/*! 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 91 A mixture of 75 mole \(\%\) meth... [FREE SOLUTION] | 91Ó°ÊÓ

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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.)

Short Answer

Expert verified
(a) The concentration of \(CO\) in stack gas is 8204 ppm. (b) Two potential costs of increasing the air fed to the furnace are increased fuel costs and lowering the efficiency.

Step by step solution

01

Determine the Composition of the Feed

The feed contains 75 mole% methane (0.75) and 25 mole% hydrogen (0.25) burned with 25% excess air. Therefore, for every 100 moles of mixture there are 75 moles of methane and 25 moles of hydrogen. Incomplete combustion of methane forms carbon monoxide (CO) and water (H2O) while hydrogen fully burns to form water (\(H_{2}O\)).
02

Find the Amount of Air Required for Combustion

Each methane molecule (\(CH_{4}\)) requires 2 mole of O2 for full combustion and each hydrogen (\(H_{2}\)) requires 0.5 mole of O2. Totaling, the stoichiometric air required for 75 mole of methane is \(75*2 =150\) mole and for 25 mole of hydrogen is \(25*0.5 = 12.5\), totally 162.5 moles. Considering 25% excess air, total air supplied is 1.25*162.5 = 203.125 moles.
03

Find the Amount of Methane and Hydrogen Reacted

90% of methane and 85% of hydrogen are reacted. So, the reacted methane is \(75*0.9 = 67.5\) mole and reacted hydrogen is \(25*0.85 = 21.25\) mole.
04

Compute the Formation of Products

95% of the methane that reacts, reacts to form CO2 and the balance to form CO. So, the formed CO2 is \(67.5*0.95 = 64.125\) mole and CO is \(67.5*0.05 = 3.375\) mole. Each reacted methane also forms 2 moles of water and each reacted hydrogen forms 1 mole of water. Thus, the total water formed is \(67.5*2+21.25*1= 156.25\) mole.
05

Calculate the Concentration of CO

The total amount of combustion product gases is the sum of the unreacted methane and hydrogen, \(CO_2\), \(CO\), \(H_{2}O\), and unreacted O2 and N2 from the air. Each mole of air contains 0.21 mole of O2 and 0.79 mole of N2. Thus, the unreacted O2 is \(203.125 - 162.5 = 40.625\) mole and unreacted N2 is \(203.125*0.79 = 160.46875\) mole. The total mole is \(75-67.5+25-21.25+64.125+3.375+156.25+40.625+160.46875=411.09375\) mole. Now, the concentration of \(CO\) is \((3.375/411.09375) * 10^6 = 8204\) ppm.
06

Think of the Costs of Decreasing CO Concentration

Increasing the percent of excess air fed to the furnace can reduce CO concentration. However, this comes with two potential costs: 1. Increased fuel costs: More air means more fuel is needed to maintain the temperature, which could drastically increase fuel costs. 2. Lowering the efficiency: The extra nitrogen in the air that is not taking part in the reaction absorbs some of the heat produced, which could decrease the efficiency of the combustion process.

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Key Concepts

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

Combustion Analysis
Combustion analysis involves understanding the chemical reactions that occur when a fuel burns in the presence of oxygen. It helps in calculating the products formed and gauging the effectiveness of the combustion process. When dealing with hydrocarbon fuels like methane and hydrogen, we typically observe reactions such as complete and incomplete combustion. Complete combustion results in products like carbon dioxide (CO2) and water (H2O). However, when there is insufficient oxygen, incomplete combustion may occur, leading to the production of carbon monoxide (CO). In the given exercise, the stoichiometry of the reactions is crucial. For methane combustion, each mole requires two moles of O2, while hydrogen needs 0.5 moles of O2. This knowledge is used to compute how much oxygen is needed for the reactions. Stoichiometry ensures the correct proportions, which affects the amounts of CO, CO2, and water produced.
Excess Air in Combustion
Excess air in combustion refers to the additional air supply beyond the stoichiometric requirement needed to ensure all the fuel burns completely. This practice is common to prevent the formation of carbon monoxide, a harmful pollutant. In the exercise, 25% excess air is introduced. This implies that 25% more air than the theoretical amount needed is supplied. It ensures complete combustion of the reactants by compensating for potential inefficiencies in air and fuel mixing, which can otherwise lead to incomplete burning and CO formation. However, using excess air comes with downsides. It can lead to energy losses as more air needs to be heated along with the fuel. The extra nitrogen from the air absorbs heat without contributing to combustion, which can reduce the system's thermal efficiency.
Fractional Conversion
Fractional conversion is the fraction of reactant that is converted into products in a chemical reaction. It indicates the efficiency with which reactants are transformed into desired products during a combustion process. In this context, with a fractional conversion of 90% for methane and 85% for hydrogen, not all fuel is converted into the desired combustion products. Instead, a portion remains unreacted or converts into less desired forms, like carbon monoxide. This concept is vital as it affects the overall efficiency and the emissions from the combustion process. By improving fractional conversion, one can optimize fuel usage and reduce pollutant emissions, hence enhancing the overall combustion efficiency and reducing harmful environmental impacts.
Pollutant Control
Pollutant control in combustion involves strategies to minimize the release of harmful substances, such as carbon monoxide, into the atmosphere. Proper management is crucial for limiting environmental and health impacts. The exercise highlights that CO can be reduced by increasing excess air. However, it is important to balance this approach to avoid increased fuel costs and efficiency losses. Alternatives to increasing air include methods like catalytic converters or fuel restructuring, which can help mitigate CO emissions without significantly altering the combustion process. In addition to technological solutions, operational changes, such as maintaining optimal flame temperature and ensuring proper mixing of fuel and air, can effectively lower CO emissions and improve overall system performance.

