/*! 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

Methanol is produced by reacting carbon monoxide and hydrogen. A fresh feed stream containing \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) joins a recycle stream and the combined stream is fed to a reactor. The reactor outlet stream flows at a rate of \(350 \mathrm{mol} / \mathrm{min}\) and contains \(10.6 \mathrm{wt} \% \mathrm{H}_{2}, 64.0 \mathrm{wt} \% \mathrm{CO},\) and \(25.4 \mathrm{wt} \% \mathrm{CH}_{3} \mathrm{OH} .\) (Notice that those are percentages by mass, not mole percents.) This stream enters a cooler in which most of the methanol is condensed. The liquid methanol condensate is withdrawn as a product, and the gas stream leaving the condenser- -which contains \(\mathrm{CO}, \mathrm{H}_{2},\) and \(0.40 \mathrm{mole} \%\) uncondensed \(\mathrm{CH}_{3} \mathrm{OH}\) vapor \(-\mathrm{is}\) the recycle stream that combines with the fresh feed. (a) Without doing any calculations, prove that you have enough information to determine (i) the molar flow rates of CO and \(\mathrm{H}_{2}\) in the fresh feed, (ii) the production rate of liquid methanol, and (iii) the single-pass and overall conversions of carbon monoxide. Then perform the calculations. (b) After several months of operation, the flow rate of liquid methanol leaving the condenser begins to decrease. List at least three possible explanations of this behavior and state how you might check the validity of each one. (What would you measure and what would you expect to find if the explanation is valid?)

Methanol is formed from carbon monoxide and hydrogen in the gas-phase reaction The mole fractions of the reactive species at equilibrium satisfy the relation where \(P\) is the total pressure (atm), \(K_{c}\) the reaction equilibrium constant (atm \(^{-2}\) ), and \(T\) the temperature (K). The equilibrium constant \(K_{c}\) equals 10.5 at 373 K, and \(2.316 \times 10^{-4}\) at \(573 \mathrm{K}\). A semilog plot of \(K_{\mathrm{c}}\) (logarithmic scale) versus 1/ \(T\) (rectangular scale) is approximately linear between \(T=300 \mathrm{K}\) and \(T=600 \mathrm{K}\) (a) Derive a formula for \(K_{\mathrm{c}}(T),\) and use it to show that \(K_{\mathrm{e}}(450 \mathrm{K})=0.0548 \mathrm{atm}^{-2}\) (b) Write expressions for \(n_{A}, n_{B},\) and \(n_{C}\) (gram-moles of each species), and then \(y_{A}, y_{B},\) and \(y_{C},\) in terms of \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0},\) and \(\xi,\) the extent of reaction. Then derive an equation involving only \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0}, P, T,\) and \(\xi_{e},\) where \(\xi_{e}\) is the extent of reaction at equilibrium. (c) Suppose you begin with equimolar quantities of CO and \(\mathrm{H}_{2}\) and no \(\mathrm{CH}_{3} \mathrm{OH}\), and the reaction proceeds to equilibrium at 423 K and 2.00 atm. Calculate the molar composition of the product ( \(y_{\mathrm{A}}\), \(\left.y_{\mathrm{B}}, \text { and } y_{\mathrm{C}}\right)\) and the fractional conversion of \(\mathrm{CO}\) (d) The conversion of CO and \(\mathrm{H}_{2}\) can be enhanced by removing methanol from the reactor while leaving unreacted CO and \(\mathrm{H}_{2}\) in the vessel. Review the equations you derived in solving Part (c) and determine any physical constraints on \(\xi_{c}\) associated with \(n_{\mathrm{A} 0}=n_{\mathrm{B} 0}=1\) mol. Now suppose that 90\% of the methanol is removed from the reactor as it is produced; in other words, only 10\% of the methanol formed remains in the reactor. Estimate the fractional conversion of CO and the total gram moles of methanol produced in the modified operation. (e) Repeat Part (d), but now assume that \(n_{\mathrm{B} 0}=2\) mol. Explain the significant increase in fractional conversion of CO. (f) Write a set of equations for \(y_{\mathrm{A}}, y_{\mathrm{B}}, y_{\mathrm{C}},\) and \(f_{\mathrm{A}}\) (the fractional conversion of \(\mathrm{CO}\) ) in terms of \(y_{\mathrm{A} 0}, y_{\mathrm{B} 0}, T,\) and \(P(\) the reactor temperature and pressure at equilibrium). Enter the equations in an equation-solving program. Check the program by running it for the conditions of Part (c), then use it to determine the effects on \(f_{\mathrm{A}}\) (increase, decrease, or no effect) of separately increasing, (i) the fraction of \(\mathrm{CH}_{3} \mathrm{OH}\) in the feed, (ii) temperature, and (iii) pressure.

