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Chapter 4: Problem 101

Fuel oils contain primarily organic compounds and sulfur. The molar composition of the organic fraction of a fuel oil may be represented by the formula \(\mathrm{C}_{\rho} \mathrm{H}_{q} \mathrm{O}_{r}\); the mass fraction of sulfur in the fuel is \(x_{\mathrm{S}}\left(\mathrm{kg} \mathrm{S} / \mathrm{kg} \text { fuel); and the percentage excess air, } P_{\mathrm{xs}},\) is defined in terms of the theoretical air required to \right. burn only the carbon and hydrogen in the fuel. (a) For a certain high-sulfur No. 6 fuel oil, \(p=0.71, q=1.1, r=0.003,\) and \(x_{S}=0.02 .\) Calculate the composition of the stack gas on a dry basis if this fuel is burned with \(18 \%\) excess air, assuming complete combustion of the fuel to form \(\mathrm{CO}_{2}, \mathrm{SO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) and expressing the \(\mathrm{SO}_{2}\) fraction as ppm (mol SO_/10^6 mol dry gas). (b) Create a spreadsheet to calculate the mole fractions of the stack gas components on a dry basis for specified values of \(p, q, r, x_{\mathrm{S}},\) and \(P_{\mathrm{xs}} .\) The output should appear as follows: (Rows below the last one shown can be used to calculate intermediate quantities.) Execute enough runs (including the two shown above) to determine the effect on the stack gas composition of each of the five input parameters. Then for the values of \(p, q, r,\) and \(x_{S}\) given in Part (a), find the minimum percentage excess air needed to keep the dry-basis \(\mathrm{SO}_{2}\) composition below 700 ppm. (Make this the last run in the output table.) You should find that for a given fuel oil composition, increasing the percentage excess air decreases the \(S O_{2}\) concentration in the stack gas. Explain why this should be the case. (c) Someone has proposed using the relationship between \(P_{\mathrm{xs}}\) and ppm \(\mathrm{SO}_{2}\) as the basis of a pollution control strategy. The idea is to determine the minimum acceptable concentration of \(\mathrm{SO}_{2}\) in the stack gas, then run with the percentage excess air high enough to achieve this value. Give several reasons why this is a poor idea.

Short Answer

Expert verified
The molar composition of the stack gas is determined by level of excess air, fraction of sulfur and the composition of the fuel. Increasing the percentage of excess air decreases the SO2 concentration. A higher \(SO_{2}\) concentration leads to more pollution. However, using excessive air as a pollution control strategy can be counterproductive due to higher energy requirements and reduced combustion temperature, potentially leading to other pollutants.

Step by step solution

01

Analysis of Fuel Combustion

Let's begin by determining the moles of air required for complete combustion. The fuel has \(C_\rho H_q O_r S\). The balanced reaction for combustion will lead to \(\rho\) C in CO2, \(q/2\) H2 in H2O, \(r/2\) O2 in CO2, and S in SO2. Thus, it's possible to write the reaction as: \(C_\rho H_q O_r S + (\rho + q/4 - r/2 ) O_2 -> \rho CO_2 + q/2 H2O + SO_2\). Here, \(\rho + q/4 - r/2\) is stoichiometric oxygen required. Therefore, air needed (79% N2 and 21% O2 by volume) is \((\rho + q/4 - r/2) *(1/0.21)\).
02

Compute Stack Gas Composition

Now, consider 18% excess air. This means the actual air supplied is \((\rho + q/4 - r/2) *(1/0.21)* 1.18\). Using the complete combustion assumption, stack gas consists of \(CO_2\), \(SO_2\), \(H_2 O\) and \(N_2\). Given the fuel is dry, the H2O is not considered. Thus, the moles of \(N_2\), \(CO_2\), and \(SO_2\) are respectively \(\rho_q*(1/0.21)*1.18*0.79\), \(\rho\) and \(2*0.02\) because 2 kg of S will give 2 moles of SO2. Finally, convert to ppm by using \((SO_2 / (N_2 + CO_2 + SO_2 ))*10^6\).
03

Spreadsheet for Different Conditions

Create an excel sheet that allows to input different values of \(p, q, r, x_{\mathrm{S}}\) and \(P_{\mathrm{xs}}\), and perform the same calculation as outlined in step 2 for each. This will help to observe the impact of each parameter on the stack gas composition.
04

Discuss Strategy for Pollution Control

The minimum percentage excess air needed to keep the dry-basis \(SO_{2}\) composition below 700 ppm can be obtained by setting the formula from step 2 less than 700 and solving for \(P_{\mathrm{xs}}\). According to the analysis, increasing excess air reduces \(SO_{2}\) concentration in the stack gas. This is because higher excess air means more \(O_2\) available per unit of fuel for combustion which ensures complete oxidation of S to \(SO_{2}\) and dilutes the resultant gases.
05

Analysis of Pollution Control Strategy

Using the relationship between percentage excess air and \(SO_{2}\) concentration may seem beneficial, but there are several disadvantages. Firstly, excess air means extra pumping and heating which is energy-intensive. Secondly, it increases the volume of stack gases, hence the size and cost of gas handling - heat recovery and exhaust systems. Also, high excess air can reduce the combustion temperature which reduces efficiency and can lead to the production of other harmful pollutants.

