condensate has a density of \(49.01 \mathrm{b}_{\mathrm{m}} / \mathrm{ft}^{3},\) and the reflux ratio is \(3 \mathrm{lb}_{\mathrm{m}}\) reflux/lb \(_{\mathrm{m}}\) overhead product. When the system is operating at steady state, the tank collecting the condensate is half full of liquid and the mean residence time in the tank (volume of liquid/volumetric flow rate of liquid) is 10.0 minutes. Determine the overhead product volumetric flow rate (ft \(^{3}\) /min) and the condenser tank volume (gal).

Chlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\right),\) an important solvent and intermediate in the production of many other chemicals, is produced by bubbling chlorine gas through liquid benzene in the presence of ferric chloride catalyst. In an undesired side reaction, the product is further chlorinated to dichlorobenzene, and in a third reaction the dichlorobenzene is chlorinated to trichlorobenzene. The feed to a chlorination reactor consists of essentially pure benzene and a technical grade of chlorine gas (98 wt\% \(\mathrm{Cl}_{2}\), the balance gaseous impurities with an average molecular weight of 25.0 ). The liquid output from the reactor contains \(65.0 \mathrm{wt} \% \mathrm{C}_{6} \mathrm{H}_{6}, 32.0 \% \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}, 2.5 \% \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},\) and \(0.5 \%\) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}_{3} .\) The gaseous output contains only \(\mathrm{HCl}\) and the impurities that entered with the chlorine. (a) You wish to determine (i) the percentage by which benzene is fed in excess, (ii) the fractional conversion of benzene, (iii) the fractional yield of monochlorobenzene, and (iv) the mass ratio of the gas feed to the liquid feed. Without doing any calculations, prove that you have enough information about the process to determine these quantities. (b) Perform the calculations. (c) Why would benzene be fed in excess and the fractional conversion kept low? (d) What might be done with the gaseous effluent? (e) It is possible to use 99.9\% pure ("reagent-grade") chlorine instead of the technical grade actually used in the process. Why is this probably not done? Under what conditions might extremely pure reactants be called for in a commercial process? (Hint: Think about possible problems associated with the impurities in technical grade chemicals.)

Liquid methanol is fed to a space heater at a rate of \(12.0 \mathrm{L} / \mathrm{h}\) and burned with excess air. The product gas is analyzed and the following dry-basis mole percentages are determined: \(\mathrm{CH}_{3} \mathrm{OH}=0.45 \%\) \(\mathrm{CO}_{2}=9.03 \%,\) and \(\mathrm{CO}=1.81 \%\) (a) Draw and label a flowchart and verify that the system has zero degrees of freedom. (b) Calculate the fractional conversion of methanol, the percentage excess air fed, and the mole fraction of water in the product gas. (c) Suppose the combustion products are released directly into a room. What potential problems do you see and what remedies can you suggest?

Draw and label the given streams and derive expressions for the indicated quantities in terms of labeled variables. The solution of Part (a) is given as an illustration. (a) A continuous stream contains 40.0 mole\% benzene and the balance toluene. Write expressions for the molar and mass flow rates of benzene, \(\dot{n}_{\mathrm{B}}\left(\operatorname{mol} \mathrm{C}_{6} \mathrm{H}_{6} / \mathrm{s}\right)\) and \(\dot{m}_{\mathrm{B}}\left(\mathrm{kg} \mathrm{C}_{6} \mathrm{H}_{6} / \mathrm{s}\right),\) in terms of the total molar flow rate of the stream, \(\dot{n}(\mathrm{mol} / \mathrm{s})\) (b) The feed to a batch process contains equimolar quantities of nitrogen and methane. Write an expression for the kilograms of nitrogen in terms of the total moles \(n(\) mol) of this mixture. (c) A stream containing ethane, propane, and butane has a mass flow rate of \(100.0 \mathrm{g} / \mathrm{s}\). Write an expression for the molar flow rate of ethane, \(\dot{n}_{\mathrm{E}}\left(\text { Ib-mole } \mathrm{C}_{2} \mathrm{H}_{6} / \mathrm{h}\right)\), in terms of the mass fraction of this species, \(x_{\mathrm{E}}\). (d) A continuous stream of humid air contains water vapor and dry air, the latter containing approximately 21 mole \(\% \mathrm{O}_{2}\) and \(79 \% \mathrm{N}_{2}\). Write expressions for the molar flow rate of \(\mathrm{O}_{2}\) and for the mole fractions of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}_{2}\) in the gas in terms of \(\dot{n}_{1}\left(\mathrm{lb}-\mathrm{mole} \mathrm{H}_{2} \mathrm{O} / \mathrm{s}\right)\) and \(\dot{n}_{2}(\text { lb- mole dry air/s })\) (e) The product from a batch reactor contains \(\mathrm{NO}, \mathrm{NO}_{2},\) and \(\mathrm{N}_{2} \mathrm{O}_{4} .\) The mole fraction of \(\mathrm{NO}\) is 0.400. Write an expression for the gram-moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) in terms of \(n(\mathrm{mol}\) mixture) and \(y_{\mathrm{NO}_{2}}\left(\operatorname{mol} \mathrm{NO}_{2} / \mathrm{mol}\right)\)
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