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 \(y_{\mathrm{NO}_{2}}\left(\operatorname{mol} \mathrm{NO}_{2} / \mathrm{mol}\right)\)

Ammonia is oxidized to nitric oxide in the following reaction: $$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$$ (a) Calculate the ratio (lb-mole \(\mathrm{O}_{2}\) react/lb-mole NO formed). (b) If ammonia is fed to a continuous reactor at a rate of \(100.0 \mathrm{kmol} \mathrm{NH}_{3} / \mathrm{h}\), what oxygen feed rate (kmol/h) would correspond to 40.0\% excess O_? (c) If \(50.0 \mathrm{kg}\) of ammonia and \(100.0 \mathrm{kg}\) of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

The hormone estrogen is produced in the ovaries of females and elsewhere in the body in men and postmenopausal women, and it is also administered in estrogen replacement therapy, a common treatment for women who have undergone a hysterectomy. Unfortunately, it also binds to estrogen receptors in breast tissue and can activate cells to become cancerous. Tamoxifen is a drug that also binds to estrogen receptors but does not activate cells, in effect blocking the receptors from access to estrogen and inhibiting the growth of breast-cancer cells. Tamoxifen is administered in tablet form. In the manufacturing process, a finely ground powder contains tamoxifen (tam) and two inactive fillers- -lactose monohydrate (lac) and corn starch (cs). The powder is mixed with a second stream containing water and suspended solid particles of polyvinylpymolidone (pvp) binder, which keeps the tablets from easily crumbling. The slurry leaving the mixer goes to a dryer, in which 94.2\% of the water fed to the process is vaporized. The wet powder leaving the dryer contains 8.80 wr\% tam, 66.8\% lac, 21.4\% cs, 2.00\% pvp, and 1.00\% water. After some additional processing, the powder is molded into tablets. To produce a hundred thousand tablets, 17.13 kg of wet powder is required. (a) Taking a basis of 100,000 tablets produced, draw and label a process flowchart, labeling masses of individual components rather than total masses and component mass fractions. It is unnecessary to label the stream between the mixer and the dryer. Carry out a degree-of-freedom analysis of the overall two-unit process. (b) Calculate the masses and compositions of the streams that must enter the mixer to make 100,000 tablets. (c) Why was it unnecessary to label the stream between the mixer and the dryer? Under what circumstances would it have been necessary? (d) Go back to the flowchart of Part (a). Without using the mass of the wet powder (17.13 kg) or any of the results from Part (b) in your calculations, determine the mass fractions of the stream components in the powder fed to the mixer and verify that they match your solution to Part (b). (Hint: Take a basis of \(100 \mathrm{kg}\) of wet powder.) (e) Suppose a student does Part (d) before Part (b), and re-labels the powder feed to the mixer on the flowchart of Part (a) with an unknown total mass ( \(m_{1}\) ) and the three now known mole fractions. (Sketch the resulting flowchart.) The student then does a degree-of-freedom analysis, counts four unknowns (the masses of the powder, pvp, and water fed to the mixer, and the mass of water evaporated in the dryer), and six equations (five material balances for five species and the percentage evaporation), for a net of -2 degrees of freedom. since there are more equations than unknowns, it should not be possible to get a unique solution for the four unknowns. Nevertheless, the student writes four equations, solves for the four unknowns, and verifies that all of the balance equations are satisfied. There must have been a mistake in the degree-of-freedom calculation. What was it?

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