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Most popular questions from this chapter

A garment to protect the wearer from toxic agents may be made of a fabric that contains an adsorbent, such as activated carbon. In a test of such a fabric, a gas stream containing \(7.76 \mathrm{mg} / \mathrm{L}\) of carbon tetrachloride (CCl_) was passed through a 7.71-g sample of the fabric at a rate of 1.0 L/min, and the concentration of \(\mathrm{CCl}_{4}\) in the gas leaving the fabric was monitored. The run was continued for \(15.5 \mathrm{min}\) with no \(\mathrm{CCl}_{4}\) being detected, after which the \(\mathrm{CCl}_{4}\) concentration began to rise. (a) How much CCl_ was fed to the system during the first 15.5 min of the run? How much was adsorbed? Using this information as a guide, sketch the expected concentration of \(\mathrm{CCl}_{4}\) in the exit gas as a function of time, showing the curve from \(t=0\) to \(t \gg 15.5\) min. (b) Assuming a linear relationship between amount of \(\mathrm{CCl}_{4}\) adsorbed and mass of fabric, what fabric mass would be required if the feed concentration is \(5 \mathrm{mg} / \mathrm{L},\) the feed rate \(1.4 \mathrm{L} / \mathrm{min},\) and it is desired that no \(\mathrm{CCl}_{4}\) leave the fabric earlier than 30 min?

A catalytic reactor is used to produce formaldehyde from methanol in the reaction $$\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{HCHO}+\mathrm{H}_{2}$$ A single-pass conversion of \(60.0 \%\) is achieved in the reactor. The methanol in the reactor product is separated from the formaldehyde and hydrogen in a multiple-unit process. The production rate of formaldehyde is 900.0 kg/h. (a) Calculate the required feed rate of methanol to the process ( \(\mathrm{kmol} / \mathrm{h}\) ) if there is no recycle. (b) Suppose the unreacted methanol is recovered and recycled to the reactor and the single-pass conversion remains 60\%. Without doing any calculations, prove that you have enough information to determine the required fresh feed rate of methanol (kmol/h) and the rates (kmol/h) at which methanol enters and leaves the reactor. Then perform the calculations. (c) The single-pass conversion in the reactor, \(X_{\mathrm{sp}},\) affects the costs of the reactor \(\left(C_{\mathrm{r}}\right)\) and the separation process and recycle line \(\left(C_{\mathrm{s}}\right) .\) What effect would you expect an increased \(X_{\mathrm{sp}}\) would have on each of these costs for a fixed formaldehyde production rate? (Hint: To get a \(100 \%\) singlepass conversion you would need an infinitely large reactor, and lowering the single-pass conversion leads to a need to process greater amounts of fluid through both process units and the recycle line.) What would you expect a plot of \(\left(C_{\mathrm{r}}+C_{\mathrm{s}}\right)\) versus \(X_{\mathrm{sp}}\) to look like? What does the design specification \(X_{\mathrm{sp}}=60 \%\) probably represent?

Methanol is synthesized from carbon monoxide and hydrogen in a catalytic reactor. The fresh feed to the process contains 32.0 mole \(\%\) CO, \(64.0 \%\) H \(_{2}\), and \(4.0 \%\) Ne. This stream is mixed with a recycle stream in a ratio 5 mol recycle/ 1 mol fresh feed to produce the feed to the reactor, which contains 13.0 mole\% \(\mathrm{N}_{2}\). A low single-pass conversion is attained in the reactor. The reactor effluent goes to a condenser from which two streams emerge: a liquid product stream containing essentially all the methanol formed in the reactor, and a gas stream containing all the \(\mathrm{CO}, \mathrm{H}_{2}\), and \(\mathrm{N}_{2}\) leaving the reactor. The gas stream is split into two fractions: one is removed from the process as a purge stream, and the other is the recycle stream that combines with the fresh feed to the reactor. (a) Assume a methanol production rate of \(100 \mathrm{kmol} / \mathrm{h}\). Perform the DOF for the overall system and all subsystems to prove that there is insufficient information to solve for all unknowns. (b) Briefly explain in your own words the reasons for including (i) the recycle stream and (ii) the purge stream in the process design.

\- 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).

A paint mixture containing \(25.0 \%\) of a pigment and the balance binders (which help the pigment stick to the surface) and solvents (which ensure that the paint stays in liquid form) sells for 18.00 dollar/kg, and a mixture containing 12.0\% sells for 10.00 dollar /kg. (a) If a paint retailer produces a blend containing \(17.0 \%\) pigment, for how much (S/kg) should it be sold to yield a 10\% profit? (b) Paint manufacturers have begun to market "low VOC" paint as a more environmentally friendly product. What are VOCs? List some ways in which paint products can be altered to lower the VOC content.

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