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Key Concepts

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

Fuel Composition
The composition of a fuel is crucial for understanding how it will behave during combustion. Fuel oils, like the one described in the problem, contain organic compounds and sulfur. The organic part of the fuel can be represented by a composition formula, such as \(\mathrm{C}_{\rho} \mathrm{H}_{q} \mathrm{O}_{r}\). This formula indicates the presence of carbon, hydrogen, and oxygen atoms in the fuel.

Identifying these elements helps in balancing the chemical equation needed to predict the products of combustion. In combustion reactions, every element in the fuel composition plays a role in determining the type and amount of products generated. For instance, carbon forms \(\mathrm{CO}_{2}\), hydrogen forms \(\mathrm{H}_{2}\mathrm{O}\), and sulfur forms \(\mathrm{SO}_{2}\). If the fuel contains sulfur \(x_{S}\), this will directly influence the \(\mathrm{SO}_{2}\) levels in the stack gas produced.

When dealing with complex fuels, having a clear understanding of \(p\), \(q\), \(r\), and \(x_{S}\) allows engineers to estimate both the fuel's energy potential and the possible environmental impact of its combustion.
Stack Gas Composition
Stack gas refers to the gases emitted from the exhaust stack of a combustion system. The composition of these gases is a direct reflection of the original fuel makeup combined with the combustion process conditions. Upon complete combustion, typical stack gases include \(\mathrm{CO}_{2}\), \(\mathrm{SO}_{2}\), \(\mathrm{H}_{2}\mathrm{O}\), and \(\mathrm{N}_{2}\).

In the exercise, since we assume a dry basis (ignoring water vapor), the stack gas primarily includes \(\mathrm{CO}_{2}\), \(\mathrm{SO}_{2}\), and \(\mathrm{N}_{2}\). The concentration of these gases in the stack gas depends on the amounts and types of elements in the fuel. For instance, a higher sulfur content will always result in a higher \(\mathrm{SO}_{2}\) percentage in the stack gas.

Monitoring stack gas composition is essential for optimizing combustion and for pollution control. High levels of harmful pollutants like \(\mathrm{SO}_{2}\) can have severe environmental impacts. That's why regulatory bodies set limits on pollutant concentrations and why plant operators strive to keep emissions within permissible levels.
Excess Air in Combustion
Excess air refers to the additional air supplied during combustion beyond the theoretical amount required to completely burn the fuel. The percentage excess air, \(P_{xs}\), is a critical parameter in combustion systems. It ensures complete combustion and affects the composition of stack gases.

When you supply excess air, it helps consume all the fuel elements thoroughly. This can notably reduce pollutants like \(\mathrm{SO}_{2}\) in the exhaust. The reason is simple 鈥� more air available means more oxygen for converting sulfur to \(\mathrm{SO}_{2}\), which dilutes the concentration of \(\mathrm{SO}_{2}\) by increasing the total volume of gases emitted. Therefore, as excess air increases, \(\mathrm{SO}_{2}\) expressed in ppm tends to decrease.

However, excessive excess air isn't always favorable. It can lead to inefficient combustion, increased energy costs due to excessive pumping, and large volumes of gases that require handling and treatment. This can increase operational costs and the environmental footprint of combustion. Optimizing the balance of air supply is key to achieving both efficient energy use and low pollutant emissions.

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Most popular questions from this chapter

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.

Ethanol can be produced commercially by the hydration of ethylene: $$\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$$ Some of the product is converted to diethyl ether in the side reaction $$2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}$$ The feed to the reactor contains ethylene, steam, and an inert gas. A sample of the reactor effluent gas is analyzed and found to contain 43.3 mole\% ethylene, 2.5\% ethanol, 0.14\% ether, 9.3\% inerts, and the balance water. (a) Take as a basis 100 mol of effluent gas, draw and label a flowchart, and do a degree-of-freedom analysis based on atomic species to prove that the system has zero degrees of freedom. (b) Calculate the molar composition of the reactor feed, the percentage conversion of ethylene, the fractional yield of ethanol, and the selectivity of ethanol production relative to ether production. (c) The percentage conversion of ethylene you calculated should be very low. Why do you think the reactor would be designed to consume so little of the reactant? (Hint: If the reaction mixture remained in the reactor long enough to use up most of the ethylene, what would the main product constituent probably be?) What additional processing steps are likely to take place downstream from the reactor?

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.)

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?)

Two aqueous sulfuric acid solutions containing \(20.0 \mathrm{wt} \% \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{SG}=1.139)\) and \(60.0 \mathrm{wt} \% \mathrm{H}_{2} \mathrm{SO}_{4}\) (SG = 1.498) are mixed to form a 4.00 molar solution (SG = 1.213). (a) Calculate the mass fraction of sulfuric acid in the product solution. (b) Taking \(100 \mathrm{kg}\) of the \(20 \%\) feed solution as a basis, draw and label a flowchart of this process, labeling both masses and volumes, and do the degree-of-freedom analysis. Calculate the feed ratio (liters 20\% solution/liter 60\% solution). (c) What feed rate of the \(60 \%\) solution (L/h) would be required to produce \(1250 \mathrm{kg} / \mathrm{h}\) of the product?
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